Ioan TOFAN - Facultatea De Matematica Iasitofan/depozit/teoria_divizibilitatii.pdf · Ioan TOFAN Aurelian Claudiu VOLF Ring Arithmetic, Field Extensions and Applications

Ioan TOFAN

Aurelian Claudiu VOLF

Ring Arithmetic, Field Extensions

and Applications

Table of Contents

Foreword..............................................................................................7 I. Arithmetic in integral domains ....................................................10

I.1 Divisibility...................................................................................10 I.2 Euclidian domains .......................................................................22 I.3 Euclidian rings of quadratic integers...........................................26 I.4 Principal ideal domains ...............................................................31 I.5 Unique factorization domains .....................................................36 I.6 Polynomial ring arithmetic..........................................................43 Exercises ...........................................................................................50

II. Modules.........................................................................................58 II.1 Modules, submodules, homomorphisms....................................58 Exercises ...........................................................................................72 II.2 Factor modules and the isomorphism theorems.........................74 Exercises ...........................................................................................80 II.3 Direct sums and products. Exact sequences ..............................81 Exercises .........................................................................................102 II.4 Free modules ............................................................................104 Exercises .........................................................................................116

III. Finitely generated modules over principal ideal domains ....118 III.1 The submodules of a free module...........................................118 Exercises .........................................................................................128 III.2 Finitely generated modules over a principal ideal domain.....130 Exercises .........................................................................................138 III.3 Indecomposable finitely generated modules ..........................140 Exercises .........................................................................................149

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III.4 The endomorphisms of a finite dimensional vector space ..... 150 Exercises ........................................................................................ 167

IV. Field extensions ........................................................................ 169 IV.1. Algebraic extensions............................................................. 169 Exercises ........................................................................................ 190 IV.2 Roots of polynomials. Algebraically closed fields ................ 193 Exercises ........................................................................................ 207 IV.3 Finite fields ............................................................................ 209 Exercises ........................................................................................ 221 IV.4 Transcendental extensions ..................................................... 223 Exercises ........................................................................................ 230

V. Galois Theory............................................................................. 232 V.1 Automorphisms ....................................................................... 233 V.2 Normal extensions................................................................... 238 Exercises ........................................................................................ 242 V.3 Separability.............................................................................. 244 Exercises ........................................................................................ 259 V.4 The Fundamental Theorem of Galois Theory ......................... 262 Exercises ........................................................................................ 271

VI. Applications of Galois Theory ................................................ 273 VI.1 Ruler and compass constructions........................................... 273 VI.2 Trace and norm ...................................................................... 289 Exercises ........................................................................................ 296 VI.3 Cyclic extensions and Kummer extensions ........................... 297 Exercises ........................................................................................ 312 VI.4 Solvability by radicals............................................................ 313 VI.5 Discriminants, resultants........................................................ 320 Exercises ........................................................................................ 326

Appendices ...................................................................................... 331 1. Prime ideals and maximal ideals................................................ 331 2. Algebras. Polynomial and monoid algebras............................... 333 3. Symmetric polynomials ............................................................. 348 4. Rings and modules of fractions.................................................. 354 5. Categories, functors.................................................................... 362 6. Solvable groups.......................................................................... 373

Index ................................................................................................ 379 Bibliography.................................................................................... 387

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Foreword

The book is aimed at undergraduate students in Mathematics, Com-puter Science or technical universities having a background of stan-dard courses of Abstract Algebra and Linear Algebra as well as at general Mathematics readers with an interest in Algebra.

The topics covered by the book are mandatory algebraic back-ground of any mathematics graduate: arithmetic in integral domains, module theory basics, the structure of the finitely generated modules over a principal ideal domain with applications in abelian groups and Jordan forms of matrices, field extensions and Galois theory with applications.

Throughout the book the reader is motivated by concrete applica-tions and exercises.

The book is reasonably self contained, in the sense that the reader is assumed to be familiar with general notions on algebraic structures (monoids, groups, rings, fields), factor rings and isomorphism theo-rems, vector spaces and bases, matrices, polynomials, permutation groups basics, elementary arithmetic of cardinals. Some topics less likely to appear in standard general Algebra courses are presendted in the Appendix.

The language of Category Theory is used beginning with the chap-ter on Modules, the reader being invited to refer to the appendix (or to a book on Categories) when categorical notions appear in the text. We

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think that the Categories are particularly useful for a better understan-ding and for unifying many algebraic concepts and proofs.

The chapters on Modules (II, III) can be read independently of the rest of the book. The section VI.1, Ruler and compass constructions, is not a prerequisite for the other sections in chapter VI.

Some notations used in the text: - | A | denotes the cardinal of the set A (the number of elements of

A, if A is finite). - x := y means x is equal by definition to y (where y is already

defined) or we denote y with x" - ! marks the end or the absence of a proof. - N is the set of natural numbers, 0, 1, 2, - N* is the set of positive natural numbers, 1, 2, - Z is the set of integers

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I. Arithmetic in integral domains

The set Z of integers, endowed with the operations of addition and multiplication, is the prototype for the familiar concept of ring. The classical divisibility theory in Z (the arithmetic of Z) can be extended with outstanding results to a large class of rings, the integral domains. Such a generalization is interesting by itself and it also illuminates and yields nontrivial results on the divisibility in Z.

After a general study of the divisibility in integral domains, three classical and important classes of domains are studied (Euclidian do-mains, principal ideal domains and unique factorization domains). The definitions of these classes of rings originate in fundamental arithmetic properties of Z. The material in this chapter is at the very basis of all Algebra, and vital in Algebraic Number Theory, Field Extensions and Galois Theory.

I.1 Divisibility

The classical definition for the relation of divisibility in the ring of integers Z generalizes easily to an arbitrary ring R:

I.1 Divisibility

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1.1 Definition. Let R be a ring and let a, b ∈ R. We say that a di-vides b in R (and write a | b) if there exists c ∈ R such that b = ac.

The fact that a | b can be also expressed by writing b ! a (read b is divisible by a). Other ways of reading a | b are: a is a divisor of b or b is a multiple of a.

One says a divides b in R because the ring R plays an essential role here. For instance, 2 | 3 in Q, but of course not in Z! We shall omit any reference to R in the notation a | b if the ring R is clear from the context. Write a - b if a does not divide b.

If the ring R lacks some natural properties, the theory of divisibility in R can be very poor or very peculiar when compared to the classical theory in Z. For instance, in a ring without identity element, an ele-ment may not divide itself; other difficulties arise if R is not commuta-tive or if R has zero divisors.

This motivates the following definition:

1.2 Definition. A ring R is called an integral domain (a domain, for short) if it has identity (denoted by 1), it is commutative and has no zero divisors: for any x, y ∈ R, xy = 0 implies x = 0 or y = 0. This can also be said: for any nonzero x, y ∈ R, we have xy ≠ 0.

In what follows, all rings we consider are domains, 0 denotes the zero element of the domain and 1 its identity element. All subrings considered are unitary (they contain the identity element of the ring).

1.3 Examples. a) Any field F (for instance Q, R, C, ...) is a do-main. Indeed, if x, y ∈ F are nonzero, then xy = 0 implies (by multiplying with x −1, which exists in F since F is a field and x ≠ 0), that y = 0, contradiction. The theory of divisibility in fields is trivial, though (cf. 1.7).

b) Every subring of a domain is itself a domain. In particular, every subring of a field is a domain. So, if d ∈ Z is squarefree (i.e.: d ≠ 0,


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d ≠ 1 and d is not divisible by the square of any integer greater than 1), the subring of C generated by 1 and d , denoted by [ ]d" , is a do-main. One easily checks that [ ]d" consists of the complex numbers of the form dba + , with a, b ∈ Z. The ring [ ]1−" is called the ring of Gauss integers.1

c) If R is a domain and n ∈ N*, then the polynomial ring in n indeterminates with coefficients in R, R[X1,..., Xn], is a domain.

In a domain, one can simplify the nonzero factors:

1.4 Proposition. Let R be a domain and let a, b, c ∈ R, with c ≠ 0. If ac = bc, then a = b.

Proof. We have ac = bc ⇔ ac − bc = 0 ⇔ (a − b)c = 0. Since R is a domain, a − b = 0 or c = 0. But c ≠ 0, so a − b = 0. !

We will develop a theory of divisibility in R, with Z as a model (and a particular case). The proof of the following properties is an easy exercise:

1.5 Proposition. Let R be a domain. Then: a) For any a ∈ R, a | a. b) For any a, b, c ∈ R such that a | b and b | c, we have a | c. c) For any a ∈ R, we have a | 0 and 1 | a. d) For any x, y ∈ R and a, b, c ∈ R such that a | b and a | c, we

have a | (bx + cy). !

Properties a) and b) say that the divisibility relation on R is reflex-ive and transitive.

1.6 Definition. The elements a, b in R are called associated in divisibility (or, simply, associated) if a | b and b | a. Notation: a ∼ b.

1 Carl Friedrich Gauss (1777 − 1855), famous German mathematician.

I.1 Divisibility

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For d, a ∈ R, d is called a proper divisor of a if d | a and d is neither invertible in R, nor associated with a.

The relation "∼" defined above is an equivalence relation on R (exercise!) and it is very important when studying the arithmetic of R: two elements associated in divisibility have exactly the same divisors and the same multiples. One can say they are indistinguishable as far as divisibility is concerned.

An invertible element u ∈ R is called a unit of R, because u ∼ 1 (so u behaves just like 1 from the divisibility standpoint). Let U(R) denote the set of all invertible elements of R:

U(R) = x ∈ R | (∃) y ∈ R such that xy = 1. U(R) is a group with respect to the ring multiplication (as a

straightforward checking shows) and is called the group of units of R.

1.7 Proposition. Let R be a domain. Then : a) For any u ∈ R, we have: u ∈ U(R) ⇔ u ∼ 1 ⇔ u | a, (∀) a ∈

R ⇔ uR = R. b) For any a, b ∈ R, we have: a ∼ b ⇔ there exists u ∈ R such that

a = bu. !

For a given domain R, knowing the group U(R) is a first step, very important, in the study of divisibility in R.

1.8 Examples. a) U(Z) = −1, 1. b) If K is a field, U(K[X]) = f ∈ K[X] | deg f = 0 = K* (we iden-

tify nonzero elements in K with the polynomials of degree 0). c) If d ∈ Z is squarefree, then:

UZ[ d ] = dba + | a, b ∈ Z, a 2 − db 2 = ±1 Proof. Let R = Z[ d ]. It is useful to define the norm N : R → Z,

by N(α) = ασ(α), where σ(α) is the conjugate of α, defined as : ( ) dbadba −=+σ , for any a, b ∈ Z.


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So ( ) 22N dbadba −=+ , ∀ a, b ∈ Z. An easy computation

shows: N(α)N(β) = N(α)N(β), ∀α, β ∈ R.

This implies: if α, β ∈ R with α | β in R, then N(α) | N(β) in Z. Let u = dba + ∈ U(R). Then N(u) = 22 dba − divides 1 in Z, so

N(u) = ±1. Conversely, if N(u) = ±1, then ( )( ) 1±=−+ dbadba , so ( )dba −± is the inverse of u. Thus:

UZ[ d ] = dba + | a, b ∈ Z, a 2 − db 2 = ±1 = = α ∈ Z[ d ] | N(α) = ±1.

We define the central concepts of greatest common divisor and least common multiple. Since an order relation like the one on N is not available, we use the divisibility relation itself to order the common divisors.

1.9 Definition. Assume R is a domain, n ∈ N* and a1, ..., an ∈ R. We call the element d ∈ R a greatest common divisor (abbreviated GCD) of the elements a1, ..., an if:

i) d | a1, ..., d | an. (d is a common divisor of a1, ..., an) ii) For any e ∈ R such that e | a1, ..., e | an, it follows that e | d (d is

the greatest among the common divisors of a1, ..., an). If 1 is a GCD of a1 and a2, we call a1 and a2 coprime or mutually

prime (or relatively prime). We call m ∈ R a least common multiple (LCM for short) of a1, ...,

an if: i') a1 | m, ..., an | m. ii') For any e ∈ R such that a1 | e, ..., an | e, it follows that m | e. We denote by (a1, ..., an) or GCD(a1, ..., an) a GCD of a1, ..., an, if it

exists. Similarly, [a1, ..., an] or LCM(a1, ..., an) denotes a LCM of a1, ...,

an, if it exists.

I.1 Divisibility

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1.10 Remarks. a) Given a1, ..., an ∈ R, if there exists a GCD for a1, ..., an, say d ∈ R, then d is uniquely determined up to association in divisibility: if e is also a GCD of a1, ..., an, then e ∼ d. Moreover, if e ∼ d, then e is a GCD of a1, ..., an.

The same remark applies to the LCM. b) When the domain R in which we work is not clear from the con-

text, we use occasionally a subscript, like in the notation (a1, ..., an)R. For a given domain R and given x, y ∈ R, a GCD(x, y) may not ex-

ist (see the Exercises for some examples). A domain R with the prop-erty that any two elements x, y ∈ R possess a GCD is called a GCD domain. For instance, Z is a GCD domain.

Writing d = (a1, ..., an) means that d is associated with a GCD of a1, ..., an. This can lead to some oddities: in Z, we can write 1 = (1, 2) = −1, but this does not imply that 1 = −1 (of course, it implies that 1 ∼ −1).

c) Note that a1, a2 are coprime if and only if all their common divi-sors are units in R.

d) For any domain R and any a ∈ R, there exists GCD(a, 0) = a. If u is a unit, then there exists GCD(a, u) = u. What can you say about the LCM in these cases?

If d ≠ 0 and d|a, a/d denotes the unique element x ∈ R with a = dx.

1.11 Proposition. Let R be a domain and let a1, ..., an, r ∈ R \0. a) If there exists d = (a1, ..., an), then a1/d,..., an/d have a GCD,

equal to 1. b) If there exists (a1, ..., an) =: d and exists (ra1, ..., ran) =: e, then

e = rd. Thus: (ra1, ..., ran) = r(a1, ..., an).

c) If there exists [a1, ..., an] = m and exists [ra1, ..., ran] =: µ, then µ = rm. Thus:

[ra1, ..., ran] = r[a1, ..., an].


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Proof. a) Let xi ∈ R such that ai = dxi, ni ,1= . If e ∈ R is a com-mon divisor of x1, ..., xn, then de is a common divisor of a1, ..., an, so de | d. This implies e | 1.

b) Since rd | rai, ni ,1= , we get rd | e. Let u ∈ R with e = rdu. It is enough to prove that u | 1. Let xi, yi ∈ R such that ai = dxi and rai = eyi,

ni ,1= . For any ni ,1= , rai = rdxi = rduyi. It follows that u is a com-mon divisor of the elements xi, whose GCD is 1, by a). So, u | 1.

c) Because rm is a common multiple of rai, ni ,1= , we have µ | rm, so rm = µt, for some t ∈ R. We can write m = aibi, µ = raixi, for some xi, bi ∈ R, ni ,1= . We have rm = raibi = µt = raixit, ni ,1= . Simplifying, bi = xit. Also, a1x1 = ... = anxn is a common multiple of a1, ..., an. Thus, m | aixi , ni ,1= . Since µ = raixi, we also get mr | µ. !

1.12 Corollary. Let R be a GCD-domain and let K be the field of quotients of R. Then every element in K can be written as a quotient

ba , with a, b ∈ R, b ≠ 0 and (a, b) = 1.

Proof. If dc ∈ K, with c, d ∈ R, d ≠ 0, then let e = GCD(c, d). Then

c = ea, d = eb, for some a, b ∈ R, with (a, b) = 1 (use the statement a)

above). Moreover, ba

dc

= . !

The next result is ubiquitous in divisibility arguments.

1.13 Corollary. Let R be a GCD domain and a, b, c ∈ R such that a | bc and (a, b) = 1. Then a | c.

Proof. (a, b) = 1 and the preceding result, part b), imply that (ac, bc) = c. Since a | ac and a | bc, the definition of the GCD ensures that a | (ac, bc) = c. !

Although the definitions of the GCD and the LCM are dual to each other, the situation is not entirely symmetric (the existence of the

I.1 Divisibility

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LCM implies the existence of the GCD, but not conversely, in gen-eral).

1.14 Proposition. Let R be a domain and x, y ∈ R. Then: a) If a LCM of x, y exists, [x, y] = m ∈ R, than a GCD of x, y exists,

(x, y), and xy ∼ [x, y](x, y).

b) If any two elements in R have a GCD, then any two elements in R have a LCM.

c) If any two elements in R have a GCD, then, for any n ∈ N, n > 1, any n elements a1, ..., an in R have a GCD and a LCM.

Proof. a) When x = 0, [0, y] exists and it is 0. Similarly, (0, y) = y. Suppose now that x and y are nonzero. The definition of the LCM im-plies m | xy. Let d, a, b ∈ R with xy = md and m = xa, m = yb. We need only prove that d = (x, y). We have xy = xad, so y = ad ⇒ d | y. Like-wise, d | x. Take e ∈ R with e | x, e | y and pick r, s ∈ R such that x = er and y = es. Then ers is a common multiple of x and y, so m | ers. Let t ∈ R such that mt = ers. We have dm = xy = temrse =2 . Simplify-ing by m, d = te, so e | d.

b) Let a, b ∈ R \ 0 and let d = (a, b). There exist x, y ∈ R with a = dx, b = dy. The element m = dxy is obviously a common multiple of a and b. Let µ be another common multiple of a and b. There exist z, t ∈ R such that µ = az = dxz and µ = bt = dyt. So m divides µy = dxyz and µx = dxyt, which means m divides also (µx, µy) = µ(x, y) = µ. This shows that m is a LCM of a and b.

c) Induction on n. (Exercise!). !

1.15 Example. In R = Z [ ]5− , x = 51 −+ and y = 2 have a GCD, but no LCM. Indeed, let d = 5−+ ba (a, b ∈ Z) be a common divisor of x and y. Using the properties of the norm N (see 1.8.c.), we get N(d ) | N(x) = 6 and N(d ) | N(y) = 4 in Z. So, N(d ) | 2 in Z. Since


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N(d ) = 22 5ba + , a case-by-case inspection leads to the conclusion that

a = ±1 and b = 0. Thus, d is invertible. We proved that any common divisor of x and y is a unit, so x and y have GCD equal to 1.

Suppose a LCM µ ∈ R of x and y exists. Then 6 | N(µ) and 4 | N(µ) in Z, so 12 | N(µ) in Z. On the other hand, 6 = 2·3= ( )( )5151 −+−− and ( )512 −+ are common multiples of x and y, so they are common multiples of µ. Thus, N(µ) divides N(6) = 36 and

( ) ( ) 24512 =−+NN in Z, so N(µ) | 12. Combining with 12 | N(µ), we get N(µ) = 12, which is impossible (the equation 125 22 =+ ba has no solutions in Z).

If R is a domain, let R° designate the set of nonzero and non-invert-ible elements in R:

R° := R \ 0\ U(R)

In Z, prime numbers play a central role in divisibility questions. Usually, the (elementary) definition for the notion of prime number is the natural number p > 1 is prime if its only divisors in N are 1 and p. The generalization to the case of a domain of this definition leads to the notion of irreducible element (also compare with the notion of prime element below).

1.16 Definition. Let R be a domain. The element p ∈ R° is called irreducible (in R) if it has no proper

divisors. In other words, any divisor of p is either a unit or is associ-ated to p: ∀d ∈ R, d | p ⇒ d ∼ 1 or d ∼ p.

The element p ∈ R° is called prime (in R) if, for any a, b ∈ R, p | ab ⇒ p | a or p | b.

We emphasize that a prime element or an irreducible element is by definition nonzero and non-invertible.

I.1 Divisibility

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A quick argument shows that, for any m ∈ N*, if p is prime and p divides a product of m factors in R, then p divides one of the factors.

1.17 Proposition. Every prime element is also irreducible. Proof. Let p ∈ R be prime. If d ∈ R is a divisor of p, there exists x

∈ R (nonzero) such that p = dx. So p | dx, which implies p | d (and we are finished) or p | x. But p | x means that p ∼ x (since x | p), so p = ux, with u a unit. So, ux = dx = p and thus u = d is a unit. !

The notions of prime element and irreducible element (which coin-cide for Z, as we will see) are not the same in general.

1.18 Example. In [ ]5−" , 2 is irreducible and it is not prime. In-deed, 2 divides ( )( )1 5 1 5 6− − + − = , but 2 divides neither factor.

On the other hand, if d is a divisor of 2, then N(d ) can only be 1, 2 or

4. An examination of the possible cases shows that d is ±1 or ±2. Thus, the notion of prime element depends heavily on the ring in

which it is considered: 2 is prime in Z, but not in [ ]5−" . The same

remark applies to the notion of irreducible element.

The GCD domains do not have the peculiarity described in the example above:

1.19 Proposition. Let R be GCD domain. Then any irreducible ele-ment in R is prime in R.

Proof. Let p ∈ R, irreducible and x, y ∈ R such that p | xy. If p - x, then the GCD of p and x (which exists!) is 1. Indeed, if d | x and d | p, we cannot have d ∼ p (we would get p | x), so d ∼ 1. Thus, p | xy and (p, x) = 1. Corollary 1.13 guarantees that p | y. !

The notion of divisibility can be translated in the language of ide-als. This approach allows extending classical results on the divisibility


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in Z to much more general classes of rings (for instance, the primary decomposition theory).

Recall that a subset I of the commutative ring R is called an ideal of R if:

a) (I, + ) is a subgroup in the additive group (R, + ): ∀x, y ∈ I ⇒ x + y ∈ I.

b) ∀x ∈ I, ∀r ∈ R ⇒ rx ∈ I. Write I ≤ R if I is an ideal in the ring R. The ideal I is proper if

I ≠ R. For a ∈ R, the ideal generated by a is the set ra | r ∈ R, denoted

by Ra or aR and is called the principal ideal generated by a. The sum of two ideals I and J of R is the ideal:

I + J := i + j | i ∈ I, j ∈ J.

1.20 Proposition. Let R be a domain, n ∈ N* and a, b, x1, ..., xn ∈ R. Then:

a) a | b if and only if Ra ⊇ Rb. b) a ∼ b if and only if Ra = Rb. c) a ∈ U(R) if and only if Ra = R. d) a is prime in R if and only if Ra is a prime ideal. e) a is irreducible in R if and only if Ra is a maximal ideal among

the principal proper ideals of R (more precisely: ∀ x ∈ R such that Ra ⊆ Rx, we have Ra = Rx or Rx = R).

f ) a is a common divisor of x1, ..., xn if and only if Rx1 +...+ Rxn is included in Ra.

g) If Rx1 +... + Rxn = Ra, then a = (x1, ..., xn).2 h) a is a common multiple of x1, ..., xn if and only if Rx1 ∩ ∩ Rxn

includes Ra.

2 The converse is false in general. For a counterexample, see the section

Principal Ideal domains.

I.1 Divisibility

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i) a = [x1, ..., xn] if and only if Rx1 ∩ ∩ Rxn = Ra. Proof. a) a | b ⇔ ∃ c ∈ R with b = ca ⇔ b ∈ Ra ⇔ Rb ⊆ Ra. b) Obvious, by a). c) If a is invertible, then ∃ c ∈ R with ca = 1. So 1 ∈ Ra ⇒ Ra = R.

Conversely, if Ra = R, then 1 ∈ Ra, so there exists c ∈ R such that 1 = ca.

d) Let x, y ∈ R. We have xy ∈ Ra ⇔ a | xy. If a is prime, then a | x or a | y, i. e. x ∈ Ra or y ∈ Ra, which shows that Ra is prime. If Ra is a prime ideal and a | xy, then xy ∈ Ra, so x ∈ Ra or y ∈ Ra ⇔ a | x or a | y.

e) Suppose a is irreducible. If Rx is a proper principal ideal of R with Ra ⊆ Rx, then x | a. Since a has no proper divisors, x is associ-ated with a or it is a unit. But x cannot be a unit because Rx ≠ R. So, x ∼ a, i.e. Rx = Ra. Suppose now Rx is maximal among principal proper ideals, and d ∈ R is a divisor of a. Then Ra ⊆ Rd, so Rd = Ra or Rd = R. This means d ∼ a or d ∼ 1.

f ) If a is a common divisor of x1, ..., xn, then a | nn xrxr ++ ...11 , for any r1, ..., rn ∈ R, so any element in the ideal Rx1 +...+ Rxn is divisible by a. The other implication is left to the reader.

g) From f ), a is a common divisor of x1, , xn. Let d ∈ R be an-other common divisor. Because a ∈ Rx1 + + Rxn, ∃ c1, , cn ∈ R with a = c1x1 + + cnxn. Since d | x1, , d | xn, we obtain d | a.

h), i) are left to the reader. !

An essential role in the arithmetic of Z is played by the theorem of division with remainder:

For any a, b ∈ Z, b ≠ 0, there exist q, r ∈ Z, such that a = bq + r and | r | < | b | or r = 0.

From this theorem, one deduces other two fundamental properties of Z:

Every ideal of Z is principal (of the form nZ, for some n ∈ Z).


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Any nonzero non-invertible integer can be written as a finite prod-uct of prime integers and this writing is unique up to the order of fac-tors and an association of the factors in divisibility. This result is called the fundamental theorem of integer arithmetic or the unique integer factorization theorem.

Abstracting these properties of Z, one obtains the notions of Euclidian domain (a domain in which a property analogous to the theorem of division with remainder holds), principal ideal domain (a domain whose every ideal is principal) and, respectively, unique factorization domain (a domain in which every nonzero non-invertible element can be written as a product of primes).

The following sections are devoted to the study of these classes of rings.

I.2 Euclidian domains

2.1 Definition. A domain R is called an Euclidian domain if there exists a mapping ϕ : R \ 0 → N satisfying:

For any a, b ∈ R, b ≠ 0, there exist q, r ∈ R such that: a = bq + r and (r = 0 or ϕ(r) < ϕ(b)).

(DRT)

We say in this case that R is an Euclidian domain with respect to ϕ.

The property (DRT) is called the Division with remainder theorem in R; q is called traditionally the quotient and r is called the remain-der of the division of a by b.

Of course, the definition above originates in the division with remainder theorems in Z (where ϕ is the absolute value | | : Z \ 0 → N), respectively in K[X], with K a field (where ϕ is the


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degree function deg : K[X] \ 0 → N). These rings are also the most important examples of Euclidian domains.

2.2 Remark. Sometimes, in the definition of an Euclidian domain, the following condition on ϕ is also required:

For any a, b ∈ R \ 0, a | b implies ϕ(a) ≤ ϕ(b). This extra condition is not essential and in fact defines the same

class of rings as the definition above (see 4.9).

2.3 Remark. If b ∈ R, b ≠ 0, and ϕ(b) = 0, then b ∈ U(R). Indeed, the remainder r of the division of 1 to b satisfies 1 = bq + r, for some q ∈ R, and (r = 0 or ϕ(r) < ϕ(b) = 0). Clearly, the natural number ϕ(r) cannot be < 0, so r = 0.

The converse is false in general (give a counterexample!).

The Euclidian domains are GCD rings. This follows from the fact that the Euclidian algorithm can be performed in these rings and guarantees the existence of GCD's:

2.4 Theorem (Euclidian Algorithm). Let R be an Euclidian domain and let a, b ∈ R, with b ≠ 0. Then a GCD of a and b exists and it can be found by the following algorithm:

Algorithm Euclid (R, a, b, d) Input: a, b ∈ R. Output: d = GCD(a, b) ∈ R.

begin if b = 0 then d := a; Stop.

else (Step 1) Find q, r ∈ R with a = bq + r and (r = 0 or ϕ(r) < ϕ(b)).

if r = 0 then d := b; Stop. else a := b, b := r; go to Step 1 end


24

Moreover, there exist (and can be algorithmically determined) u, v ∈ R such that

d = au + bv. Proof. The algorithm3 above implies the following sequence of

divisions with remainder performed in R: (1) a = bq1 + r1 with r1 = 0 or ϕ(r1) < ϕ(b); (2) b = r1q2 + r2 with r2 = 0 or ϕ(r2) < ϕ(r1); (3) r1 = r2q3 + r3 with r3 = 0 or ϕ(r3) < ϕ(r2); ... (n − 2) rn − 4 = rn − 3qn − 2 + rn − 2 with rn − 2 = 0 or ϕ(rn − 2) < ϕ(rn − 2); (n − 1) rn − 3 = rn − 2 qn − 1 + rn − 1 with rn − 1 = 0 or ϕ(rn − 1) < ϕ(rn − 2); (n) rn − 2 = rn − 1 qn + rn with rn = 0.

The existence of the elements qi, ri ∈ R with the properties above is guaranteed by the condition (1) in the definition of the Euclidian do-main. The strictly decreasing sequence of natural numbers ϕ(b) > ϕ(r1) > ϕ(r2) > ... must terminate, so there exists n ∈ N* with rn = 0 (which means that the algorithm terminates in a finite number of steps). We must prove that rn − 1 (the last nonzero remainder) is a GCD of a and b.

The relation (n) shows rn − 1 | rn − 2. The relation (n − 1) implies rn − 1 | rn − 3. Using relations (n − 2), ..., (3), (2), (1), we obtain (by induction) that rn − 1 | b and rn − 1 | a. Now let e ∈ R be a common divi-sor of a and b; then e also divides r1 = a − bq1. Using (2), e divides

3 We hope that the algorithm is clear to the reader. We do not want to present a

rigorous pseudo-programming language or use strict syntax rules from a particular language. Also, this algorithm is intended to serve a theoretical purpose; for instance, finding the elements q, r at step 1 does not imply a description of a concrete procedure (such procedures can be given for particular rings as Z, Q[X], ) and merely uses the fact that these elements exist. Moreover, implementing this algorithm must take into account computer representations of the elements of R, decision algorithms of the equality of two elements in R, addition and multiplication algorithms in R etc. These important issues are not discussed here.


25

r2 = b − r1q2. By induction, e | ri for any i < n, so e | rn − 1. Thus, rn − 1 = (a, b).

In order to obtain d = rn − 1 written as au + bv, we note that r1 = a − bq1 is a linear combination of a and b; replacing r1 in (2) with a − bq1, r2 can be written as au' + bv' and so on. By induction, rn − 1 is of the form au + bv. The following version of Euclid's algorithm (the extended Euclidian algorithm) outputs d, u, v with d = (a, b) and d = au + bv. Note that the variables u and v are such that the last remainder in the computation is always au + bv) :

Input: a, b ∈ R. Output : d = GCD(a, b) ∈ R and u, v ∈ R such that d = au + bv.

begin if b = 0 then d := a; u := 1; v := 0; stop. else u1 := 1; v1 := 0; u := 0; v := 1. Step 1. Find q, r ∈ R with a = bq + r and r = 0 or ϕ(r) < ϕ(b). if r = 0 then d := b; Stop. else a := b; b := r; u1 := u1 − qu; v1 := v1 − qv; t := u; u := u1; u1 := t; t := v; v := v1; v1 := t; go to Step 1

end !

2.5 Examples. a) Z is an Euclidian domain with respect to the absolute value. Here is a proof. Let b ∈ Z, b > 0. We shall prove (by induction) that for every a ∈ N, there exist q, r ∈ N satisfying a = bq + r and r = 0 or r < b. If a < b, q = 0, r = a satisfy these condi-tions. If a ≥ b, suppose that the claim is true for any n < a and we prove it for a. Since a − b < a, by induction exist q, r ∈ N such that a − b = bq + r and r = 0 or r < b. This implies a = b(q + 1) + r and the claim is proven. If a < 0, − a ∈ N, so there exist q, r ∈ N such that − a = bq + r and r = 0 or r < b; so a = b(−q) + (−r), with r = 0 or |−r | < b. We leave the case b < 0 to the reader.


26

Given a, b ∈ Z, the quotient q and remainder r of the division of a by b are not unique: for instance, 3 = 2·1 + 1 = 2·2 + (−1). If we re-quire that the remainder should be positive, then q and r are unique.

b) The ring K[X] of polynomials in one indeterminate X with coefficients in the field K is Euclidian with respect to deg : K[X] \ 0 → N. The proof originates in the procedure of long polynomial division taught in school. Let f = a0 + + an X

n and g = b0 + + bm X

m ∈ K[X], with g ≠ 0 (i.e. bm ≠ 0). We prove by induction on n = deg f. If n < m, set q = 0, r = f. If n ≥ m, h = gabf nm

1−− has degree less than n (the degree n terms cancel out) and, by induction, there exist q, r ∈ K[X], h = gq + r, with deg r < m. Thus, f = ( ) rXabqg mn

nm ++ −−1 . Note that q and r are unique (prove!).

For further reference, we recall:

2.6 Definition. Let n ∈ N be fixed. Two integers a, b ∈ Z are congruent modulo n if n | a − b. We denote this fact by a ≡ b (mod n).

We have a ≡ b (mod n) ⇔ a − b ∈ nZ ⇔ a and b yield the same re-mainder when divided by n. Recall that the relation defined here is an equivalence relation on Z and the set of equivalence classes is struc-tured as a ring (which is in fact the factor ring Z/nZ), called the ring of integers modulo n, often denoted by Zn.

I.3 Euclidian rings of quadratic integers

Besides Z and the polynomial ring in one indeterminate with coefficients in a field, important examples of Euclidian domains are among the rings of the form

Z[ d ] = a + b d | a, b ∈ Z.


27

A somewhat surprising fact is that the ring Z[ ( ) 21 d+ ] = a + b ( ) 21 d+ | a, b ∈ Z has sometimes better arithmetic properties than Z[ d ]. This fact is closely connected to the theory of the rings of quadratic integers, an important topic in (algebraic) number theory. In the following, some elementary facts on these rings are presented. The unproven statements below are pro-posed as exercises (some of them in the next chapters). For a system-atic treatment of the theory of algebraic integers, see for instance LANG [1964].

3.1. Definition. A subfield of C that has dimension 2 viewed as a vector space over Q is called a quadratic number field.

Using elementary tools of field extensions theory, one can readily show that any quadratic number field is of the form Q[ d ] = a + b d | a, b ∈ Q, where d ∈ Z is squarefree.

3.2. Definition. A complex number that is the root of a monic4 polynomial in Z[X] is called integral over Z (or algebraic integer). For instance, 2 is integral over Z, but 1/2 is not. Sometimes, for avoiding confusions, the numbers in Z are called rational integers.5

An element of a quadratic number field that is integral over Z is called a quadratic integer. One can prove that: a quadratic integer is a root of a monic polynomial in Z[X] of degree 2.

Fix d ∈ Z, squarefree. If α = a + b d ∈ Q[ d ], the element a b dα = − is called the conjugate of α. Define the norm

N : [ ]dQ → Q and the trace Tr : [ ]dQ → Q, N(α) := αα = a2 − db2

4 A polynomial is called monic if the coefficient of its monomial of maximum

degree is 1. 5 Because every algebraic integer over Z which is rational (in Q) must be in Z.

Prove!


28

Tr(α) := α +α = 2a for any α = a + b d ∈ Q[ d ] (a, b ∈ Q).

The norm N is multiplicative and the trace Tr is additive: for any α, β ∈ Q[ d ],

N(αβ) = N(α)N(β), Tr(α + β) = Tr(α) + Tr(β).

One can prove that, for any x ∈ [ ]dQ : x is integral over Z (x is a quadratic integer) ⇔ Tr(x) ∈ Z and N(x) ∈ Z.

The quadratic integers in [ ]dQ form a ring, called the ring of integers of [ ]dQ . This ring is traditionally called a ring of quadratic integers (imaginary if d < 0, respectively real if d > 0). We have the following (for the proof, see the Exercises):

3.3 Proposition. The ring of integers of [ ]dQ is Z[θ ], where Z[θ ] = a + bθ | a, b ∈ Z and

θ = ⎩⎨⎧

d if d ≡ 2 or 3 (mod 4)( ) 21 d+ if d ≡ 1 (mod 4). !

In what follows, for a fixed d ∈ Z, squarefree, θ denotes the num-ber above.

We remark also that Q[ d ] = Q[θ ] = a + bθ | a, b ∈ Q.

If d ≡ 1(mod 4), then Z[θ ] = a + b ( ) 21 d+ | a, b ∈ Z can also be described as the set of complex numbers of the form (u + v d )/2, with u, v ∈ Z having the same parity.

According to the above, Z[i], [ ] ( )[ ] ( )[ ]251 ,231,2 ++ ZZZ i are examples of rings of quadratic integers.

The norm N : [ ]dQ → Q has the property that N(α) ∈ Z, ∀α ∈ Z[θ ], as we saw above. We obtain a mapping:

|N| : Z[θ ] → N, |N|(α) = |N(α)|, ∀α ∈ Z[θ ]. A natural problem arises:


29

For which d ∈ Z, squarefree, Z[θ ] is Euclidian with respect to |N|?

We remark first that, since the norm N : [ ]dQ → Q is multiplica-tive (N(xy) = N(x)N(y), ∀x, y ∈ [ ]dQ ), |N| : Z[θ ] → N is also multiplicative.

If d < 0, the representation of the complex numbers in Z[θ ] in the plane yields a grid (rectangular if d ≡ 2 or 3 (mod 4) and oblique if d ≡ 1(mod 4)); moreover, for any x, y ∈ [ ]dQ , |N|(x − y) is the Euclidian distance between the points x and y.

If d > 0, Z[θ ] is a (dense) subset of R and no geometric interpreta-tion is available.

3.4 Lemma. Let d ∈ Z, squarefree. Then R := Z[θ ] is Euclidian with respect to |N| if and only if for any x ∈ [ ]dQ , there exists γ ∈ R such that |N|(x − γ) < 1.

Proof. ⇒ [ ]dQ is the field of quotients of R, i.e.: ∀x ∈ [ ]dQ , ∃α, β ∈ R, β ≠ 0 such that x = α/β. Since R is Euclidian with

respect to |N|, ∃γ, δ ∈ R such that α = βγ + δ, with |N|(δ ) < |N|(β) or δ = 0. So,

x = α/β = γ + δ/β, with |N|(x − γ) = |N|(δ /β) = |N|(δ )/(β) < 1.

d ≡ 2 or 3 (mod 4) d ≡ 1 (mod 4)


30

⇐ Let α, β ∈ R. For x = α/β ∈ [ ]dQ , ∃γ ∈ R with |N|(x − γ) < 1. Let δ = x − γ ∈ [ ]dQ . Thus, α = βγ + βδ , with βδ = α − βγ ∈ R and |N|(βδ ) = |N|(β)·|N|(δ ) < 1. !

If d < 0, because [ ]dQ is dense6 in C, the lemma can be re-phrased as follows:

The ring Z[θ ] is Euclidian with respect to |N| if and only if any point in the complex plane is situated at a distance smaller than 1 to some point in the grid Z[θ ].

3.5 Proposition.7 Let d ∈ Z, d < 0, squarefree. a) If d is congruent with 2 or 3 (mod 4), [ ]dZ is Euclidian with

respect to |N| if and only if d = −1 or d = −2. b) If d ≡ 1 (mod 4), ( )[ ]21 d+Z is Euclidian with respect to |N| if

and only if d ∈ −3, −7, −11.

Proof. a) [ ]dZ is Euclidian with respect to |N| if and only if all points inside a rectangle of the grid are situated at a distance less than 1 from some vertex of the rectangle. The greatest distance to the verti-ces is attained at the intersection of the diagonals, at the distance of

21 d− to any vertex. We have 21 d− < 1 if and only if d > −3, i.e. d = −1 or d = −2.

b) In this case, the grid forms isosceles triangles with base 1 and height 2d− . The points inside the triangle have distance less than 1 to some vertex if and only if the circles of radius 1 centered in base vertices intersect in a point P situated at a distance less than 1 to the third vertex. An easy geometric reasoning shows that the distance from P to the third vertex is ( ) 23−− d . We must have then ( ) 23−− d < 1 ⇔ d > −7 − 34 ⇔ d ∈ −3, −7, −11. !

6 For any z ∈ C and any ε ∈ R, ε > 0, there exists x ∈ [ ]dQ such that |z − x| < ε. 7 This result was obtained in 1923 by L. E. Dickson.

I.4 Principal ideal domains

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We obtain that the following rings are Euclidian with respect to |N|:

Z[i], [ ] ( ) ( ) ( )2 , 1 3 2 , 1 7 2 , 1 11 2i i i i⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦" " " " .

One can prove that these are all the imaginary Euclidian rings of quadratic integers (not necessarily with respect to |N|). The real case d > 0 has no geometric interpretation and is considerably more diffi-cult.


4.1 Definition. A domain R is called a principal ideal domain (PID) if any ideal of R is principal. In other words, for every ideal I of R, there exists a ∈ R such that I = Ra.

Any field is a PID8. The most important examples of PIDs are given by the following result.

4.2 Theorem. Any Euclidian domain is a principal ideal domain. Proof. Let R be Euclidian with respect to ϕ and let I be a nonzero

ideal of R. The set of natural numbers ϕ(x) | x ∈ I, x ≠ 0 contains a minimum element: ∃a ∈ I, a ≠ 0, such that ϕ(a) = minϕ(x) | x ∈ I, x ≠ 0 (a may not be unique). We claim that a is a generator of the ideal I. Obviously, Ra ⊆ I. In order to prove the opposite inclusion, suppose by contradiction that there exists b ∈ I \ Ra. Applying the division with remainder in R, we obtain elements q, r ∈ R such that b = aq + r, r ≠ 0 (since b ∉ Ra) and ϕ(r) < ϕ(a). But a, b ∈ I, so r ∈ I and ϕ(r) < ϕ(a) contradict the choice of a. !

8 Sometimes the fields are not considered principal ideal domains by definition.


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For instance, if K is a field, K[X] is a PID; if I ≠ 0 is an ideal in K[X], a generator of I is a polynomial g ∈ I of minimum degree among the degrees of nonzero polynomials in I.

The principal ideal domains are GCD-domains; for any a, b ∈ R, there exists their GCD, namely any generator of the ideal aR + bR:

4.3 Proposition. Let R be a PID and let a, b ∈ R. Then: a) A GCD d of a and b exists and, for some u, v ∈ R, d = au + bv .

Moreover, the element d ∈ R is a GCD of a and b if and only if dR = aR + bR.

b) The element m ∈ R is a LCM of a and b if and only if mR = aR ∩ bR.

Proof. a) Since R is a PID, ∃d ∈ R such that the ideal aR + bR = ax + by | x, y ∈ R is generated by d. Since d ∈ aR + bR, there exist u, v ∈ R such that d = au + bv. We have a, b ∈ dR, so d | a, d | b. If e ∈ R is such that e | a, e | b, then e | ax + by, ∀x, y ∈ R. In particular, e | d. Thus, any generator d of aR + bR is a GCD of a and b.

Conversely, if d is a GCD of a and b, d divides a and b, so dR ⊇ aR and dR ⊇ bR. Thus, dR ⊇ aR + bR. If e is a generator of aR + bR, this means d | e. But e is also a common divisor of a and b, so d is associated to e. Thus, dR = eR = aR + bR.

b) The proof is similar to the one above and is proposed as an exer-cise. !

Recall that, for a, b ∈ R, the notation (a, b) is used to designate the ideal generated by a and b, aR + bR. The proposition above shows that this notation is consistent to the fact that it designates also the GCD of a and b (which is a generator of the ideal aR + bR).

4.4 Example. Let R be a domain that is not a field and take r ∈ R nonzero and non invertible. Then the ideal (r, X) in R[X] is not princi-


33

pal, so R[X] is not a PID. In particular, the rings Z[X], K[X, Y], with K a field, are not PID's.

Indeed, suppose that for some f ∈ R[X], we have ( f ) = (r, X). Then f | r. We obtain deg f = 0, so f ∈ R. Since f | X, there exists g ∈ R[X] with X = fg, so f is invertible in R. Thus, the GCD of r and X is 1, so f = 1. But the ideal generated by r and X does not contain 1: if h, q ∈ R[X] are such that 1 = hr + qX, setting X = 0 in this equality of polynomials, it follows that 1 = h(0)·r, which means r is invertible, a contradiction.

Using Proposition 1.19 and the fact that any PID is a GCD-domain, we obtain:

4.5 Proposition. In a PID, an element is irreducible if and only if it is a prime element. !

Thus, the domains containing irreducibles that are not primes are not principal ideal domains. such a ring is Z[ 5− ], as Example 1.18 shows.

4.6 Corollary. In a PID R, the prime nonzero ideals are maximal ideals. Any maximal ideal is of the form pR, where p is irreducible in R. Moreover, p ∈ R is irreducible if and only if pR is a maximal ideal.

Proof. It is sufficient to remark that any nonzero prime ideal is principal, generated by a prime p (Prop. 1.20.d)). The element p is irreducible, so (Prop. 1.20.e)) the ideal pR is maximal. The other state-ments are obvious. !

The case R = Z of the following proposition is known as the the fundamental theorem of integer arithmetic. Recall that

R° = x ∈ R | x is nonzero, non-invertible.


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4.7 Theorem. Let R be a PID. Then every nonzero non-invertible element r in R is a finite product of prime elements.9

Proof. Since R is a PID, the primes in R coincide with the irreduci-bles in R. Suppose by contradiction that there exists r0 ∈ R° such that r0 cannot be written as a finite product of irreducibles. In particular, r0 is reducible, so r0 = r1s1, with r1, s1 ∈ R°, not associated with r0. If r1 and s1 are finite products of irreducibles, then r0 is too, false. So, at least one of them (say r1) is not a product of irreducibles. We obtain thus r1∈ R° with r1 | r, r1 ¿ r and r1 is not a product of irreducibles. This reasoning applies to r1, so we get r2 ∈ R°, r2 | r1, r2 ¿ r1, and r2 is not a product of irreducibles. By induction, there exists a sequence (rn)n ≥0 of elements in R, with the property that for any n ∈ N, rn + 1 is a proper divisor of rn. In other words, we obtain a strictly increasing se-quence of ideals r0R ⊂ r1R ⊂ ⊂ rnR ⊂ rn+1R ⊂ . This is impossi-ble in a PID, as the following lemma shows.

4.8 Lemma. Let R be a PID and let (In)n ≥ 0 be a sequence of ideals in R such that In ⊆ In + 1, for any n ∈ N. Then there exists m ∈ N such that Im = Im + i , for any i ∈ N. (In other words, every ascending se-quence of ideals is stationary).

Proof. Let I be the union of all ideals In, n ∈ N. Since the sequence (In)n ≥ 0 is ascending, I is an ideal in R: if x, y ∈ I, then, for some i, j ∈ N, x ∈ Ii, y ∈ Ij. So x, y ∈ It, where t = max(i, j), hence x + y ∈ It ⊆ I. If r ∈ R, rx ∈ Ii ⊆ I. But R is a PID, so there exists a ∈ R such that I = aR. Since a ∈ I, there exists m ∈ N such that a ∈ Im, so aR ⊆ Im ⊆ I = aR. So aR = I = Im + i, ∀i ∈ N. !

A ring R satisfying the ascending chain condition (ACC): every as-cending sequence of ideals of R, I0 ⊆ I1 ⊆ , is stationary (there ex-

9 The products may contain a single factor (i.e., the element itself is a prime).


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ists m ∈ N such that Im = Im + i, ∀i ∈ N) is called a Noetherian ring.10 We have shown that every PID is Noetherian. Note that the proof above can be slightly modified to obtain the theorem: A ring in which every ideal is finitely generated is Noetherian. The converse is also true. Noetherian rings are an important topic in Commutative Algebra.

We have remarked that sometimes in the definition of the Euclidian domain R with respect to ϕ : R \ 0 → N, an extra condition is re-quired: for any a, b ∈ R*, a | b implies ϕ(a) ≤ ϕ(b). We show that this condition is not essential.

4.9 Proposition. Let R be a Euclidian domain with respect to ϕ : R*→ N. Then there exists ψ : R*→ N such that R is Euclidian with respect to ψ and, for any a, b ∈ R*, a | b implies ψ(a) ≤ ψ(b).

Proof. Let ψ : R* → N, defined by ψ(x) = minϕ(y) | y ∼ x, ∀x ∈ R. Clearly, x ∼ y implies ψ(x) = ψ(y). We claim that R is Euclid-ian with respect to ψ. Let a, b ∈ R, b ≠ 0 and let b0 be associated with b such that ψ(b) = ϕ(b0). There exist q, r ∈ R such that a = b0q + r and r = 0 or ϕ(r) < ϕ(b0). Since b0 = bu, for some unit u, we have a = b(uq) + r and r = 0 or ψ(r) ≤ ϕ(r) < ϕ(b0) = ψ(b). Now let a, b ∈ R* with a | b. As seen in the proof of 4.2, a generator of the ideal aR is an element c (which must be associated with a) such that ψ(c) = minψ(x) | x ∈ aR, x ≠ 0. So ψ(a) =ψ(c) ≤ ψ(b), since b ∈ aR. !

Are there any principal ideal domains that are not Euclidian? The answer is affirmative, but the examples are not easy to find. Such an

example is Z ⎥⎦⎤

⎢⎣⎡ +

2191 i , as shown by Dedekind in 1894.

10 Emmy Noether (18821935), German.


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I.5 Unique factorization domains

5.1 Definition. A domain R with the property that every nonzero and non-invertible element is a finite product11 of prime elements is called a unique factorization domain (UFD).

Theorem 4.7 shows that any principal ideal domain (thus also any Euclidian domain) is a UFD. Every field is a UFD, because it has no nonzero and non-invertible elements.

5.2 Proposition. Every irreducible element in a UFD is also a prime.

Proof. Let R be a UFD and let p be irreducible in R. Since p ∈ R°, p is a product of primes. But this product can have only one factor, otherwise p would not be irreducible. So, p is itself a prime. !

The next Proposition justifies and explains the epithet unique in the name of a UFD.

5.3 Proposition. Let R be a domain and r ∈ R°. If r has a prime de-composition, then this decomposition is unique up to an ordering of the factors and an association in divisibility. This means that: if r = p1pn = q1qm are two prime decompositions of r, then m = n and there exists a permutation σ of the set 1, , n such that pi ∼ qσ(i), ∀i ∈ 1, , n.

Proof. Let r = p1 pn be a prime decomposition of r. Call the natural number n the length of the decomposition. Define l(r) as the smallest n ∈ N* such that there exists a prime decomposition of r of length n.

11 Such a product is also called a prime decomposition of the element. Products

may have a single factor.


37

We prove the claim by induction on l(r). If l(r) = 1, then let r = p1 = q1 qm, with p1, q1, , qm prime. Since

r is a prime and r divides the product q1 qm, r divides one of the factors, say q1 (relabel if necessary). But q1 is irreducible, so we have r ∼ q1. Thus r = q1u, with u invertible. We must prove that m = 1. If m ≥ 2, simplifying by q1 in q1u = q1 qm, we obtain q2 qm = u, so q2, , qm are invertible, contradiction.

Suppose the claim is true for any x ∈ R° with l(x) < n. Let r ∈ R, r = p1 pn = q1 qm, with p1, , pn, q1, , qm primes. Because pn is prime, ∃i ∈ 1, , n such that pn | qi. But qi is irreducible, so pn ∼ qi, that is, vpn = qi, with v a unit. Simplifying by pn, we obtain the relation p1 pn−1 = vq1 qi−1qi+1 qm. The induction hypothesis applied to p1 pn−1 shows that n − 1 = m − 1 and p1, , pn − 1 are associated with q1, , qi − 1, qi + 1, , qm, possibly in other order. !

A UFD is also a GCD-domain. This is an important property of the UFD's. Several other characterizations are collected in the next theo-rem.

5.4 Theorem. Let R be a domain. The following statements are equivalent:

a) R is a UFD. b) Every element in R° is a finite product of irreducibles and every

irreducible is a prime. c) Every element in R° has a decomposition in irreducible factors,

unique up to a reordering of the factors and an association in divisibility.

d)Every element in R° has a decomposition in irreducible factors and every two elements have a GCD.

Proof. a)⇒b) It follows from Prop. 5.2. b)⇒c) It follows from Prop. 5.3.


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c)⇒d) Let a, b ∈ R° (if a, b ∈ 0 ∪ U(R), it is trivial to exhibit a GCD of a and b). A GCD of a and b can be determined by taking the common prime factors, at the smallest power. To be precise, let P be a system of representatives of the equivalence classes of the irreduci-ble elements in R (with respect to the association in divisibility). This means that every irreducible in R is associated to exactly one element in P. Property c) assures that there exist and are unique distinct p1, , pn ∈ P, s1, , sn, t1, , tn ∈ N, u, v ∈ U(R) such that

a = upp nsn

s …11 and b = vpp ntn

t …11 . The uniqueness claim follows the uniqueness assertion in c). Let ri

= min(si, ti) and define d = nrn

r pp …11 . We have d | a, d | b. If e | a,

e | b, then every irreducible c ∈ P that divides e divides also a and b. This implies c ∈ p1, , pn. Indeed, otherwise a (or b) would possess two decompositions in irreducible factors, one containing c and the other one not, contradicting the uniqueness property. So e = qpp nw

nw …11 , with w1, , wn ∈ N, q ∈ U(R). From e | a it follows

that wi ≤ si, and e | b implies wi ≤ ti, i = n,1 . So wi ≤ ri and e | d. d)⇒a) Prop. 1.19 makes sure that any irreducible is a prime, since

R is a GCD domain. The implication is now obvious. !

Note that in a UFD any two elements a and b have a LCM. Given two prime decompositions of a and b, their LCM can be determined by taking all prime factors in the decompositions, at the greatest exponent; with the notations in the proof above, define qi = max(si, ti); the element m = nq

nq pp …11 is then a LCM of a and b.

5.5 Example. Z [ ]5− is not a UFD, since 2 is irreducible and not a prime (cf. Ex. 1.18). The reader can also check that 6 has two decompositions as a product of irreducibles in Z [ ]5− :

6 = 2·3 = ( )( )5151 −−−+ , and 2 is not associated with 51 −+ or with 51 −− .


39

5.6 Proposition. Let R be a UFD, n ∈ N* and a, b1, , bn ∈ R. If (a, bi) = 1, for any 1 ≤ i ≤ n, then (a, b1bn) = 1.

Proof. It is enough to show that there is no prime p that simultane-ously divides a and b1bn. If p is such an element, then there exists j, 1 ≤ j ≤ n such that p | bj. Since p | a, we have p | (a, bj) = 1. Therefore, p is invertible, contradiction. !

The UFDs have many characterizations. The following, due to Kaplansky, is inspired from Commutative Algebra techniques.

5.7 Theorem. Let R be a domain. Then R is a UFD if and only if any prime nonzero ideal of R contains a prime element.

Proof. Suppose R is a UFD and P ≠ 0 is a prime ideal. Take a ∈ P, a ≠ 0. Since P ≠ R, a is not a unit, so a decomposes in prime factors: a = p1 pn. Because P is a prime ideal, there exists i such that pi ∈ P.

Suppose now that any nonzero prime ideal in R contains a prime element. Consider

S = a ∈ R | a ∈ U(R) or a is a product of prime elements. If S = R 0, then R is a UFD. Suppose, by contradiction, that

there exists a ∈ R, nonzero, with a ∉ S. We note that S is a multiplica-tive system (1 ∈ S and ∀a, b ∈ S, ab ∈ S). Then there exists a prime ideal P in R containing a, with P ∩ S = ∅. This fact, together with the hypothesis, implies the existence of a prime element p ∈ P. But p ∈ S, contradicting P ∩ S = ∅. So we must have S = R 0.

We have to prove the existence of P. We will prove that any ideal I, maximal with the properties I ∩ S = ∅ and aR ⊆ I, is prime. Let us prove first that such an ideal exists. The idea is to use Zorn's Lemma. Consider the set of ideals (ordered by inclusion):

J = I ideal in R | I ∩ S = ∅ and aR ⊆ I . First, J is not empty, since aR ∈ J. Indeed, if ar ∈ S for some r ∈

R, then ar is a unit or is a product of primes. If ar is a unit, then a is also a unit, contradiction. If ar is a product of primes, we show that


40

a ∈ S, by induction on the number of factors. If ar = p, with p a prime implies p ∼ r (so a is a unit) or p ∼ a (so a ∈ S). If ar = p1 pn with p1, , pn primes, then p1| a or p1| r. If p1 | a, then a = bp1; simplifying, br = p2 pn. The induction hypothesis shows that b ∈ S, so a ∈ S. If p1| r, let r = cp1, for some c, so ac = p2 pn. Again by induction, a ∈ S.

J is inductively ordered: every chain in J, indexed by some set L, (Il)l∈L, is bounded above in J by ∪l∈L Il. The checking is straightfor-ward and is left to the reader.

Zorn's Lemma guarantees now that some maximal element P ∈ J exists.

The ideal P is prime: let x, y ∈ R, with x ∉ P and y ∉ P. If, by ab-surd, xy ∈ P, consider the ideals P + Rx and P + Ry, which strictly in-clude P. Since P is maximal, there exist some elements s ∈ S ∩(P + Rx) and t ∈ S ∩(P + Ry); then st ∈ S ∩ P, contradiction. !

The following important result shows the property of R being a factorization domain is inherited by the polynomial ring R[X]. This has far from trivial consequences: for instance, it is not obvious that every polynomial in n indeterminates with coefficients in Z decom-poses uniquely in irreducible factors.

5.8 Theorem. If R is a UFD, then the polynomial ring in one indeterminate R[X] is a UFD.

The proof of this theorem needs some preparations, which are also interesting on their own. First, we determine the units of R[X].

5.9 Proposition. Let R be a domain. Then U(R[X]) = U(R) and R[X]° = R° ∪ f ∈ R[X] | deg f ≥ 1.

In particular, if K is a field, U(K[X]) = K* and K[X]° = f ∈ K[X] | deg f ≥ 1.


41

Proof. U(R) ⊆ U(R[X]) is evident. If f is a unit R[X], there is some g such that fg = 1. Since R is a domain, deg f + deg g = 0, so deg f = 0 = deg g, which means that f, g ∈ R. From fg = 1 we deduce that f ∈ U(R).

If K is a field, then U(K) = K* and the other claims follow. !

5.10 Definition. Let R be a UFD and let f = a0 + a1X + + anX n ∈

R[X]. The GCD of the coefficients a0, a1, , an is called the content of f, denoted c f . A polynomial with content (associated with) 1 is called a primitive polynomial. Note that f is primitive if and only if there exists no prime p in R such that p divides all the coefficients of f. Any f ∈ R[X] can be written

f = c f ·g, for some primitive g ∈ R[X]. Also, if f = a·g, with a ∈ R and g primitive, then c f = a.

5.11 Proposition. a) Let R be a domain. If p is prime in R, then p is prime in R[X], too.

b) Let R be a UFD and let f, g be primitive in R[X]. Then the prod-uct fg is also primitive.12

c) Let R be a UFD and let f, g ∈ R[X]. Then c f g = c f ·c(g). Proof. a) Note first that: p divides a polynomial in R[X] if and only

if p divides all the coefficients of the polynomial. Let f = a0 + a1X + + anX

n, g = b0 + b1X + + bmX m ∈ R[X] such that p - f and p - g. Let us prove that p - fg. From p - f we deduce that there exists i, 0 ≤ i ≤ n, such that p - ai . Take i minimal with p - ai. Similarly, let j be minimal such that p - bj. Then the coefficient of X i + j in fg is

∑+=+ jilk

lkba

In this sum, aibj is not divisible by p and the other terms are divisi-ble by p (as products of two factors, at least one of which is divisible

12 This is called Gauss' Lemma.


42

by p). So, the coefficient of X i + j is not divisible by p and neither is the polynomial fg .

b) If fg is not primitive, there exists p ∈ R, prime, such that p | fg. The previous paragraph implies p | f or p | g, contradiction.

c) Let f = c f ·f1, g = c(g)·g1, where f1 and g1 are primitive. Then fg = c f c(g)·f1·g1,

with f1g1 primitive by b). It is clear now that c fg = c f c(g). !

5.12 Proposition. Let R be a UFD and let K be its field of quo-tients. Then a polynomial f in R[X] of degree ≥ 1 is irreducible in R[X] if and only if f is primitive and irreducible in K[X].

Proof. Let f be irreducible in R[X]. Then f is clearly primitive. Let us prove that f is irreducible in K[X]. If f = gh, with g, h ∈ K[X], by multiplying with the LCM of the denominators of the coefficients of g and h, we get something like af = g1h1, for some polynomials g1, h1 ∈ R[X] and some a ∈ R. Computing the contents, we have a = c(g1)c(h1), since c f = 1. We have g1 = c(g1)·g2, h1 = c(h1)·h2, with primitive g2, h2 ∈ R[X]. So af = c(g1)·c(h1)·g2·h2; simplifying by a( = c(g1)c(h1)), we obtain f = g2h2. Since f is irreducible in R[X], deg g2 = 0 or deg h2 = 0. But deg g = deg g1 = deg g2 and similarly for h, so deg g = 0 or deg h = 0.

Conversely, if f is irreducible in K[X], it has no proper divisors (of degree ≥ 1) in K[X]; so it does not have divisors of degree ≥ 1 in R[X]. Since f is primitive, f has no non-invertible factors of degree 0. !

This result is also important from a practical point of view: in order to prove that some polynomial with coefficients in R is irreducible in K[X], it is sufficient to prove it is irreducible in R[X], which is often easier to accomplish.

We can give now the Proof of the Theorem 5.8. We will use the characterization in 5.4.b). Le R be a UFD, K its field of quotients and f ∈ R[X], irreducible. Let us prove that f is a prime. If f | gh, with g, h ∈

I.6 Polynomial ring arithmetic

43

R[X], because f is irreducible in K[X] (and thus also prime in K[X]), we have f | g or f | h in K[X]. Say f | g in K[X]; so there exist a ∈ R, q ∈ R[X], such that ag = fq. This implies a·c(g) = c(q). We can write q = c(q)·q1, g = c(g)·g1, with q1, g1 primitive in R[X]. Thus ac(g)·g1 = f·c(q)·q1; simplifying by c(q), we have g1 = fq1, so f | g in R[X].

Next, we must show that any nonzero non-invertible polynomial f ∈ R[X] is a product of irreducibles. We prove this by induction on the degree of f. If deg f = 0 and f is not a unit in R[X], then f ∈ R° and it has a decomposition in irreducible factors in R, factors that are also irreducible in R[X]. If deg f > 0, write f = c f f1, with f1 primitive. It is sufficient to prove the existence of a decomposition for f1. If f1 is irreducible, we are finished; if not, f1 has a proper divisor in R[X], which must be a polynomial of degree strictly less than deg f ( f1 has no proper divisors in R, being primitive). Thus, f1 = gh, with g, h ∈ R[X], having degrees smaller than f. Applying the induction for g and h, we infer that f1 is a product of irreducibles in R[X]. !

Thus, the following rings are unique factorization domains:

Z[X], Z[X1, , Xn], K[X1, , Xn], where K is a field.

An analogous result holds for Noetherian rings: if R is a commuta-tive Noetherian ring, then R[X] is Noetherian. (David Hilbert's Basis-satz).


In this section, R designates a domain and K its field of quotients. Recall that R° denotes the set of nonzero and non-invertible elements in R.


44

The problem of deciding the irreducibility of a polynomial in R[X] is important and often nontrivial. A rich collection of irreducibility criteria is thus very useful in such problems.

6.1 Remark. Given f ∈ R[X], the following simple facts are worth remembering:

- if deg f = 0, then f ∈ R. In this case, f is irreducible in R[X] if and only if it is irreducible in R. If R = K (R is a field), then f is invertible and thus reducible.

- if deg f = n > 0, then f is irreducible in R[X] if and only if f has no non invertible divisors of degree 0 and there are no decompositions f = gh, with g, h ∈ R[X] and 1 ≤ deg g, deg h < n.

The fact that f has no non-invertible divisors of degree 0 amounts to saying that the GCD of the coefficients of f exists and is 1. In practice, when R is a UFD, this condition reads f is primitive. Recall that, in this case, f is irreducible in R[X] ⇔ f is primitive and f is irreducible in K[X].

If R is a domain and not a field, R[X] is not principal (and certainly not Euclidian), so the theorem of division with remainder does not hold in R[X]. However, if f, g ∈ R[X], and g has as leading coefficient a unit, the argument of the proof of the division with remainder theo-rem for K[X] (K a field) still holds (see Example2.5.b)). The proof of the following result is left to the reader:

6.2 Proposition. (Integer division theorem) Let f, g ∈ R[X]. If the leading coefficient of g is a unit in R, then there exist q, r ∈ R[X] such that f = gq + r, with r = 0 or deg r < deg f . !

An important consequence is the Theorem of Bézout:

6.3 Theorem. Let f ∈ R[X] and a ∈ R. The following statements are equivalent:


45

a) a is a root of f. b) the polynomial X − a divides f in R[X]. Proof. There exist q, r ∈ R[X] such that f = (X − a)q + r, where

deg r = 0 or r = 0. Note that X − a divides f if and only if r = 0. But f (a) = (a − a)q(a) + r(a) = r, so f (a) = 0 is equivalent to r = 0. !

This theorem is the basis of the notion of multiple root:

6.4 Definition. If f ∈ R[X] and a ∈ R, a is called a multiple root of multiplicity m for f if (X − a) m | f and (X − a) m + 1- f ; the natural num-ber m is called the multiplicity of the root a. A root of multiplicity 1 is called a simple root.

6.5 Corollary. Let f ∈ R[X], deg f > 1. If f has a root a ∈ R, then f is reducible in R[X] (it is divisible with X − a). !

The converse of this statement is false: (X 2 + 1)2 has no roots in Q, yet it is obviously reducible in Q[X]. Nevertheless, we have:

6.6 Proposition. Let K be a field. Then a polynomial f of degree 2 or 3 in K[X] is irreducible if and only if f has no roots in K. In particu-lar, if R is a UFD, a primitive polynomial of degree 2 or 3 in R[X] is irreducible in R[X] if and only if it has no roots in K.

Proof. Since f is irreducible in K[X] and deg f > 1, f has no roots in K. Conversely, if f is reducible and has degree 2 or 3, then, by looking at the degrees of the factors in a decomposition of f, one concludes that f has a divisor of degree 1, which has a root in K. The remaining claims follow from f is irreducible in R[X] if and only if f is primitive and irreducible in K[X]. !

If the ring R is not a UFD, then the criterion above may not work:

6.7 Examples. a) f = (2X + 1)2 is reducible in Z[X], but has no roots in Z. Of course, f has roots in Q, the field of quotients of Z.


46

b) Let R = a + 2bi | a, b ∈ Z. A quick check shows that R is a subring in Z[i], so it is a domain. The polynomial X 2 + 1 is irreducible in R[X] (prove!), but has the roots i, −i in Q[i], the field of quotients of R. This means that a polynomial of degree 2 or 3 in R[X] that has roots in Q(R) is not necessarily reducible in R[X].

The following criterion is widely used to find all rational roots of a polynomial in Z[X] (also see Exercise 26).

6.8 Proposition. Let R be a UFD and f = a0 + a1X + + anX n ∈ R[X]. If p/q ∈ K is a root of f, with p, q ∈ R, ( p, q) = 1, then p | a0 and q | an.

Proof. We remark first that every element of K can be written as p/q, with p and q coprime. Writing that f(p/q) = 0 and multiplying

with q n, we have: −a0q

n = a1 pq n−1 + + an p n,

so p | a0q n. Since ( p, q) = 1, we also have ( p, q n) = 1 (R is a UFD), so

p | a0. A similar proof can be given for q | an. !

6.9 Example. Let f = X 3 − X + 2 ∈ Z[X]. If p/q ∈ Q is a root of f, (p, q) = 1, then p | 2 and q | 1. So, the rational roots of f belong to the set 1, −1, 2, −2. By direct testing, we get that none of these numbers is a root. So, f has no rational roots. Since f has degree 3, f is irreduci-ble in Q[X] (also in Z[X], being primitive).

Here are some general tricks that may prove useful in irreducibility problems.

6.10 Proposition. Let f = a0 + a1X + + anX n ∈ R[X], f ≠ 0. a) Let c, d ∈ R, where c is a unit in R. Then f is irreducible if and

only if f (cX + d) is irreducible. b) Suppose f(0) = a0 ≠ 0. Then f is irreducible if and only if

r f = an + an − 1X + + a0X n


47

(the reciprocal of f ) is irreducible. c) Suppose f has no divisors of degree 0 other than units. If S is a

commutative ring and ϕ : R → S is a unitary ring homomorphism such that ϕ(an) ≠ 0 and ϕ(a0) + ϕ(a1)X + + ϕ(an)X n is an irreducible polynomial in S[X], then f is irreducible in R[X].

d) (Eisenstein criterion) Let R be a UFD. If there exists a prime ele-ment p ∈ R such that p | ai, ∀ i < n, p - an , p

2-a0, then f is irreducible in K[X] (thus f is irreducible in R[X] if it is primitive).

Proof. a) Let ϕ : R[X] → R[X] be the unique homomorphism of R-algebras (that is, ψ is a ring homomorphism and ψ |R = ϕ) satisfying ϕ(X) = cX + d. In other words, ϕ f is obtained by replacing the indeterminate X in f with cX + d. The element c is a unit if and only if ϕ is an isomorphism of R-algebras (the homomorphism of R-algebras ψ : R[X] → R[X] that takes X to c −1X − c −1d is the inverse of ϕ). Therefore, f = gh ⇔ ϕ f = ϕ(g)ϕ(h), ∀g, h ∈ R[X]. Since ϕ preserves the degrees and ϕ | R = idR, one obtains that: f is irreducible if and only if ϕ f is irreducible.

b) If g, h ∈ R[X], with nonzero g(0) and h(0), then r(gh) = r(g)r(h).

Indeed, note that ( ) ⎟⎠⎞

⎜⎝⎛ 1

=X

fXfr n , where n = deg f (for a rigorous

argument, consider the equality in K(X), the field of quotients of K[X]). So, if deg g = m, deg h = p, we have

( ) ( ) ( ) ( )hrgrX

hXX

gXX

ghXghr pmpm =⎟⎠⎞

⎜⎝⎛ 1

⎟⎠⎞

⎜⎝⎛ 1

=⎟⎠⎞

⎜⎝⎛ 1

= + .

Because r preserves the degrees and, for any d ∈ R, we have d | f ⇔ r(d) | r f , we get the conclusion.

c) Let ψ : R[X] → S[X] be the unique homomorphism of R-algebras such that ψ(X) = X. We must prove that ψ f is irreducible implies f is irreducible. Suppose f = gh, with g, h ∈ R[X]. Thenψ f = ψ(g)ψ(h) ; the condition ϕ(an) ≠ 0 ensures that degψ(g) + degψ(h) = degψ f = n. Since degψ(q) ≤ deg q, ∀q ∈ R[X], we obtain that degψ(g) = deg g and


48

degψ(h) = deg h. Butψ f is irreducible, so ψ(g) (to make a choice) is a unit, of degree 0. So, 0 = degψ(g) = deg g. Since f has no 0 degree non-invertible divisors, g ∈ U(R).

d) Write f = c f ·f1, with f1 primitive. We have that f and f1 are associated in K[X]. By replacing f with f1, we can assume f is primi-tive. It is sufficient now to prove that f is irreducible in R[X]. If f were reducible, then: f = a0 + a1X ++ anX

n = (b0 + b1X ++ bmX m) (c0 + c1X ++ cpX p),

where m > 0, p > 0, b0, b1, , bm, c0, c1, , cp ∈ R, bm ≠ 0, cp ≠ 0. We have b0c0 = a0, so p | b0c0 and p2 - b0c0 ; thus p divides exactly one of b0 and c0. Suppose p | b0 and p - c0. Because p - an, p does not divide all the bi's; thus there exists some i, 1 ≤ i ≤ m, such that p - bi and p | bj, ∀j < i. Then p - bic0 and

ai = bic0 + ∑1−

1=−

i

jjijcb

is not divisible by p, contradicting the hypothesis. !

6.11 Remark. If f ∈ R[X] is a monic reducible polynomial, there exists a decomposition of f of the form f = gh, where g, h ∈ R[X] are monic, of degrees > 1. The proof is proposed as an exercise. This sim-ple remark is useful in reducibility issues.

A few instances of using the above criteria on, concrete cases will give an idea on the strategies of approaching the problem of irreduci-bility of a polynomial. The exist algorithms that decide if a given polynomial in Z[X] is irreducible. Such an algorithm (due to Kronecker) that also outputs a factor of the polynomial if it is not irre-ducible is described in Exercise 31. This algorithm, applied repeat-edly, yields a factorization algorithm (producing a decomposition in irreducible factors) for any polynomial in Z[X] or Q[X]. The modern symbolic computation software (Maple, Mathematica, Macaulay, Axiom, etc.) have powerful routines that decide polynomial irreduci-


49

bility, including polynomials in several indeterminates, polynomials with coefficients in algebraic extensions of Q or in a finite field. One can prove that, if there exists a factorization algorithm for K[X], with K a field, then there exists one or L[X], where L is a finitely generated extension of K. For details and developments, see SPINDLER [1994], WINKLER[1996].

6.12 Examples. a) The polynomial 6X 9 + 13X 2 + 26 is irreducible in Q[X] (and in Z[X], it is primitive), by the Eisenstein criterion with p = 13.

b) For any prime p and any n ∈ N*, X n − p is irreducible in Q[X] and in Z[X] (use Eisenstein again).

c) Let p be a prime number and let f = X p−1 + X p−2 + + X + 1 ∈ Z[X].

The Eisenstein criterion cannot be applied directly to f. Consider the polynomial

g = ( ) ( ) ( ) 1

1

1 11

1 1

p pi

i

pi

Xf X X

X−

=

+ −+ = =

+ − ∑

Remark that the Eisenstein criterion can be applied to g, since p di-vides all the binomial coefficients ( )p

i , with 1 ≤ i < p. Thus, g is

irreducible and, by a) above, f is irreducible. d) The polynomial f = Y 9 + X 9Y 7 − 3X 2Y + 2X is irreducible in

Z[X, Y]. Indeed, consider f as a polynomial in Y with coefficients in Z[X], a UFD. Apply Eisenstein with p = X (X is irreducible in Z[X]). Note that Z can be replaced with any UFD of characteristic ≠ 2.

e) Consider f = X 5 + X 2 + 1 ∈ Z2[X]. The polynomial f has no roots in Z2 (easy check: Z2 has 2 elements, 0 and 1), so the proper divisors of f can be of degrees 2 or 3. A decomposition of f may look only like:

X 5 + X 2 + 1 = (X 3 + aX 2 + bX + 1)(X 2 + cX + 1),


50

with a, b, c ∈ Z2. Identifying the coefficients, we obtain a system of equations in a, b, c, which is readily seen to have no solutions in Z2. Hence, f is irreducible in Z2[X].

f) A typical use of 6.10.c) for a polynomial f with integer coeffi-cients is to reduce the coefficients modulo n. More precisely, for some n ∈ N conveniently chosen, consider the unique ring homomorphism ϕ : Z → Zn and investigate the polynomial f reduced modulo n, (denoted with ψ f in the Proof). Take, for instance, f = 7X 5 + 4X 3 − X 2 + 6 X + 9 ∈ Z[X]. The polynomial f reduced modulo 2 is X 5 + X 2 + 1 ∈ Z2[X], which is irreducible. The conditions in 6.10.c) are satisfied, so f is irreducible in Z[X] (also in Q[X]).

g) 10X 7 + 5X 2 + 2 is irreducible in Z[X]: its reciprocal is 2X 7 + 5X 5 + 10, irreducible by Eisenstein with p = 5.

Exercises

Throughout the exercises, R is a domain and K is its field of quo-tients (unless specified otherwise). 1. Let R be a commutative unitary ring that has zero divisors and let a ∈ R be a zero divisor. If b ∈ R \ 0 is not a zero divisor, prove that ax + b = 0 has no solutions in any ring S that includes R as a subring. 2. Let R be a finite domain. Prove that R is a field. 3. Let R be an infinite unitary ring. Prove that the set R° of the non-zero non-invertible elements in R is infinite. (Hint: If R° is finite, then U(R) is infinite. Let S(R°) be the set of all bijections from R° to R°. The mapping ϕ : U(R) → S(R°), x & ϕx, ϕx(a) = xa, ∀a ∈ R°, is injec-tive, contradiction.)

Exercises

51

4. Let R be a GCD domain. Then any element in K can be written as a/b, (b ≠ 0), with a, b ∈ R, coprime. What can you say about the uniqueness of such a representation? 5. Show that a commutative ring R is a domain if and only if R[X] is a domain. 6. Let p ∈ R°. Prove that the ideal generated by p in R[X], pR[X], is prime if and only if p is a prime element in R. (Hint: R[X]/(pR[X]) ≅ (R/pR)[X]). Deduce a new proof for 5.11.a). 7. Let d ∈ Z, d < 0 be squarefree. Determine U(Z[ d ]). 8. The elements 6 and 2 + 5− have no GCD (and thus no LCM) in Z[ 5− ]. 9. Show that Z[ 2 ] is Euclidian. (Hint. Use 3.4.) 10. Let θ = ( ) 251+ . Show that the norm N : Z[θ ] → Z is N(a + bθ) = a2 + ab − b2, ∀a, b ∈ Z. Prove that ( )[ ]251+Z is Euclidian. (Hint. Use 3.4 and Q[ 5 ] = Q[θ ]). 11. Let θ = ( ) 231 i+ . Write a formula for the norm N : Z[θ ] → Z and determine U(Z[θ ]). 12. Let d ∈ Z be squarefree.

a) Any element in [ ]dQ can be uniquely written as a + b d , with a, b ∈ Q.

b) Suppose known that any quadratic integer is a root of a monic polynomial of degree 2 with integer coefficients. Show that: x = a + b d ∈ [ ]dQ (a, b ∈ Q) is a quadratic integer ⇔ Tr(x) = 2a ∈ Z and N(x) = a 2 − db 2 ∈ Z.

c) Show that x = a + b d (with a, b ∈ Q) is a quadratic integer ⇔ 2a ∈ Z, 2b ∈ Z and 4a 2 − 4b 2d ≡ 0 (mod 4).

d) Show that R := α ∈ [ ]dQ | α is a quadratic integer is a subring of [ ]dQ and R = Z[θ ], where θ is given by Prop. 3.3.

e) For d < 0, determine explicitly U(R).


52

13. Let d ∈ Z be squarefree and α, β ∈ Z[ d ] such that αβ ∈ Z. Show that there is some γ ∈ Z[ d ] and a, b ∈ Z such that α = aγ and β = bγ

⎯ (γ

⎯ is the conjugate of γ).

14. The purpose of the exercise is to determine all primes in Z[i]. As a bonus, we find the primes in Z that can be written as a sum of two squares.

a) Prove: for any d ∈ Z and any a + bi ∈ Z[i], d | a + bi in Z[i] ⇔ d | a and d | b in Z.

b) Suppose that p ∈ Z is a prime in Z and in Z[i]. Show that the equation x2 + 1 = 0 has no solutions in Zp ( = Z/pZ, the field of inte-gers mod p).

c) Let p be prime in Z. Then: (∃) a, b ∈ Z such that p = a2 + b2 (p can be written as a sum of two squares) ⇔ the equation x2 + 1 = 0 has solutions in Zp.

d) If p is a prime in Z and p ≡ 3(mod 4), then p cannot be written as a sum of two squares.

e) If p is a prime in Z and p ≡ 1(mod 4), then ( )pp mod1!2

1−≡⎟

⎠⎞⎜

⎝⎛ −

(Hint. Use Wilson's Theorem: (p − 1)! ≡ −1(mod p)). f) Prove that a prime p ∈ Z is also prime in Z[i] if and only if

p ≡ 3(mod 4). g) Prove that all the prime elements in Z[i] (up to association in

divisibility) are: 1 + i, 1 − i and the primes p in Z with p ≡ 3(mod 4). h) Prove that a prime p ∈ Z can be written as a sum of two squares

⇔ p ≡ 1(mod 4).

15. Let R be a unitary subring of the commutative ring S. An element of S is called integral over R if it is a root of a nonzero monic polyno-mial in R[X]. Prove that, if R is a GCD-domain, K is its field of quo-tients and x ∈ K is integral over R, then x ∈ R. (A domain R with this property is called integrally closed).

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53

16. Let R be a PID and let S be multiplicatively closed system in R. Then the ring of quotients S −1R is a PID. 17. Let R be an Euclidian domain with respect to the function ϕ and let S be a multiplicatively closed system in R. Then S −1R is an Euclid-ian domain. (Hint. Take S saturated. Use exercise 4.) 18. Let R be a UFD and let S be a multiplicatively closed system in R. Then S −1R is a UFD. 19. Does the property of a domain R of being Euclidian (respectively a PID, a UFD) is inherited by the unitary subrings of R? 20. Let R be Euclidian with respect to ϕ. Show that there exists u ∈ R nonzero and not a unit with the property: ∀x ∈ R, ∃q ∈ R such that x − qu is a unit or 0. Find such a u for R = Z, K[X]. (Ind. min ϕ(v) | v ∈ R° = ϕ(u) for some u ∈ R°.) 21. Let d ≤ −13, d squarefree and denote by R the ring of integers of

[ ]dQ (so, R = Z[θ ], with θ as in Prop. 3.3). Show that U(R) = −1, 1. Show that R is not Euclidian. Is this result a particular case of Proposition 3.5? (Ind. If R is Euclidian, let u ∈ R given by the preced-ing exercise. Then, ∀x ∈ R, we have u | x or u | x ± 1. Take x = 2 and deduce that u ∈ Z and u = ±2 or ±3. Then find y ∈ R such that u does not divide any of y or y ± 1). 22. Let d ∈ Z be squarefree, d ≡ 1 (mod 4). Then Z[ d ] is not a GCD-domain. (Hint. 2 is irreducible and not a prime.) 23. Show that Z contains an infinity of prime elements not associated in divisibility to each other. 24. Let R be a UFD that is not a field, such that the group of units U(R) is finite. Then R contains an infinity of prime elements not associated in divisibility to each other. (Hint. If p1, , pn are all the primes up to association in divisibility , then there exists m ≥ 1 such that 1 + (p1pn)

m ∈ R°.)


54

25. Let R be a UFD and let p ∈ R be a prime. Using the canonical homomorphism π : R → R/pR and its extension to a homomorphism ψ : R[X] → (R/pR)[X], give a new proof for the Eisenstein criterion. (Hint. If f = a0 + a1 X + + an X

n satisfies the hypothesis of the crite-rion and f = gh, thenψ f = π(an)X

n = ψ(g)ψ(h). If deg g, deg h ≥ 1, then g(0) and h(0) are divisible by p.) 26. Let R be a UFD and f = a0 + a1 X + + an X

n ∈ R[X]. a) If p/q ∈ K is a root of f, where p, q ∈ R and ( p, q) = 1, then

p | a0, q | an and (p − qr) | f (r), ∀r ∈ R. Write down explicitly the conclusions for an = 1.

b) Let g = nnn

nn

nn XXaXaaaa ++++ −

−−− 1

112

01 … . Then ( )1n

na f X− = g(anX). What connection is between the roots of g and the roots of f?

c) Find the rational roots of 2X 3 + 5X 2 + 9X − 15 and 4X 3 − 7X 2 − 7X + 15. 27. Let K be a field. Prove that any nonzero polynomial f ∈ K[X] has at most deg f roots in K (every root is counted with its multiplicity). 28. Let R be a commutative ring. Prove that the following statements are equivalent:

a) Any nonzero polynomial f ∈ R[X] has at most deg f roots in R. b) Any polynomial of degree 1 has at most one root in R. c) R is a domain. (Hint. Consider the field of quotients of R and use the previous

problem). 29. Let R be a commutative ring. If f ∈ R[X], define the polynomial function f~ : R → R, defined by: ∀x ∈ R, f~ (x) = f (x) (the value of f in x). Prove that, if R is an infinite domain, then the mapping ϕ : R[X] → RR, ϕ f = f~ , ∀f ∈ R[X], is injective. Is the conclusion still valid if one does not assume that R is infinite? 30. (The Lagrange interpolation polynomial) Let K be a field, n ≥ 1 an integer, fix n + 1 distinct elements x0, , xn ∈ K and (not necessarily

Exercises

55

distinct) y0, , yn ∈ K. Prove that there exists a unique polynomial L ∈ K[X] satisfying: deg L ≤ n and L(xi) = yi, 0 ≤ i ≤ n. 31. Let p ∈ Z[X], primitive, deg p = n and let m = the largest integer ≤ n/2.

a) Show that p is reducible in Z[X] ⇔ p has a divisor of degree be-tween 1 and m.

b) Choose m + 1 distinct integers, (x0, , xm) ∈ Z m + 1. Show that the following algorithm terminates in a finite number of steps and out-puts a nontrivial factor of p of degree ≤ m or proves that p is irreduci-ble:

1. If ∃i with p(xi) = 0, then X − xi is a factor of p and STOP. If not, go to 2.

2. Let D = d = (d0, , dm) ∈ Z m + 1 | di| p(xi), ∀i. D is a finite set. For any d ∈ D, let Ld ∈ Q[X] be the Lagrange interpolation polynomial with Ld(xi) = di, ∀i, and deg Ld ≤ m. If there exists d ∈ D with Ld ∈ Z[X] and Ld | p, then Ld is a factor of p and STOP. If not, then p is irreducible.

c) Deduce an algorithm of deciding the irreducibility of polynomi-als in Q[X].

d) Suppose m = 2. Propose a choice for (x0, , xm). e) Suppose there is available a factorization algorithm for R (an

algorithm that produces a decomposition of any element in R° in prime factors). What properties should R have in order to adapt the algorithm above to R[X]?

f) Suppose R is a UFD and there exists a factorization algorithm for R[X]. Then there exists a factorization algorithm for K[X]. 32. Decide the irreducibility of X 4 + X 2 + 2X − 1 ∈ Z[X]. 33. Show that a0 + a1 X + + an X

n ∈ Z2[X] (an ≠ 0) has no roots in Z2 if and only if a0(a0 + a1 + + an) ≠ 0.


56

34. Using the equality (in Z2[X]): X 5 + X + 1 = ( X

3 + X 2+ 1)( X 2 + X + 1),

prove that X 5 − X − 1 ∈ Q[X] is irreducible. 35. a) Let f ∈ R[X], deg f = m. If f has at least m + 1 roots in R, then f = 0.

b) Let g ∈ R[X1,, Xn], with R infinite. If g(a1, , an) = 0, ∀(a1, , an) ∈ R n, then g = 0. Deduce that two polynomials in R[X1,, Xn] are equal if and only if the associated polynomial func-tions are equal.

c) Give an example of a finite field K and distinct polynomials in K[X] that have the same associated polynomial function.

d) (Generalization of b) Let R be infinite and let g ∈ R[X1,, Xn], with deg(g, Xi) = mi. Suppose that (∃) S ⊆ R with |S| > m1; (∀) a1 ∈ S, (∃) S(a1) ⊆ R with |S(a1)| > m2; (∀) a1 ∈ S, (∀) a2 ∈ S(a1), (∃) S(a1, a2) ⊆ R with |S(a1, a2)| > m3 and so on. If g(a1, , an) = 0, ∀a1 ∈ S, ∀a2 ∈ S(a1), ∀a3 ∈ S(a1, a2), , ∀an ∈ S(a1, , an), then g = 0. 36. Let R be a domain. A polynomial in n indeterminates p ∈ R[X1,, Xn] =: R[ X ] is called homogeneous of degree q if all the monomials in p have total degree q (see the Appendix). Show that:

a) Any p ∈ R[ X ] can be written uniquely as: p = p0 + p1 + + pm, with pi ∈ R[ X ], homogeneous of degree i. b) The product of two homogeneous polynomials of degrees a,

respectively b, is homogeneous of degree a + b. c) p is homogeneous of degree q ⇔ p(TX1,, TXn) = T qp(X1,, Xn)

(equality in R[X1,, Xn, T] = R[X1,, Xn][ T ]). d) Let p ∈ R[ X ] be homogeneous. Then any divisor of p in R[ X ]

is homogeneous. (Hint. For p = gh, write g = ga + + gm, where gi is homogeneous of degree i, 0 ≤ a ≤ m and ga ≠ 0 ≠ gm. It suffices to prove that a = m.)

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57

e) If R is infinite and p ∈ R[ X ], then p is homogeneous of degree q ⇔ p(tx1,, txn) = t qp(x1,, xn), ∀t, x1,, xn ∈ R.

f) Is it true that any symmetric polynomial in R[ X ] has all its divi-sors symmetric polynomials in R[ X ]? 37. Let K be a field of characteristic not equal to 2 (1 + 1 ≠ 0 in K) and p a homogeneous polynomial of degree 2 in K[X, Y], i.e. p = aX 2 + bXY + cY 2, with a, b, c ∈ K. Prove that p is reducible in K[X, Y] ⇔ b 2 − 4ac is a square in K ⇔ b 2 − 4ac = 0 or there exist α, β ∈ K, (α, β) ≠ (0, 0), with p(α, β) = 0. 38. Assume K is a field and p = a0Y

n + a1 Y n − 1X + + an X

n is a homogeneous polynomial of degree n in K[X, Y]. Let p(X, 1) = a0 + a1 X + + an X

n ∈ K[X]. Prove that, for any g ∈ K[X, Y], g | p in K[X, Y] if and only if g is homogeneous and g(X, 1) | p(X, 1) in K[X]. 39. Let K be a field and let p ∈ K[X, Y] be homogeneous. Prove that p is irreducible in K[X, Y] ⇔ p(X, 1) is irreducible in K[X]. 40. Write a decomposition in irreducible factors for X1

3 + X23 ∈ K[X1, X2]. (Hint. The cases char K = 3 and char K ≠ 3

should be treated separately). 41. Let K be a field, char K ≠ 3 and f = X1

3 + + Xn3 ∈ K[X1,, Xn].

Show that f is irreducible if and only if n ≥ 3. Generalization. (Hint. For n = 3, apply Eisenstein to f ∈ K[X1, X2][X3]. Use then an induction on n.) 42. Consider n 2 indeterminates Xij, 1 ≤ i, j ≤ n, and consider the n×n matrix A = (Xij)1 ≤ i, j ≤ n ∈ Mn(Z[Xij;1 ≤ i, j ≤ n]). Then the polynomial:

det A = ∑X1σ(1) Xnσ(n) | σ ∈ Sn is irreducible in Z[Xij;1 ≤ i, j ≤ n]. 43. Prove that a commutative ring R is Noetherian if and only if every ideal in R is finitely generated.

58

II. Modules

Module theory can be seen as a generalization of the classic linear algebra (which studies vector spaces over an arbitrary field1). The the-ory is fundamental in many areas of mathematics: commutative alge-bra, algebraic number theory, group representation theory, algebra structure theorems, homological algebra etc. Also, module theory illustrates and uses concepts of category theory (we use some elemen-tary notions from category theory in this chapter, notions that can be found in the Appendix). Module theory language and results are indispensable throughout most of modern algebra.

II.1 Modules, submodules, homomorphisms

The notion of a module over a ring can be obtained by replacing the word field with the word ring in the definition of the vector space.

1 Although the volume Algèbre linéaire (1961) of the famous Bourbaki series

Eléments de Mathématique begins with the definition of the module.


59

1.1 Definition. Let R be a ring with identity (not necessarily commutative) and (M, +) an Abelian group. We say that M is a left R-module (or left module over R) if there exists an external operation on M with operators in R2, i.e. a function

µ : R × M → M (notation: µ (r, x) =: rx, ∀r ∈ R, ∀x ∈ M), satisfying, for any r, s ∈ R and x, y ∈ M:

i) r(x + y) = rx + ry; ii) (r + s)x = rx + sx; iii) (rs)x = r(sx); iv) 1x = x,

1.2 Remark. The addition in R is denoted by + , as the addition in M. Also, the zero element in R is denoted by 0, like the zero element in M. For instance, in axiom ii), the + in the LHS denotes the addition in R, whereas the + in the RHS denotes the addition in R. This nota-tional abuse (which is widely used) should not confuse the reader.

If the axiom iii) is replaced by: iii') (rs)x = s(rx), ∀r, s ∈ R, ∀x ∈ M,

we say that M is a right R-module. The usual notation for the scalar multiplication in the case of right R-modules is with the scalars on the right, i.e. the scalar multiplication is a function µ : M × R → M, with the notation µ (x, r) = xr, ∀r ∈ R, ∀x ∈ M. The axioms for the right R-module become in this case:

i') (x + y)r = xr + yr; ii') x(r + s) = xr + xs; iii') x(rs) = (xr)s; iv') x1 = x,

2 Also called multiplication of elements in M with scalars in R.

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60

for any r, s ∈ R and x, y ∈ M. There is a handy notation for the fact that M is a left R-module,

namely R M. M is a right R-module is denoted by MR. If R is commutative, then the notions of left R-module and right

R-module are the same (look at axiom iii'). If R is an arbitrary ring and M is a right R-module, then M becomes

a left Rop-module, where (Rop, +, *) is the opposite of the ring R (Rop and R have the same underlying Abelian group R, but the multiplica-tion * in Rop is defined by r*s = sr, ∀r, s ∈ R).

The construction above shows that a result that holds for any ring R and any right R-module is valid also for any left R-module, and con-versely. In the same way, all definitions for left modules have a natu-ral correspondent for right modules.

1.3 Examples. a) If K is a field, a K-module is exactly a K-vector space.

b) If R is a ring with identity, R has a (canonical) structure of left R-module, denoted R R. Indeed, (R, +) is an Abelian group; the exter-nal operation R × R → R is the ring multiplication: (r, s) & rs, ∀r, s ∈ R. Similarly, R is canonically a right R-module, denoted RR.

c) Any Abelian group (A, +) is canonically a Z-module. For n ∈ Z and a ∈ A, na is defined as the multiple of a in the additive group A (if n ∈ N, na = a + + a (n terms); if n < 0, na = (−a) + + (−a) (n terms)). This is the only external operation that endows A with a Z-module structure (exercise!). The theory of Abelian groups is thus a particular case of module theory.

d) Let R be a ring and let n ∈ N*. The n-fold Cartesian product R n = (x1, , xn) | xi ∈ R, 1 ≤ i ≤ n becomes an R-module if the addi-tion and the scalar multiplication are defined component-wise:

(x1, , xn) + (y1, , yn) = (x1 + y1, , xn + yn), ∀(x1, , xn), (y1, , yn) ∈ R n.


61

r(x1, , xn) = (rx1, , rxn), ∀r ∈ R, ∀(x1, , xn) ∈ R n. e) If R is a ring and m, n ∈ N*, the set Mm,n(R) of m×n matrices with

entries in R is an Abelian group endowed with usual matrix multipli-cation and becomes an R-module by defining the multiplication of matrices with scalars: for r ∈ R, A = (aij) ∈ Mm, n(R), r(aij) := (raij) (multiply every entry of the matrix with r).

f) Let R := M2(Z) (the ring of 2×2 matrices with entries in Z) and M := M2, 1(Z) (the Abelian group of 2×1 matrices with entries in Z). M has a natural structure of left R-module: ∀A ∈ M2(Z) = R, ∀U ∈ M2, 1(Z) = M, AU ∈ M is the usual matrix product. Checking the module axioms is straightforward and it boils down to the known properties of matrix operations. Can you generalize this example? Can M be endowed with a natural structure of a right R-module ?

g) Let ϕ : R → S be a unitary ring homomorphism. If M is a left S-module, then M has a structure of left R-module by restriction of scalars: ∀r ∈ R, ∀x ∈ M, rx := ϕ(r)x. In particular, S becomes a left R-module (and also a right R-module). This example generalizes a situation often encountered in field extensions: any field S is a vector space over any subfield R.

1.4 Remark. For a ring R and an Abelian group M, defining a left R-module structure on M amounts to defining a unitary ring homomorphism λ : R → End(M), where (End(M), + ,) is the endomorphism ring of the Abelian group M, defined as follows: + is homomorphism addition: ∀u, v ∈ End(M), (u + v)(x) := u(x) + v(x), ∀x ∈ M, and is the usual map composition: ∀u, v ∈ End(M), (uv)(x) = u(v(x)), ∀x ∈ M. 3

3 Sometimes this ring is called the ring of left endomorphisms of M, emphasizing

that the functions are written at the left of the argument, like u(x); this forces the definition of the composition of functions in the usual manner defined above. But

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62

Indeed, if M is a left R-module, define λ : R → End(M) by λ(r)(x) = rx, ∀r ∈ R, ∀x ∈ M. Conversely, given λ : R → End(M), de-fine scalar multiplication R × M → M by (r, x) & rx := λ(r)(x), ∀r ∈ R, ∀x ∈ M. The reader should check the details. What corresponds to a structure of right R-module of M?

Throughout this section, R denotes a ring with identity. All modules are left R-modules (unless specified otherwise).

1.5 Proposition. Let M be an R-module. The, for any x ∈ M and r ∈ R, we have:

a) 0x = r0 = 0. b) r(−x) = (−r)x = −(rx). Proof. a) 0x = (0 + 0)x = 0x + 0x. Since M is a group, by simplifica-

tion we deduce that 0x = 0. The same method shows r0 = 0. b) 0 = r0 = r(x + (−x)) = rx + r(−x). So −(rx), the opposite of rx in

(M, +), is r(−x). !

We define the natural notion of submodule:

1.6 Definition. a) Let M be a left R-module. A non-empty subset L of M is called left R-submodule of M if

i) L is a subgroup in (M, +): ∀x, y ∈ L ⇒ x − y ∈ L; ii) ∀r ∈ R, ∀x ∈ L ⇒ rx ∈ L.

Usually one says L is a submodule of M if no confusions can occur. Notation: L ≤ R M (or, simpler, L ≤ M ). The fact that L is a right R-submodule of M is written L ≤ MR.

if one writes (x)u for the value of u at x, then the composition of u and v is defined as (x)(uv) = ((x)u)v. With this multiplication, End(M) is called the ring of right endomorphisms of M and it is the opposite of the ring of left endomorphisms of M.


63

The definition above is the natural generalization of the notion of vector subspace.

1.7 Proposition. Let M be an R-module and ∅ ≠ L ⊆ M. The following statements are equivalent:

a) L ≤ RM: (∀r ∈ R, ∀x, y ∈ L ⇒ x − y ∈ L and rx ∈ L). b) For any r, s ∈ R and x, y ∈ L, it follows that rx + sy ∈ L. c) For any n ∈ N*, r1,, rn ∈ R, x1,, xn ∈ L, it follows that

r1x1 + + rnxn ∈ L. ! Proof. a)⇒ b) If r, s ∈ R and x, y ∈ L, then a) implies that rx and

(−s)y ∈ L, so L contains also rx − (−s)y = rx − (−sy) = rx + sy. b)⇒a) If r ∈ R and x, y ∈ L, then rx + 0y = rx ∈ L and

1x + (− 1)y = x − y ∈ L. b)⇒c) Induction on n (exercise). !

An element of the form r1x1 + + rnxn, with r1, , rn ∈ R and x1, , xn ∈ M, is called a linear combination of x1, , xn (r1, , rn are called the coefficients of the linear combination). Thus, L ≤ R M iff any linear combination of elements in L is still in L.

Since any submodule L of a module M is a subgroup of the additive group (M, +), we have 0 ∈ L. Also, if L ≤ RM, L has a structure of left R-module: the external operation is the restriction at R × L of the external operation of M.

1.8 Examples. a) For any R-module M, 0 is an R-submodule in M, denoted simply by 0. Also, M is an R-submodule of M. A submod-ule of M, not equal to M, is called a proper submodule of M.

b) The left submodules of the canonical module R R are exactly the left ideals of the ring R; the notation I ≤ R R means I is a left ideal of R.

c) If M is an Abelian group (= Z-module), a Z-submodule of M is the same thing with a subgroup of M.

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64

1.9 Proposition. Let (Mi)i∈I be a family of submodules of RM. Then the intersection of this family, ∩ i∈I Mi , is a submodule of M. !

This simple result allows us to define the notion of submodule generated by a subset:

1.10 Definition. Let M be an R-module and let X be a subset of M. a) The intersection of all submodules of M that include X is a

submodule of M, called the submodule generated by X and denoted by R < X > (or simply < X >). If L ≤ R M and < X > = L, one says also that X is a system of generators for L (or that X generates L).

b) Define the set of linear combinations of elements in X with coefficients in R as the set RX, where

RX := r1x1 + + rnxn| n ∈ N, r1, , rn ∈ R, x1, , xn ∈ X. If X = ∅, define R∅ = 0.

1.11 Proposition. Let M be an R-module and X ⊆ M. Then: a) < X > is the smallest (inclusion-wise) submodule of M that in-

cludes X. b) < X > = RX, that is, the submodule generated by X is the same as

the set of linear combinations of elements in X with coefficients in R. Proof. a) Evidently, < X > is a submodule and includes X. If L is a

submodule in M that includes X, then < X > ⊆ L because L is a mem-ber of the family of submodules the intersection of which is L.

b) We show that X ⊆ RX and that RX is the smallest submodule that includes X. The case X = ∅ is trivial. If X ≠ ∅ and x ∈ X, then x is a linear combination, x = 1x ∈ RX. So X ⊆ RX. The difference of two linear combinations in RX and the product of any r ∈ R with a linear combination is still in RX. So, RX is a submodule. If L is a submodule that includes X, 1.7.c) implies RX ⊆ L. !

1.12 Definition. For any a ∈ RM, the submodule generated by a is Ra = ra | r ∈ R and it is also called the cyclic submodule


65

generated by a. The R-module M is called finitely generated if there exists a finite system of generators for M, i.e. a finite subset F of M such that < F > = M.

While the intersection of a family of submodules is a submodule, the union of a family of submodules is not a submodule in general.

1.13 Definition. Let M be an R-module and let E, F be submodules in M. The submodule generated by E ∪ F is called the sum of the submodules E and F, and is denoted E + F. Thus, E + F is just another notation for < E ∪ F >.

For an arbitrary family (Ei)i∈I of submodules of M, the submodule generated by ∪i∈I Ei is called the sum of the family of submodules (Ei)i∈I, denoted ∑i∈I Ei or ∑I Ei.

The sum of the submodules E1, , En is denoted E1 + + En

or ∑=

n

iiE

1.

The sum of the family of submodules (Ei)i∈I is the smallest submod-ule of M including all submodules Ei.

1.14 Proposition. a) If E, F are submodules of RM, then the sum of E and F is

E + F = e + f | e ∈ E, f ∈ F, b) If E1, , En are submodules of M, then

E1 + + En = e1 + + en | e1 ∈ E1, , en ∈ En. Proof. a) Let S := e + f | e ∈ E, f ∈ F. A straightforward verifica-

tion shows that S is a submodule. If L is a submodule containing E and F, then e + f ∈ L, ∀e ∈ E, ∀f ∈ F. Thus, S ⊆ L and S is the smallest submodule including E ∪ F. !

In order to formulate a similar result for the case of the sum of an arbitrary (possibly infinite) family of submodules, we introduce the following notion: for a set I (seen as a set of indexes), an R-module M

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66

and a family of elements4 (ei)i∈I, with ei ∈ M, ∀i ∈ I, define the sup-port of the family (denoted Supp((ei)i∈I)):

Supp((ei)i∈I) := i ∈ I | ei ≠ 0 For any family (ei)i∈I having finite support J ⊆ I, its sum is defined

as ∑i∈I ei := ∑i∈J ej.

1.15 Proposition. If (Ei)i∈I is a family of submodules of M, then ( ) , , having finite supporti i i i i Iii I i I

e e E i I eE ∈∈ ∈

∈ ∀ ∈= ∑∑ =

= 1 1 11, , , , , ,n n ni i n i i i ie e n i i I e E e E+ + ∈ ∈ ∈ ∈… ' … … .

Proof. Let S = ( ) , , having finite supporti i i i i Ii I

e e E i I e∈

∈

∈ ∀ ∈∑ .

As above, we show that S is a submodule: if r ∈ R, and e = ∑∈Ii

ie ∈ S,

with ei ∈ Ei, ∀i ∈ I, and Supp(ei)i∈I finite, then re = ∑∈Ii

ire ∈ S, since Ei

is a submodule. Likewise, if e, f ∈ S, then e − f ∈ S. On the other hand, clearly Ei ⊆ S, ∀i ∈ I. If L is another submodule of M that in-cludes all submodules Ei, then S ⊆ L. So S = < ∪i∈I Ei>. !

1.16 Remark. The set LR(M) of all submodules of the R-module M is ordered by inclusion; furthermore, (LR(M), ⊆) is a complete lattice: 5 for any subset F of LR(M) (i.e., any family of submodules of M), sup F is the sum of the family F, and inf F = ∩F.

4 Note that a family (ei)i ∈ I of elements of M is in fact a function f : I → M

(denoting f (i) = ei, ∀i ∈ I ). 5 An ordered set A is called a complete lattice if, for any B ⊆ A, there exists

sup B (the smallest upper bound of B) and inf B (the largest lower bound of B) in A.


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1.17 Remark. The sum of the submodules I, J of the left canonical module R R is the same with the sum of the left ideals I and J. If R is a field, the submodules of an R-module M are the vector subspaces of M sum of submodules means sum of vector subspaces.

The lattice LR(M) always has a greatest element (M itself) and a smallest element (the submodule 0). For this reason, the notions of maximal submodule and minimal submodule are defined as follows.

1.18 Definition. The submodule L of the R-module M is called a maximal submodule if L is maximal among the proper (distinct from M) submodules, i.e.:

∀E ≤ R M with E ≠ M, L ⊆ E implies L = E. The submodule L ≤ M is called a minimal submodule if L is mini-

mal among the nonzero submodules: ∀E ≤ R M with E ≠ 0, E ⊆ L implies E = L.

The following theorem is very important.

1.19 Theorem. Let M be a nonzero finitely generated R-module. Then any proper submodule of M is included in some maximal submodule. In particular, M has a maximal submodule.

Proof. Let L ≤ R M, L ≠ M and let x1, , xn be a finite generator set of M. Let P be the set of proper submodules of M that include L. P is ordered by inclusion; its maximal elements (if they exist!) are ex-actly the maximal submodules of M that include L. We use Zorn's lemma to prove that maximal elements exist in P. First, note that P ≠ ∅ since L ∈ P. Take a chain (Ei)i∈I, with Ei ∈ P, ∀i ∈ I. This chain of submodules is bounded above in P by ∪i∈I Ei =: E. Indeed, E is a submodule6: if x, y ∈ E, then, for some i, j ∈ I, x ∈ Ei and y ∈ Ej;

6 Here is a (singular) situation when the union of a family of submodules is a

submodule.

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(Ei)i∈I being a chain, we have Ei ⊆ Ej or Ej ⊆ Ei. So x − y ∈ Ej (because Ej ≤ R M) or x − y ∈ Ei. Anyway, x − y ∈ E. Similarly, ∀r ∈ R, ∀x ∈ E, we have rx ∈ E. So, E ≤ M and E includes L.

We must also prove that E ≠ M. Suppose E = M. Then x1, , xn ⊆ E = ∪i∈I Ei, so, ∀t ∈ 1, , n, there exists it ∈ I such that

xt ∈ tiE . But (Ei)i∈I is a chain, so ∃j ∈ i1, , in such that tiE ⊆ Ej,

∀t ∈ 1, , n. Thus x1, , xn ⊆ Ej. We deduce M = < x1, , xn > ⊆ Ej, contradicting Ej ≠ M (since Ej ∈ P).

Zorn's Lemma provides us with a maximal element of P. Taking L = 0, the existence of a maximal submodule in M is

proven. !

1.20 Corollary (Krull's Lemma7) Let R be a ring with identity. Then every left proper ideal of R is included in some maximal left ideal. In particular, R has a maximal left ideal.

Proof. The canonical left R-module R R is finitely generated (by 1). !

1.21 Definition. Let M and N be left R-modules. A function ϕ : M → N is called a left R-module homomorphism (or simply module homomorphism) or R-homomorphism if it preserves the module opera-tions:

ϕ(x + y) = ϕ(x) + ϕ(y), ∀x, y ∈ M; ϕ(rx) = rϕ(x), ∀x ∈ M, ∀r ∈ R.

Other names: R-linear application, linear transformation 8 , R-morphism. An R-module homomorphism ϕ : M → M is called an endomorphism of M.

7 Wolfgang Adolf Ludwig Helmuth Krull (1899-1971), German mathematician. 8 A geometric terminology, used mainly for vector space homomorphisms.


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The first condition in the definition of the module homomorphism ϕ : M → N means that ϕ is an Abelian group homomorphism. Thus,

ϕ(0) = 0 (0 denotes the zero element of M, as well as the zero element of N ). Also:

ϕ(− x) = − ϕ(x), ∀x ∈ M.

1.22 Remark. Let L ⊆ M. If L has an R-module structure such that the canonical inclusion ι : L → M is a module homomorphism, then L ≤ M. Conversely, if L ≤ M, then ι : L → M is a module homomorph-ism.

1.23 Examples. a) For any R-modules M and N, 0 : M → N, 0(x) = 0, ∀x ∈ M, is an R-module homomorphism, called the zero homomorphism. The identity application idM : M → M, idM(x) = x, ∀x ∈ M, is also a homomorphism, the identity homomorphism of M.

b) If M is an R-module and x ∈ M, the multiplication by x, rx : R R → M defined by rx(a) = ax, ∀a ∈ R, is a module homomorph-ism. Indeed,

rx(a + b) = (a + b)x = ax + bx = rx(a) + rx(b); rx(ba) = (ba)x = b(ax) = brx(a), ∀a, b ∈ R.

c) If R is commutative ring and r ∈ R, then the multiplication by r, λr : M → M, λr(x) = rx, ∀x ∈ M, is a homomorphism: it is obvi-ously additive and

λr(ax) = r(ax) = (ra)x = (ar)x = a(rx) = aλr(x), ∀x ∈ M, ∀a ∈ R. Note that the commutativity of R is effectively used.

1.24 Definition. For any R-modules M, N, we denote HomR(M, N) := ϕ | ϕ : M → N, ϕ is an R-module homomorphism;

EndR(M) := HomR(M, M). We denote sometimes: Hom(R M, N), respectively End(R M) - for left modules;

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Hom(MR, N), respectively End(MR) - for right modules.

1.25 Remark. HomR(M, N) is always nonempty; it contains at least the zero homomorphism 0 : M → N, 0(x) = 0, ∀x ∈ M. Moreover, HomR(M, N) is an Abelian group with respect to homomorphism addi-tion, defined below:

1.26 Proposition. a) Let E, F be R-modules and let ϕ : E → F, η : E → F be R-module homomorphisms. Then the sum ϕ +η : E → F, defined by:

(ϕ +η)(x) := ϕ(x) + η(x), ∀x ∈ E, is an R-module homomorphism. HomR(E, F) is an Abelian group with respect to homomorphism addition; the zero element is the 0 homomorphism, the opposite of ϕ is (−ϕ), (−ϕ)(x) = −ϕ(x), ∀x ∈ E.

b) Let E, F, G be R-modules and let ϕ : E → F, ψ : F → G be R-module homomorphisms. Then their composition ψ ϕ : E → G is also an R-module homomorphism. As a consequence, (EndR(E), + , ) is a ring, the unity element being the identity homomorphism idE.

c) Let E be a submodule of the R-module F and let ψ : F → G be an R-module homomorphism. Then the restriction of ψ to E,ψ|E : E → G, is an R-module homomorphism. !

A homomorphism is perfectly determined by its values on a generating set:

1.27 Proposition. Let E, F be R-modules, S a system of generators of E and ϕ, ψ : E → F module homomorphism. Then ϕ = ψ iff ϕ|S = ψ|S.

Proof. Suppose ϕ|S = ψ|S. If x ∈ E, there exist x1, , xn ∈ S and r1, , rn ∈ R such that x = r1x1 + + rnxn. So,

ϕ(x) = ϕ(r1x1 + + rnxn) = r1ϕ(x1) + + rnϕ(xn) = r1ψ(x1) + + rnψ(xn) = ψ(x). !


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1.28 Definition. An R-module homomorphism ϕ : E → F is called an R-module isomorphism if there exists a homomorphism ψ : F → E such that ϕ ψ = idF and ψ ϕ = idE. The R-modules E and F are called isomorphic if there exists an R-module isomorphism ϕ : E → F. We write in this case E ≅ R F (or E ≅ F if it is clear that it is an R-module isomorphism ). An isomorphism ϕ : E → E is called an automorphism of E.

1.29 Proposition. An R-module homomorphism is an isomorphism iff it is bijective.

Proof. Let ϕ : E → F be a homomorphism. If ϕ is an isomorphism, then it is an invertible map, so it is a bijection. Suppose ϕ is a bijective homomorphism. Then the inverse of the map ϕ exists, ψ : F → E. We must prove that ψ is a homomorphism. Recall that, ∀y ∈ F, ψ(y) = x, where x is the unique element in E with ϕ(x) = y. Let y1, y2 ∈ F and x1, x2 ∈ E such that ψ(y1) = x1 and ψ(y2) = x2. We have ψ(y1 + y2) = x1 + x2, since ϕ(x1 + x2) = ϕ(x1) + ϕ(x2) = y1 + y2. So, ψ(y1 + y2) = x1 + x2 = ψ(y1) + ψ(y2). Similarly, ∀r ∈ R, ∀y ∈ F, ψ(ry) = rψ(y). !

1.30 Definition. Let ϕ : M → N be an R-module homomorphism. Consider the subset of M defined by:

Kerϕ := x ∈ M |ϕ(x) = 0 = ϕ −1(0) Kerϕ is called the kernel of ϕ. Let Imϕ denote the image of ϕ:

Imϕ := y ∈ N| ∃x ∈ M such that y = ϕ(x) = ϕ(x)| x ∈ M.

1.31 Proposition. If ϕ : M → N is an R-module homomorphism and M' ≤ M, N' ≤ N, then ϕ −1(N') is a submodule of M, and ϕ(M) is a submodule of N. In particular, Kerϕ is a submodule of M, and Imϕ is a submodule of N.

Proof. Let x, y ∈ ϕ −1(N'). Then ϕ(x − y) = ϕ(x) − ϕ(y) ∈ N', so x − y ∈ ϕ −1(N'). If r ∈ R, then ϕ(rx) = rϕ(x) ∈ N', so rx ∈ ϕ −1(N')

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Let z, t ∈ ϕ(M') and r ∈ R. Then there exist x, y ∈ M' such that z = ϕ(x), t = ϕ(y). We have z − t = ϕ(x) − ϕ(y) = ϕ(x − y) ∈ ϕ(M'), and rz = rϕ(x) = ϕ(rx) ∈ ϕ(M'). !

1.32 Proposition. Let ϕ : M → N be an R-module homomorphism. Then:

a) ϕ is injective iff Kerϕ = 0. b) ϕ is surjective iff Imϕ = N. !

Exercises

1. Prove that the commutativity of the addition of a module is a conse-quence of the other axioms of the definition of the module. 2. Let K be a field. Which of the following subsets of the K-module K[X] is a K-submodule?

a) The set of all polynomials of degree 7. b) The set of all polynomials of degree at most 7. c) The set of all monic polynomials. d) The set of all polynomials of degree 1 that have the root 1. e) The set of all polynomials of even degree. Which of the sets above is a K[X]-submodule?

3. Let E and F be submodules of the R-module M. Show that E ∪ F is a submodule of M iff E ⊆ F or F ⊆ E. 4. Let M be an R-module. Study the properties of the operations + and ∩ on LR(M ) (such as commutativity, associativity, distributivity, exis-tence of neutral elements). 5. Let M be an R-module and let A, B, C ≤ M. If B ≤ A, then A∩(B + C) = B + A∩C (one says that LR(M ) is a modular lattice).


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6. Determine all submodules of the R-module R2. 7. Give an example of a module M and A, B, C ≤ M such that A∩(B + C) ≠ A∩B + A∩C (i.e., LR(M ) is not distributive). (Hint. Try a vector space.) 8. Let (G, +) be an Abelian group and n ∈ N* such that na = 0, ∀a ∈ G. Then G has a canonical structure of Zn-module. Is the con-verse true? Generalization. 9. Identify the Euclidian plane with R2, seen as an R-vector space. Which of the following transformations of the plane is a linear transformation (an R-module homomorphism) from R2 to R2?

a) The rotation of angle α around (0, 0). b) The rotation of angle α around (0, 1). c) The translation by the vector v = (x, y). d) The symmetry with respect to a line. e) The projection on a line.

10. Let R be a ring, let S be a set and R S := ϕ : S → R. Define a left R-module structure on R S. More generally, let M be a left R-module and let S be a set. Define a left R-module structure on M S = ϕ : S → M. 11. Let R be a ring. Show that, for any R M, R M', HomR(M, M') ⊆ HomZ(M, M'). Give examples of R, M, M' such that the inclusion is strict. 12. Let u : M → M' be an R-module homomorphism, A, B ≤ R M and A', B' ≤ R M'. Study the validity of the statements:

a) u(A + B) = u(A) + u(B). b) u(A ∩ B) = u(A) ∩ u(B). c) u −1(A' + B') = u −1(A') + u −1(B'). d) u −1(A' ∩ B') = u −1(A') ∩ u −1(B').

13. Let R be a commutative ring. Show that EndR(R) ≅ R. More gener-ally, for any R M, HomR(R, M) ≅ M.

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14. Let R be a commutative ring. Show that R[X] is not a finitely generated R-module. 15. Let K be a field, n ∈ N* and v1, , vn ∈ K n, where vi = (vi1, , vin). Find necessary and sufficient conditions for v1, , vn to be a system of generators for the K-module K n. 16. Write 4 distinct homomorphisms from the R-module R2 to R. Which is the general form of such a homomorphism? Generalization. 17. Let A be an R-module. Prove that A = 0 ⇔ HomR(A, B) = 0, ∀RB ⇔ HomR(B, A) = 0, ∀RB. 18. Does the Z-module Q have minimal submodules? 19. Let V be a finite dimensional K-vector space and let u ∈ EndK(V). Then: u is injective ⇔ u is surjective ⇔ u is an isomorphism. 20. Give examples of an R-module M and of an endomorphism ϕ : M → M such that:

a) ϕ is injective, but not surjective. b) ϕ is surjective, but not injective.

II.2 Factor modules and the isomorphism theorems

The method used to construct the factor ring (given a ring and an ideal in the ring) can be applied to modules, with minor modifications.

Let M be a left R-module and let L be a submodule in M. Consider-ing only the Abelian group structure on M, L is a subgroup in M, so we can construct the factor group M/L, which is also an Abelian group. The Abelian group M/L can be endowed with a natural R-module structure, inherited from the R-module structure on M.


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We briefly recall the construction of the factor group M/L. Define on M the equivalence relation (also called modulo L congruence) by:

∀x, y ∈ M : x ∼ y (mod L) ⇔ x − y ∈ L. An easy check shows that this is indeed an equivalence relation and

that the equivalence class of x ∈ M, i.e. the set y ∈ M | x ∼ y (mod L) is x + L, where

x + L := x + l |l ∈ L (called the class of x modulo L) Define now the set M/L as the set of all equivalence classes:

M/L = x + L | x ∈ M Make M/L into an Abelian group by putting:

(x + L) + (y + L) := (x + y) + L, ∀x, y ∈ M. One proves that: the operation above is well defined (it does not de-

pend on the representatives of the classes modulo L) and (M/L, +) is an Abelian group; the neutral element is 0 + L (equal to l + L, ∀l ∈ L); − (x + L) = (− x) + L, ∀x ∈ M.

Getting back to the module case, define an external operation R × (M/L) → M/L: ∀r ∈ R, ∀x + L ∈ M/L, set

r(x + L) := rx + L. This definition is correct: if r ∈ R and x, y ∈ M, with x + L = y + L,

then x − y ∈ L, so r(x − y) ∈ L (since L ≤ R M), i.e. rx − ry ∈ L. Thus, rx + L = ry + L.

A routine exercise shows that M/L becomes a left R-module. For instance, ∀r, s ∈ R, ∀x ∈ M: (r + s)(x + L) = (r + s)x + L = (rx + sx) + L = (rx + L) + (sx + L) = r(x + L

) + s(x + L).

2.1 Definition. The R-module M/L defined above is called the fac-tor module of M with respect to L. The map π : M → M/L, π(x) = x + L, ∀x ∈ M, is a surjective module homomorphism, called the canonical homomorphism or the canonical surjection.

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2.2 Examples. If L ≤ R M and π : M → M/L is the canonical homo-morphism, then

Kerπ = x ∈ M | x + L = 0 + L = L and Imπ = M/L. So, any submodule is the kernel of some homomorphism.

2.3 Proposition. Let M be an R-module and let L be a submodule. Then there is a natural one-to-one increasing map between the lattice of the submodules of M that include L and the lattice of submodules of M/L, given by

ϕ : A ≤ R M | L ⊆ A → LR(M/L), ϕ(A) := a + L | a ∈ A, ∀A ≤ R M, L ⊆ A.

Proof. We remark that ϕ(A) = π(A), the image, through the canoni-cal homomorphism π : M → M/L, of the submodule A. So, ϕ(A) ≤ R M/L. If B ≤ R M/L, then π −1(B) = x ∈ M | x + L ∈ B is a sub-module of M that includes L, and ϕ(π −1(B)) = B. So, ϕ is surjective. The rest of the proof is left to the reader. !

2.4 Example. Let n ∈ N. Let us find the submodules of the Z-module Z/nZ (also known as Zn). The above says that L(Z/nZ) is in one-to-one correspondence to the submodules of Z that include nZ, which are mZ | m ∈ N, m|n. To mZ ⊇ nZ corresponds mZ/nZ ≤ Z/nZ. So, L(Z/nZ) = mZ/nZ | m ∈ N, m|n. Also, mZ/nZ = < m + nZ > is the unique submodule having n/m elements in Z/nZ:0 + nZ, m + nZ, , (n/m − 1)m + nZ.

The following notions are general concepts in category theory, ap-plied to the category R-Mod of the left R-modules:

2.5 Definition. Let ϕ : M → N be an R-module homomorphism. a) ϕ is called a monomorphism if, for any R A and any homomor-

phisms u, v : A → M, ϕ u = ϕ v implies u = v. b) ϕ is called an epimorphism if, for any R A and any homomor-

phisms u, v : N → A, uϕ = vϕ implies u = v.


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2.6 Proposition. Let ϕ : M → N be a homomorphism. Then: a) ϕ is a monomorphism if and only if ϕ is injective. b) ϕ is an epimorphism if and only if ϕ is surjective. Proof. a) Suppose ϕ injective. Let R A and u, v ∈ HomR(A, M) such

that ϕu = ϕv. So, ∀a ∈ A, ϕ(u(a)) = ϕ(v(a)). Since ϕ is injective, we deduce u(a) = v(a). Suppose now ϕ is a monomorphism and let A = Kerϕ. Consider ι : Kerϕ → M the inclusion homomorphism and 0 : Kerϕ → M the zero homomorphism. For any a ∈ Kerϕ, ϕ ι(a) = ϕ(a) = 0 = ϕ 0(a); so, ι = 0 ⇔ Kerϕ = 0.

b) Let ϕ be an epimorphism. Consider the factor module N/Imϕ and π : N → N/Imϕ (canonical surjection) and 0 : N → N/Imϕ. We have π ϕ (x) = ϕ(x) + Imϕ = 0 + Imϕ = 0ϕ (x), ∀x ∈ M. So, π ϕ = 0ϕ, i.e. π = 0. This means N/Imϕ = 0⇔N = Imϕ. The converse is left to the reader. !

2.7 Theorem. (The fundamental isomorphism theorem) Let

ϕ : M → N be an R-module homomorphism. ThenϕKer

M ≅ Imϕ.

More precisely, there exists a unique isomorphism ψ : M/Kerϕ → Imϕ such that the following diagram is commutative:

ϕϕ ψιπ

ϕ

ImKer ⎯→⎯↑↓

⎯→⎯

M

NM

i.e. ϕ = ι ψ π, where ι is the inclusion and π is the canonical surjec-tion. ψ is given by ψ(x + Kerϕ) = ϕ(x), ∀x ∈ M.

Proof. ψ : M/Kerϕ → Imϕ is well defined: if x, y ∈ M are such that x + Kerϕ = y + Kerϕ, then x − y ∈ Kerϕ ⇔ ϕ(x − y) = 0 ⇔ ϕ(x) = ϕ(y). Thus, ψ(x + Kerϕ) does not depend on x, but only on the equivalence class x + Kerϕ. Checking that ψ is a homomorphism is simple.

ψ is surjective: Imψ = ψ(x + Kerϕ)| x ∈ M = ϕ(x)| x ∈ M = Imϕ. For the injectivity, we show that Kerψ = 0 + Kerϕ: if x ∈ M

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with ψ(x + Kerϕ) = 0, then ϕ(x) = 0, so x ∈ Kerϕ, ⇔ x + Kerϕ = 0 + Kerϕ.

Also, ι ψ π (x) = ψ(x + Kerϕ) = ϕ(x), ∀x ∈ M. If η : M/Kerϕ → Imϕ is a homomorphism such that ϕ = ι η π,

then, ∀x ∈ M, η(x + Kerϕ) = ιηπ(x) = ϕ(x), so η = ψ. !

2.8 Remark. A typical use for the isomorphism theorem is as fol-lows: suppose B ≤ R A and we want to prove that A/B ≅ C. One looks for a surjective homomorphism ϕ : A → C, with Kerϕ = B. The isomorphism theorem provides then the required isomorphism. The following corollaries illustrate this technique (which is used also for groups, rings, algebras, ).

2.9 Corollary. Let R M and E, F ≤ M such that E ⊆ F. Then F/E is a submodule of M/E and:

FM

EFEM

≅ (R-module isomorphism).

Proof. Since F/E = x + E | x ∈ F, F/E ⊆ M/E = x + E | x ∈ M. Let ϕ : M/E → M/F, ϕ(x + E) = x + F, ∀x ∈ M. The map ϕ is well de-fined: if x, y ∈ M, with x + E = y + E, then x − y ∈ E. So, x − y ∈ F and x + F = y + F. It easy to see that ϕ is a surjective module homomorphism. Kerϕ = x + E | x ∈ M, x + F = 0 + F = F/E. Apply now the fundamental isomorphism theorem. !

2.10 Corollary. Let R M and E, F ≤ R M. Then

FEF

EFE

∩≅

+ .

Proof. Let ϕ : F → (E + F)/E, ϕ(x) = x + E, ∀x ∈ F. The mapping ϕ is a module homomorphism, being the restriction of the canonical homomorphism E + F → (E + F)/E to the submodule F of E + F. Moreover, ϕ is surjective: ∀(e + f ) + E ∈ (E + F)/E, with e ∈ E, f ∈ F, (e + f ) + E = f + E = ϕ( f ).


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Kerϕ = x ∈ F | x + E = 0 + E = x ∈ F | x ∈ E = E ∩ F. Apply the fundamental isomorphism theorem, we get F/(E ∩ F) ≅ (E + F)/E.!

The following result is often used in module theory arguments:

2.11 Proposition. Let ϕ : E → F and ψ : E → G be R-module homomorphisms, with ϕ surjective. If Kerϕ ⊆ Kerψ, then:

a) ψ factorizes through ϕ: there exists a unique homomorphism η : F → G such that ψ = η ϕ .

b) η is injective if and only if Kerϕ = Kerψ.

Proof. a) Let y ∈ F. Since ϕ is surjective, there exists x ∈ E with ϕ(x) = y. Define η(y) := ψ(x). The definition is independent on the choice of x ∈ E with ϕ(x) = y. Indeed, if x, x' ∈ E such that ϕ(x) = ϕ(x') = y, then x − x' ∈ Kerϕ ⊆ Kerψ, so ψ(x) = ψ(x'). Also, η is a homomorphism (standard check) and, ∀x ∈ E, η(ϕ(x)) = ψ(x). If η´ ∈ HomR(F, G) has the property that η'ϕ = ψ, then η(ϕ(x)) = η'(ϕ(x)), ∀x ∈ E. Since ϕ is surjective, η = η'.

b) Suppose η is injective. Let x ∈ E with ψ(x) = 0. Then η(ϕ(x)) = 0, so ϕ(x) = 0, i.e. x ∈ Kerϕ. Therefore Kerϕ = Kerψ. Sup-pose now that Kerϕ = Kerψ and take y ∈ F with η(y) = 0. Then ∃x ∈ E with ϕ(x) = y; so, η(ϕ(x)) = ψ(x) = 0, which means x ∈ Kerψ. Since Kerϕ = Kerψ, x ∈ Kerϕ, so ϕ(x) = y = 0. !

η FE ⎯→⎯ϕ

G ψ

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Exercises

1. Prove that every R-module homomorphism u : M → N can be writ-ten as u = vw, where v is an injective homomorphism and w is a surjective homomorphism. 2. Let M be a finitely generated R-module. Show that there exists n ∈ N and a surjective homomorphism ϕ : R n → M. Deduce that, if M is cyclic (generated by one element), then M is isomorphic to a R-module of the form R/I, where I ≤ R R. 3. Prove that every factor module of a finitely generated module is finitely generated. 4. Let m, n ∈ N*. Prove that:

a) HomZ(Zn, Z) = 0 and HomZ(Z, Zn) ≅ Zn. b) HomZ(Zm, Zn) ≅ Zd, where d = GCD(m, n).

5. Let m, n ∈ N. Under what conditions the Abelian group (Zm, +) has a Zn-module structure? Does there exist a Zn-module structure on (Z, +)? 6. Determine all the submodules in: Z6, Z8, Z24. Generalize the result.

7. Prove the following version of the fundamental isomorphism theo-rem: Let ϕ : M → N be an R-module homomorphism and let L ≤ R M such that Ker ϕ ⊆ L. Then there exists a canonical isomorphism

( )( )LM

LM

ϕϕ

≅ .

8. An R-module is called simple if it has no submodules other than zero or itself. Show that M is simple if and only if it is isomorphic to a module of the type R/I, where I is a maximal left ideal in R. Write all (up to isomorphism) simple Z-modules and all simple K[X]-modules, where K is a field.

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9. Is it true that any intersection of finitely generated submodules is a finitely generated submodule? (Hint. Let K be a field, S = K[Xn; n ∈ N] and T = S/I, where I is the ideal of S generated by (X1 − X2)Xi | i ≥ 3. Let ξ1, ξ2 be the images of X1, X2 in T. Then Tξ1∩Tξ2 is not a finitely generated T-module.)


The methods of constructing new structures (from a given collec-tion of structures) are extremely important. In the case of modules (and also groups, rings) some of the most used constructions are the direct product and the direct sum.

We begin with the case of two R-modules M1 and M2. The Carte-sian product M1 × M2 = (x1, x2) | x1 ∈ M1, x2 ∈ M2 is made into an Abelian group by defining:

(x1, x2) + (y1, y2) := (x1 + y1, x2 + y2), ∀ (x1, x2), (y1, y2) ∈ M1 × M2. M1 × M2 is in fact the direct product of the groups (M1, +) and

(M2, +). The Abelian group M1 × M2 becomes a left R-module by defining:

r(x1, x2) := (rx1, rx2), ∀r ∈ R, ∀(x1, x2) ∈ M1 × M2. It is trivial to check the R-module axioms. Define the homomorphisms π1 : M1 × M2 → M1 and

π2 : M1 × M2 → M2, by: π1(x1, x2) := x1 and π2(x1, x2) := x2, ∀(x1, x2) ∈ M1 × M2.

The R-module M1 × M2, together with the homomorphisms π1 and π2, is called the direct product of M1 and M2. The homomorphisms π1 and π2 are called the canonical projections of the direct product

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M1 × M2. The direct product satisfies the following universality prop-erty:

3.1 Theorem. (the universality property of the direct product) Let M1 and M2 be R-modules. Then, for any R-module E and any homomorphisms v1 : E → M1, v2 : E → M2, there exists a unique homomorphism v : E → M1 × M2 such that v1 = π1v and v2 = π2v (the diagram below is commutative, for i ∈ 1,2): 9

Proof. We prove first the uniqueness. Let v : E → M1 × M2 be a

homomorphism with the required property. Let e ∈ E and suppose v(e) = (x1, x2) ∈ M1 × M2. We have x1 = π1(v(e)) = (π1v)(e) = v1(e) and, likewise, x2 = v2(e). So, if there exists v as required, then v(e) = (v1(e), v2(e)), ∀e ∈ E. On the other hand, the map v defined in this way is clearly a module homomorphism. !

The construction above can be generalized for an arbitrary family (possibly infinite) of modules. Let I be a set (seen as a set of indices) and let (Mi)i∈I be a family of R-modules indexed by I. Recall that the

Cartesian product of this family is ∏i∈I Mi = f : I → ∪i∈I Mi | f (i) ∈ Mi, ∀i ∈ I.

An element f ∈ ∏i∈I Mi is usually written as (xi)i∈I or (xi)I (where we denoted f (i) = xi ∈ Mi, ∀i ∈ I). Define an operation of addition on ∏i∈I Mi by:

∀f, g ∈ ∏i∈I Mi, ( f + g)(i) := f(i) + g(i), ∀i ∈ I;

9 So, v1 factorizes through π1 and v2 factorizes through π2.

vi

ii MMM ⎯→⎯× π

21

E

v


83

With the alternate notation: ∀(xi)I, (yi)I ∈ ∏i∈I Mi, (xi)I + (yi)I := (xi + yi)I.

∏i∈I Mi is an Abelian group with respect to this operation. Define the external operation: ∀r ∈ R, ∀f ∈ ∏i∈I Mi,

(rf )(i) := rf (i), ∀i ∈ I; With the alternate notation: ∀r ∈ R, ∀(xi)I ∈ ∏i∈I Mi, r(xi)I := (rxi)I. Proving that ∏i∈I Mi becomes an R-module with the operations

above is routine: for instance, ∀r ∈ R, ∀f, g ∈ ∏i∈I Mi, ∀i ∈ I, we have

(r( f + g))(i) = r·( f + g)(i) = r( f (i) + g(i)) = rf (i) + rg(i) = (rf + rg)(i), which shows that r( f + g) = rf + rg.

For any j ∈ I, define the homomorphisms πj : ∏i∈I Mi→ Mj by: πj( f ) := f ( j), ∀f ∈ ∏i∈I Mi;

using the notation (xi)I : πj((xi)I) := xj, ∀(xi)I ∈ ∏i∈I Mi. The R-module ∏i∈I Mi, together with the family of homomor-

phisms (πi)i∈I, is called the direct product of the family of modules (Mi)i∈I. The homomorphisms πi, i ∈ I, are called the canonical projec-

tions. The direct product satisfies a universality property:

3.2 Theorem. (the universality property of the direct product) Let (Mi)i∈I be family of R-modules. Then, for any R-module E and any

family of homomorphisms indexed by I, vi : E → Mi, i ∈ I, there exists a unique homomorphism v : E → ∏i∈I Mi, such that vi = πiv, ∀i ∈ I.

ii

Ii i MM ⎯→⎯∏ ∈π

viE

v

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Proof. It is basically the same as for two modules. Suppose there exists a homomorphism v : E → ∏i∈I Mi with the desired property. Let e ∈ E and denote v(e) by f ∈ ∏i∈I Mi. For any i ∈ I, f (i) = πi( f ) = πi(v(e)) = (πiv)(e) = vi(e). So, if such a homomorphism v exists, then v(e)(i) = vi(e), ∀e ∈ E. On the other hand, v defined this

way is a module homomorphism. !

The direct product of the family (Mi)i∈I is denoted also ×i∈I Mi. For

a finite family M1, , Mn, the notation is M1 ×× Mn or ∏=

n

iiM

1. If the

modules of the family (Mi)i∈I are all equal to some module M, the di-rect product of the family is denoted by M I (or M n if I is finite, having n elements).

A remarkable fact is that the universality property of the direct product characterizes the direct product up to a unique isomorphism:

3.3 Theorem. Let (Mi)i∈I be a family of R-modules. Suppose that an R-module P, and the family of homomorphisms indexed by I, pi : P → Mi, i ∈ I, satisfies the property:

For any R-module E and any family of homomorphisms indexed by I, vi : E → Mi, i ∈ I, there exists a unique homomorphism v : E → P such that vi = piv, ∀i ∈ I. 10 (U)

If the R-module Q, together with the family of homomorphisms qi : Q → Mi, i ∈ I, satisfies also property (U) above, then there exists a

unique isomorphism ϕ : P → Q such that pi = qi ϕ, ∀i ∈ I. Proof. In (U), set E = Q and vi = qi, ∀i ∈ I. We obtain a (unique)

homomorphism ψ : Q → P such that qi = piψ. Since Q satisfies also 10 This is exactly the universality property of the direct product.


85

(U), apply (U) for P with the homomorphisms (pi)i ∈ I, there exists a

unique homomorphism ϕ : P → Q, such that pi = qiϕ. We show that ϕ and ψ are inverse one to each other. Indeed, ϕ ψ : Q → Q satisfies qi(ϕ ψ) = (qiϕ)ψ = piψ = qi, ∀i ∈ I; but the homomorphism idQ

: Q → Q has the same property: qiidQ = qi, ∀i ∈ I. By uniqueness, guaranteed by (U), we get ϕ ψ = idQ. Similarly, ψ ϕ = idP. !

3.4 Remark. The universality property of the direct product is articulated only in terms of objects (modules, in this case) and homomorphisms, without involving any elements of the underlying sets of the modules. For this reason, the universality property above is taken as a definition of the direct product in a category C (replace module with object in C and module homomorphism with morphism in C). The direct product of a family of modules in C (if it exists) is uniquely determined up to an isomorphism: the proof of the proposition above is valid in any category. Theorem 3.2 says that in R-Mod the direct product of an arbitrary family of objects (Mi)i∈I exists, namely (∏i∈I Mi, (πi)i∈I).

Dualizing (reversing the arrows) in the universality property of the direct product, one obtains the universality property of the direct sum (taken as a definition of this notion):

3.5 Definition. Let (Mi)i∈I be a family of R-modules. An R-module S, together with a family of homomorphisms (σi)i∈I, σi : Mi → S, is called a direct sum of the family (Mi)i∈I if, for any module E and any family of homomorphisms (vi)i∈I, vi : Mi → E, there exists a unique homomorphism v : S → E such that vσi = vi, ∀i ∈ I (the diagram be-low is commutative, ∀i ∈ I):

vi

SM ii ⎯→⎯σ

E

v

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86

The homomorphisms (σi)i∈I are called the canonical injections of the direct sum (S, (σi)i∈I).

3.6 Remark. The canonical injections are indeed injective. For a fixed j ∈ I, in the definition above set E = Mj and vi : Mi → Mj defined as follows: vi = 0 if i ≠ j and vj = idMj. The homomorphism v : S → Mj,

given by the definition, satisfies v σj = id, so σj is injective.

The direct sum is also called coproduct (a categorical name, emphasizing the duality with the product).

We now construct an object satisfying the definition of the direct sum.

If f ∈ ∏i∈I Mi, recall that the support of f is the set Supp f := i ∈ I | f (i) ≠ 0. Consider the following subset of ∏i∈I Mi:

(i∈I Mi := f ∈ ∏i∈I Mi | Supp f is finite. We claim that (i∈I Mi is a submodule of ∏i∈I Mi. Indeed, if f,

g ∈ (i∈I Mi, then Supp( f + g) ⊆ Supp f ∪ Supp g, so f + g has finite support. If r ∈ R, then Supp(rf ) ⊆ Supp f. Identify f ∈ (i∈I Mi with the family of elements (xi)i∈I, where xi = f (i) ∈ Mi, ∀i ∈ I; the family (xi)i∈I has finite support.

Define, ∀j ∈ I, σj : Mj → (i∈I Mi by: ∀x ∈ Mj,

(σj(x))(i) := ⎩⎨⎧

=≠

jixji

if

,if,0

. In other words, if σj(x) = (xi)i∈I, then xj = x,

and xi = 0, ∀i ∈ I, i ≠ j. It is easy to check that σj is a homomorphism, ∀j ∈ I. Note that ∀(xi)I ∈ (i∈I Mi, we have

(xi)I = ∑i∈I σi(xi) (the sum is finite, i runs only on Supp((xi)I), which is finite).


87

3.7 Theorem. Let (Mi)i∈I be a family of R-modules. Then the mod-ule (i∈I Mi, together with the homomorphisms (σi)i∈I, is a direct sum of the family (Mi)i∈I.

Proof. Let E be an R-module and let vi : Mi → E, ∀i ∈ I. If v : (i∈I Mi → E is a homomorphism with vσi = vi, ∀i ∈ I, then, ∀(xi)I ∈ (i∈I Mi, we have:

v((xi)I) = v∑i∈I σi(xi) = ∑i∈I (vσi)(xi) = ∑i∈I vi(xi). This shows that the homomorphism v is uniquely determined by the

condition vσi = vi, ∀i ∈ I. If we define v in this way, v is indeed a homomorphism: ∀(xi)I, (yi)I ∈ (i∈I Mi, ∀r ∈ R,

v((xi)I + (yi)I) = v((xi + yi)I) = ∑i∈I vi(xi + yi) = ∑i∈I vi(xi) +vi( yi) = v((xi)I) + v((yi)I),

v(r(xi)I) = v((rxi)I) = ∑i∈I vi(rxi) = ∑i∈I rvi(xi) = r∑i∈I vi(xi) = rv((xi)I).!

The module (i∈I Mi is often denoted ⊕i∈I Mi. For a finite family

M1, , Mn, the notation is M1 ⊕⊕ Mn or in

iM⊕

=1. If the modules in

the family (Mi)i∈I are all equal to the same module M, ⊕i∈I Mi is de-

noted by M ( I ). In order to avoid the confusion with the notion of inter-

nal direct sum of submodules (see 3.11), (i∈I Mi is sometimes called

external direct sum.

As in the case of the direct product, the universality property characterizes the direct sum of a family of modules up to an isomor-phism. The reader is invited to formulate this precisely and prove it, as in the case of the direct product.

3.8 Remark. If the set of indices I is finite, the module direct sum ⊕i∈I Mi = (i∈I Mi coincides with the module direct product ∏i∈I Mi in 3.2. Nevertheless, the direct sum is a couple (⊕i∈I Mi, (σi)i∈I) and is not the same with the direct product ∏i∈I Mi, (πi)i∈I.

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3.9 Example. Let R be a commutative ring and let R[X] be the polynomial ring in the indeterminate X. As an R-module, R[X] is isomorphic with the direct sum R(N) of a countable family of copies of R R.

We defined the notion of sum of a family (Li)i∈I of submodules of R M. The direct sum of the family (Li)i∈I, where Li are seen are modules by themselves can be also constructed. When are these sums the same?

3.10 Proposition. Let M be an R-module and let (Li)i∈I be a family of submodules of M. Let L denote their sum, L := ∑i∈I Li, and let ηi : Li → L be the inclusion homomorphisms. The following statements are equivalent:

a) L is the external direct sum of the modules (Li)i∈I, and the (ηi)i∈I are the canonical injections.

b) For any x ∈ L, there exists a unique family (xi)i∈I, with xi ∈ Li, ∀i ∈ I, having finite support, such that x = ∑i∈I xi.

c) For any j ∈ I, Lj ∩ \

ii I j

L∈

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

∑ = 0.

d) For any family (xi)i∈I, with xi ∈ Li, ∀i ∈ I, having finite support,

such that ∑i∈I xi = 0, we have xi = 0, ∀i ∈ I. Proof. a)⇒b) Let (i∈I Li be the direct sum constructed as in 3.7,

with (σi)i∈I the canonical injections; let ϕ : (i∈I Li → L be the unique isomorphism with ϕ σi = ηi, ∀i ∈ I. For any x ∈ L, there exists (xi)i∈I ∈ (i∈I Li such that ϕ((xi)i∈I) = x. So,

x = ϕ((xi)i∈I) = ϕ∑i∈I σi(xi) = ∑i∈I (ϕ σi)(xi) = ∑i∈I ηi(xi) = ∑i∈I xi, which means that x is the sum of the family (xi)i∈I, xi ∈ Li, ∀i ∈ I, hav-ing finite support. If (yi)i∈I is another family having finite support, with yi ∈ Li, ∀i ∈ I, such that ∑i∈I xi = ∑i∈I yi, then ϕ((xi)i∈I) = ϕ((yi)i∈I). Since ϕ is an isomorphism, (xi)i∈I = (yi)i∈I.


89

b)⇒c) Let xj ∈ Lj ∩

⎟⎟⎠

⎞⎜⎜⎝

⎛∑

∈ jIiiL

\. Then there exists a family of finite

support, (xi)i∈I \ j, xi ∈ Li, ∀i ∈ I \ j, such that xj = ∑i∈I \ j xi. We obtain that 0 is the sum of the finite support family (yi)i∈I, where

yi = xi, ∀i ≠ j and yj = −xj. Since 0 has a unique writing as a sum of a finite support family (evidently, 0 is the sum of the family (0)i∈I), we

obtain xi = 0, ∀i ∈ I. So, xj = 0. c)⇒d) Let (xi)i∈I, xi ∈ Li, ∀i ∈ I, be such that ∑i∈I xi = 0. Let j ∈ I.

Then xj ∈ Lj; since xj = ∑i∈I \ j(−xi), we get xj ∈

⎟⎟⎠

⎞⎜⎜⎝

⎛∑

∈ jIiiL

\. So, xj = 0.

d)⇒a) We show that (L, (ηi)i∈I) satisfies the definition 3.5. Let RE and let a family of homomorphisms (vi)i∈I, vi : Mi → E. If x ∈ L, there exists a family of finite support (xi)i∈I (xi ∈ Li, ∀i ∈ I), such that

x = ∑i∈I xi. From d), we obtain that this family is unique: if ∑i∈I xi = ∑i∈I yi, with yi ∈ Li, ∀i ∈ I, then ∑i∈I (xi − yi) = 0, so xi = yi,

∀i ∈ I. Define the homomorphism ϕ : L → E by: ∀x ∈ L, put

ϕ(x) := ∑i∈I vi(xi), where (xi)i∈I is the unique family of finite support

with xi ∈ Li, ∀i ∈ I, and x = ∑i∈I xi. The map ϕ thus defined is an R-module homomorphism: if x, y ∈ L, and (xi)i∈I, (yi)i∈I are the unique

families of finite support with xi, yi ∈ Li, ∀i ∈ I and such that x = ∑i∈I xi, y = ∑i∈I yi, then x + y = ∑i∈I xi + ∑i∈I yi = ∑i∈I(xi + yi), where (xi + yi)i∈I has finite support and xi + yi ∈ Li, ∀i ∈ I. So,

ϕ(x + y) = ∑i∈I vi(xi + yi) = ∑i∈I vi(xi) + ∑i∈I vi(yi) = ϕ(x) + ϕ(y). Similarly one sees that ϕ(rx) = rϕ(x), ∀r ∈ R, ∀x ∈ L. If j ∈ I and xj ∈ Lj, then (ϕ ηj)(xj) = ϕ(xj) = vj(xj), which shows that ϕ ηj = vj,

∀j ∈ I. We must show that ϕ is unique with this property. Let

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ψ : L → E be a homomorphism with ψ ηi = vi, ∀i ∈ I, and let x ∈ L. Then x = ∑i∈I xi, xi ∈ Li, ∀i ∈ I and:

ψ(x) = ψ(∑i∈I xi) = ∑i∈Iψ( xi) = ∑i∈Iψ(ηi( xi)) = ∑i∈I vi(xi) = ϕ(x). !

3.11 Definition. Let (Li)i∈I be a family of submodules of an R-module M. We say that the sum of the family (Li)i∈I is a direct (internal) sum if and only if one of the equivalent statements of the proposition above is satisfied. In this situation, one also says that (Li)i∈I is an independent family of submodules. A submodule A of M is called a direct summand of M if there exists 0 ≠ B ≤ R M such that M = A ⊕ B (a submodule B having this property is called a comple-ment of A).

3.12 Corollary. Let M be an R-module and let A, B ≤ R M. The following statements are equivalent:

a) M = A ⊕ B. b) M = A + B and A ∩ B = 0. c) ∀x ∈ M, there exists a unique pair (a, b) ∈ A × B such that

x = a + b. !

3.13 Examples. a) If V is a K-vector space, then any vector sub-space U of V is a direct summand in V.

b) If R is a domain, then R R has no nonzero submodule that is a di-rect summand. Indeed, for any nonzero submodules A and B of R R, we have A ∩ B ≠ 0: for 0 ≠ a ∈ A and 0 ≠ b ∈ B, 0 ≠ ab ∈ A ∩ B.

3.14 Remark. a) The notions of (internal) direct sum of submod-ules and external direct sum are very close: if S is a direct sum of the modules (Mi)i∈I, with canonical injections (σi)i∈I, then S is the internal direct sum of its submodules (Imσi)i∈I, and Mi ≅ Imσi, ∀i ∈ I. It is usual to identify Mi with Imσi and call S the direct sum of the modules Mi, i ∈ I.


91

b) If M = ⊕i∈I Mi is the direct sum of the modules (Mi)i∈I, with canonical injections (σi)i∈I, then, ∀j ∈ I, the mapping pj : M → Mj, x & xj, (where x = ∑i∈I σi(xi) , with xi ∈ Mi, ∀i ∈ I) is a module homo-

morphism, called the canonical projection on Mj. Considering M as a submodule of the direct product ∏i∈I Mi, pj is exactly the restriction to M of the canonical projection πj of the direct product. Furthermore, pjσj is the identity of Mj : j

pij MMM jj ⎯→⎯⎯→⎯ ∏σ .

The direct summands of an R-module M are closely related with the idempotents of the ring EndR(M). (An element e of a ring S is called an idempotent if e 2 = e).

3.15 Proposition. Let M be an R-module. If M = A ⊕ B, then p := iApA (where pA : M → A is the canonical projection and iA : A → M is the canonical injection) is an idempotent of EndR(M) and Im p = A, Ker p = B.

Conversely, let p ∈ EndR(M) be an idempotent. Then M = Im p ⊕ Ker p.

Proof. For any x ∈ M, there exist unique a ∈ A, b ∈ B, such that x = a + b. Then p(x) = a. We have: p(p(x)) = p(a) = a = p(x). So, pp = p2 = p. We have p(x) = 0 iff x = 0 + b, with b ∈ B, i.e. Ker p = B. Evidently, Im p = A.

Suppose now that p ∈ EndR(M) and p2 = p. For any x ∈ M, write x = p(x) + (x − p(x)).

We have p(x) ∈ Im p and p(x − p(x)) = p(x) − p2(x) = 0, so x − p(x) ∈ Ker p. Thus, M = Im p + Ker p. If x ∈ Im p ∩ Ker p, then p(x) = 0 and x = p(y), with y ∈ M. Therefore, x = p(y) = p2(y) = p(x) = 0 and Im p ∩ Ker p = 0, which means that M = Im p ⊕ Ker p. !

3.16 Proposition. Let M and N be R-modules and let u : M → N and v : N → M be homomorphisms such that vu = idM. Then

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92

N = Im u ⊕ Ker v. In particular, M is isomorphic to a direct summand of N.

Proof. Let p = uv : N → N. Since p2 = uvuv = uidv = uv = p, N = Im p ⊕ Ker p by 3.15. Because v is surjective, Im p = Im uv = Im u. Also, Ker uv = x ∈ N | u(v(x)) = 0 = x ∈ N | v(x) = 0 = Ker v, since u is injective. !

3.17 Definition. Let vi : Mi → Ni (i ∈ I) be a family of R-module homomorphisms. We define the direct product and the direct sum of the family of homomorphisms (vi)i∈I.

Let (∏ Mi, (πi)i∈I) (respectively (∏Ni, (ρi)i∈I)) be the direct product of the family (Mi)i∈I (respectively (Ni)i∈I). For any j ∈ I,

vjπj : ∏ Mi → Nj is a homomorphism. The universality property of the direct product (3.2), produces a unique homomorphism

v : ∏ Mi → ∏ Ni such that ρjv = vjπj, ∀j ∈ I: The homomorphism v is called the direct product of the family of

homomorphisms (vi)i∈I and is usually denoted by ∏i∈I vi or ×i∈I vi. If

I = 1, , n, the notations are ∏ =ni iv1 or v1 ×× vn.

For any x = (xi)i∈I ∈ ∏ Mi, ∏i∈I vi(x) = (vi(xi))i∈I ∈ ∏ Ni.

Similarly one defines the direct sum of the family of homomor-phisms (vi)i∈I. Let (⊕ Mi, (σi)i∈I) (respectively (⊕Ni, (τi)i∈I)) be the di-rect sum of the family (Mi)i∈I (respectively (Ni)i∈I). For any j ∈ I, τjvj : Mj → ⊕Ni is a homomorphism. The universality property of the

jj

i MM ⎯⎯→⎯∏π

jj

i NN ⎯⎯→⎯∏ρ

vj v


93

direct sum (3.5) yields a unique homomorphism w : ⊕ Mi → ⊕ Ni such that wσj = τjvj, ∀j ∈ I.

The homomorphism w is called the direct sum of the family (vi)i∈I and is denoted by (i∈I vi or ⊕i∈I vi. If I = 1, , n, w is denoted by

ini v1=⊕ or v1⊕⊕vn. If x = ∑i∈I xi ∈ ⊕ Mi, where xi ∈ Mi and the family (xi)i∈I has finite

support11, then (⊕i∈I vi)(x) = ∑i∈I vi(xi) ∈ ⊕ Ni.

3.18 Proposition. Let vi : Mi → Ni (i ∈ I) be a family of module homomorphisms. Then:

a) If vi is injective (surjective), then ∏i∈I vi and ⊕i∈I vi are injective (surjective).

b) If ui : Ni → Pi (i ∈ I) are homomorphisms, then ∏i∈I ui∏i∈I vi = ∏i∈I uivi and ⊕i∈I ui⊕i∈I vi = ⊕i∈I uivi. !

Let R-Mod be the category of the left R-modules and Ab the cate-gory of Abelian groups.

3.19 Definition. (The Hom functors) For any A ∈ R-Mod, define the (covariant) functor hA : R-Mod → Ab:

∀ B∈ R-Mod, hA(B) := HomR(A, B), (note that HomR(A, B) is an Abelian group with respect to homomorphism addition);

11 We identify xi with its image through the canonical injection σi(xi).

ij

j MM ⊕⎯⎯→⎯σ

ij

j NN ⊕⎯→⎯τ

w vj

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94

∀u : B → B' morphism in R-Mod, hA(u) : HomR(A, B) → HomR(A, B') is defined as

hA(u)(g) := ug, ∀g ∈ HomR(A, B). It is immediate that hA(u) is a morphism in Ab, that hA(1B) = 1hA(B)

and that hA(vu) = hA(v)hA(u), for any R-module B and any R-module homomorphisms u : B → B' and ∀v : B' → B". So, hA is a functor, also denoted by HomR(A, -).

In a similar manner one defines the contravariant functor hA : R-Mod → Ab. For any B ∈ R-Mod, hA(B) := HomR(B, A); for any u : B → B' in R-Mod, hA(u) : HomR(B', A) → HomR(B, A) is given by hA(u)(g) := gu, ∀g ∈ HomR(B', A). The functor hA is denoted also by HomR(-, A).

We study the behavior of HomR(A, -) and HomR(-, A) with respect to direct products and direct sums.

3.20 Proposition. Let A be an R-module and let (Mi)i∈I be a family of R-modules. Then there exist canonical isomorphisms in Ab:

HomR(A, ∏i∈I Mi) ≅ ∏i∈I HomR(A, Mi), HomR(⊕i∈I Mi, A) ≅ ∏i∈I HomR(Mi, A).

Proof. We prove the second isomorphism (the first is proposed as an exercise).

Let S = ⊕i∈I Mi and let σi : Mi → S be the canonical injections. We give two proofs, a direct proof and a category proof. The

direct proof uses the form of the elements in the concrete construc-tions of the direct sum and the direct product and defines a natural homomorphism from HomR(S, A) to ∏i∈I HomR(Mi, A); we then show this is an isomorphism. The category proof shows that HomR(S, A) is a direct product in Ab of the family (HomR(Mi, A))i∈I (more precisely, it satisfies the universality property of the direct product) and applies then 3.3.

Direct proof. Define α : HomR(S, A) → ∏i∈I HomR(Mi, A) as fol-


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lows: for any homomorphism g : ⊕i∈I Mi → A, α(g) =: (gσi)i∈I ∈ ∏i∈I HomR(Mi, A).

It is routine to check that α is a homomorphism. If α(g) = (gσi)i∈I = (0)i∈I, then g(∑i∈I σi(xi)) = ∑i∈I g(σi(xi)) = 0, for any family having finite support (xi)i∈I, with xi ∈ Mi, so g = 0. This proves that α is injective. If (gi)i∈I ∈ ∏i∈I HomR(Mi, A), the universality prop-erty of the direct sum S = ⊕i∈I Mi (definition 3.5) supplies a unique homomorphism g : S → A such that gσi = gi, ∀i ∈ I. So, α is surjec-tive.

Category proof: For any i ∈ I, hA(σi) : hA(S) → hA(Mi) is a mor-phism in Ab. We show that hA(S) is a direct product of the Abelian groups hA(Mi)i∈I, with canonical projections hA(σi). This means that we have to prove that, for any X ∈ Ab and any morphisms τi : X → hA(Mi) in Ab, i ∈ I, there exists a unique morphism τ : X → hA(S) in Ab such that τi = hA(σi)τ. So, ∀x ∈ X, τi(x) : Mi → A is a morphism in R-Mod. By the universality property of the direct sum, ∃! τ(x) : S → A homomorphism such that τi(x) = τ(x)σi. We ob-tain τ : X → HomR(S, A) = hA(S), x & τ(x), which is a morphism in Ab. Indeed, ∀x, y ∈ X, τ(x) + τ(y) : S → A is an R-homomorphism satisfy-ing

(τ(x) + τ(y))σi = τ(x)σi + τ(y)σi = τi(x) + τi(y) = τi(x + y). Since τ(x + y) is the only homomorphism with this property,

τ(x + y) = τ(x) + τ(y). If ϕ : X → hA(S) is a homomorphism with τi = hA(σi)ϕ, then, ∀x ∈ X, ϕ(x) : S → A is a homomorphism with τi(x) = hA(σi(ϕ(x)) = ϕ(x)σi. But τ(x) is unique with this property, so τ(x) = ϕ(x), ∀x ∈ X, i.e. τ = ϕ. !

Many facts in Algebra have a convenient form in the language of exact sequences (used intensively Homological Algebra, for instance).

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3.21 Definition. Consider a (finite or infinite) sequence of R-modules and module homomorphisms:12 (S): ⎯→⎯⎯→⎯⎯→⎯⎯→⎯ GFE vu ,

The sequence (S) is called semiexact at F if vu = 0. This may be rephrased as Im u ⊆ Ker v. The sequence (S) is called exact at F if Im u = Ker v. The sequence (S) is called semiexact (respectively exact) if it is semiexact (respectively exact) at any term. A semiexact se-quence of modules is called a complex of modules.

3.22 Remarks. a) Often, the homomorphisms that are uniquely determined (or clear from the context) are not marked on the arrows. For instance, write 0 → L instead of L⎯→⎯00 , since the only homo-morphism defined on the module 0 is the zero homomorphism.

b) A sequence of the form FE u⎯→⎯→0 is exact if and only if u is a monomorphism. Indeed, Ker u = 0 ⇔ Ker u = Im 0.

c) A sequence of the form 0→⎯→⎯ FE u is exact if and only if u is an epimorphism.

d) The sequence 00 →⎯→⎯→ FE u is exact ⇔ u is an isomor-phism.

3.23 Examples. a) If ϕ : E → F is a homomorphism, then the following sequence is exact:

0ImKer0 →⎯→⎯⎯→⎯⎯→⎯→ ϕϕ πϕι FFE The inclusion homomorphism is denoted by ι and the canonical

surjection by π. The module F/Imϕ is called the cokernel of ϕ and is denoted by Cokerϕ. Thus, we have an exact sequence:

0CokerKer0 →⎯→⎯⎯→⎯⎯→⎯→ ϕϕ πϕι FE . b) If A ≤ R B, then the sequence

12 Called in the sequel sequence of modules.


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00 →⎯→⎯⎯→⎯→ ABBA πι is exact.

3.24 Definition. An exact sequence of the form 00 →⎯→⎯⎯→⎯→ CBA vu is called a short exact sequence. For such a sequence, A is isomorphic to the submodule Im u = Ker v of B, and B/Im u = B/Ker v is isomor-phic to C. This is the reason B is also called an extension of A by C (cf. Example b) above).

3.25 Example. If A and C are two R-modules, then the sequence 00 →⎯→⎯⊕⎯→⎯→ CCAA CA πι is short exact; ιA : A → A ⊕ C is the canonical injection and πC : A ⊕ C → C is the canonical projection (recall that A ⊕ C = A × C). So A ⊕ C is an extension of A by C. The problem of finding all extensions of A by C is highly nontrivial. It is natural in this sense to consider the following definition:

3.26 Definition. Let A and C be two modules and consider two extensions of A by C: 00 →⎯→⎯⎯→⎯→ CBA vu , 00 →⎯→⎯′⎯→⎯→ ′′ CBA vu .

We say that the extensions are equivalent if there exists a homomorphism g : B → B' such that gu = u' and v'g = v. In other words, the diagram below is commutative:

00

00

→⎯→⎯′⎯→⎯→↓↓↓

→⎯→⎯⎯→⎯→

′′ CBA

CBA

vu

g

vu

(the vertical unmarked homomorphisms are identity homomorphisms). A homomorphism g that makes the diagram commutative is called a homomorphism of extensions.

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3.27 Proposition. In the definition 3.26, g is an isomorphism. In particular, the relation defined at 3.26 is an equivalence relation on the class of all extensions of A by C.

Proof.13 Let us show that Ker g = 0. Let b ∈ B with g(b) = 0. Then v'g(b) = 0 = v(b), so b ∈ Ker v = Im u. Thus there exists a ∈ A with u(a) = b; we obtain 0 = g(b) = gu(a) = u'(a); since u' is injective, a = 0, so b = u(a) = 0.

Let b' ∈ B'; then v'(b') ∈ C and the surjectivity of v implies that, for some b ∈ B, v(b) = v'(b'). Since v'g(b) = v(b), v'(g(b)) = v'(b'), which means g(b) − b' ∈ Ker v' = Im u'. There exists a ∈ A with u'(a) = g(b) − b'; since u' = gu, g(b) − b' = u'(a) = gu(a). Thus b' = g(b − u(a)) ∈ Im g. !

3.28 Definition. Consider two complexes: E: …… →⎯→⎯⎯⎯→⎯→ +−

−11

1i

ui

ui EEE ii

F: …… →⎯→⎯⎯⎯→⎯→ +−−

111

iv

iv

i FFF ii , A homomorphism of complexes from E to F is a sequence of mod-

ule homomorphisms g := (gi)i∈Z, gi : Ei → Fi, such that gi + 1ui = vigi, ∀i ∈ Z. In other words, the following diagram commutes:

……

……

→⎯→⎯⎯⎯→⎯→↓↓↓

→⎯→⎯⎯⎯→⎯→

+−

+−

−

+−

−

11

11

1

11

1

iv

iv

i

gggi

ui

ui

FFF

EEE

ii

iii

ii

The homomorphism g is called an isomorphism if gi is an isomor-phism, ∀i ∈ Z. In this case, (gi

−1)i∈Z is also a homomorphism of com-plexes. The complexes E and F are called isomorphic if there exists an isomorphism from E to F.

13 The technique used is called diagram chasing and it is used extensively in

arguments involving diagrams of module homomorphisms.


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3.29 Examples. a) The exact sequence 00 →⎯→⎯⎯→⎯→ CBA vu is isomorphic (in the sense given by the previous definition) to the se-quence 0ImIm0 →⎯→⎯⎯→⎯→ uBBu πι , where ι is the inclusion and π the canonical surjection. The homomorphisms that compose this isomorphism are u0 : A → Im u, u0(a) = u(a), ∀a ∈ A; idB : B → B; in order to define w : C → B/Im u, observe that the surjective homomorphism v induces a canonical isomorphism v0 : B/Ker v → C, v0(b + Ker v) = v(b), ∀b ∈ B (cf. the fundamental isomorphism theo-rem). Set w = v0

−1. Checking the commutativity of the diagram is left to the reader.

b) Any extension of Z2 by Z2 is isomorphic either to Z2⊕Z2 or to Z4.

The extension direct sum of A by C, 00 →⎯→⎯⊕⎯→⎯→ CCAA CA πι ,

has the remarkable property that by reversing the arrows we obtain an-other exact sequence: 00 ←⎯⎯←⊕⎯⎯←← CCAA CA ιπ . The exten-sions of A by C that are isomorphic to the extension direct sum A ⊕ C are characterized in the next proposition:

3.30. Proposition. Let (S): 00 →⎯→⎯⎯→⎯→ CBA vu be a short exact sequence of modules (i.e., B is an extension of A by C). The following statements are equivalent:

a) B is isomorphic (as an extension of A by C) with 00 →⎯→⎯⊕⎯→⎯→ CCAA CA πι , the direct sum extension, where

ιA, ιC are the canonical injections of the direct sum A ⊕ C and πA, πC are the canonical projections.

b) Im u = Ker v is a direct summand in B. c) There exists a homomorphism u' : B → A such that u'u = idA. d) There exists a homomorphism v' : C → B such that vv' = idC. Proof. c)⇒b) and d)⇒b) follow from 3.16.

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a)⇒c), d) Let ϕ : B → A ⊕ C be an isomorphism such that the dia-gram

00

00

→⎯→⎯⊕⎯→⎯→↓↓↓

→⎯→⎯⎯→⎯→

CCAA

CBA

CA

vu

πι

ϕ

commutes. Define u' : B → A, u' = πAϕ. We have u'u = πAϕ u = πAιA = idA (see Remark 3.14). This proves c). Defining v' : C → B, v' = ϕ −1ιC, we have vv' = vϕ −1ιC = πCιC = idC, so d) holds.

b)⇒a) Let B = Im u ⊕ D; thus, ∀b ∈ B, ∃! a ∈ A and d ∈ D such that b = u(a) + d. Define then α(b) = a ∈ A. We obtain a module homomorphism α : B → A. Indeed, if b = u(a) + d, b' = u(a') + d' ∈ B, with a, a' ∈ A, d, d' ∈ D, then

b + b' = u(a) + d + u(a') + d' = u(a + a') + d + d'; so, α(b + b') = a + a' = α(b) + α(b').

In the same way is shown that α preserves multiplication with sca-lars. Note that αu : A → A is idA.

Consider ϕ : B → A⊕C, ϕ = ιAα + ιCv. We have ϕu = (ιAα + ιCv)u = ιAαu + ιCvu = ιA

πCϕ = πC (ιAα + ιCv) = πCιAα + πCιCv = v. So, ϕ is a homomorphism (and isomorphism) of extensions. !

3.31 Definition. A short exact sequence 00 →⎯→⎯⎯→⎯→ CBA vu is called split (or we say, the sequence

splits) if it satisfies the equivalent conditions of the previous proposi-tion.

A monomorphism BA u⎯→⎯→0 is called split if the short exact sequence 0Im0 →→⎯→⎯→ uBBA u splits (⇔ Im u is a direct summand in B).

An epimorphism 0→⎯→⎯ CB v is called split if the short exact se-quence 0Ker0 →⎯→⎯→→ CBv v splits (⇔ Ker v is a direct sum-mand in B).


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The behavior with respect to short exact sequences is primordial in the study of functors defined on categories of modules. The next proposition says the functor Hom is left exact:

3.32 Proposition. Let R M be a module and let 00 →⎯→⎯⎯→⎯→ CBA vu be a short exact sequence. Then the following sequence ( ) ( ) ( ) ( ) ( )ChBhAh MvhMuhM MM

⎯⎯ →⎯⎯⎯ →⎯→0 is exact in Ab. In particular, if u : A → B is an R-module monomor-phism, then h M(u) is an Abelian group monomorphism.

Proof. We have to prove that hM(u) is monomorphism and Im hM(u) = Ker hM(v). Let ϕ ∈ hM(A) (i.e. ϕ : M → A is a homomorph-ism) such that hM(u)(ϕ) = uϕ = 0. Since u is monomorphism, ϕ = 0. So, Ker hM(u) = 0.

Since hM(v)hM(u) = hM(vu) = hM(0) = 0, Im hM(u) ⊆ Ker hM(v). The reverse inclusion also holds: if ψ ∈ hM(B) (i.e. ψ : M → B) is in Ker hM(v), then vψ = 0, so Imψ ⊆ Ker v = Im u. But u is injective, so u0 : A → Im u (u0(a) = u(a), ∀a ∈ A) is an isomorphism; let u' : Im u → A be the inverse of u0. Then u'ψ : M → B is well defined and it is a homomorphism. Also, hM(u)(u'ψ) = uu'ψ = ψ.

If u is monomorphism, consider the exact sequence: 0Im0 →⎯→⎯⎯→⎯→ uBBA u and write that hM is left exact to obtain that h M(u) is monomorphism.!

The contravariant functor hM is also left exact. We invite the reader to prove:

3.33 Proposition. Let R M be an R-module and let 00 →⎯→⎯⎯→⎯→ CBA vu be a short exact sequence. Then the sequence ( ) ( ) ( ) ( ) ( )AhBhCh M

uhM

vhM

MM ⎯⎯ →⎯⎯⎯ →⎯→0

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is exact in Ab. In particular, if v : B → C is an R-module epimor-phism, then hM(v) is a monomorphism. !

Exercises

In the exercises, R denotes a fixed ring with identity. 1. Prove that the R-module S is a direct sum of the family of R-modules (Mi)i∈I, with canonical injections (σi)i∈I, if and only if for any x ∈ S, there exists a unique family of elements (xi)i∈I, with xi ∈ Mi, ∀i ∈ I, having finite support, such that x = ∑i∈I σi(xi). 2. Let A and B be R-modules. Prove that there exists a (canonical) isomorphism A⊕B ≅ B⊕A. Generalization. 3. Prove that the R-module P is a direct product of family of R-modules (Mi)i∈I, with canonical projections (πi)i∈I, if and only if for any family (xi)i∈I, with xi ∈ Mi, ∀i ∈ I, there exists a unique x ∈ P such that πi(x) = xi, ∀i ∈ I. 4. Let R be a ring and let I be a left ideal of R. For any module R A, de-fine the submodule of A:

IA := <ia | i ∈ I, a ∈ A>. a) Prove that IA = i1a1 + + inan | n ∈ N, i1, , in ∈ I, a1, ,

an ∈ A. b) If A = ⊕s∈S As (internal direct sum), then IA = ⊕s∈S IAs. c) If A = ∏s∈S As, then IA ⊆ ∏s∈S IAs (we identify As with its image

in ∏s∈S As; the same applies to IAs). If I is finitely generated, then the inclusion is in fact an equality. Prove that he inclusion may be strict. 5. Taking the universality property of the direct product as its defini-tion, prove that the canonical projections are surjective.


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6. Find homomorphisms u, v, w such that the following sequence is exact:

0 → Z3 399 ZZZ ⎯→⎯⎯→⎯⎯→⎯ wvu → 0. 7. Let R be a PID. Prove that the R-module M is cyclic (generated by one element) ⇔ there exists an exact sequence of the form 0 → R → R → M → 0. 8. Let R be a ring and let M be an R-module. Then any R-epimorphism ϕ : M → R splits. Is it true that for any ring R and for any R-module M, any R-monomorphism ψ : R → M splits? 9. Suppose that the following diagram of R-modules commutes and has exact rows:

CBA

CBA

vu

vu

′⎯→⎯′⎯→⎯′↓↓↓

⎯→⎯⎯→⎯

′′

γβα

Prove, using diagram chasing, that: a) If α, γ and u' are monomorphisms, then β is monomorphism. b) If α, γ and v are epimorphisms, then β is an epimorphism. c) If β is a monomorphism and α, v are epimorphisms, then γ is an

epimorphism. d) If β is an epimorphism and u', γ are monomorphisms, then α is a

monomorphism. 10. Give an example of an R-module M with a direct summand S in M that has an infinity of complements. 11. Show that any composition of split monomorphisms (epimorphisms) is a split monomorphism (epimorphism). Give an example of monomorphisms u, v such that uv splits, but v does not split.

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II.4 Free modules

Many nice properties of the vector spaces are consequences of the fact that any vector space has a basis. The existence of a basis is not guaranteed for a module, and this increases considerably the difficulty in the study of modules in comparison to the vector spaces. It is natu-ral to define and to study the class of modules that have a basis.

In this section, R is a ring with identity and all modules are left R-modules.

4.1. Definition. Let M be an R-module, n ∈ N* and let x1, , xn be a family of elements of M. We say that the family x1, , xn is linearly independent (over R)14 if, for any r1, , rn ∈ R, r1x1 + + rnxn = 0 implies r1 = = rn = 0. In other words, any linear combination of x1, , xn is 0 if and only if all its coefficients are 0.

A family of elements in M that is not linearly independent is called linearly dependent. A relation of the form r1x1 + + rnxn = 0, with r1, , rn ∈ R, not all zero, is called a relation of linear dependence of the family x1, , xn.

4.2. Remarks. a) If there exists i ≠ j with xi = xj, then the family x1, , xn is linearly dependent: the linear combination xi − xj is 0. Thus, in studying linear independence we may suppose that x1, , xn are distinct. On the other hand, the notion of linear independence does not depend on the indexing of x1, , xn. This is the reason we can speak about a linearly dependent (finite) subset of M.

b) The set x (containing a single element x ∈ M ) is linear independent if and only if ∀r ∈ R, rx = 0 implies r = 0. This suggest the following definition: the annihilator of x in R is

14 The reference to the ring R is often omitted.

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AnnR(x) := r ∈ R | rx = 0 (sometimes denoted by lR(x)). Since AnnR(x) is exactly the kernel of the R-module homomorphism ρx : R → M, ρx(r) = rx, ∀r ∈ R, AnnR(x) is a left ideal in R and (by the isomorphism theorem) the submodule Rx = Imρx is isomorphic as a left R-module to R/AnnR(x). Summarizing: x is linearly independent ⇔ AnnR(x) = 0 ⇔ ρx is an isomorphism from R to Rx.

4.3. Definition. Let X be a subset of R M. We say that X is linearly independent over R (or free over R) if any finite subset of X is linearly independent over R in the sense of the remark above. If X is not line-arly independent, we say that X is linearly dependent over R. Thus, X is linearly dependent if and only if there exists a finite linearly dependent subset of X.

4.4. Remarks. a) The empty subset of M is linearly independent. b) Let xii∈I be a family of elements of the R-module M. The family xii∈I is linearly independent if and only if, for any fam-

ily rii∈I of elements of R, with finite support, ∑i∈I rixi = 0 implies ri = 0, ∀i ∈ I.

It follows that the sum of the family of submodules Rxii∈I is di-rect (see 3.11). In fact, xii∈I is linearly independent if and only if the sum of the family of submodules Rxii∈I is direct and AnnR(xi) = 0,∀i ∈ I (prove this!).

c) Every subset of an linearly independent set is linearly independ-ent.

4.5. Examples. a) The set 1, containing only the unity of the ring R (seen as a left R-module) is linearly independent. More generally, ∀r ∈ R, r is linearly dependent ⇔ r is a right zero divisor in R (∃s ∈ R, s ≠ 0, such that sr = 0).

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b) The Z-module Z3 has no Z-linearly independent subsets: for any x ∈ Z3, 3x = 0 . Can you generalize this? Of course, the set 1 is linearly independent over Z3, as seen at example a).

c) If R is a domain, in the R-module R[X] the set X n| n ∈ N is linearly independent. This amounts to saying that a polynomial a0 + a1X + + an X

n (a0, a1, , an ∈ R) is 0 if and only if a0 = = an = 0. More generally, if fn | n ∈ N is a family of polynomials such that deg fn ≠ deg fm if m ≠ n, then the family fn | n ∈ N is linearly independent.

4.6. Definition. A subset B of an R-module M is called a basis of M if it is simultaneously linearly independent and a system of generators for M. The module M is called a free module if it has a basis.

4.7 Proposition. Let M be an R-module and let B be a subset of M. Then: B is a basis of M if and only if any x ∈ M is written uniquely as a linear combination of elements of B.

Proof. Let B = eii∈I be a basis of R M and let x ∈ M. Since B generates M, there exists a family rii∈I of elements in R with finite support, such that x = ∑i∈I riei. If sii∈I is another family of elements in R, with x = ∑i∈I siei, then ∑i∈I (si − ri)ei = 0. The linear independ-ence of B implies si = ri, ∀i ∈ I.

Conversely, if any element in M is written uniquely as a linear combination of elements of B, then B is a system of generators of M. Writing the fact that 0 has a unique writing as a linear combination of elements of B, we obtain that B is linearly independent. !

With the notations above, for x ∈ M, the elements rii∈I in R with x = ∑i∈I riei are called the coordinates of x in the basis B = eii∈I.

4.8. Examples. a) ∅ is a basis (the only one!) of the module 0. b) 1 is a basis of R R. More generally, r is a basis of R R if and

only if r is right invertible in R.

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c) If I is a set, the R-module R( I ) (the direct sum of | I | copies of R, with canonical injections σi : R → R( I )) is free, a basis being eii∈I, where ei = σi(1). This basis is called the canonical basis of R( I ); R( I ) is also called the free R-module on the set I (or the free R-module of ba-sis I ). If I = 1, , n, the canonical basis of the free R-module R n are e1 = (1, 0,, 0), e2 = (0, 1,, 0), , en = (0, 0,, 1).

A module homomorphism is determined by its values on a generat-ing set. But, for an arbitrary generating set, there may be no homomorphism that takes prescribed values on the elements of the generating set. In the privileged case of free modules, for any choice of the values on the elements of a basis, a unique homomorphism takes the respective values on the elements of the basis:

4.9 Proposition. (The universality property of the free module of basis B) Let L be a free R-module of basis B and let i : B → L be the inclusion map. The pair (L, i) has the following universality property:

For any pair (M, u), where M is an R-module and u : B → M is a map, there exists a unique homomorphism v : L → M with vi = u.

If B = (ei)i∈I, this can be formulated: For any module M and any family (yi)i∈I of elements of M, there ex-

ists a unique homomorphism v : L → M such that v(ei) = yi, ∀i ∈ I. Moreover, we have: a) v is injective ⇔ u(B) is linearly independent. b) v is surjective ⇔ u(B) is a system of generators for M. c) v is an isomorphism ⇔ u(B) is a basis.

Proof. Uniqueness claim: suppose v, w ∈ HomR(L, M) satisfy v(ei) = yi = w(ei), ∀i ∈ I. Let x ∈ L; since (ei)i∈I is a basis, there exists a unique family (ri)i∈I of elements in R such that x = ∑i∈I riei. We have v∑i∈I riei = ∑i∈I riv(ei) = ∑i∈I riyi. The same thing is obtained for w.

For the existence claim, we have to prove that

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v(x) := ∑i∈I riyi, for any x = ∑i∈I riei with (ri)i∈I family of elements in R, having a finite support,

defines a module homomorphism. Since, for any x ∈ L there exists a unique family (ri)i∈I of elements in R such that x = ∑i∈I riei, v is well defined. Let x = ∑i∈I riei and y = ∑i∈I siei be elements in L, with (ri)i∈I, (si)i∈I families of elements in R. For any a, b ∈ R, we have:

v(ax + by) = v∑i∈I ariei + ∑i∈I bsiei = v∑i∈I (ari + bsi)ei = ∑i∈I (ari + bsi)yi =

= ∑i∈I ariyi + ∑i∈I bsiyi = av(x) + bv(y). We prove a): Ker v = ∑i∈I riei | (ri)i∈I ∈ R( I ), ∑i∈I riyi = 0. It is

clear that Ker v = 0 ⇔ (yi)i∈I is linearly independent. The rest of the proof is left to the reader. !

The R-modules of the type R( I ) are all free R-modules:

4.10 Proposition. Let L be a free R-module and let (xi)i∈I be a basis of L. Then R L ≅ R R

( I ). Proof. Let (ei)i∈I be the canonical basis in R( I ). By the result above,

there exists a unique homomorphism u : L → R( I ) with u(xi) := ei, ∀i ∈ I. Since Im u includes eii∈I, which generates R( I ), it follows that u is surjective. If x = ∑i∈I rixi ∈ Ker u (with (ri)i∈I family of ele-ments in R), then 0 = u(x) = ∑i∈I riei. Since eii∈I is a basis, ri = 0, ∀i ∈ I, so x = 0. !

A direct sum of free modules is also free:

4.11 Proposition. Let (Mi)i∈I be a family of free R-modules. Then the (external) direct sum ⊕i∈I Mi is a free R-module.

Proof. Let M := ⊕i∈I Mi. For any j ∈ I, let σj : Mj → M be the canonical injection and let Bj be a basis of Mj. We show that B := ∪i∈I σi(Bi) is a basis of M. By 3.10, M is the (internal) direct sum

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of the family of submodules (Im σi)i∈I. Since σi(Bi) generates Im σi,

∀i ∈ I, ∪i∈I σi(Bi) generates M. The linear independence of B is easy

to show (only the notations are complicated): let Bi = (eit)t∈Ti the basis in Mi and let yit = σi(eit). If rit | i ∈ I, t ∈ Ti is a finite support family of elements of R, with ∑i∈I∑t∈Ti rityit = 0, then ∑t∈Tk rktykt = −∑i∈I \k∑t∈Ti rityit, ∀k ∈ I. But Mk ∩∑i∈I \k Mi = 0, so

∑t∈Tk rktykt = 0. But (ykt)t∈Tk is a basis in Mk, so rkt = 0, ∀k ∈ I, ∀t ∈ Tk.!

4.12 Proposition. Every module is (isomorphic to) a factor module of a free module. More precisely, if M is an R-module and S is a sys-tem of generators of M, then there exists an epimorphism ϕ : R(S) → M. Hence M ≅ R(S)/Kerϕ.

Proof. Let (es)s∈S be the canonical basis in R(S). Define ϕ : R(S) → M by ϕ(es) = s, ∀s ∈ S (see 4.9). Since the submodule Imϕ includes S, and S generates M, we obtain Imϕ = M. !

This simple fact is very important: if one knows the structure of the submodules and of the factor modules of free modules, the structure of an arbitrary module is known. This method will be used to study the structure of finitely generated modules over a PID. Here is another application of this principle:

4.13 Example. If M is an R-module, then there exists an exact se-quence: → Ln+1 → Ln → → L1 → L0 → M → 0, where Ln is free, ∀n ∈ N. Such a sequence is called a free resolution of M. For the proof, let L0 be free and let u0 : L0 → M be an epimor-phism (as given by the Proposition above). Then we have an exact se-quence: 0Ker0 0

00 →⎯→⎯⎯→⎯→ MLu ui . Apply again the proposi-tion for Ker u0 and obtain a free module L1 and an epimorphism u1 : L1 → Ker u0. We have now the exact sequence

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0Ker0 01011 →⎯→⎯⎯⎯→⎯→→ MLLu uui)

since Ker u0 = Im iu1. One continues by induction on n such that ex-ists an exact sequence of the form: 0 → Kn → Ln → → L1 → L0 → M → 0, with Li free, 1 ≤ i ≤ n.

In a vector space, any two bases have the same cardinal. In the case of a free module over an arbitrary ring, this fact is not guaranteed. Nevertheless, the free modules over a commutative ring have this property; in fact, the proof of this result reduces the problem to the case of vector spaces.

4.14 Proposition. Let R be a commutative ring and let M be a free R-module. Then any two bases of M have the same cardinal.

Proof. Let us try to connect the R-module M to a vector space over a field. Recall that, if I is a maximal ideal in R, then R/I is a field.

A maximal ideal I of R exists (see 1.20). Then R/I is a field (see Appendix, Prime and maximal ideals). The module M has no natural structure of R/I-vector space: the natural multiplication given by (r + I)x = rx, for r ∈ R, x ∈ M, is not well defined unless ix = 0, ∀i ∈ I. This is the reason we need to kill off the elements x of M for which ix ≠ 0 for some i ∈ I, as follows:

Let IM := a1x1 + + anxn | n ∈ N, ai ∈ I, xi ∈ M, ∀i = 1, , n. A straightforward check shows that IM is an R-submodule of M and that the factor module M/IM is also an R/I-module (i.e. an R/I-vector space!) with respect to the external operation:

(r + I)(x + IM) := rx + IM, ∀r ∈ R, ∀x ∈ M. Let π : M → M/IM be the canonical surjection, π(x) = x + IM,

∀x ∈ M. We prove that: if (eα)α ∈ A is a basis of R M, then (π(eα))α ∈ A is a basis in the R/I-vector space M/IM.

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Indeed, (π(eα))α ∈ A generates the R-module M/IM, since π is surjec-tive and (eα)α ∈ A generates M. It easily seen that it is also generates M/IM as an R/I-module.

Let us show that (π(eα))α ∈ A is R/I-linear independent. Let (rα + I)α ∈ A be a family of elements of R/I (where rα ∈ R, ∀α ∈ A), of finite support, such that

∑α∈A (rα + I)π(eα ) = 0 + IM. This means: ∑α∈A rα eα ∈ IM. Note that, for any x ∈ IM, x = a1x1 +

+ amxm ∈ IM, for some ai ∈ I, xi ∈ M, ∀i = 1, , m. Writing every xi as a linear combination of (eα)α ∈ A, it follows that x is of the form ∑α∈A bα eα , where bα ∈ I, ∀α ∈ A (since ai ∈ I and I is an ideal). Thus,

∑α∈A rα eα = ∑α∈A bα eα But (eα)α ∈ A is a basis, so rα = bα ∈ I, ∀α ∈ A. Thus, rα + I = bα + I

= 0 + I, ∀α ∈ A. Now, taking two bases (eα)α ∈ A and (vβ)β ∈ B of R M, we obtain that

(π(eα))α ∈ A and (π(vβ))β ∈ B are bases in the R/I-vector space M/IM. Since two bases of a vector space have the same cardinal, we obtain |A| = |B|. !

4.15 Definition. Let R be a commutative ring and let L be a free R-module. The cardinal of a basis of R L is called the rank of L, de-noted rankR L (or, simply, rank L). The previous result ensures that the definition is independent of a choice of the basis in R L. If R is a field, the rank of a vector space is the same as its dimension.

We study now the homomorphisms between free modules of finite rank. The idea is to mimic the vector space situation: for two fixed bases in the modules, we associate a matrix to a homomorphism. In this manner, the operations with (and the properties of) module homomorphisms translate in matrix language (and conversely). We

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suppose that the ring R is commutative, although some results hold for any ring with identity.

4.16 Definition. Let E and F be free R-modules and let ϕ : E → F be a homomorphism. Fix e = (e1, , em) an ordered basis in E and f = (f1, , fn) an ordered basis15 in F. For any i ∈ 1, , m, there exist and are unique aij ∈ R, ( j ∈ 1, , n) such that

ϕ(ei) = ai1 f1 + + ain fn. The matrix (aij)1≤i≤m, 1≤j≤n ∈ Mm, n(R) is denoted by Me, f(ϕ) and it is

called the matrix of the homomorphism ϕ (in the pair of bases (e, f)). In other words, the i-th row of Me, f(ϕ) is made up of the coordinates ai1, , ain of ϕ(ei) in the basis f. If E = F, one usually takes e = f and Me, f(ϕ) is denoted simply Me(ϕ). The importance of ordering the bases is quite clear now: a permutation of the basis (e1, , em) leads to an-other matrix of ϕ, whose rows are a permutation of the rows of the ini-tial matrix.

4.17 Remark. This is the algebraic convention of writing the ma-trix Me, f(ϕ). Of course, one can agree to the geometric convention: write the coordinates of ϕ(ei) in the basis f on the i-th column (which means considering the transpose matrix tMe,f(ϕ)). Choosing this rule leads to changing property b) in the next sentence into: tMe, f(ψ ϕ ) = tMf, g(ψ )·tMe, f(ϕ ).

4.18 Proposition. a) The mapping: Me, f : HomR(E, F) → Mm, n(R),

ϕ & Me, f(ϕ) is an R-module isomorphism.

15 We consider the bases as being totally ordered (the place of the element in the

basis matters).

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113

b) If G is a free R-module, g = (g1,, gp) is a basis in G, and ψ : F → G is an R-module homomorphism, then Me, g(ψ ϕ) = Me, f(ϕ)·Mf, g(ψ).

c) The mapping Me : EndR(E) → Mm(R) is an R-algebra anti-isomorphism (i.e. Me : EndR(E)op → Mm(R) is an R-algebra isomor-phism).

Proof. a) Let η : E → F be another homomorphism and let (bij)1≤i≤m, 1≤j≤n = Me, f(η). Then

(ϕ +η)(ei) = ϕ(ei) +η(ei) = ∑j aij fj + ∑j bij fj = ∑j (aij + bij) fj So, Me, f(ϕ +η) = (aij + bij)1≤j≤n, 1≤i≤m = Me, f(ϕ) + Me, f(η), so Me, f is

an Abelian group homomorphism. If Me, f(ϕ) = 0 (the matrix 0 ∈ Mm, n(R)), then ϕ(ei) = 0, 1≤ i ≤ m. Since (ei)1≤i≤m is a basis in E, ϕ = 0; this shows that Me, f is injective. For the surjectivity, let A = (aij) ∈ Mm, n(R). The universality property of the free module yields a unique homomorphism ϕ ∈ HomR(E, F) such that ϕ(ei) = ai1 f1 + + ain fn, i.e. Me, f(ϕ) = A. If r ∈ R and ϕ ∈ HomR(E, F), a sim-ple computation shows that Me, f(rϕ) = rMe, f(ϕ).

b) Let Mf, g(ψ) = (bjk)1≤j≤n, 1≤k≤p. We have (ψ ϕ)(ei) = ψ ∑j aij fj = ∑j aijψ( fj ) = ∑j aij∑k bjk gk = ∑k ∑j aijbjk gk.

So, Me, g(ψ ϕ) = Me, f(ϕ)·Mf, g(ψ). c) The statement follows from a) and b): a) implies that Me is an

R-module homomorphism and b) implies that Me : EndR(E)op → Mm(R) is a ring homomorphism. !

In the setting above, if e = (e1,, en) and f = (f1,, fn) are ordered bases of the free module E, then every element of the basis f is written uniquely as a linear combination of ei's:

∑=

=n

jjiji esf

1, sij ∈ R

We obtain the basis change matrix Te, f = (sij) ∈ Mn(R). If g is an-other basis in E, then a straightforward computation yields:

Te, g = Tf, g·Te, f

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In particular, In = Te, e = Te, f ·Tf, e, i.e. the basis change matrix is invertible in Mn(R). When is a matrix in Mn(R) invertible?

If A ∈ Mn(R), the adjoint matrix of A is A* (the entry (i, j) of A* is (−1)i+jdet(Aji), where Aji is the matrix in Mn−1(R) obtained by suppress-ing the row j and the column i of the matrix A). Then A·A* = A*·A = det(A)·In. This shows that, if det(A) ∈ U(R), then A is invertible in Mn(R) and A−1 = (det A)−1A*. Conversely, if A is invert-ible, A·A−1 = In implies (taking determinants) that det(A·A−1) = det(A)det(A−1) = 1, so det(A) ∈ U(R). Thus, we have:

4.19 Proposition. Let R be a commutative ring and let n ∈ N*. a) A matrix S ∈ Mn(R) is invertible in Mn(R) if and only if det(S) is

invertible in R. b) Let E be a free R-module of rank n and let e = e1,, en, f, g be

bases in E. Then: Te, g = Tf, g·Te, f

In particular, any basis change matrix Te, f is invertible. Con-

versely, if S = (sij) is invertible in Mn(R) and ∑=

=n

jjiji esf

1, 1 ≤ i ≤ n,

then f1,, fn is a basis in E. !

The facts not already proven in the proposition above have proofs that can be taken word for word from the case of the vector spaces.

The multiplicative group of the n×n invertible matrices with entries in R, U(Mn(R)) = S ∈ Mn(R) | ∃T ∈ Mn(R) such that ST = TS = I = S ∈ Mn(R) | det S ∈ U(R) is also denoted GL(n, R) or GLn(R) and is called the linear general group of degree n over R.

The following formalism is useful in many situations involving vectors and matrices. Let R be a commutative ring and let E be an R-module. If m, n, p are positive integers, let Mn, p(E) denote the set of n×p matrices with entries in E. It is evident that Mn, p(E) is an Abelian

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group with respect to usual matrix addition. For any A = (aij) ∈ Mm, n(R) and any X = (xjk) ∈ Mn, p(E), define the product AX ∈ Mm, p(E) by

AX = (yik) ∈ Mm, p(E), yik = ∑=

n

jjkij xa

1, ∀i ∈ 1, , m, ∀k ∈ 1, , p,

Of course, aijxjk is the product given by the R-module structure of E.

We obtain an external operation · : Mm, n(R)×Mn, p(E) → Mm, p(E),

with the following properties (the proof of which is straightforward): (A + B)X = AX + BX, A(X + Y) = AX + AY, ∀ A, B ∈ Mm, n(R), ∀ X,

Y ∈ Mn, p(E). Moreover, if q ∈ N*, then for any A ∈ Mq, m(R), B ∈ Mm, n(R),

X ∈ Mn, p(E) we have: (AB)X = A(BX).

If I is the identity matrix in Mn(R), then IX = X, ∀X ∈ Mn, p(E). Thus, Mn, p(E) is a left module over the ring Mn(R). As an application, let us see how the matrix of a homomorphism of

free modules changes when we change bases in the modules.

4.20 Proposition. Let E and F be free R-modules, rank E = n, rank F = m and let e = (e1,, en), e' = (e'1,, e'n) be bases in E, f = f1,, fm, f' = ( )mff ′′…,1 bases in F. Let S = (sij) ∈ Mn(R) be the basis change matrix from e to e' and let T = (tij) ∈ Mm(R) be the basis change matrix from f to f'. If ϕ : E → F is a homomorphism, then:

Me, f(ϕ) = S·Me', f'(ϕ)·T −1. Proof. Let Me, f(ϕ) = A = (aij) ∈ Mn, m(R), i.e.:

ϕ(ei) = ∑j aij fj, ∀i ∈ 1, , n. We view e = t(e1, , en) as a column matrix, e ∈ Mn, 1(E) and, simi-

larly, f ∈ Mm, 1(F). The relations above can be written as ϕ(e) = A·f,

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where ϕ(e) = t(ϕ(e1), , ϕ(en)). In the same way, if B = Me', f'(ϕ), then ϕ(e') = B·f'. Also we have e' = S·e and f' = T·f, or f = T −1·f'. Since ϕ is an R-module homomorphism, we have ϕ(S·e) = S·ϕ(e) (prove this!). Thus:

ϕ(e') = ϕ(S·e) = S·ϕ(e) = S·(A·f) = (SA)·f = (SA)·(T −1·f') = (SAT −1)·f', which says that the matrix of ϕ in the bases e' and f' is SAT −1. !

Exercises

1. Let R be a ring and let L be a free R-module. Then any R-epimorphism ϕ : M → L splits. In particular, if K is a field, any short exact sequence of K-linear spaces 0 → U → V → W → 0, is split. 2. a) Let R be a domain and g1, , gn ∈ R[X], with deg gi ≠ deg gj if i ≠ j. Then g1, , gn are linearly independent in the R-module R[X].

b) Let K be a field of characteristic 0 (i.e. n·1 ≠ 0, ∀n ∈ N*, where 1 is the identity of K) and let a ∈ K. Then 1, X − a, (X − a) 2, , (X − a) n, is a basis in the K-vector space K[X]. If p ∈ K[X], com-pute the coordinates of p in this basis. (Hint. Recall the Taylor series expansion, applied to polynomials.) 3. Prove that in the Z-module Q any subset having at least two ele-ments is linearly dependent and that Z À ZQ. Deduce that ZQ is not free. 4. Let G be a finite Abelian group. Can G be a free Z-module? 5. Characterize the ideals I of the commutative ring R that are free R-modules.

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6. If M is a free R-module of basis B, express |M| as a function of |R| and |B|. 7. Let M be an R-module with the following property:

There exists a subset B of M such that, for any R N and any function ϕ : B → N, there exists a unique R-homomorphism ψ : M → N with ψ|B = ϕ.

Then M is free and B is a basis of M. (the converse of Prop. 4.9). 8. Let R be a commutative ring and let I, J be ideals in R. Consider the statements:

(i) I ≅ J (as R-modules). (ii) R/I ≅ R/J (as R-modules). (iii) R/I ≅ R/J (as rings). Which are the valid implications between the above statements?

9. Let W be a finite dimensional K-vector space and let U, V ≤ KW. a) Show that dim(W/U) = dim W − dim U. b) Using the isomorphism theorems, prove that

dim(U + V) = dim U + dim V − dim(U ∩ V). 10. Is it true that any submodule (resp. factor module) of a free module is still a free module?

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III. Finitely generated modules over principal ideal domains

In the case of finitely generated modules over PID's very precise structure theorems are available. Applying these theorems for Z, a complete description of finitely generated Abelian groups is obtained. Another important application is to the problem of invariant subspaces of an endomorphism of a finitely dimensional vector space: we obtain the existence of a basis in which the given endomorphism has a sim-ple matrix (the Jordan canonical form).

III.1 The submodules of a free module

Any module is a factor module of a free module. In order to build a factor module we need a submodule. It is therefore natural to study first the submodules of a free module. Recall that, if R is commuta-tive, any free R-module E has the property that any two bases have the same cardinal (called the rank of E).

In this section, unless otherwise specified, R denotes a principal ideal domain.


119

1.1 Theorem. Let E be a free R-module of rank n and let F be a submodule of E. Then F is free and rank F ≤ n.

Proof. Evidently, we may suppose F ≠ 0. We use an argument by induction on n. If n = 1 let e be a basis of

E. Then E = Re ≅ R. In this case, the statement of the theorem be-comes: any submodule ( = ideal) of R is free, of rank ≤ 1. Since R is a PID, this is true : any nonzero ideal of R is of the form Ra, with a ∈ R; so a is a basis of Ra.

Suppose that, for any free R-module H of rank n − 1, any submod-ule of H is free, of rank ≤ n − 1. Take R E, free of rank n, e1,, en a basis in E and F ≤ R E. Let L := Re2 + + Ren and G := F ∩ L. Then L

is free, of basis e2,, en, so its rank is n − 1, and G = F ∩ L ≤ L. By

induction, G is free of rank m ≤ n − 1. If G = F, we are done. If not, note that F + L ≤ R E, so

LE

LLF

FLF

GF ≤+≅= ∩

Accordingly, 0 ≠ F/G is isomorphic to a submodule of E/L. But E/L is free, of basis e1 + L (easy check) and thus F/G is free of rank 1 (by the case n = 1, already proven). Let f1,, fm be a basis in G and f0 + G a basis in F/G. We claim that B = f0, f1,, fm is a basis

in F. Indeed, if a0, a1,, am ∈ R, with a0 f0 + a1 f1 + + am fm = 0, then

a0 f0 + G = 0 + G (since a1 f1 + + am fm ∈ G); but f0 + G is a basis in F/G, so a0 = 0. We get a1 f1 + + am fm = 0, which implies f1 = = fm = 0 (f1,, fm being a basis in G). Therefore, B is linearly independent.

Let us show that it is a generating system. Let x ∈ F. Then x + G ∈ F/G, so there is some a0 ∈ R such that x + G = a0 f0 + G. This means that x − a0 f0 ∈ G, so x − a0 f0 = a1 f1 + + am fm, for some ai ∈ R. Thus, x = a0 f0 + a1 f1 + + am fm. !


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Let R E and F ≤ R E, as above. If e1,, en is a basis of E and f1,, fm is a basis of F, any fi is written uniquely as a linear combination of e1,, en:

fi = ∑j aijej, with aij ∈ R. We obtain a matrix A = (aij) ∈ Mm, n(R). Can we choose the bases in

E and F such that A has a simple (e.g., diagonal) form?1. The next theorem says the answer is Yes.

1.2 Theorem. Let R be a principal ideal domain, E a free R-module of rank n and F a submodule of E. Then there exists a basis e = (e1,, en) in E and a basis f = ( f1,, fm) in F, such that m ≤ n, fi = diei, ∀i ∈ 1,, m, where d1,, dm ∈ R and d1|d2||dm.

(Beginning of) Proof. For any basis e' = (e'1,, e'n) in E and any basis f' = ( )mff ′′…,1 in F, there exists a unique matrix A = (aij) ∈ Mm, n(R) such that f 'i = ∑j aije'j, ∀i ∈ 1, , m. If we see e' as a column matrix, e' ∈ Mn, 1(E), (see the discussion preceding 4.20) and f' as a matrix in Mm, 1(R), then these relation are written shortly:

f' = Ae'. Suppose the problem is solved: we have two bases e = (e1,, en)

and f = f1,, fm as in the statement of the theorem. Then the unique matrix D = (dij) ∈ Mm, n(R) with f = De has the properties : dij = 0 if i ≠ j and d11|d22||dmm. Let V ∈ GLn(R) the basis change matrix from basis e to basis e' and U ∈ GLm(R) the basis change matrix from basis f to basis f'. Thus, e' = Ve and f' = Uf. Replacing in f' = Ae', we have Uf = A(Ve) = (AV)e; so

f = U −1(AV)e. Comparing with f = De, we obtain D = U −1AV. We see that the existence of the bases e and f with the desired prop-

erty is equivalent to the following: given A ∈ Mm, n(R), there exist 1 What about the case of vector spaces?


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invertible matrices U and V such that U −1AV = D = (dij) ∈ Mm, n(R) is a diagonal matrix (i ≠ j implies dij = 0), with the additional condition d11|d22||dmm.

We reduced the statement to a problem concerning matrices with elements in R. The following definitions are helpful:

1.3 Definitions. Let m, n ∈ N*. a) A matrix D = (dij) ∈ Mm, n(R) is called a diagonal matrix if its en-

tries not situated on the main diagonal are zero: i ≠ j implies dij = 0. If r = min(m,n) and d1, , dr ∈ R,

diag(d1, , dr) := ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00

001

rd

d

…!!*!

…

denotes the diagonal matrix (dij) ∈ Mm, n(R) with dii = di, ∀i ∈ 1, , r.

b) We call a diagonal matrix D = diag(d1, , dr) ∈ Mm, n(R) (where r = min(m,n)) canonically diagonal (or in Smith normal form) if d1|d2||dr.

c) The matrices A, B ∈ Mm, n(R) are called arithmetically equivalent (denoted A ∼ B) if there exist invertible matrices U ∈ Mm(R) and V ∈ Mn(R) such that B = UAV. This is an equivalence relation on Mm, n(R) (exercise!).

We finish the proof of theorem 1.2 by proving the next theorem, which is interesting on its own: the proof is in fact an algorithm used to compute the Smith normal form of a matrix.

1.4 Theorem. Let R be a PID and let m, n ∈ N*. Then any matrix A ∈ Mm, n(R) is arithmetically equivalent to a matrix in Smith normal form. Moreover, the Smith normal form of A is unique in the following sense: if D = diag(d1,, dr) and D' = diag(d'1,, d'r) are in Smith nor-


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mal form and are arithmetically equivalent, then d1 ∼ d'1,, dr ∼ d'r (" ∼ " means here associated in divisibility).

If D is in Smith normal form and is arithmetically equivalent to A, D is called the Smith normal form of A; it is uniquely determined up to an association in divisibility of the entries on the diagonal.

Before proving the theorem, let us review the transformations that can be performed on the matrix A, such that the resulting matrix is arithmetically equivalent to A. We shall see that these are the transformations that arise when computing determinants: swapping rows (or columns), addition to a row (column) of another row (column) multiplied by some element.

1.5 Definitions. Let m ∈ N* and let I be the identity matrix of the ring Mm(R). An elementary matrix is a square matrix in Mm(R) which is of one of the following types:

- Type I : Tij(a), where a ∈ R, i, j ∈ 1, , m, i ≠ j. Tij(a) is ob-tained from I by adding to the row i the row j multiplied by a.

i j

1 ! ! 0

* ! !

i + + 1 + a Tij(a) = ! * !

j + + 0 + 1

* 0 1

- Type II : Pij, where i, j ∈ 1, , m, i ≠ j. Pij is the matrix ob-tained from I by swapping row i with row j. - Type III : Di(u), where i ∈ 1, , m and u ∈ U(R). Di(u) is the matrix obtained from I by multiplying row i by u.


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i j i

1 ! ! 0 1 0 ! 0

* ! ! 0 1 ! 0

i ++ 0 + 1 * !

Pij = ! * ! i +++ u

j ++ 1 + 0 *

*

Di(u) =

0 0 + 0 1 0 1

If A ∈ Mm, n(R), and Tij(a), Pij, Di(u) are elementary matrices

in Mm(R) as above, a direct calculation shows that: - Tij(a)A is obtained from A by adding to the row i the row j

multiplied by a. - Pij A is obtained from A by swapping the row i with the row j. - Di(u)A is obtained from A by multiplying the row i by u.

The transformations described above are called elementary transformations of the rows of A (of type I, II, respectively III). If we take elementary matrices in Mn(R) and multiply A on the right with these matrices, we obtain the elementary transformations of the col-umns of A:

- Type I: ATij(a) is obtained from A by adding to the column i the column j multiplied by a.

- Type II: APij is obtained from A by swapping the column i with the column j.

- Type III: ADi(u) is obtained from A by multiplying the column i by u.

All elementary matrices are invertible. This follows from the following relations, easy to prove:

Tij(a)Tij(b) = Tij(a + b), ∀a, b ∈ R; so Tij(a)Tij(−a) = Tij(0) = I.


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PijPij = I; Dij(u)Dij(v) = Dij(uv), ∀u, v ∈ U(R); so Dij(u)Dij(u

−1) = Dij(1) = I. In other words, the inverse of an elementary matrix exists and is

also an elementary matrix (of the same type). Looking at the definition of the relation of arithmetic equivalence

between matrices, we obtain:

1.6 Proposition. For any A ∈ Mm, n(R), any matrix obtained from A by elementary transformations of rows and/or columns is arithmeti-cally equivalent to A. !

We also need:

1.7 Definition. a) Let R be an UFD and a ∈ R, a ≠ 0. Define the number l(a), called the length of a, as follows: if a is a unit, set l(a) = 0; if a is nonzero and not a unit, l(a) is the number of prime fac-tors (not necessarily distinct) of a decomposition of a in prime factors. For example, in Z, l(1) = 0; l(8) = 3; l(24) = 4. By convention, l(0) = −∞.

b) If A = (aij) ∈ Mm, n(R), define the length of A: l(A) := minl(aij) | i ∈ 1, , m, j ∈ 1, , n.

We can finally prove theorem 1.4: The proof is by induction on m. More precisely, we prove that P(m)

holds for any m, where:

P(m): For any matrix A ∈ Mm, n(R), there exists d1 ∈ R and a matrix A' ∈ Mm−1, n−1(R) such that d1 divides all the entries of A' and A is arithmetically equivalent to the matrix (written in block

form): ⎥⎦

⎤⎢⎣

⎡′ −−−

−

1,11,1

1,11

00

nmm

n

Ad

, where the subscripts indicate the

dimensions of the matrix. We prove P(m) by induction on l(A).


125

If A = 0 (l(A) = −∞), A is in Smith normal form. Case 1. A has an entry that is a unit (⇔ l(A) = 0). Let aij be a unit. Swap the rows i and 1, then swap the columns j

and 1. The matrix obtained is arithmetically equivalent to A, with aij in the position (1,1). To simplify notations, we may suppose thus from the beginning that a11 is invertible. We now obtain a matrix that has 0 on the column 1 (except for a11): for each i, 2 ≤ i ≤ m, add to the row i the row 1 multiplied by (−a11

−1ai1) (which amounts to multiply A on the left with Ti1(−a11

−1ai1)). Similarly, we make 0's on the first row and obtain a matrix of the form:

B = ⎥⎦

⎤⎢⎣

⎡'0

011

Aa

with B ∼ A. The condition that a11 divides all entries of A' is fulfilled (a11 is invertible!).

Case 2. l(A) ≥ 1. Let aij be the entry of A for which l(aij) = l(A). As in case 1, we may

suppose (swapping some rows and/or some columns) that l(a11) = l(A). Subcase 2.1. a11 divides all the entries of A. This case is essentially Case 1. Indeed, since a11|ai1, ∃ bi ∈ R with

ai1 = a11bi. For 2 ≤ i ≤ m, add to the row i the row 1 multiplied by (−bi); the matrix obtained has 0 on the first column, except a11. Clearly, a11 divides the entries of the new matrix (these entries are lin-ear combinations of the entries of A). In the same way we can annihi-late the entries of the first row. Thus, A is arithmetically equivalent to a matrix of the form B, as in case 1.

Subcase 2.2. a11 does not divide all the entries of A. In this case we prove that A is arithmetically equivalent to a matrix

C ∈ Mm, n(R) with l(C) < l(A), which finishes the proof of P(m) by induction on l(A).

We may suppose that a11 does not divide an entry of the first row or of the first column. Indeed, otherwise a11 divides all entries of the first


126

row and of the first column. Working as in subcase 2.1, we obtain a matrix with a11 the only nonzero entry of the first row and of the first column. If a11 divides all the entries of the matrix, we are done! If a11 does not divide all the entries of the matrix, there exists an entry aij (with i, j > 1) such that a11 - aij. By adding to the column 1 the column j, we obtain an entry on the column 1, not divisible by a11.

To fix the ideas, suppose that a11 does not divide an element on the first column (the proof in the other case is similar, the difference being that one multiplies on the right with adequate invertible matrices see below). Suppose a11-a21 (if not, swap rows to achieve this). Let d = GCD(a11, a21). We cannot have d ∼ a11, because this implies a11|a21, false. Since d |a11, l(d) < l(a11) = l(A). We exhibit a matrix C, arithmetically equivalent to A, whose (1, 1) - entry is d (thus l(C) ≤ l(d) < l(A)). Let a11 = da, a21 = db, for some a, b ∈ R. Since (a, b) = 1 and R is a PID, there exist u, v ∈ R such that au + bv = 1. So, dau + dbv = a11u + a21v = d. Consider the matrix (written in block form):

u v 0-b a 0U =0 0 I

where I is the (m − 2)×(m − 2) identity matrix (if m = 2, then I does not appear anymore, i.e. U is a 2×2 matrix). The matrix U is invertible (det U = ua + vb = 1), so C := UA ∼ A. But C has as the (1, 1)-entry ua11 + va21 = d, so l(C) ≤ l(d) < l(a11) = l(A).

This finishes the proof of the existence part. We prove now the uniqueness part in theorem 1.4. For any

A ∈ Mm,n(R) and 1 ≤ k ≤ min(m, n), let ∆k(A) be the GCD of the minors of order k of the matrix A (Recall that a minor of order k of A is the

determinant of a matrix obtained as follows: choose k rows and k col-umns of A and retain only the entries located at the intersection of the


127

chosen rows and columns. There are ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

kn

km

minors of order k in an

m×n matrix). Note that, if U ∈ Mm(R), then ∆k(A)|∆k(UA). Indeed, the rows of UA

are linear combinations (with coefficients in R) of the rows of A. Thus, the rows of a minor of order k of UA (corresponding to the choice of columns i1, , ik of UA) are linear combinations of the rows of A (truncated to contain only the entries on the columns i1, , ik). Apply-ing the fact that the determinant of a matrix is a multilinear function of the rows of the matrix2, it follows that a minor of order k of UA is a linear combination of minors of order k of A. The claim now follows.

Similarly, if V ∈ Mn(R), then ∆k(A)|∆k(AV). So, if A ∼ B, then ∆k(A)|∆k(B) and, by symmetry, ∆k(B)|∆k(A), i.e. ∆k(A) ∼ ∆k(B). If D = diag(d1, , dr) is in Smith normal form, an easy check show that ∆k(D) ∼ d1dk. So, if A ∼ D, d1, , dr are determined (up to associa-tion in divisibility) by ∆1(A), , ∆r(A) and

d1 ∼ ∆1(D)∼ ∆1(A), dk ∼ ∆k(A)/∆k−1(A) , for k ≥ 2. These relations indicate another method to compute effectively d1,

, dr (although the amount of computation is prohibitive if m, n are not small). !

1.8 Remark. The existence part of the proof above is in fact an algorithm to find the Smith normal form of a given matrix. In practice, R is an Euclidian ring. In this case, the function length defined at 1.7 (whose computation is costly, involving the prime factorizations of the

2 By denoting (l1, , lk) the matrix having the rows l1, , lk, the following

relation holds: det(al1 + bl'1, , lk) = adet(l1, , lk) + bdet(l'1, , lk), ∀a, b ∈ R (similarly for any row li).


128

elements of the matrix) can be advantageously replaced by the func-tion ϕ in the definition of the Euclidian ring.

To be precise, if R is Euclidian with respect to ϕ, and A = (aij) ∈ Mm,n( R), define

ϕ(A) = minϕ(aij) | i ∈ 1, , m, j ∈ 1, , n. The proof above is rewritten word for word (replace everywhere l

with ϕ); at the subcase 2.2, replace the matrix U with T21(− q), where a21 = a11q + r, with ϕ(r) < ϕ(a11) = minϕ(aij) (i.e., subtract from the row 2 of A the row 1 multiplied by q, the quotient in the division with remainder of a21 to a11). This places r on the entry (2, 1) of the matrix UA, which has thus ϕ(UA) < ϕ(A). We invite the reader to check the details and apply the algorithm in the proof in concrete cases (see also the exercises).

Exercises

In the exercises, R denotes a principal ideal domain, unless other-wise specified. 1. Find the Smith normal form of the matrix:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

126012105962

∈ M3(Z).

If L is the submodule of Z3 generated by v1 = (2, 6, 9), v2 = (5, 10, 12), v3 = (0, 6, 12), determine a basis of L, rank L and the factor module Z3/L.


129

2. Let a, b ∈ R. Show that the Smith normal form of ⎥⎦⎤

⎢⎣⎡

ba0

0 is

⎥⎦⎤

⎢⎣⎡

md0

0, where d = GCD(a, b), m = LCM(a, b). (Hint. Use the invari-

ants ∆k.)

3. Find the Smith normal form of a diagonal matrix diag (a1, , an) ∈ Mn(R).

4. Find the Smith normal form of a row matrix (a1, , an) ∈ M1, n(R).

5. Determine all subgroups of (Z×Z, +). 6. Let n ∈ N*, x1, , xn ∈ R and d = GCD(x1, , xn). Show that there exists V ∈ GL(n, R) such that (x1, , xn)V = (d, 0, , 0). (Ind. Con-sider the Smith normal form of the row matrix (x1, , xn)). 7. Let n ∈ N* and let a1, , an ∈ R. Show that: there exists V ∈ GL(n, R) such that the first row of V is (a1, , an) if and only if GCD(a1, , an) = 1. 8. Let K be a field and let A ∈ Mm, n(K). Then the Smith normal form of A is diag(1,,1, 0,, 0), where 1 appears r times (r is the rank of the matrix A). 9. Let m, n ∈ N* and let ϕ : R n → R m be an R-homomorphism whose matrix is A ∈ Mm, n(R) (in the canonical bases). Let U ∈ GL(m, R) and V ∈ GL(n, R) such that UAV is in Smith normal form, namely diag(d1,, dr, 0,, 0), with r ≤ min(m, n) and d1,, dr nonzero. Show that a basis in Kerϕ is (vr+1, , vn), where vi is the column i of the ma-trix V (vi is seen as an element in R n). 10. Suppose L is a free R-module of rank n, (e1, , en) is a basis in L and f1, , fm ⊆ L. Show that a basis in F = < f1, , fm > can be ob-tained as follows:


130

Let A = (aij) ∈ Mm, n(R) such that fi = ∑j aijej (1 ≤ i ≤ n) and let U ∈ GLm(R), V ∈ GLn(R), with UAV in Smith normal form. Let gi := ∑j uij fj ∈ F. Then a basis in F is gi | 1 ≤ i ≤ m, gi ≠ 0. 11. Let m, n ∈ N*, A ∈ Mm, n(R) and b ∈ Mm, 1(R). Consider the equa-tion

Ax = b, x ∈ Rn (a linear system of m equations and n unknowns). Consider the ex-tended matrix A = (A, b) ∈ Mm, n + 1(R) (the first n columns of A are the columns of A, the last column is b). Show that: Ax = b has solu-tions x in R n if and only if the Smith normal form of A is (D, 0), where D is the Smith normal form of A and 0 is the zero column in Mm, 1(R). Note that if R is a field, this is the Kronecker-Capelli theo-rem. (see also exercise 8.)

III.2 Finitely generated modules over a principal ideal domain

We are now ready to describe the structure of finitely generated modules over a principal ideal domain. Since any module is a factor of a free module, and a detailed description of any submodule of a finitely generated free module is available (Theorem 1.2 ), the task is now easy. We need only some preparations:

2.1 Definition. If R is a PID, M is an R-module and x ∈ M, then AnnR(x) (= r ∈ R | rx = 0) is a principal ideal of R. A generator of the ideal AnnR(x) is called an order of x, denoted by o(x). Therefore, the order of an element in M is defined up to an association in divisibility. Thus, we have


131

AnnR(x) = Ro(x); Rx ≅ R/Ro(x). This notion generalizes the usual concept of order of an element in

an Abelian group. If ϕ : E → F is an R-module isomorphism and x ∈ E, then

o(x) = o(ϕ(x)) since AnnR(x) = AnnR(ϕ(x)).

2.2 Lemma. Let R be a ring and let M be a left R-module such that M is the direct sum of a family of submodules (Mi)i∈I, M = ⊕I Mi. If Ni ≤ R Mi, ∀i ∈ I, then the sum of the submodules (Ni)i∈I is direct and we have a canonical isomorphism :

i

iIiiIi

iIi

NM

N

M

∈∈

∈ ⊕≅⊕

⊕.

Proof. Let πj : ⊕I Mi → Mj be the canonical surjections. Define

ϕ : M →i

iIi N

M∈⊕ by ϕ(x) = (πi(x) + Ni)i∈I, ∀x ∈ M. One easily checks

that ϕ is a surjective homomorphism (in fact, ϕ is the direct sum of the family of homomorphisms ηiπi : M → Mi/Ni, where ηi : Mi → Mi/Ni is the canonical surjection). We have Kerϕ = x ∈ M | πi(x) ∈ Ni, ∀i ∈ I. Since x = ∑i∈I πi(x), it follows that Kerϕ = ∑i∈I Ni = ⊕I Mi. Apply now the isomorphism theorem. !

We state now the following important theorem, which determines the structure of finitely generated modules over a principal ideal do-main. Recall that R° is the set of nonzero noninvertible elements of R.

2.3 Theorem. (Invariant factors theorem) Let R be a principal ideal domain and let M be a finitely generated R-module. Then M is a direct sum of a finite number of cyclic submodules.

More precisely, there exist n, m ∈ N, with m ≤ n, and x1, , xn ∈ M such that : M = Rx1⊕ ⊕ Rxm ⊕ Rxm+1⊕ Rxn, (D)


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Moreover, o(xi) =: di ∈ R satisfy the conditions: di ∈ R°, ∀i ∈ 1, , m and d1|d2||dm ; dm+1 = = dn = 0.

The numbers n, m ∈ N and the orders o(xi) ∈ R, i = 1 n with the above properties are uniquely determined, in the following sense: if n', m' ∈ N, with m' ≤ n', and y1, , yn' ∈ M such that : M = Ry1⊕ ⊕ Rym' ⊕ Rym' +1⊕ Ryn', (D') and o(yi) =: ei satisfy : ei ∈ R°, ∀i ∈ 1, , m' and e1|e2||em' ; em'+1 = = en' = 0, then

m = m', n = n' and di ∼ ei, ∀i ∈ 1, , n. The orders o(xi) are called the invariant factors of the module M. Proof. The existence part: if S ⊆ M is finite and generates M, then

there is an isomorphism ψ : R(S)/F → M, where F is the kernel of ϕ, given by ϕ(es) = s, ∀s ∈ S ((es)s∈S is the canonical basis of R(S)). Let n = |S|, E = R(S). Theorem 1.2 provides us with a basis e = (e1,, en) in E and a basis f = (f1,, fm) in F (with m ≤ n) such that fi = diei, with di ∈ R, di ≠ 0 and d1|d2||dm. There exists k ∈ N (0 ≤ k ≤ m) such that d1,, dk ∈ U(R) and dk+1,, dm ∉ U(R). Then Rfi = Rdiei = Rei, 1 ≤ i ≤ k, and we can write (applying lemma 2.2):

M≅E/F ≅ nmm

m

k

k

k

k eReRRf

eRRf

eReReR

eReR

⊕⊕⊕⊕⊕⊕⊕⊕ ++

+ ……… 11

1

1

1

11

1

1 1

1

k mm n

k m

k m m n

k m

Re Re Re ReRf Rf

Re Re Re ReRf Rf

++

+

+ +

+

= ⊕ ⊕ ⊕ ⊕ ⊕

⊕ ⊕ ⊕ ⊕ ⊕≅

⊕ ⊕

… …

… ……

Thus, the structure of M is determined by dk+1,, dm, ek+1,, en, fk+1,, fm. Changing notations if necessary, we may suppose from the beginning that d1,, dm are non invertible. Let ψ(ei + F) = xi ∈ M,

1 ≤ i ≤ n. We have M = Rx1⊕⊕Rxn because ei + F |1 ≤ i ≤ n generates

E/F, so xi |1 ≤ i ≤ n generates M. Besides,


133

E/F = R(e1 + F)⊕⊕R(en + F), since ∑1 ≤ i ≤ n ri(ei + F) = 0 + F ⇔ ∃si ∈ R such that ∑1 ≤ i ≤ n riei = ∑1 ≤ i ≤ m sidiei ⇔ ri = sidi, for any 1 ≤ i ≤ m and ri = 0, for m + 1 ≤ i ≤ n ⇔ ri(ei + F) = 0 + F, 1 ≤ i ≤ n.

Also, o(xi) = o(ei + F) = di, if 1 ≤ i ≤ m and o(xi) = 0 if m + 1 ≤ i ≤ n. Indeed, for any r ∈ R, rxi = 0 ⇔ rei ∈ F ⇔ ∃si ∈ R such that ∑1 ≤ i ≤ n riei = ∑1 ≤ i ≤ m sidiei, so r = 0 if m + 1 ≤ i ≤ n and r = sidi if 1 ≤ i ≤ m.

This finishes the proof of existence of a decomposition (D) of M.

For the uniqueness part, we need some invariants.

2.4 Definitions. Let R be a domain and let M be an R-module. a) Define t(M) := x ∈ M | ∃r ∈ R \0 with rx = 0. t(M) is called

the torsion submodule of M). The elements of t(M) are called torsion elements. If M = t(M), M is called a torsion module; if t(M) = 0, M is called a torsion-free module.

b) Let p ∈ R be a prime element. Define tp(M) := x ∈ M | ∃n ∈ N cu pnx = 0 (called the p-torsion submodule of M or the p-submodule of M).

If d ∈ R and k ∈ N, pk||d means pk|d and pk + 1-d (if R is a UFD, pk||d ⇔ k is the exponent of p in the prime factor decomposition of d).

c) If a ∈ R, let za(M) := x ∈ M | ax = 0 (called the annihilator of a in M, also denoted sometimes by AnnM(a) or rM(a)).

d) AnnR(M) := r ∈ R | rx = 0, ∀x ∈ M is called the annihilator of the module M.

We need to collect the basic proprieties of these invariants.

2.5 Proposition. Let R be a domain and let M be an R-module. a) t(M) is a submodule of M and t(M/t(M)) = 0; if M ≅R N, then

t(M) ≅ t(N). b) If M = ⊕i∈I Mi, then t(⊕i∈I Mi) = ⊕i∈I t(Mi).


134

c) Let R be a PID. If M is finitely generated, then t(M) is a direct summand in M and there exists L ≤ R M, free, such that M = t(M)⊕L. In particular, a finitely generated torsion-free module is free.

Proof. a) Let x, y ∈ t(M) and let r, s ∈ R. Then there exist a, b ∈ R, nonzero, such that ax = by = 0. We have ab ≠ 0 and ab(rx + sy) = 0, so rx + sy ∈ t(M).

Let x + t(M) ∈ t(M/t(M)): there exists 0 ≠ a ∈ R with ax ∈ t(M). Then there exists 0 ≠ b ∈ R with bax = 0, so x ∈ t(M) (since ba ≠ 0). Thus, x + t(M) = 0 + t(M).

b) Let (xi)i∈I have finite support, with xi ∈ Mi, ∀i ∈ I, such that ∑i∈I xi := x ∈ t(M). Then there exists 0 ≠ a ∈ R such that ax = ∑i∈I axi = 0. From M = ⊕i∈I Mi it follows that axi = 0, i.e. xi ∈ t(Mi), ∀i ∈ I. So, t(M) ⊆ ⊕i∈I t(Mi).

If xi ∈ t(Mi) such that J := supp((xi)i∈I) is finite, there is some family (rj)j∈J, with 0 ≠ rj ∈ R and rjxj = 0, ∀j ∈ J. Let r := ∏j∈J rj (well de-fined, since J is finite). Obviously, r ≠ 0 and r∑i∈I xi = 0. This shows that ⊕i∈I t(Mi) ⊆ t(M).

c) Take a decomposition (D) of M. We show that t(M) = Rx1⊕⊕Rxm. Evidently, x1, xm ∈ t(M), since dixi = 0 and di ≠ 0, so Rx1⊕⊕Rxm ⊆ t(M). If x ∈ t(M), there is some 0 ≠ a ∈ R and r1,, rn ∈ R such that x = r1x1 + + rmxm + + rnxn and ax = ar1x1 + + armxm + + arnxn = 0. Because the sum of the submodules Rxi is direct, arixi = 0, i = 1 n . If i > m, then ari ∈ AnnR(xi) = 0, so ri = 0. We deduce that x ∈ Rx1⊕⊕Rxm.

It is now clear that, denoting by L the free module ⊕i > m Rxi, M = t(M)⊕L. !

2.6 Proposition. Let R be a PID, let M be an R-module and p ∈ R a prime element.

a) tp(M) is a submodule in M and tp(M/tp(M)) = 0; if M ≅R N, then tp(M) ≅ tp(N).

b) If M = ⊕i∈I Mi, then tp(⊕i∈I Mi) = ⊕i∈I tp(Mi).


135

c) Let d ∈ R and k ∈ N such that pk||d. Then tp(R/Rd) ≅ R/Rpk. In particular, if x ∈ M and o(x) = d, then tp(Rx) ≅ R/Rpk; if p-d, then tp(Rx) = 0.

Proof. a) and b) have similar proofs with 2.5 and are proposed as an exercise.

c) Let b ∈ R such that d = pkb. We claim that tp(R/Rd) = Rb/Rd. In-deed, let r + Rd ∈ tp(R/Rd). There exist s ∈ N such that psr ∈ Rd, i.e. psr = dc = pkbc, with c ∈ R. So b| p

sr and (b, p) = 1, which imply b|r. So, r ∈ Rb and tp(R/Rd) ⊆ Rb/Rd. The other inclusion is obvious.

We have R/Rpk ≅ Rb/Rd by the isomorphism theorem applied to ϕ : R → Rb/Rd, ϕ(r) = rb + Rd, ∀r ∈ R. !

2.7 Proposition. Let M be an R-module and let r ∈ R. a) zr(M) is a submodule in M; if M ≅R N, then zr(M) ≅R zr(N). b) If M = ⊕i∈I Mi , then zr(M) = ⊕i∈I zr(Mi). c) If R is a PID, p ∈ R is a prime element and d ∈ R, then

zp(R/Rd) = 0 if p-d and zp(R/Rd) ≅ R/Rp if p|d. In particular, if x ∈ M, then zp(Rx) = 0 if p-o(x) and zp(Rx) ≅ R/Rp if p|o(x).

Proof. a) Exercise. b) Let x = ∑i∈I xi, with xi ∈ Mi, ∀i ∈ I. We have rx = 0 ⇔

∑i∈I rxi = 0 ⇔ rxi = 0, ∀i ∈ I (since rxi ∈ Mi, and the sum is direct) ⇔ xi ∈ zr(Mi), ∀i ∈ I.

c) Let p - d and r ∈ R. We have p(r + Rd) = 0 + Rd ⇔ pr ∈ Rd ⇔ d | pr. Since (d, p) = 1, this implies d | r, so r + Rd = 0.

Let p|d and a ∈ R with d = pa. For any r ∈ R, pr ∈ Rd = Rpa ⇔ r ∈ Ra. So, zp(R/Rd) = Ra/Rd. Consider the surjective homomorphism ϕ : R → Ra/Rd, ϕ(r) = ra + Rd, ∀r ∈ R. We have Kerϕ = Rp, so Ra/Rd ≅ R/Rp. !

We can proceed now to prove the uniqueness part in 2.3. The idea is to apply the invariants t, zp, tp to the decompositions (D) and (D').

Since t(M) = Rx1⊕⊕Rxm = Ry1⊕⊕Rym' (see 2.5.c)), we get:


136

M/t(M) ≅ Rxm+1⊕⊕Rxn ≅ Rym'+1⊕⊕Ryn'. But xm+1, , xn and ym'+1, , yn' are bases in the free module

M/t(M), so they have the same number of elements: n − m = n' − m'. It remains to prove that m = m' and di ∼ ei, 1 ≤ i ≤ m. To this end,

we suppose in the sequel that M = t(M) = Rx1⊕⊕Rxm = Ry1⊕⊕Rym'.

Since Rxi ≅ R/Rdi, Ryi ≅ R/Rei, we have to prove that, if M ≅ R/Rd1⊕⊕R/Rdm, with d1|d2||dm, di ∈ R°, 1 ≤ i ≤ m, (*) M ≅ R/Re1⊕⊕R/Rem', with e1|e2||em', ei ∈ R°, 1 ≤ i ≤ m', (**) then m = m' and di ∼ ei, 1 ≤ i ≤ m.

Note first that dm ∼ em'. Indeed, AnnR(M) = r ∈ R| rx = 0, ∀x ∈ M is an ideal in R; by (*), AnnR(M) = Rdm (prove!). By (**), AnnR(M) = Rem', so dm ∼ em'.

Let p be a prime divisor of d1. Then p|di, 1 ≤ i ≤ m. Consider zp(M). By 2.7 and (*), we can write the R-module isomorphisms:

zp(M) ≅ zp(Rx1⊕⊕Rxm) ≅ zp(Rx1)⊕⊕zp(Rxm) ≅ R/Rp⊕⊕R/Rp (m terms). (we used that p|di, 1 ≤ i ≤ m). Use now (**); let k be the number of those indices i, with 1 ≤ i ≤ m', such that p divides ei. We have :

zp(M) ≅ zp(Ry1)⊕⊕zp(Rym') ≅ R/Rp⊕⊕R/Rp (k terms). So, (R/Rp)m ≅ (R/Rp)k (R-module isomorphism). It is clear that this

is also an R/Rp-module isomorphism. But R/Rp is a field (p is prime and R is a PID), so m = k, since two isomorphic R/Rp-vector spaces have the same dimension. Of course, k ≤ m', so m ≤ m'. By symmetry, m' ≤ m, so m = m'.

Let us show that di ∼ ei, 1 ≤ i ≤ m. Let p be an arbitrary prime factor of dm and let αi (respectively βi) be the exponent of p in the decomposition of di (respectively ei), 1 ≤ i ≤ m. Since d1|d2||dm, we have α1 ≤ α2 ≤ ≤ αm (and also β1 ≤ β2 ≤ ≤ βm). It is sufficient to show that αi = βi, 1 ≤ i ≤ m. Suppose that is not so: then there exists j,


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1 ≤ j ≤ m, minimal with the property that αj ≠ βj (thus αi = βi if i < j). Say, for instance, that αj < βj. Apply tp. By (*) and 2.6, we have:

tp(M) ≅ tp(Rx1)⊕⊕tp(Rxm) ≅ mRpRRpR αα ⊕⊕…1 By (**), tp(M) ≅ tp(Ry1)⊕⊕tp(Rym) ≅ mRpRRpR ββ ⊕⊕…1 .

Multiply tp(M) by jpα (i.e., consider the submodule ( )Mtp pjα ). We

obtain from (*):

( ) ( ) ( )1

1

j j j m

j j m j

pp t M p R Rp p R Rp

R Rp R Rp

α α α αα

α α α α+ − −

≅ ⊕ ⊕

≅ ⊕ ⊕

…

…

We have used the proprieties (whose proof is immediate): if M = ⊕i∈I Mi and r ∈ R, then rM = ⊕i∈I rMi ; pαR/Rpβ ≅ 0 if β ≤ α; pαR/Rpβ ≅ R/Rpβ −α if β > α.

From (**), with a similar argument, we have : ( ) ( ) ( )1j j j m

j j m j

pp t M p R Rp p R Rp

R Rp R Rp

α α α ββ

β α β α− −

≅ ⊕ ⊕

≅ ⊕ ⊕

…

…

Therefore, jmjjjmjj RpRRpRRpRRpR αβαβαααα −−−− ⊕⊕≅⊕⊕+ ……1

These are decompositions of the type (*), as one easily sees. Let k be the number of indices i for which αi > αj (evidently, 0 ≤ k ≤ m − j). In the left hand side we have k nonzero terms and in the right hand side there are m − j + 1 nonzero terms. The first part of the proof shows that k = m − j + 1, contradicting k ≤ m − j. Thus, we must have αi = βi, 1 ≤ i ≤ m. !

2.8 Remark. An R-module M can have several decompositions of type (D). But the sequence of ideals

AnnR(x1) ⊇ AnnR(x2) ⊇ ⊇ AnnR(xn) is uniquely determined. The generators (uniquely determined up to an association in divisibility) d1, d2, , dn ∈ R of these ideals are called the invariant factors of the R-module M. Sometimes this name is


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given to the sequence of ideals above. If in the principal ideal domain R there is a natural way to choose a generator of an ideal, the invariant factors are the natural generators of the annihilator ideals above. For example, in Z the positive generator is chosen, in K[X] (K a field) the monic polynomial that generates the ideal is chosen.

2.9 Example. Let Z2 = 1,0 , Z6 = 5 , ,1 ,0 … and the Z-module Z2×Z6. We have the following decompositions:

Z2×Z6 = Zx1⊕Zx2 = Zy1⊕Zy2, where x1 = ( 0,1 ), x2 = ( 1,0 ), y1 = ( 3,1 ), y2 = ( 5,0 ) (check this!). Of course, these are distinct decompositions, but AnnR(x1) = AnnR(y1) = 2Z and AnnR(x2) = AnnR(y2) = 6Z. The se-quence of invariant factors of the Z-module Z2×Z6 is: 6, 2.

Exercises

1. Give an example of a finitely generated Z-module such that its tor-sion submodule has at least two complements. (cf. Prop. 2.5.) 2. Let R be a domain and let u : M → N be an R-module homomorph-ism. Then u(t(M)) ⊆ t(N). State and prove a similar propriety for tp(M) (R is a PID and p is a prime element in R) and zr(M) (where r ∈ R). 3. Assume R is a domain, M is an R-module and L ≤ R M. Then M/L is torsion-free if and only if t(M) ⊆ L. 4. This exercise gives an example of a Z-module M such that t(M) is not a direct summand in M. (Thus, by 2.5.c), M cannot be finitely generated). Let P = pn |n ≥ 1 be the set of prime natural numbers and let M := ∏p∈P Zp. Show that:


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a) t(M) = ⊕p∈P Zp ( = (ap)p∈P | ∀p ∈ P, ap ∈ Zp and supp((ap)P) fi-nite).

b) ∀n ∈ Z, n ≠ 0, n·(M/t(M)) = M/t(M). c) ∀x ∈ M, ∃p ∈ P such that x ∉ pM. d) If M = t(M)⊕S, with S ≤ R M, then S ≅ M/t(M). e) t(M) is not a direct summand in M.

5. Let p be a prime natural number and let G = Zp×Zp. Prove that the subgroups of G coincide with the Zp-vector subspaces of G. How many vector subspaces of dimension 1 are there in G? In how many ways can ZG be written as a direct sum of two proper submodules? (Hint: the sum of any two nonzero distinct subspaces is direct, equal to G). 6. State a structure theorem for finitely generated Abelian groups. 7. Give an example of an Abelian group G that is neither cyclic, nor a direct sum of cyclic groups. (Hint: G cannot be finitely generated. An example is Q). 8. Determine all Abelian groups with 100 elements. 9. Let G be a finite Abelian group whose order is squarefree (not divisible with the square of any prime). Then G is cyclic. 10. If G is a finite Abelian group and n is the exponent of G ( = GCD of the orders of the elements of G), then G contains an element of or-der n. (Hint: use the theorem of invariant factors). Give an example of a group G such that n is a proper divisor of the order of G. 11. Let G be a finite Abelian group. Show that, for any divisor d of |G|, there exists a subgroup of G having d elements. 12. For n ∈ N*, let gn denote the number of isomorphism types of Abelian groups having n elements. For what n ∈ N* we have gn = 1? Determine n ∈ N | n ≤ 100, gn ≥ 4. 13. Using the structure theorem of finite Abelian groups, show that any finite subgroup G of the multiplicative group K* (where K is a


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field) is cyclic. (Hint. If G has more than one invariant factor and d is the greatest invariant factor, then any element of G is a root of a X d − 1 and d < |G|). 14. Determine the finitely generated Abelian groups with the property that their lattice of subgroups is a chain ( = totally ordered with respect to inclusion).

III.3 Indecomposable finitely generated modules

Theorem 2.3 says that any finitely generated R-module (if R is a PID) is a direct sum of cyclic submodules. Can such a decomposition be refined? In other words, can we decompose further these submod-ules as direct sums of proper submodules?

3.1 Definition. Let R be a ring (not necessarily commutative). An R-module M is called indecomposable if M ≠ 0 and M has no proper direct summands (if L, N ≤ R M are such that M = L⊕N, then L = 0 or N = 0). An R-module that is not indecomposable is called decomposa-ble.

3.2 Examples. a) The Z-module Z6 is decomposable: Z6 = 2Z6⊕3Z6.

b) If R M is isomorphic to a direct product of modules of the type A×B, with A, B nonzero, then M is decomposable.

b) If K is a field, a K-vector space V is decomposable if and only if dim V > 1. (Besides, any proper subspace of a vector space is a direct summand).

c) The Z-module Z2 is indecomposable (it has no proper submod-ules altogether).


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3.3 Proposition. If the lattice of submodules of the R-module M is a chain (it is totally ordered) with respect to inclusion, then M is indecomposable.

Proof. Suppose M = A⊕B, with A, B ≤ R M. But the lattice of submodules of M is a chain, so A ⊆ B or B ⊆ A. Thus, A ∩ B = 0 im-plies A = 0 or B = 0. !

3.4 Corollary. Let R be a PID, p ∈ R a prime element and k ∈ N. Then the cyclic module R/Rpk is indecomposable.

Proof. Suppose k ≥ 1. It is enough to check that the lattice of ideals of R that include Rpk is a chain (see II.2.3). But the ideal I includes Rpk if and only if I = Ra (I is principal), where a ∈ R, a|pk. The divisors of pk are pt, 0 ≤ t ≤ k, so a ∼ pt for some t ≤ k. This says that the ideals that include Rpk are chained: Rpk ⊆ Rpk−1 ⊆ ⊆ Rp2 ⊆ Rp ⊆ R.

If k = 0, we have to show that R R is indecomposable. This is true for an arbitrary domain: the intersection of any two nonzero submod-ules ( = ideals) I, J of R is nonzero: if a ∈ I, b ∈ J are nonzero, then 0 ≠ ab ∈ I ∩ J. !

If R is a PID, the cyclic modules (isomorphic to) R/Rpk, with p prime in R and k ≥ 0, are all finitely generated indecomposable R-modules. This follows from the following result, valid in any commutative ring.

3.5 Theorem. (Chinese remainder theorem) Let R be a commuta-tive ring, n ≥ 2 and I1,, In ideals of R.

a) If Ii + Ij = R for i ≠ j,3 then the product 4 I1··In is equal to the intersection I1 ∩∩ In and there is a natural isomorphism of rings (and of R-modules) η:

3 The ideals Ii şi Ij are called in this case comaximal. For example, the ideals Za

and Zb of Z are comaximal if and only if a and b are coprime.


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nnn IR

IR

IIR

IIR

××≅=⋅⋅

…∩…∩… 111

,

η(r + I1 ∩∩ In) = (r + I1, , r + In), ∀r ∈ R.

b) Conversely, if the homomorphism ϕ : R →nI

RIR

××…1

,

ϕ(r) = (r + I1, , r + In), ∀r ∈ R is surjective (inducing an isomor-

phism nn I

RIR

IIR

××≅ …∩…∩ 11

as above), then Ii and Ij are comaxi-

mal ideals, for any i ≠ j. Proof. a) We prove by induction on n that I1··In = I1 ∩∩ In and

that η is an isomorphism. For n = 2, I1 + I2 = R implies the existence of x ∈ I1, y ∈ I2 such that x + y = 1. Let z ∈ I1 ∩ I2. Then z = z·1 = zx + zy, with zx, zy ∈ I1·I2, so I1 ∩ I2 ⊆ I1I2. Thus, I1 ∩ I2 = I1I2.

Let ϕ : R →21 I

RIR

× , ϕ(r) = (r + I1, r + I2), ∀r ∈ R. ϕ is a ring (and

an R-module) homomorphism (it is the direct product of the canonical surjections R → R/Ij). We have:

Kerϕ = r ∈ R| (r + I1, r + I2) = (0 + I1, 0 + I2) = I1 ∩ I2

The isomorphism theorem implies that R/I1 ∩ I2 ≅ Imϕ and the

isomorphism is precisely η. Let us prove that ϕ is surjective (which

will finish the proof). Let (r1 + I1, r2 + I2) ∈21 I

RIR

× . We must exhibit

r ∈ R with r − r1 ∈ I1, r − r2 ∈ I2. Such an element is r = r1 y + r2 x. In-deed,

r − r1 = r1y + r2x − r1x − r1y = (r2 − r1)x ∈ I1.

4 Recall that the product IJ of two ideals I and J is the ideal generated by all the

products ij, i ∈ I, j ∈ J. The product of ideals is associative and always IJ ⊆ I ∩ J.


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Similarly, r − r2 ∈ I2. Suppose now that for any k < n and any ideals I1,, Ik, pairwise co-

maximal , we have I1··Ik = I1 ∩∩ Ik and η is an isomorphism. Take n pairwise comaximal ideals I1,, In. Since Ij + In = R, 1 ≤ j ≤ n − 1, there exist aj ∈ Ij, bj ∈ In, such that aj + bj = 1. Multiply these n − 1 equalities:

( )∏−

=+

1

1

n

jjj ba = a1··an−1 + b = 1, where b ∈ In, a1··an−1 ∈ I1··In−1.

So, I1··In−1 + In = R. Applying the case n = 2 to the comaximal ideals I1··In−1 and In,

I1··In−1·In = (I1··In−1) ∩ In = (I1 ∩∩ In−1) ∩ In We also used the induction hypothesis I1··In−1 = I1∩∩ In−1. Also,

case n = 2 says that

( ) nnnn IR

IIR

IIIR

×⋅⋅

≅⋅⋅⋅ −− 1111 ……

,

r + I1··In & (r + I1In−1, r + In), ∀r ∈ R. On the other hand, by induction, we have the isomorphism:

1111 −−

××≅⋅⋅ nn I

RIR

IIR ……

,

r + I1··In−1 & (r + I1, r + In−1), ∀r ∈ R. Combining these isomorphisms, we obtain the result. b) We prove that I1 and I2 are comaximal. Let (1 + I1, 0 + I2, ,

0 + In) ∈ nI

RIR

××…1

. There exists y ∈ R such that (y + I1, y + I2, ,

y + In) = (1 + I1, 0 + I2, , 0 + In), i.e. y ∈ I2 and y − 1 =: x ∈ I1. So, 1 = −x + y ∈ I1 + I2, which means I1 + I2 = R. !

3.6 Corollary. a) Let R be a PID and let a1, , an ∈ R. If (ai, aj) = 1, ∀i ≠ j, then:


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nn RaR

RaR

aRaR

××≅ …… 11

r + Ra1an & (r + Ra1, , r + Ran), ∀r ∈ R. b) If M is a cyclic R-module M = Rx (x ∈ M), with

o(x) = d = a1··an ∈ R° and (ai, aj) = 1, ∀i ≠ j, then there exist xi ∈ M, 1 ≤ i ≤ m, such that o(xi) = ai and

M = Rx = Rx1⊕⊕Rxm c) Any cyclic R-module can be written as a direct sum of

indecomposable submodules. Proof. a) If a, b ∈ R, then (a, b) = 1 if and only if Ra and Rb are co-

maximal. Indeed, the ideal generated by GCD(a, b) is Ra + Rb. So, (a, b) = 1 ⇔ Ra + Rb = R. Apply now the Chinese remainder theorem for Ra1, , Ran.

b) We have M ≅ R/Rd. By a), we have R/Ra1 ×× R/Ran ≅ R/Rd. So, there is an isomorphism ϕ : R/Ra1 ×× R/Ran → M. Let yi := (0 + Ra1,, 1 + Rai, , 0 + Ran) and xi := ϕ(yi). Obviously, R/Ra1 ×× R/Ran = Ry1⊕⊕Ryn, so, (applying ϕ) M = Rx1⊕⊕Rxn. Also, o(xi) = o(yi) = ai, 1 ≤ i ≤ n.

c) This follows from a): let M = Rx, with x ∈ M and let d = o(x). If d = 0, then M ≅ R is indecomposable. If d ≠ 0, let tk

tk ppd …11= the

prime factor decomposition of d (where p1, , pt are distinct primes in

R). Clearly, ikip and jk

jp are coprime if i ≠ j; applying b), there exist

xi ∈ M such that M = Rx1⊕⊕Rxt, with o(xi) = ikip . So Rxi is

indecomposable, being isomorphic to ikiRpR . !

3.7 Corollary. Let R be a PID and let M be a finitely generated R-module. Then:

- M is indecomposable if and only if: either M is cyclic, isomorphic to R (in this case M is torsion-free) or M is cyclic, isomorphic to some


145

R/Rp k, where p ∈ R is prime and k ∈ N* (in this case M is a torsion module).

Proof. The proof of the if part is easy: we saw that the modules R/Rp k are indecomposable.

Let M be indecomposable finitely generated. There exists a decomposition (D), as in theorem 2.3. Keeping the notations in 2.3, we see that M is indecomposable only if m = n = 1 or m = 0 and n = 1. If m = 0, n = 1,then M is free of rank 1, and thus isomorphic to R.

If m = n = 1, then M = Rx, with o(x) = d. So, M ≅ R/Rd. Let tk

tk ppd …11= be the prime decomposition of d (p1, , pt are distinct

primes in R). We claim that t = 1. If t > 1, ikip and jk

jp are coprime if

i ≠ j, so R/Rd ≅ tkt

k RpRRpR ××…11 , which is clearly decomposable. !

3.8 Proposition. If M is a finitely generated torsion module over a PID R, then M is the direct sum of its p-torsion submodules.

More precisely, if P is a system of representatives of the equiva-lence classes (with respect to association in divisibility) of the prime elements in R, then p ∈ P | tp(M) ≠ 0 is finite and M is the finite di-rect sum:

M = ⊕p∈P tp(M) Proof. We know that M is a direct sum of cyclic submodules:

M = Rx1⊕⊕Rxm, o(xi) = di ∈ R°. From 2.6 we deduce that, if p ∈ P and p-di, for all 1 ≤ i ≤ m, then tp(M) = 0. So, p ∈ P | tp(M) ≠ 0 ⊆ p ∈ P | ∃i, p|di, which is finite.

If p ∈ P, then tp(M) ∩ ∑tq(M) |q ∈ P, q ≠ p = 0. Indeed, if x be-

longs to the intersection, then pkx = 0 for some k ∈ N; also, x = ∑q≠p yq, with yq ∈ tq(M), ∀q ∈ P, q ≠ p (only a finite number of yq

are nonzero). So, ∀q ≠ p for which yq ≠ 0, ∃kq ∈ N such that qk yq q = 0.

Let ∏ ≠= 0q

qy

kqa (finite number of factors!). Evidently, ax =


146

a∑q≠p yq = 0. Also (pk, a) = 1, since p is not associated to any prime q

occurring in the decomposition of a. So, there exist u, v ∈ R such that upk + va = 1. Then x = (upk + va)x = upkx + vax = 0.

Let x ∈ M. We prove that x ∈ ∑ p∈P tp(M). Let d = o(x) ∈ R and tk

tk ppd …11= be its prime decomposition. By 3.6.b),

Rx = Rx1⊕⊕Rxt, with o(xi) = ikip , i.e. xi ∈ ( )Mt

ip . !

The invariant factor theorem and the list of finitely generated indecomposable modules allow us to formulate the following structure theorem:

3.9 Theorem. Let R be a PID and let M be a finitely generated R-module. Then M is a finite direct sum of indecomposable submod-ules. The following property of uniqueness holds: if

M = A1⊕⊕Am = B1⊕⊕Bn

are decompositions of M as direct sums of indecomposable submod-ules, then m = n and there exists a permutation σ ∈ Sn such that Ai ≅ Bσ(i), 1 ≤ i ≤ m.

Proof. The existence part is proven as follows: from theorem 2.3, M is a finite direct sum of cyclic submodules, and 3.6 says that every cyclic submodule is a finite direct sum of indecomposable submod-ules.

As in the proof of uniqueness in theorem 2.3, note that the free module M/t(M) has rank equal to the cardinal of the set i | 1 ≤ i ≤ m, Ai torsion free (and equal to the cardinal of the set j | 1 ≤ j ≤ n, Bj torsion free). We may suppose then that

M = t(M) = Rx1⊕⊕Rxm = Ry1⊕⊕Ryn, with xi, yj ∈ M and Rxi, Ryj indecomposable (by 3.7, o(xi), o(yj) are prime powers in R). Let p be a prime in R. We have tp(Rxi) = 0 if p-o(xi) and tp(Rxi) = Rxi if o(xi) is a power of p (see 2.6). Thus

tp(M) = ⊕Rxi | 1 ≤ i ≤ m, p|o(xi) = ⊕Ryj | 1 ≤ j ≤ n, p|o(yj).


147

Relabeling if necessary, let i | 1 ≤ i ≤ m, p|o(xi) =: 1, , r and j | 1 ≤ j ≤ n, p|o(yj) =: 1, , s, such that o(xi) = ikp and o(yj) = jlp , with k1 ≤ ≤ kr and l1 ≤ ≤ ls. Then

tp(M) = Rx1⊕⊕Rxr = Ry1⊕⊕Rys, o(x1)| |o(xr), o(y1)| |o(ys). The uniqueness part of theorem 2.3 says that r = s and o(xi) = o(yi)

(so, Rxi ≅ Ryi), 1 ≤ i ≤ r. Since M = ⊕p∈P tp(M) (see 3.8), the proof is finished. !

3.10 Remark. If M is a torsion module and Rx1⊕⊕Rxm is o decomposition of M as a direct sum of indecomposables, then the fam-ily of ideals (AnnR(xi))1≤ i ≤ m is independent of the chosen decomposi-tion. The family of elements (o(xi))1≤ i ≤ m is thus uniquely determined (up to an association in divisibility and a permutation) by the module M and is called the family5 of the elementary divisors of M. The elementary divisors are powers of prime elements in R, by 3.7.

3.11 Examples. a) The elementary divisors of the Z-module Z6⊕Z24 ≅ Z2⊕Z3⊕Z8⊕Z3 are (2, 23, 3, 3). Its invariant factors are (6, 24).

b) If G is an Abelian group with n elements, and d1||dm is the se-quence of its invariant factors, then G ≅

mdd ZZ ⊕⊕…1

, so

n = d1··dm . Thus, any Abelian group of order n is perfectly deter-mined (up to an isomorphism) by an m-uple (d1, , dm) of natural numbers (m ≥ 1), such that: d1 ≥ 2, d1··dm = n and d1||dm. For example, for n = 60, the possible choices for (d1, , dm) are (60), (2, 30). Thus there are two types of isomorphism6 of a Abelian groups

5 We avoid the term set, because in a set all elements are distinct, while the

elementary divisors can occur more than once. 6 The class of all groups isomorphic to a given group G is called the type of

isomorphism of G. (This definition can be generalized to any type of algebraic


148

with 60 elements: Z60 ≅ Z4⊕Z3⊕Z5 (the elementary divisors are (22, 3, 5)) and Z2⊕Z30 ≅ Z2⊕Z2⊕Z3⊕Z5 (the elementary divisors are (2, 2, 3, 5)).

c) If the invariant factors (d1, , dm) are given, the elementary divi-sors can be obtained by decomposing in a product of powers of primes every di and writing down all the powers that occur, as many times that they arise.

Conversely, if the family of the elementary divisors is given, the invariant factors are obtained as follows: write a product containing (only once) all primes in the family of the elementary divisors, at the greatest power. The product obtained is dm (the largest invariant fac-tor divisibility-wise). Erase from the family of the elementary divi-sors the powers written in the product and repeat the procedure with what is left. Continue until the elementary divisors are exhausted.

For example, if the family of the elementary divisors of a Z-module is (2, 2, 22, 3, 33, 5), following the procedure above we obtain succes-sively: 22·33·5, 2·3, 2, which are the invariant factors of the Z-module. Which is this Z-module?

structures: Abelian groups, rings, modules, fields, ordered sets). For a given type of algebraic structure, the description of all types of isomorphism of the structure is a most important (and hard to attain, in general) objective, called classification. For instance, the theorem of invariant factors yields a classification of finitely generated Abelian groups. The classification of the finite simple groups (having no normal proper subgroups) is one of the great successes of group theory, accomplished in the 1980's.


149

Exercises

1. Give an example of a ring R and an indecomposable R-module that has a decomposable factor module. 2. Prove that R M is indecomposable ⇔ the ring EndR(M) contains no idempotents other than 0 and 1M. 3. Give an example of a ring R an indecomposable R-module that has decomposable submodules. Can R be a PID? (Hint. Let K be a field, R = K[X, Y], I = (XY) = XYK[X, Y] and M = R/I. The idempotents of EndR(M) are 0 and 1M. XK[X, Y] + YK[X, Y]/I ≤ M and is decomposa-ble). 4. Let R be a PID, M an R-module and x1, , xn ∈ M, with (o(xi), o(xj)) = 1, ∀i ≠ j. Then o(x1 + + xn) = o(x1)··o(xn). (This general-izes the known fact: if a, b ∈ G, (G, +) an Abelian group, and (ord x, ord y) = 1, then ord(x + y) = ord(x)·ord(y)). 5. Let R be a PID, M an R-module with AnnR(M) = Rr, r ≠ 0. Then o(x)| r, ∀x ∈ M. 6. Let R be a PID and M an R-module (not necessarily finitely gener-ated) with AnnR(M) ≠ (0). We want to prove that M is a direct sum of cyclic submodules. Prove that:

a) For any N ≤ R M, AnnR(M/N) ≠ (0). b) There exists y ∈ M such that AnnR(y) = AnnR(M). (Hint. Let

AnnR(M) = Rr, r = nan

a pp …11 , with pi prime in R and ai ∈ N*. Let bi = max b ∈ N | ∃x ∈ M with pi

b|o(x) (bi ≤ ai). For any i, ∃xi ∈ M with o(xi) = ib

ip . Set y =∑ xi. o∑ xi = ∏o(xi) by ex. 4. Also, r = ∏ ib

ip , so bi = ai) c) Let C = C ≤ M | C is a direct sum of cyclic submodules and

satisfies condition (*), where the condition (*) is: ∀s ∈ R, ∀x ∈ M, if sx ∈ C, then ∃ x0 ∈ C with sx = sx0. (*) C is nonempty, inductively ordered, so it has a maximal element F.


150

d) ∀C ∈ C, if C ≠ M, then there exists D ∈ C such that C ( D. (Ind. Apply b) to M/C and obtain y ∈ M such that AnnR(M/C) = AnnR(y) =: µ. So µy ∈ C and let y0 ∈ C with µy = µy0, given by (*). Then set D := C + R(y − y0).)

e) F = M.

III.4 The endomorphisms of a finite dimensional vector space

Throughout this section, K is a field, K[X] is the polynomial ring in the indeterminate X with coefficients in K, V is a finite dimensional K-vector space and u : V → V is a K-endomorphism (also called linear transformation or operator).

The idea of the theory we develop here is the following: The endo-morphism u defines a natural structure of K[X]-module on V. Because K[X] is a PID, we can apply the invariant factors theorem to obtain information on this K[X]-module (therefore on the endomorphism u). Furthermore, keeping in mind the connection between endomorphisms and matrices, the results obtained translate into matrix language.

4.1 Definition. Let u ∈ EndK(V). We endow the Abelian group (V, +) with a structure of K[X]-module (depending on u): ∀f = a0 + a1X + + anX

n ∈ K[X], ∀v ∈ V, define: f·v := a0v + a1u(v) + + anu

n(v) ∈ V, where u n = uu (n times).

In other words, f·v is the image of the vector v by the endomorph-ism f(u) := a0id + a1u + + an u

n, where id is the identity automor-phism of V.


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Checking the module axioms is simple. It is worth mentioning that the axioms ( f·g)·v = ( f )·(g·v) and ( f + g)·v = f·v + g·v ∀f, g ∈ K[X], ∀v ∈ V, are a consequence of the fact that:

( f·g)(u) = f(u)g(u) and ( f + g)(u) = f(u) + g(u), ∀f, g ∈ K[X] This means that the evaluation map in u from K[X] to EndK(V),

f & f(u), ∀f ∈ K[X], is a ring homomorphism (more exactly, a K-algebra homomorphism).

The external operation defined above extends to K[X] × V the external operation (defined on K × V) of the K-vector space V.

Note that the K[X]-module structure defined on V depends strongly on the endomorphism u.

Let Vu denote the K[X]-module V defined by the endomorphism u, as above.

4.2 Remark. Let u ∈ EndK(V). The universality property of the polynomial ring K[X] ensures the existence of a unique K-algebra homomorphism η : K[X] → EndK(V) such that η(X) = u. If f = a0 + a1X + + anX

n∈ K[X], η( f ) = f (u) = η(a0 + a1X + + anX

n) = a0idV + a1u + + anu n.

The homomorphism η defines a K[X]-module structure on V (see remark II.1.4) that is the same as the one defined above.

Can distinct endomorphisms of V define the same K[X]-module structure on V?

4.3 Proposition. Let u, w ∈ EndK(V). Then the K[X]-modules Vu and Vw are isomorphic if and only if there exists a K-automorphism ϕ of V such that ϕuϕ −1 = w.

Proof. Let ϕ : Vu → Vw be a K[X]-isomorphism. Let ·u denote the external operation of the K[X]-module Vu. For any v ∈ Vu, X ·u v = u(v) and, ∀v ∈ Vw, X ·w v = w(v). We have ϕ(X ·u v) = X ·w ϕ(v), so ϕ(u(v)) = w(ϕ(v)), i.e. ϕu = wϕ. It is clear that ϕ is also a K-automorphism.


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If ϕ : V → V is a K-isomorphism with ϕu = wϕ, then ϕ is a K[X]-isomorphism from Vu to Vw. Indeed, ϕu = wϕ is written ϕ(X ·u v) = X ·w ϕ(v). From this we deduce that ϕ(X n·u v) = X n ·w ϕ(v), ∀n ∈ N; since ϕ is K-linear, ϕ((a0 + a1X + + anX

n)·u v) = a0·wϕ(v) + a1X·wϕ(v) + + anX n·wϕ(v),

for any a0 + a1X + + anX ∈ K[X]. !

4.4 Definition. a) The endomorphisms u, w ∈ EndK(V) are called similar if there exists ϕ ∈ AutK(V) such that ϕ uϕ

−1 = w. Notation: u ≈ w.

b) Let n ∈ N and A, B ∈ Mn(K). We call the matrices A and B simi-lar (denoted by A ≈ B) if there exists an invertible matrix U ∈ Mn(K) such that B = UAU −1.

4.5 Remark. Let KV be finite dimensional and let v = (v1, , vn) be a basis of V. The endomorphisms u, w ∈ EndK(V) are similar if and only if the matrices Mv(u) and Mv(w) are similar. To see that, recall that the rings EndK(V) and Mn(K) are (anti-)isomorphic by the mapping u & Mv(u).

A typical example of similar matrices is given by the matrices of an endomorphism in various bases of V.

The similarity relation on EndK(V) and the similarity relation on Mn(K) are equivalence relations.

4.6 Definition. Let u ∈ EndK(V) and W ≤ KV. The K-subspace W of V is called invariant relative to u (or u-invariant) if u(W) ⊆ W. We say that V is indecomposable relative to u (or u-indecomposable) if V can-not be written as a direct sum of proper u-invariant subspaces.

4.7 Proposition. Let W be a nonempty set of V. Then: W is an u-invariant subspace ⇔ ∀w ∈ W, X·w ∈ W ⇔ W is a K[X]-submodule of Vu.


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Proof. u(W) ⊆ W if and only if ∀w ∈ W, u(w) ∈ W ⇔ X·w ∈ W. By induction, X n·w ∈ W, ∀n ∈ N; since W is a linear subspace in V, this implies (a0 + a1X + + anX

n)·w ∈ W, ∀ai ∈ K. The converse is an easy exercise. !

The following statements (whose proofs are left to the reader) are further examples of translations from vector space language to K[X]-module language.

V is the direct sum of the u-invariant subspaces V1, , Vm ⇔ K[X]Vu is the direct sum of the K[X]-submodules V1, , Vm.

There exists v ∈ V such that V is generated by ui(v) | i ∈ N ⇔ Vu is a cyclic K[X]-module (generated by v). (In this case, u is called a cyclic endomorphism).

V is u-indecomposable ⇔ Vu is an indecomposable K[X]-module.

4.8 Proposition. a) The K[X]-module Vu is a finitely generated tor-sion module.

b) V is u-indecomposable if and only if there exists v ∈ Vu such that Vu = K[X]v, with o(v) = pk, where p is an irreducible polynomial in K[X] and k ∈ N*.

Proof. a) Any system of generators KV is also a system of genera-tors for the K[X]-module Vu. If dimKV = n ∈ N*, then, ∀v ∈ V, the vec-tors v, u(v), , un(v) cannot be linearly independent. Consequently, there exist a0, a1, , an ∈ K, not all zero, such that a0v + a1u(v) + + anu

n(v) = 0 ⇔ (a0 + a1X + + anX n)·v = 0.

b) V is u-indecomposable if and only if Vu is an indecomposable K[X]-module. But Vu is a torsion, finitely generated K[X]-module,. Apply now Prop. III.3.7. !

We apply to the K[X]-module Vu the structure theorems from III.2 and III.3:


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4.9 Proposition. If dimK V = n ∈ N*, then there exists m ∈ N* and v1, , vm ∈ Vu, such that Vu is a direct sum of u-invariant subspaces

Vu = K[X]v1⊕⊕K[X]vm, cu o(v1)||o(vm). The natural number m and the monic polynomials o(v1),,

o(vm) ∈ K[X] with the properties above are uniquely determined (o(v1),, o(vm) are the invariant factors of the K[X]-module Vu).

Vu is also a direct sum of indecomposable submodules: Vu = K[X]w1⊕⊕K[X]wt,

where wi ∈ Vu and o(wi) are powers of irreducible polynomials in K[X], 1 ≤ i ≤ t. The natural number t and the monic polynomials o(w1),, o(wt) ∈ K[X] are uniquely determined (o(w1),, o(wt) are the elementary divisors of the K[X]-module Vu). !

4.10 Definition. If dimKV = n and u ∈ EndK(V), the monic polyno-mial that generates AnnK[X](Vu) is called the minimal polynomial of the endomorphism u (and is denoted by µu). The monic polynomials that are the invariant factors (respectively the elementary divisors) of the K[X]-module Vu are called the invariant factors (respectively the elementary divisors) of the endomorphism u. The same terminology is used for matrices: choosing a basis v in V, the invariant factors of a matrix A ∈ Mn(K) are the invariant factors of the unique endomorph-ism u with Mv(u) = A (similarly for the minimal polynomial, the elementary divisors).

4.11 Proposition. Two endomorphisms (matrices) are similar ⇔ they have the same invariant factors ⇔ they have the same elementary divisors.

Proof. Let u, w ∈ EndK(V). We have: Mv(u) ≈ Mv(w) ⇔ u ≈ w ⇔ Vu ≅ K[X]Vw (by 4.3) ⇔ Vu and Vw have the same invariant factors ⇔ Vu and Vw have the same elementary divisors. !


155

4.12 Remark. With the notations in 4.9, µu is o(vm), the highest de-gree invariant factor of u. If f ∈ K[X] is monic, the following proper-ties are equivalent:

a) f = µu. b) f (u) = 0 and ∀g ∈ K[X], g(u) = 0 implies f |g. c) f (u) = 0 and ∀g ∈ K[X], g ≠ 0, g(u) = 0, implies deg f ≤ deg g. The proof is easy, using the definitions. Note that, unlike the mini-

mal polynomial of an algebraic element in a field extension, the mini-mal polynomial of an endomorphism is not necessarily irreducible.

The next result translates in matrix language the fact that V is a di-rect sum of u-invariant subspaces. 4.13 Proposition. a) Let V = V1⊕⊕Vm, where V1, , Vm are u-invariant subspaces. If vi is a basis in Vi, 1 ≤ i ≤ m, then v1∪∪vm =: v is a basis7 in V. If Ai is the matrix of the restriction of u to Vi, in the basis vi, 1 ≤ i ≤ m, then the matrix of u in the basis v is (written on blocks):

Mv(u) = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

mA

A

0

01

* ,

b) Conversely, if the matrix of u in a basis v is of the form above, then the rows of the block Ai correspond to a set of vectors in v that generate an u-invariant subspace Vi (1 ≤ i ≤ m) and V = V1⊕⊕Vm.

Proof. a) It is clear that v is a basis in V (see also II.4.11). To keep notations manageable, suppose m = 2 and v1 = (e1, , ep), v2 = ( f1, , fq), p + q = n = dimV. Then v = (e1, , ep, f1, , fq). Since V1, V2 are u-invariant, u(ei) is a linear combination of e1, , ep, and u( fj) is a linear combination of f1, , fq. Writing the matrix of u in the basis v,

7 We totally order the vectors in the basis v, by sequencing the elements of the

bases v1, , vm, in this order.


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Mv(u) = ⎥⎥⎦

⎤

⎢⎢⎣

⎡

2

1

0

0

A

A.

b) The task of detailing the proof is left to the reader. !

We want to find a basis v of V such that Mv(u) has as simple a form as possible. Since Vu is a direct sum of indecomposable submod-ules (theorem 4.9) the previous result allows us to study the restriction of u to each of the u-invariant subspaces in the direct sum. It is thus natural to study first the case in which Vu is indecomposable: Vu = K[X]v, for some v ∈ V, o(v) = pk, p irreducible in K[X], k ∈ N*.

4.14 Definition. If p ∈ K[X], p = X r − ar −1 X r −1 − − a1 X − a0,

define the r×r matrices with entries in K:

Cp =

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−1210

1000

01000010

raaaa ……*!……

, N =

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

00010000

00000000

……

………

The matrix Cp is called the matrix companion of the polynomial p. Define the rk×rk matrix (written in block form):

Cp N 0 0 0 0 Cp N 0 0 * * J pk = * *

∈ Mrk(K)

0 0 0 Cp N 0 0 0 0 Cp

J pk is called the Jordan cell 8 associated to the polynomial p k.

8 Camille Jordan (1838-1922), French mathematician.


157

A matrix (written in block form) having on the diagonal Jordan cells (and 0 elsewhere), i.e. a matrix of the form

( )( )

( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

tkt

k

k

pJ

pJpJ

0

02

1

2

1

*,

where p1, , pt are monic irreducible polynomials in K[X], is called a Jordan canonical matrix9 over K.

4.15 Proposition. a) Suppose Vu is a indecomposable K[X]-module and v ∈ Vu is such that Vu = K[X]v, with o(v) = µu = pk, where p = X r − ar−1X

r−1 − − a1X − a0 ∈ K[X] is irreducible of degree r and k ∈ N*. Then dimKV = rk and there exists a basis v of KV such that the matrix of u in v is the Jordan cell J p

k associated to pk. b) In the general case, let tk

tk pp ,,11 … be the elementary divisors of

u, with p1, , pt irreducible and monic in K[X]. Then there exists a ba-sis of V in which the matrix of u is the Jordan canonical matrix:

J =

( )( )

( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

tkt

k

k

pJ

pJpJ

0

02

1

2

1

*.

Proof. a) We exhibit a basis e = (e0, , ekr−1) such that Me(u) = J pk :

9 In other texts this matrix is called a rational canonical matrix, the name Jordan

canonical matrix being given only if pi are polynomials of degree 1.


158

e0 = v; e1 = X·v = u(e0); ; er−1 = X r−1·v = u(er−2); er = p·v; er+1 = Xp·v = u(er); ; e2r−1 = X r−1p·v = u(e2r−2); e(k −1)r = pk −1·v; e(k −1)r+1 = Xp·v = u(e(k −1)r); ; ekr−1 = X r−1pk −1·v = u(ekr−2).

The next lemma proves that e = (e0, , ekr−1) is a basis:

Lemma. If u ∈ EndK(V) is an endomorphism such that Vu = K[X]v for some v ∈ V, and f = o(v), deg f = n, then, for any g0, , gn−1 ∈ K[X], with deg gi = i, 1 ≤ i ≤ n, the vectors

g0·v, , gn−1·v form a basis of V.

Proof of the lemma. Vu ≅ K[X]/(f ) (K[X]-module isomorphism, so also a K-vector space isomorphism), hence dimKV = dimK K[X]/(f ) = deg f = n. The vectors g0·v, , gn−1·v are linearly independent: if a0g0·v + + an−1gn−1·v = 0, with ai ∈ K, then h·v = 0, where h = a0g0 + + an−1gn−1. Since o(v) = f, and deg h < n, we have f |h, hence h = 0. But the polynomials g0, , gn−1 are linearly independent in the K-vector space K[X], being of distinct degrees, so a0 = = an−1 = 0. The n elements g0·v, , gn−1·v are thus linearly independent in V, whose dimension is n, which means they are a basis.

We get back to proving that Me(u) = J pk. If 1 ≤ i < k, we have: u(eir−1) = X(X r−1p i−1·v) = X rp i−1·v = (p + a0 + a1X + + ar−1X

r−1)pi−1·v = pi·v + a0p

i−1·v + a1Xp i−1·v + + ar−1X r−1p i−1·v

= eir + a0e(i −1)r + a1e(i −1)r+1 + + ar−1e(i −1)r+r−1 If i = k, u(ekr−1) = a0e(k −1)r + a1e(k −1)r+1 + + ar−1e(k−1)r+r−1, since

pk·v = 0. These equalities, together with the relations (0), , (k − 1), prove

the claim. b) Decompose Vu as a direct sum of u-indecomposable u-invariant

subspaces (see 4.9). By a), each such subspace has a basis in which the restriction of u has the matrix of the form J pk, with pk elementary divisor of u. Apply now Prop. 4.13. !


159

4.16 Corollary. Any matrix A ∈ Mn(K) is similar to a Jordan canonical matrix. If the elementary divisors of A are tk

tk pp ,,11 … , then

A ≈ J, where J is the matrix at 4.15.b). !

How do we find the elementary divisors of an endomorphism u (of a matrix A)? This amounts to finding its invariant factors. The next theorem says: take the matrix XI − A and find its Smith normal form. The non constant polynomials on the diagonal are then the invariant factors of A.

4.17 Theorem. Let v = (v1,, vn) be a basis in V such that Mv(u) =: A = (aij) ∈ Mn(K). Then the invariant factors of u (of the ma-trix A) are the polynomials of degree > 0 on the diagonal of the diago-nally canonic matrix D ∈ Mn(K[X]), arithmetically equivalent to the matrix

XI − A =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−−−−

nnnn

n

n

aXaa

aaXaaaaX

……

……

21

22221

11211

∈ Mn(K[X]).

Proof. Recall the proof of the invariant factors theorem (2.3): since (v1, , vn) generates K[X]Vu, take a K[X]-module E, free of basis e = (e1, , en) and the K[X]-homomorphism ϕ : E → Vu, ϕ(ei) = vi, 1 ≤ i ≤ n. The homomorphism ϕ is surjective, Kerϕ =: F is a free submodule in E and E/F ≅ Vu. Since Vu is torsion, rank F = n and there exist two bases ε = (ε1, , εn) in E and φ = (φ1, , φn) in F, such that φi = diεi, di ∈ K[X], d1|d2||dn. The invariant factors of Vu are the noninvert-ible polynomials among d1, d2,, dn. (see the proof of 2.3)

How do we find d1, d2,, dn? For any basis f in F, there exists a matrix B ∈ Mn(K[X]) such that f = Be. Then the diagonally canonical matrix D ∈ Mn(K[X]), arithmetically equivalent to B, is exactly D = diag (d1, d2,, dn). Thus, we need such a matrix B. Let

fi =: Xei − (ai1e1 + ai2e2 + + ainen) ∈ E, 1 ≤ i ≤ n.


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If f = (f1, f2,, fn), then these relations can be written f = (XI − A)e. Lemma. f =: ( f1, f2,, fn) is a basis in F = Ker ϕ. Proof of the lemma. For any 1 ≤ i ≤ n,

u(vi) = ai1v1 + ai2v2 + + ainvn Thus,

ϕ(fi) = Xvi − (ai1v1 + ai2v2 + + ainvn) = u(vi) − (ai1v1 + ai2v2 + + ainvn) = 0.

This shows that fi ∈ Kerϕ = F. Let us prove that f is a system of generators for F. Note that: Xei = fi + ai1e1 + ai2e2 + + ainen, 1 ≤ i ≤ n. (*)

So, X 2ei = Xfi + ai1 Xe1 + ai2 Xe2 + + ain Xen. Using (*), we obtain: X 2ei = ∑j qj fj + ∑i rjej, for some qj ∈ K[X], rj ∈ K, 1 ≤ j ≤ n. By induc-tion, one easily sees that, ∀m ∈ N*, X mei is expressed as: X mei = ∑j qj fj + ∑i rjej, for some qj ∈ K[X], rj ∈ K, 1 ≤ j ≤ n. (**)

From (**) we deduce that, ∀g ∈ K[X], gei can be written in the same form. So, if y = ∑i giei ∈ F, with gi ∈ K[X], 1 ≤ i ≤ n, then y = ∑i giei = ∑i qi fi + r, for some qj ∈ K[X], and r ∈ E is of the form ∑i ciei, with ci ∈ K. But r = y − ∑i qi fi ∈ F, so ϕ(r) = ϕ(∑i ciei) = ∑i civi = 0. Since v is basis in V, ci = 0, 1 ≤ i ≤ n, hence r = 0. Finally, we obtain y = ∑i qi fi.

Let us prove the linear independence. Suppose ∑i gi fi = 0, with gi ∈ K[X], 1 ≤ i ≤ n. Using (*), it follows that ∑i gi Xei = ∑i gi ∑j aij ej = ∑j ∑i aijgi ej. Since (e1, , en) is a basis,

gi X = ∑j aji gj, 1 ≤ i ≤ n. Let g1 be the polynomial of maximal degree among the gi (relabel if

necessary). So, deg gi ≤ deg g1, 1 ≤ i ≤ n. If g1 ≠ 0, g1 X = ∑j aj1 gj implies

deg∑j aj1 gj ≤ max j(deg(aj1 gj)) ≤ max j(deg gj) < 1 + deg g1 = deg g1X, contradiction. This shows that ( f1, f2,, fn) is linearly independent.


161

We continue the proof of the theorem. The relations fi = Xei − (ai1e1 + ai2e2 + + ainen) show that f = (XI − A)e. If XI − A ∼ D = diag(d1, d2,, dn), with d1|d2||dn, then the invariant factors of Vu (of the endomorphism u) are dk, dk + 1, , dn, where k = mini | di noninvertible. !

4.18 Definition. Let A ∈ Mn(K). The polynomial fA := det(XI − A) ∈ K[X] is called the characteristic polynomial of the matrix A. If u is an endomorphism of the K-vector space V and A is the matrix of u (in some basis), then the characteristic polynomial fu of u is by definition fA. This definition is correct: Two similar matrices A and B have the same characteristic polynomial: if B = SAS −1, for some S ∈ GLn(K), then: fB = det(XI − SAS −1) = det(S·(XI − A)S −1) =

= det(S)·det(XI − A)·det(S −1) = fA, We used the fact that the matrix XI commutes with any matrix in

Mn(K[X]).

4.19 Remark. Let A = (aij) ∈ Mn(K) and let fA = det(XI − A) = X n − c1 X

n − 1 + c2 X n − 2 + + ( − 1) ncn,

with c1, , cn ∈ K. Writing the definition of det(XI − A) and arranging the terms by the

like powers of X, the coefficients c1, , cn are: c1 = a11 + a22 + + ann =: Tr(A) (called the trace of A),

cn = det(A). More generally, ck (1 ≤ k ≤ n) is the sum of the minors of order k of

A on the main diagonal (i.e., the minors obtained by selecting k rows

i1, , ik and the columns with the same indices i1, , ik of the

matrix A). There are ⎟⎠⎞

⎜⎝⎛kn

such minors, one for each choice of a subset

of k indices from 1, 2, , n.


162

The same terminology applies for an endomorphism u ∈ EndK(V), dimV = n, whose matrix is A (in some basis of V). Its characteristic polynomial is fu = fA, and the coefficients c1, , cn are uniquely de-fined by u, as above. c1 =: Tr(u) = Tr(A) is called the trace of u and cn =: det(u) = det(A) is called the determinant of u.

4.20 Proposition. Let A, B ∈ Mn(K). Then A and B are similar matrices if and only if XI − A and XI − B are arithmetically equivalent matrices in Mn(K[X]).

Proof. If A ≈ B, then there exists S ∈ GL(n, K) such that B = S −1AS. Then XI − B = S −1(XI − A)S, so XI − A ∼ XI − B, since obviously S ∈ GL(n, K[X]).

Suppose now that XI − A ∼ XI − B and let D ∈ Mn(K[X]) be the canonical matrix with D ∼ (XI − A) ∼ (XI − B). So, A and B have the same invariant factors: the polynomials of degree > 0 on the diagonal of D, according to 4.17. Thus, A and B have the same elementary divi-sors. By 4.15, A and B are similar with the same Jordan canonical ma-trix. !

Let n ∈ N*. The following results are about matrices in Mn(K), but they can be translated in statements on endomorphisms of a K-vector space of dimension n.

4.21 Proposition. The characteristic polynomial of a matrix A is the product of the invariant factors of A (and it is equal to the product of the elementary divisors of A).

Proof. The matrix XI − A is arithmetically equivalent to the matrix in Smith normal form D = diag(1, , 1, d1, , dm) ∈ Mn(K[X]), where d1, , dm are the invariant factors of A. There exist S, T ∈ U(Mn(K[X])) (i.e. detS, detT ∈ K*) such that XI − A = SDT. We have

d1··dm = detD = det(S(XI − A)T) = detS·fA·detT


163

This means that the monic polynomials d1··dm and fA differ by the factor detS·detT ∈ K*, which shows that they are equal. On the other hand, it is clear that the product of the elementary divisors equals the product of the invariant factors. !

4.22 Corollary. a) (The Cayley-Hamilton theorem)10 Any matrix A ∈ Mn(K) is a root of its characteristic polynomial: fA(A) = 0.

b) (The Frobenius theorem)11 The characteristic polynomial and the minimal polynomial of a matrix A ∈ Mn(K) have the same irreducible factors in K[X].

Proof. Let d1, , dm ∈ K[X] be the invariant factors of A. a) The minimal polynomial of A is dm (the invariant factor of high-

est degree). So, dm(A) = 0 and dm| fA, thus fA(A) = 0. b) This is a consequence of d1··dm = fA and d1||dm. !

Given a matrix A, is the Jordan canonical matrix that is similar to A uniquely determined? Of course, by reordering the elementary divi-sors, diverse Jordan canonical matrices are obtained. The next proposition shows that these are all the Jordan canonical matrices similar to A.

4.23 Proposition. a) Let p ∈ K[X] be monic and irreducible and let k ∈ N*. The Jordan cell J p k has only one elementary divisor, namely p k.

b) Let A be a Jordan canonical matrix whose diagonal is made up by the Jordan cells ( )ik

ipJ , where pi ∈ K[X] are monic and irreduci-ble, 1 ≤ i ≤ t. Then the elementary divisors of A are ik

ip , 1 ≤ i ≤ t.

10 Arthur Cayley (1821-1895), Sir William Rowan Hamilton (1805-1865), British

mathematicians. 11 Ferdinand Georg Frobenius (1849-1917), German mathematician.


164

c) Let A, B ∈ Mn(K) be Jordan canonical matrices such that A ≈ B. Then A and B have the same Jordan cells, perhaps in different order.

Proof. a) Let V := K[X]/ pk. V is a K[X]-module (being a factor module of K[X]) and a K-vector space of dimension k·deg p =: n. Let u ∈ EndK(V), u(y) = X·y, ∀y ∈ V (the dot denotes the external K[X]-module operation of V). The endomorphism u defines on V a structure of a K[X]-module, Vu, as in definition 4.1. It is easy to see that these two structures of K[X]-module coincide and that V = K[X]v, where v := 1 + pk ∈ V. So, V = Vu is an indecomposable K[X]-module, and its only elementary divisor is o(v) = pk. There exists a basis (as in 4.15) in which u has the matrix J pk. So J pk has the same elementary divisors as Vu, namely only pk.

b) There exists KV and u ∈ EndK(V), whose matrix is A (in some ba-sis). Then V is written as a direct sum of u-invariant subspaces: Vu = V1⊕⊕Vt, Vi being the u-invariant subspace corresponding to the Jordan cell ( )ik

ipJ . Let ui be the restriction of u to Vi, 1 ≤ i ≤ t. The matrix of ui is exactly ( )ik

ipJ and the first part of the proof shows that ik

ip is the only elementary divisor of ui. The elementary divisors of u are obtained writing down all the elementary divisors of the restric-tions ui, 1 ≤ i ≤ t, namely ik

ip , 1 ≤ i ≤ t. c) If A ≈ B, then A and B have the same elementary divisors (4.11).

By b), A and B have the same Jordan cells. !

4.24 Definition. If A ∈ Mn(K) and J is a Jordan canonical matrix such that J ≈ A, then J is called the Jordan canonical form of A. The Jordan canonical form of A is uniquely determined up to an order of the Jordan cells on the diagonal.

The classical notions of eigenvector and eigenvalue of an endomorphism are closely connected to its invariant subspaces of dimension one.


165

4.25 Definition. If u ∈ EndK(V), an element λ ∈ K is called an eigenvalue of u if there exists a vector v ∈ V, v ≠ 0, such that

u(v) = λv. Each such vector v is called an eigenvector of u for the eigenvalue

λ.

4.26 Proposition. Let v ∈ V, u ∈ EndK(V) and λ ∈ K. The following statements are equivalent:

a) v is an eigenvector of u for the eigenvalue λ. b) Considering v ∈ K[X]Vu, o(v) = X − λ ∈ K[X]. c) dimK (K[X]v) = 1 (the submodule of Vu generated by v has

K-dimension 1). Proof. a)⇒b) We have u(v) = λv. In K[X]-module language for Vu,

this means X·v = λv, thus (X − λ)·v = 0. So, o(v)| X − λ, which is irreducible, hence o(v) = X − λ. (o(v) = 1 is impossible, since it implies v = 0).

b)⇒a) o(v) = X − λ implies u(v) = λv. b)⇔c) This follows from dimK K[X]v = deg o(v). !

4.27 Proposition. λ ∈ K is an eigenvalue of u if and only if λ is a root of fu, the characteristic polynomial of u.

Proof. Suppose v ∈ V is an eigenvector of u for the eigenvalue λ. Then o(v) = X − λ; since the minimal polynomial of u, µu, is in AnnK[X](v) = (X − λ), we have X − λ|µu. Also µu | fu, so X − λ| fu, which means that λ is a root of fu. Conversely, if fu(λ) = 0, then X − λ| fu. Be-cause fu and µu have the same irreducible factors, X − λ|µu, so µu = (X − λ)g for some g ∈ K[X]. By 4.9, there exists v ∈ V such that o(v) = µu. Then o(g·v) = X − λ, i.e. g·v ∈ V is an eigenvector for the eigenvalueλ. !

The reader is invited to give an alternate proof, using facts from the theory of systems linear equations.


166

Note that if fu has no roots in K, then u has no eigenvalues and no eigenvectors.

We describe now the Jordan cells in the important cases K = C and K = R.

If K is an algebraically closed field (in particular, K = C), then the irreducible monic polynomials in K[X] are of the form X − a, a ∈ K . Thus the Jordan cell J(X − a)k is:

a 1 0 0 0

0 a 1 0 0

* * J(X − a)k = * *

∈ Mk(K).

0 0 0 a 1

0 0 0 0 a

If K = R, the monic irreducible polynomials in R[X] are X − a, a ∈ R (and the Jordan cell J(X − a)k is the one above), or X 2 − bX − c, with b, c ∈ R and b2 + 4c < 0, in which case the Jordan cell J(X 2 − bX − c)k is:

J(X 2 − bX − c)k =

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

bc

bc

bc

bc

100010010

010010

0010010

********

∈ M2k(R).


167

Exercises

In the exercises, K is a field, V is a finite dimensional K-vector space and u is an endomorphism of KV. 1. Prove that the eigenvectors corresponding to distinct eigenvalues of u are linearly independent. 2. Give an example of three matrices in M3(Q) whose only eigenvalue is 2 and any two matrices are not similar. Can you exhibit four such matrices? Generalization. 3. Determine the endomorphisms u ∈ EndK(V) whose minimal polynomials are of degree 1. 4. Give an example of two matrices having the same minimal polyno-mial and the same characteristic polynomial, but are not similar. 5. Let A ∈ Mn(K) such that the characteristic polynomial of A splits in factors of degree 1 in K[X] (one says that A has all its eigenvalues in K). Then A is similar to an upper triangular matrix T = (tij) ∈ Mn(K) (tij = 0 if i > j). In this case, A is called trigonable. Is the converse true? 6. Let A ∈ Mn(K) and let p ∈ K[X]. If A has all its eigenvalues λ1, , λn in K, then p(A) has the eigenvalues p(λ1), , p(λn). (Ind. Let A ≈ T, with T upper triangular. The diagonal of T is λ1, , λn. Com-pute p(T).) 7. Compute the characteristic polynomial and the minimal polynomial of the following matrices:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

1000011010100101

,4267921337

,

2000001100102101

8. Let A ∈ Mn(K). Then tA is similar to A. Moreover, there exists U ∈ GL(n, K), symmetric, such that tA = U −1AU.


168

9. Let R be a commutative ring with identity and let n ∈ N*. General-ize the relevant notions and prove the Cayley-Hamilton theorem: if E is a free R-module of rank n and u ∈ EndR(E), then u is a root of the characteristic polynomial: fu(u) = 0. (Hint: let A the matrix of u in a ba-sis (e1, , en) and XI − A = B = (bij) ∈ Mn(R[X]); fu = det B. In the R[X]-module Eu the relations ∑j bijej = 0 hold, ∀i. Let Bik ∈ R[X] be the algebraic complement of bik in the matrix B. For any fixed k, multiply relation i with Bik and sum after i to obtain fu(X)·ek = 0 = fu(u)(ek).) 10. Let R be a commutative ring with identity and let A ∈ U(Mn(R)) be an invertible matrix. Then A is a R-linear combination of I, A, , A n − 1. (Hint. Use the Cayley-Hamilton theorem.) 11. Let u ∈ EndK(V). Then V has no u-invariant proper subspaces ⇔ the K[X]-module Vu is simple ⇔ the characteristic polynomial of u is irreducible in K[X]. 12. Let u ∈ EndK(V) be an endomorphism having the eigenvalue 0. Then V = U ⊕ W, with U, W u-invariant subspaces and dim U = 1. 13. Let u ∈ EndK(V) be a nilpotent endomorphism (∃r ≥ 1 such that u r = 0). Then Tr u k = 0, ∀k ≥ 1. Conversely, if char K = 0 and Tr u k = 0, ∀k ≥ 1, then u is nilpotent. (Ind. Let f be the characteristic polynomial of u; in the relation f(u) = 0 apply Tr and deduce that 0 is an eigenvalue of u. So, u has an invariant subspace of dimension dim V − 1.) 14. Let V = U ⊕ W (direct sum of subspaces). Then any v ∈ V can be uniquely written as v = u + w, with u ∈ U, w ∈ W. Define π, ρ : V → V by: ∀u ∈ U, ∀w ∈ W, π(u + w) = u (π is called the projec-tion on U along W) and ρ(u + w) = u − w, (ρ is called the symmetry with respect to U along W). Show that π and ρ are K-endomorphisms of V and find their minimal polynomials.

169

IV. Field extensions

This chapter contains the basic concepts and results from the theory of field extensions. Standard facts about rings and vector spaces are a prerequisite: polynomial rings, factor rings, ring isomorphism theo-rems, bases and dimension in vector spaces, prime and maximal ide-als. Knowledge of polynomial ring arithmetic is recommended (as provided in the chapter Arithmetic in integral domains). Some elementary properties of cardinals and Zorn's Lemma are used in the proof of existence of the algebraic closure of a field. Most of these facts can be found in the Appendices; a more detailed treatment is found in most Abstract (Modern) Algebra introductory texts.

IV.1. Algebraic extensions

Recall that a field is a commutative ring (K, + , ·) with identity 1 (with 1 ≠ 0), with the property that every nonzero element is invertible with respect to multiplication. Any field has at least two elements: 0 and 1. A field K has no zero divisors (in other words, K is a domain): for any x, y ∈ K, x ≠ 0 and y ≠ 0 implies xy ≠ 0.


170

All rings and all ring homomorphisms considered are supposed to be unitary. Thus, if K, L are rings (with identity), the map σ : K → L is a homomorphism if and only if, ∀x, y ∈ K:

σ(x + y) = σ(x) + σ(y) σ(xy) = σ(x)σ(y)

σ(1) = 1.

The objects we study are field extensions: if L is a field and K is a subfield in L, we also say that L is an extension of K.

Recall that the nonempty set K of the field L is a subfield in L if it is closed under addition and multiplication and becomes a field with the induced operations. A widely used characterization is:

K is a subfield of L if and only if for any x, y ∈ K with y ≠ 0, we have x − y and xy−1∈ K.

The subfield K of L is called proper if K ≠ L. This notion of field extension is too restrictive. For instance, it is

natural to consider that C is an extension of R. But C is usually con-structed as the set of all couples (a, b) with a, b ∈ R, endowed with an addition and a multiplication that make it a field. In this setting, R is not a subset of C, but R can be identified with the set of the couples (a, 0), a ∈ R. In fact, the rigorous interpretation is the following: one defines the field homomorphism ϕ : R → C, ϕ(a) = (a, 0) and then identifies R with its image ϕ(R), which is a subfield of C.

More generally, if σ : K → L is a field homomorphism, then σ is injective. Indeed, ∀x ∈ K, x ≠ 0 implies ∃x −1 ∈ K, so σ(x)σ(x −1) = σ(xx −1) = σ(1) = 1, so σ(x) is nonzero and thus Kerσ is 0. The homomorphism σ : K → L being injective, we may identify the field K with its image σ(K), which is a subfield in L. It is thus natural to consider the following definition:

1.1 Definition. a) Let K, L be fields. If σ : K → L is a field homo-morphism, we call the triple (K, L, σ) a field extension of K. We ex-


171

press this by writing K ⊆ L is a field extension, L/K is a field extension or L is an extension of K. For any element a ∈ K, we identify σ(a) ∈ L with a ∈ K. For example, if a ∈ K and x ∈ L, we write a·x instead of σ(a)·x. This identification allows to consider K as a subfield of L. The extension K ⊆ L is called proper if the inclusion is strict (i.e. σ(K) ⊂ L). In this context, we call an intermediate field (or subextension) of the extension K ⊆ L any subfield E of L that includes K. The intermediate field E is called proper if E ≠ K and E ≠ L.

Note that L is an extension of K if and only if L is a field also struc-tured as a K-algebra (the structural homomorphism is precisely σ).

b) If K ⊆ L and K ⊆ E are extensions of K, a mapping ϕ : L → E is called a K-homomorphism if ϕ is a ring homomorphism and ϕ|K = idK (the identity function of K).

If one considers the general definition above, i.e. there exist field homomorphisms σ : K → L, τ : K → E, we call ϕ : L → E a K-homomorphism if ϕ is a field homomorphism with ϕ σ = τ (in other words, ϕ is a K-algebra homomorphism). Let HomK(L, E) denote the set of all K-homomorphism from L to E.

1.2 Examples. a) The real numbers field R is an extension of Q, the field of rational numbers.

b) The field C of complex numbers is an extension of R. c) The set [ ] 2 2 ,a b a b= + ∈ ∈, - , is a field, and it is an

extension of Q. d) If K is a field, then K(X), the field of rational fractions with

coefficients in K (K(X) is the quotient field of the polynomial ring K[X]) is an extension of K.

e) If n is a natural number, the ring Zn of the integers modulo n is a field if and only if n is a prime. For any prime p, let Fp denote the field of integers modulo p. We shall construct an extension of F2 at 1.24.


172

f) A method to construct subfields, very important in Galois the-ory39, is the following: let K be a field and let H be a set of field endomorphisms of K (homomorphisms defined on K with values in K). The set K H := x ∈ K|σ(x) = x, ∀σ ∈ H is a subfield of K (called the fixed subfield of H). Indeed, if x, y ∈ K H and σ ∈ End(K) then σ(x − y) = σ(x) − σ(y) = x − y, so x − y ∈ K H. Likewise, xy−1 ∈ K H if y ≠ 0.

A basic tool is the concept of degree of an extension.

1.3 Definition. If K ⊆ L is an extension, then L is canonically a K-vector space: the multiplication of a scalar in K with a vector in L is their multiplication in L. The dimension of L as a K-vector space is called the degree 40 of the extension K ⊆ L and is denoted by [L : K] or (L : K). An extension is called finite if its degree is finite. In exam-ple b) above, 1, i is a basis of C over R, so [C : R] = 2. What are the degrees of the other extensions?

1.4 Definition. The fields that have no proper subfields are called prime fields.

We determine now all prime fields. Recall the notion of characteristic of a ring R with identity. Let 1

be the identity element of R and let n ∈ N; n ·1 denotes the multiple 1 + + 1 (n terms); the characteristic of R, denoted char R, is defined as follows:

- if, for any n ∈ N*, n ·1 ≠ 0, then char R = 0;

39 Evariste Galois (1811-1832), French mathematician. 40 One can define the degree of an extension of skew fields (division rings): if L

is a skew field and K is a subfield (skew) of L, then L is naturally a K-vector space and one defines the degree [L : K] as dimKL.


173

- if there exists n ∈ N* such that n ·1 = 0, then char R is the smallest n ∈ N* such that n ·1 = 0.

1.5 Remark. The characteristic of R can be defined also as follows: there exists a unique ring homomorphism ϕ : Z → R (prove!). Then char R is the natural generator of the ideal Kerϕ. The proof of the equivalence of these definitions is left to the reader.

For example, char Z = char Q = 0; char F2 = 2; if R is a unitary subring of S, then char R = char S.

1.6 Proposition. Let R be a domain. Then the characteristic of R is 0 or a prime. In particular, the characteristic of a field is 0 or a prime.

Proof. Suppose n = char R ≠ 0. If, by contradiction, n = ab, with a, b ∈ N*, a < n, b < n, then 0 = n ·1 = (a·1)·(b·1). But R has no zero divi-sors, so a·1 = 0 or b·1 = 0, contradicting the minimality of n. !

A related notion is the characteristic exponent of a field: the characteristic exponent of the field K is 1 if char K = 0 and is p if char K = p > 0.

1.7 Proposition. Let K be a prime field. If char K = 0, then there exists a unique isomorphism K ≅ Q. If char K = p > 0, then there exists a unique isomorphism K ≅ Fp.

Proof. Define the ring homomorphism ϕ : Z → K, ϕ(n) = n·1, ∀n ∈ Z, where 1 is the identity of K.

Suppose char K = 0. Since for any n ∈ Z, n·1 ≠ 0, n·1 is invertible in K. Define the homomorphism ψ : Q → K that extends ϕ, ψ(a/b) = ϕ(a)ϕ(b)−1, ∀a, b ∈ Z, b ≠ 0. Since ψ(Q) is a subfield in K, we have ψ(Q) = K, so ψ is an isomorphism.

If char K = p, Kerϕ = pZ; applying the isomorphism theorem, Z/pZ = Fp ≅ Imϕ, which is a subfield in K. Since K has no proper sub-fields, Imϕ = K.

The reader is invited to prove the uniqueness part. !


174

1.8 Lemma. The intersection of any family of subfields of a field is a subfield.

Proof. Use the characterization: K is a subfield of L if and only if for any x, y ∈ K with y ≠ 0, we have x − y and xy−1∈ K. If (Li)i∈I is a family of subfields of the field L and x, y ∈ ∩Li, then x, y ∈ Li, ∀i ∈ I, so x − y ∈ Li, ∀i ∈ I, i.e. x − y ∈ ∩Li. The same argument works for xy−1 if y ≠ 0. !

1.9 Theorem. Any field K contains a unique prime subfield. In par-ticular:

- if char K = 0, then K is an extension of Q; - if char K = p > 0, then K is an extension of Fp. Proof. The intersection of all subfields of K is a subfield P, which

cannot have proper subfields (any subfield F of P would be a subfield in K, so F includes P, absurd!). If Q is a prime subfield of K, then P ∩ Q is a subfield both in P and in Q, so it is equal to both of them; this shows that P is the unique prime subfield in K. The rest of the statement follows from the list of prime subfields. !

1.10 Application. Finite fields. If F is a field with a finite number of elements, then char F cannot be 0 (otherwise F is an extension of Q, which is infinite!), so char F = p > 0 (for some prime p) and F is an extension of the field Fp. If n is the degree [F : Fp], n is finite and F is isomorphic (as an Fp-vector space) to (Fp)

n. Thus, the cardinal of a fi-nite field is p n, for some prime p and some n ∈ N*. This means there do not exist fields with 10 elements, for instance.

The following definition describes a basic construction in field extensions.

1.11 Definition. Let K ⊆ L be a field extension and let S be a sub-set of L. The intersection of all subfields of L that include K ∪ S is a subfield of L, called the field generated by K and S, and denoted by


175

K(S). One also says that K(S) is obtained by adjoining to K the ele-ments in S.

We denote by K[S] the subring of L generated by K ∪ S (the intersection of all subring of L that include K ∪ S).41

It is easy to see that K(S) (respectively K[S]) is the smallest subfield (respectively subring) of L that includes K ∪ S. Obviously, the ring K[S] is a domain (it is a subring of the field L) and K[S] ⊆ K(S). The field of fractions of the domain K[S] is canonically isomorphic to K(S) (see the universality property of the field of fractions).

If S = x1,, xn, then K(S) is denoted by K(x1,, xn) and K[S] by K[x1,, xn]. An extension K ⊆ L with the property there exists a finite subset S of L such that L = K(S) is called a finitely generated exten-sion. Do not confuse with the concept of finite extension (which means that its degree is finite)!

If there exists α ∈ L such that L = K(α), the extension K ⊆ L is called a simple extension, and α is called a primitive element. A primi-tive element need not be unique: for instance, K(α) = K(α + 1).

1.12 Definition. Let K ⊆ L be a field extension and let E, F be extensions of K, included in L. The composite of the fields E and F (denoted by EF) is the subfield of L generated by E and F: EF = K(E∪F) = E(F) = F(E). In general, for a family (Ei)i∈I of exten-sions of K, included in L, the composite of the fields (Ei)i∈I is the sub-field of L generated by ∪i∈I Ei, K(∪i∈I Ei).

41 This notation is used also for ring extensions: if R is a subring of the ring A,

and S is a subset of A, R[S] is the subring generated by R ∪ S in A (R[S] coincides

with the R-subalgebra of A generated by S).


176

1.13 Remark. The set IF(L/K) of all intermediate fields of a given extension L/K is ordered by inclusion and IF(L/K) is a (complete) lat-tice 42: for any E, F ∈ IF(L/K), supE, F = EF, infE, F = E ∩ F.

1.14 Theorem. Let L/K be an extension and let S, S1, S2 be subsets of L. Then:

a) K[S] is the set of all polynomial expression in the elements of S,

with coefficients in K, namely:

K[S] =

( )( )1

1 1

1

*1 1 1

, ,

, , , , , , ,n

n nn

n

ii' ni i n n i i n

i i

a x x n x x S a K i i∈

∈ ∈ ∈ ∀ ∈⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

∑ … …… '

… ' … … '

where ∑' means the sums are finite, i.e. the set ( ) 0,,

11 ≠∈nii

nn aii …… N is finite.

b) K(S) = αβ −1|α, β ∈ K[S], β ≠ 0. c) K[S1∪ S2] = K[S1][S2] = K[S2][S1] and K(S1∪ S2) = K(S1)(S2) =

K(S2)(S1). d) K(S) = ∪ K(T) | T ⊆ S, T finite. Proof. a) Let T be the set of all polynomial expression in the ele-

ments of S, with coefficients in K. One checks directly that T is a subring of L, that includes K and S. (Alternatively, T is the image in L via the evaluation homomorphism of polynomial ring in S indetermi-nates). On the other hand, any subring of L that includes K and S must also include T. Thus, T is the smallest subring of L that includes K and S.

42 An ordered set (S, ≤) is called a lattice (resp. a complete lattice) if any subset

with two elements (resp. any subset) of S has a least upper bound and a greatest lower bound in L.


177

b) Since K[S] ⊆ K(S) and K(S) is a field, any element of the form αβ −1, with α, β ∈ K[S], β ≠ 0, is in K(S). But the set of all these ele-ments is a subfield in L (standard check).

c), d) Exercise. !

In the particular case S = x1, , xn, with x1, , xn ∈ L, : K[x1, , xn] = ( ) [ ] nn XXKfxxf ,,, 11 …… ∈

K(x1, , xn) = ( )( )

[ ] ( )⎭⎬⎫

⎩⎨⎧ ≠∈ 0,,,,,

,,

111

1nn

n

n xxgXXKgfxxgxxf ……

…… .

If S = a ∈ L: K[a] = f (a) | f ∈ K[X] and K(a) = f (a)/g(a) | f, g ∈ K[X], g(a) ≠ 0.

Thus, K[a] = Im eva, where eva : K[X] → L is the unique K-algebra homomorphism with the property that eva(X) = a; eva is called the homomorphism of evaluation in a. If f ∈ K[X], f = b0 + b1 X + + bn X

n, then eva( f ) = b0 + b1a + + bna n ∈ L. The usual notation for

eva( f ) is f (a), called the value of f in a. If f (a) = 0, we say a is a root of f.

The following notion is central in all the theory we describe.

1.15 Definition. Let K ⊆ L be a field extension and let x ∈ L. We say that is algebraic over K if there exists a nonzero polynomial f ∈ K[X] such that f (x) = 0.

In other words, x is algebraic over K if and only if the evaluation homomorphism evx : K[X] → L is not injective.

If the element x is not algebraic over K (i.e. evx is injective), x is called transcendental over K. Thus, x is transcendental over K if and only if evx induces a K-isomorphism between the polynomial ring K[X] and the subring K[x] generated by K and x in L.

1.16 Examples. a) In the extension Q ⊆ R, the element 2 is alge-braic over Q, as a root of X 2 − 2 ∈ Q[X].


178

b) Any element of K is algebraic over K. c) Let K(X) be the field of rational fractions in the indeterminate X

with coefficients in the field K. The element X in the extension K ⊆ K(X) is transcendental over K. Indeed, the evaluation homomorphism evX : K[X] → K(X) is the canonical inclusion, and is injective.

However, X is obviously algebraic over K(X). This underscores the importance of identifying the field over which the element is alge-braic.

d) A complex number that is algebraic (respectively transcenden-tal) over Q is called by tradition algebraic number (respectively transcendental number), without any reference to Q.

e) The real numbers e (the sum of the series ( )∑ ≥−

01!n n ) and π (the

length of a circle of diameter 1) are transcendent (over Q). These facts were proven by Hermite in 1873 and Lindemann in 1882.

With the notations above, x is algebraic over K if and only if Ker evx ≠ 0. The generator of the ideal Ker evx (K[X] is a PID!) is a nonzero polynomial, which is fundamental in describing the extension K ⊆ K(x).

1.17 Theorem. Let K ⊆ L be a field extension and let x ∈ L, alge-braic over K. Let f be a monic polynomial with coefficients in K. The following statements are equivalent:

a) f (x) = 0 and deg f is minimal among the nonzero polynomials that vanish in x:

deg f= mindeg g | g ∈ K[X], g(x) = 0, g ≠ 0. b) f (x) = 0 and f is irreducible. c) f is a generator of the ideal Ker evx = g ∈ K[X] | g(x) = 0. d) f (x) = 0 and, for any g ∈ K[X], g(x) = 0 implies f |g. Proof. a)⇒b) If f were reducible, then f = gh, for some g, h ∈ K[X],

with 1≤ deg h, deg g < deg f. Since g(x)h(x) = f (x) = 0, x is a root of g


179

or of h, whose degrees are less than deg f, contradicting the hypothe-sis.

b)⇒c) Clearly, ( f ) ⊆ Ker evx. In the PID K[X], the ideal generated by f is maximal, because f is irreducible. So, Ker evx = f or Ker evx = K[X]. But obviously 1 ∉ Ker evx (⇔ Ker evx ≠ K[X]), so f = Ker evx.

c)⇔d) is evident. d)⇒a) Let g ∈ K[X] with g(x) = 0, g ≠ 0. By the hypothesis, f |g, so

deg f ≤ deg g.

1.18 Definition. Let L/K be a field extension and let x ∈ L be alge-braic over K. We call minimal polynomial of x over K the monic polynomial in K[X] that satisfies one of the equivalent properties in the previous proposition. The minimal polynomial of x over K is de-noted by Irr(x, K) or min(x, K).

1.19 Examples. a) Irr( 4 2 , Q) = X 4 2 since X 4 2 ∈ Q[X] is monic, 4 2 is a root and it is irreducible in Q[X] (by Eisenstein's crite-rion).

b) Irr( 4 2 , R) = X − 4 2 . More generally, for any field K and any a ∈ K, Irr(a, K) = X − a.

c) Irr( 4 2 , Q( 2 )) = X 2 − 2 . The fact X 2 − 2 that has minimum degree among the polynomials with coefficients in Q( 2 ) that vanish in 4 2 is equivalent to 4 2 ∉ Q( 2 ). (prove this, using the form of the elements in Q( 2 )).

1.20 Theorem (characterization of algebraic elements in an exten-sion). Let K ⊆ L be a field extension and let x ∈ L. The following statements are equivalent:

a) x is algebraic over K. b) K[x] is a field. c) K[x] = K(x).


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d) The extension K ⊆ K(x) is finite. Besides, if x is algebraic over K, then the degree of the extension

K ⊆ K(x) is the degree of the minimal polynomial Irr(x, K)): [K(x) : K] = deg Irr(x, K)

If n = deg f, a K-basis of the extension K(x) is 1, x, , xn − 1. Proof. a)⇒b) Let f = Irr(x,K) ∈ K[X] and let evx : K[X] → L be the

evaluation homomorphism in x. The ideal Ker evx of K[X] is generated by f. The isomorphism theorem implies K[X]/ f ≅ Im vx = K[x]. Since f is irreducible in K[X], the ideal f is maximal and K[X]/ f is a field. Then K[x], isomorphic to K[X]/ f , is also a field.

It is evident the b)⇔c). c)⇒a) If x = 0, then all is evident. Suppose x ≠ 0 and let x−1 = a0 +

a1x + + anx n ∈ K[x] be the inverse of x. Multiplying by x, we get a0x + a1x 2 + + anx n+1 − 1 = 0, so x is a root of a nonzero polyno-mial with coefficients in K.

d)⇒a) The infinite family x i | i ∈ N of elements of the finite di-mensional K-vector space K(x) is linearly dependent. So, there exists a linear dependence relation, a0·1 + a1x + + anx n = 0, for some n ∈ N and a0, a1, , an ∈ K, not all zero. This shows that x is algebraic over K.

a)⇒d) At a)⇒b) we saw that we have a K-isomorphism of fields K[X]/ f ≅ K(x). Evidently, this is also a isomorphism of K-vector spaces. Let n = deg f. In the K-vector space K[X]/ f , the classes of 1, X, ..., X n − 1 form a basis. If

a01 + f + a1 X + f + + an − 1 X n −1 + f = 0 + f ,

where a0, a1, , an − 1 ∈ K, then g = a0 + a1X + + an − 1X n −1 ∈ f ,

so f |g. Since deg f = n, we have g = 0, i.e. a0, a1, , an − 1 are 0. On the other hand, the division with remainder theorem shows that any class modulo f of a polynomial h ∈ K[X] has a un representative of degree less than n. This means that h + f is a linear combination with coefficients in K of the classes of 1, X, ..., X n −1.


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So, dimK K(x) = dimK K[X]/ f = n. The isomorphism K[X]/ f ≅ K(x) takes the basis 1 + f , X + f ,

, X n −1 + f in the basis 1, x, , x n − 1. !

1.21 Remark. Keep the conventions above. The proof of a)⇒b) above uses essentially the fact that if f is irreducible in K[X], then K[X]/ f is a field. This result can be proven in a more elementary way, as follows. We must show that any nonzero element in K[X]/ f has a multiplicative inverse. Take g ∈ K[X], g + f ≠ 0 + f . This implies f - g, so GCD( f, g) = 1. By the extended Euclid algorithm, there exist (and can be effectively computed) u, v ∈ K[X] such that uf + vg = 1. Taking classes modulo ( f ), this relation shows that v + f is the inverse of g + f in K[X]/ f .

This also suggests a way to compute the inverse of an arbitrary ele-ment in K[x] expressed in the basis 1, x, , xn − 1: for any element y = a0 + a1 X + + an − 1 X

n − 1 =: g(x), where g ∈ K[X], g ≠ 0, deg g < n, then y −1 is v(x), where v, u ∈ K[X] are such that uf + vg = 1.

Let us find, in the extension Q ⊆ Q ( )3 2 , the inverse of y = 33 421 ++ . Let x = 3 2 . We have Irr(x, K) = f = X 3 − 2 and y = g( 3 2 ), where g = 1 + X + X 2. Using the extended Euclid algo-rithm for X 3 − 2 and 1 + X + X 2, we find that 1 = ( − 1)(X 3 − 2) + (X − 1)(1 + X + X 2). Evaluating in 3 2 , we obtain 1 = ( 3 2 − 1)(1 + 3 2 + 3 4 ).

1.22 Definition. Let K be a field and let x be algebraic over K. The degree of the extension K ⊆ K[x] (equal to deg Irr(x, K)) is called the degree of x over K.

The factor ring K[X]/ f appears also in the following proof.


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1.23 Proposition. Let K be a field and let f ∈ K[X], deg f ≥ 1. Then there exists an extension of K in which f has a root 43.

Proof. In the UFD K[X], the polynomial f is a product of irreduci-ble polynomials. Any root of a factor of f is also a root of f. Replacing, if needed, f with one of its irreducible factors, it is sufficient to prove the claim for the case when f is irreducible.

Consider the factor ring K[X]/ f =: L, which is a field, since f is irreducible. Moreover, L is an extension of K, because the canonical mapping ϕ : K → K[X]/ f , ϕ(a) = a + f , ∀a ∈ K, is a field homo-morphism. In L, α = X + f is a root of f. Indeed, suppose that

f = a0 + a1X + + anX n

Then f (α) = a01 + f + a1X + f + + an X n + f = a0 + a1X

+ + anX n + f = f + f = 0 + f . !

1.24 Examples. The importance of this result is not just theoretical. Concrete fields can be constructed following the procedure in the proof. Here are two examples.

a) Let f = X 2 + 1 ∈ R[X]. The polynomial f is irreducible in R[X]: it has degree 2 and has no roots in R. So, R[X]/(X 2 + 1) is a field, exten-sion of R. Let g be the class of the polynomial g ∈ R[X] modulo (X 2 + 1). The class of X modulo (X 2 + 1), denoted by i, is a root of f. Any element of this field is written uniquely as the class of a polyno-mial of degree at most 1 modulo (X 2 + 1), i.e. is of the form a bX a b X a b i+ = + ⋅ = + ⋅ , where a, b ∈ R. Identifying the real

43 This result, whose proof is simple for a modern mathematician, was known

intuitively for a long time and often tacitly accepted in 17th to 19th centuries mathematical arguments (of course, only polynomials with numerical coefficients were considered). Around 1629, Albert Girard, states without proof that an equation of degree n has n roots, that can be complex numbers or other similar numbers. In 1792, Pierre Simon de Laplace gives an elegant proof of the Fundamental Theorem of Algebra any nonconstant polynomial with complex coefficients has a complex root admiting though that the roots exist somewhere.


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number a with its class a , we can uniquely write a generic element of R[X]/(X 2 + 1) in the form a + bi, with a, b ∈ R.

In this field we have i 2 = −1. Thus, the multiplication of two ele-ments is given by:

(a + bi)·(c + di) = (ac − bd) + (ad + bc)i, ∀a, b, c, d ∈ R This field is isomorphic to the complex number field C. In fact, this

construction can be taken as a definition of C. b) Construction of a field with p n elements. Let p be a prime num-

ber and let n ∈ N*. If f ∈ Fp[X] is an irreducible polynomial of degree n, then Fp[X]/ f is a field with p n elements.

For instance, the polynomial f = X 2 + X + 1 is irreducible in F2[X] (it has degree 2 and has no roots in F2). So, K := F2[X]/ f is a field with 4 elements. If α denotes the class of X modulo f, any element in K is written uniquely as a + bα, with a, b ∈ F2 (we identify any a ∈ F2 with its image a + f in F2[X]/ f ). Since α is a root of f, we have α 2 = α + 1. Thus, the addition and multiplication rules in K are given by the following rules:

(a + bα) + (c + dα) = (a + c) + (b + d)α, (a + bα)·(c + dα) = ac + (ad + bc)α + bdα2 =

ac + (ad + bc)α + bd(α + 1) = (ac + bd) + (ad + bc + bd)α, for any a, b ∈ F2.

Conversely, if L is a field with p n elements and α ∈ L is a genera-tor of (L*, ·) (see 3.4), then Irr(α, Fp) is an irreducible polynomial of degree n in Fp[X].

1.25 Corollary. Let K be a field and let f ∈ K[X], deg f = n ≥ 1. Then there exists an extension L of K such that f is written as a prod-uct of polynomials of degree 1 in L[X] (one also says f splits over L or f has n roots in L).

Proof. Induction by the degree of f. If deg f = 1, take L = K. Sup-pose that, for any field F and any g ∈ F[X] with deg g < n, there exists an extension of F in which g splits. Let f ∈ K[X], deg f = n. There ex-


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ists an extension E of K in which f has a root a. So X − a divides f in E[X], hence f = (X − a)g, for some g ∈ E[X]. Apply now the induction hypothesis for the field E and g ∈ L[X] to obtain an extension field L of E in which g splits. Obviously, f splits also in L. !

For a given f ∈ K[X], the minimal extension of K over which f splits is called the splitting field of f over K. We will study this con-cept in IV.2.14.

We saw that x is algebraic over K is equivalent to K(x) is a finite dimensional K-vector space. This translates the property of an ele-ment of being algebraic into a Linear Algebra property, concerning the dimension of a vector space. The following theorem and its conse-quences illustrate the strength of this translation.

1.26 Theorem (The transitivity of finite extensions) a) Let K ⊆ L and L ⊆ M be a tower of finite field extensions. Then K ⊆ M is a finite extension and the degree is multiplicative: [M : K] = [M : L] [L : K]

Moreover, if x1, , xm is a K-basis of L and y1, , yn is a L-basis of M, then x1 y1, , x1yn, , xm y1,, xmyn is a K-basis of M.

b) If K ⊆ M is a finite extension of fields and L is an intermediate field, then K ⊆ L and L ⊆ M are finite extensions.

Proof. a) It is sufficient to prove the claim on bases. Let us show that xiyj | 1≤ i ≤ m, 1 ≤ j ≤ n generates the K-vector space L. If z ∈ M, because y1, , yn is an L-basis, there exists b1, , bn ∈ L such that z = b1y1 + + bnyn. Each bj (1 ≤ j ≤ n) is of the form bj = aj1x1 + + ajmxm, for some aji ∈ K (1≤ i ≤ m). Using this in the expression for z, we obtain that z is a linear combination of xiyj | 1≤ i ≤ m, 1 ≤ j ≤ n with coefficients in K. Let us prove that xiyj | 1≤ i ≤ m, 1 ≤ j ≤ n is K-linearly independent.


185

If 01,1

=∑≤≤≤≤ njmi

jiij yxa , with aij ∈ K, then 01 1

=⎟⎟⎠

⎞⎜⎜⎝

⎛∑ ∑

≤≤ ≤≤mij

njiij yxa ,

where ∑≤≤ nj

iij xa1

∈ L. The L-linear independence of y1, , yn implies

that ∑≤≤ nj

iij xa1

= 0, for any i. Hence aij = 0 for any i and j, because x1,

, xm are K-linearly independent. !

1.27 Remarks. a) The theorem is also true for division rings, since the proof does not use the commutativity of multiplication.

b) More generally, if the extensions are not necessarily finite, sup-pose that (xi)i∈I is a K-basis of L and (yj)j∈J is an L-basis of M. Then (xiyj)(i,j)∈I×J is a K-basis of M.

1.28 Examples. a) Consider ( )3 2Q /Q. Irr ( 3 2 ,Q) = X 3 − 2, using

Eisenstein's criterion (or observing that X 3 − 2 has no rational roots).

Thus, the degree of the extension is 3 and a Q-basis of ( )3 2Q is

33 4,2,1 . This means that any element of ( )3 2Q is written

uniquely as 33 42 cba ++ , with a, b, c ∈ Q. The same argument

shows that for any n ∈ N*, the extension Q ⊆ ( )n 2Q has degree n and

a Q-basis is n nn 12,,2,1 −… .

b) The extension Q ⊆ ( )3,2Q has degree 4 and a Q-basis is 1, 6,3,2 . for the proof, consider the tower of extensions Q ⊆ ( )2Q ⊆ ( )( )32Q = ( )3,2Q . The first extension has de-gree 2 (since Irr ( 2 , Q) = X 2 − 2) and a basis is 1, 2 . The exten-sion ( )2Q ⊆ ( )( )32Q has also degree 2, because Irr ( )( )2,3 Q = X 2 − 3. Indeed, X 2 − 3 is irreducible over ( )2Q , because

3 ∉ ( )2Q : if, by absurd, 3 = a + b 2 , with a, b ∈ Q, then 3 = a 2 + 2b 2 + 2ab 2 . If ab ≠ 0, then 2 ∈ Q, contradiction. If ab = 0, then 3 = a 2 or 3 = 2b 2, with a, b ∈ Q (again a contradiction).


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Thus, a basis of the extension ( )2Q ⊆ ( )( )32Q is 1, 3 . Applying the method in the theorem of transitivity of finite exten-sions, a basis of the extension Q ⊆ ( )3,2Q is 1, 6,3,2 . Note that this extension has the proper intermediate fields ( )2Q ,

( )3Q , ( )6Q . These are all the proper intermediate fields. At this stage, this is difficult to prove, but it is an immediate consequence of Galois theory.

The element 32 + =: α is a primitive element: ( )3,2Q = ( )32 +Q . The inclusion ( )32 +Q ⊆ ( )3,2Q is evident. Let

us show that 2 , 3 ∈ Q(α). We have α 2 = 5 + 2 6 , so 6 ∈ Q(α). Thus 32236 +=α =: β ∈ Q(α). The system

⎩⎨⎧

=+

=+

α

β

323223

,

where α, β ∈ Q(α), shows that 2 , 3 ∈ Q(α). Let us find Irr(α, Q). The polynomial Irr(α, Q) has degree equal to

[Q(α) : Q] = [Q(α) : ( )2Q ]·[ ( )2Q : Q] = 2·2 = 4

We have α 2 = 5 2 6+ ⇒ ( ) ( )222 562 −= α ⇒ ( ) 024522 =−−α .

Thus, α is the root of ( ) 24522 −−X , which is monic, in Q[X] and has

degree 4, so it must be equal to Irr(α, Q). c) Here is a nontrivial example of a finite extension. Let K be a

field, K(T) the field of rational fractions with coefficients in K, u ∈ K(T), u ∉ K and f, g ∈ K[T] with u = f /g and ( f, g) = 1. Then the extension K(u) ⊆ K(T) is finite and [K(T) : K(u)] = max (deg f, deg g).

Let h = g(X)u − f (X) ∈ K(u)[X] (where g(X) is the polynomial in X obtained by replacing T with X in the polynomial g). We have h(T) = 0, which shows that T is algebraic over K(u), so K(u) ⊆ K(T) is finite.

Since K(T) = K(u)(T), [K(T) : K(u)] = deg Irr(T, K(u)). We claim that Irr(T, K(u)) is associated to h = g(X)u − f (X) ∈ K(u)[X].


187

We have deg h = max (deg f, deg g). This is clear if deg f ≠ deg g; if deg f = deg g, then the leading coefficient of h is ub − a, where a, respectively b, are the leading coefficients of f, respectively g. We have ub − a ≠ 0, because otherwise u = ab−1 ∈ K, contradicting the as-sumption u ∉ K.

We have to prove that h = g(X)u − f (X) ∈ K(u)[X] is irreducible. First, note that u is transcendental over K. Indeed, K ⊆ K(T) is infinite (T is transcendental over K) so K ⊆ K(u) must also be infinite (otherwise, considering the tower of extensions K ⊆ K(u) ⊆ K(T), the theorem of transitivity of finite extensions would imply that K ⊆ K(T) is finite). Thus, there is a K-isomorphism K[Y] ≅ K[u] (where Y is an indeterminate). The irreducibility of g(X)u − f (X) ∈ K(u)[X] is thus equivalent to the irreducibility of r = g(X)Y − f(X) ∈ K(Y)[X]. Since K[Y] is a UFD and K(Y) is its field of quotients, this is equivalent to r being irreducible in K[Y][X] ≅ K[Y, X]. Suppose r = pq, with p, q ∈ K[Y, X]. Let degY r be the degree of r, seen as a polynomial in Y with coefficients in K[X]. We have 1 = degY r = degY p + degY q, so we may suppose degY p = 1 and degY q = 0. So, q ∈ K[X] and p = cY + d, for some c, d ∈ K[X] and r = g(X)Y − f(X) = (cY + d )q. Identifying the coefficients of the powers of Y, we obtain that q|g and q| f in K[X], implying that q ∈ K*, since ( f, g) = 1. Thus, h is irreducible in K[Y][X].

The polynomial h is irreducible and vanishes in T, so it is associ-ated to Irr(T, K(u)). Proposition 1.20 says that [K(T) : K(u)] = deg Irr(T, K(u)) = deg h = max (deg f, deg g).

It is interesting to remark that any intermediate field K ⊆ L ⊆ K(T) with K ≠ L is of the form L = K(u), for some u ∈ K(T), u ∉ K. This fact is known as Lüroths Theorem and is of significance in Algebraic Geometry (see MORANDI [1996], WALKER [1950]).

1.29 Definition. A field extension K ⊆ L is called algebraic exten-sion if each element in L is algebraic over K. The extension K ⊆ L is


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called transcendental if it is not algebraic (it contains at least one transcendental element over K).

1.30 Proposition. Any finite extension is algebraic and finitely generated.

Proof. Let K ⊆ L be finite and let x ∈ L. Then K ⊆ K(x) is finite (as an intermediate field of K ⊆ L) and Prop. 1.20 shows that x is alge-braic over K. If x1, , xm is a K-basis of L, then certainly L = K(x1, , xm), so L is a finitely generated extension of K. !

1.31 Proposition. Suppose K ⊆ L is a field extension, n ∈ N* and x1, , xn ∈ L are algebraic over K. Then K ⊆ K(x1, , xn) is a finite extension. In particular, any algebraic finitely generated extension is a finite extension.

Proof. We prove the claim by induction on n ∈ N*. For n = 1, the conclusion follows from Prop. 1.20. If n > 1, the induction hypothesis says that K ⊆ K(x1,, xn − 1) is finite. Since xn is algebraic over K, xn is also algebraic over K(x1, , xn − 1); the extension K(x1, , xn − 1) ⊆ K(x1, , xn − 1)(xn) is thus finite. The conclusion follows by applying the transitivity of finite extensions to the tower K ⊆ K(x1, , xn − 1) ⊆ K(x1, , xn − 1, xn). !

An easy argument by induction on n ∈ N* shows that, if K ⊆ L is an extension and x1, , xn ∈ L are algebraic over K, then K[x1, , xn] = K(x1, , xn) (see exercise 1.17). The converse of this result is nontrivial and it is known as "Zariskis Lemma (see for instance SPINDLER [1994], Prop. 12.35, p. 232).

1.32 Theorem. (transitivity of algebraic extensions) If K ⊆ L and L ⊆ M are algebraic extensions, then K ⊆ M is an algebraic exten-sion.


189

Proof. Let x ∈ M. In order to prove that x is algebraic over K, we show that x is contained in a finite extension of K and apply then Prop. 1. 20.

Since x is algebraic over L, there exists 0 ≠ f ∈ L[X] such that f (x) = 0. Let a0, , an ∈ L be the coefficients of f. Then x is algebraic over K(a0, , an) and K ⊆ K(a0, , an) is finite, by Prop. 1.31. We have now the tower of finite extensions K ⊆ K(a0, , an) ⊆ K(a0, , an)(x). By transitivity, K ⊆ K(a0, , an)(x) is also finite. !

1.33 Example. The extension Q ⊆ L = Q ( )*2 N∈nn is algebraic and infinite. Q ⊆ L is algebraic because L is the union of its intermedi-ate fields of the form Q ( )mnn ≤2 , m ∈ N*, and each of these is algebraic (they are even finite). But Q ⊆ L is not finite: if, by contradiction, [L : Q] = n ∈ N*, then each intermediate field would have the degree a divisor of n. But Q ⊆ Q ( )1 2+n has degree n + 1.

1.34 Definition. If K ⊆ L is a field extension, the set of all elements in L that are algebraic over K is denoted by K'L and is called the alge-braic closure of K in L. Clearly, K ⊆ K'L.

1.35 Proposition. Let K ⊆ L be a field extension. Then K'L is a sub-field of L and an extension of K. In particular, the sum, the difference, the product and the quotient (if it exists) of two algebraic elements over K are also algebraic elements over K.

Proof. If a, b ∈ L, with b ≠ 0, are algebraic over K, then K ⊆ K(a, b) is finite, so it is an algebraic extension. Thus a + b, a − b, ab, ab −1 are algebraic over K, as elements of K(a, b). This shows that K'L is a subfield of L. !

1.36 Example. Q'C is a subfield in C, an infinite algebraic exten-sion of Q. Why?


190

Exercises

1. Let K ⊆ L be an extension. Show that L is a K-vector space with re-spect to: the addition of L and the scalar multiplication given by: ∀α ∈ K, ∀x ∈ L, α·x = αx (the multiplication in L). Prove that [L : K] = 1 if and only if K = L. 2. If S is a subset of the field L and K is a subfield in L, then the field of fractions of the domain K[S] is canonically isomorphic to K(S). 3. Let K ⊆ L be an extension. Consider the linear system with coeffi-cients aij, bi ∈ K:

(S):

⎩⎪⎨⎪⎧

a11x1 + + a1nxn = b1

a21x1 + + a2nxn = b2 am1x1 + + amnxn = bm

.

Prove that: the system (S) has a solution in E ⇔ (S) has a solution in K. Can you prove other similar properties of (S)? 4. Let K ⊆ L be an extension and let p ∈ L[X] \ K[X], deg p = n. Prove that | p(K) ∩ K | ≤ n. (Hint. Suppose p(αi) = βi, 1 ≤ i ≤ n + 1, with αi, βi ∈ K. Interpret this as a system of n + 1 equations, the unknowns be-ing the coefficients of the polynomial.) 5. Let K ⊆ L be an extension and let α ∈ L, algebraic over K. If deg Irr(α, K) = n, then deg Irr(β, K) divides n, for any β ∈ K(α). 6. Let K ⊆ L be an extension and let x ∈ L. Prove that x is transcendental over K if and only if K[x] is K-isomorphic to the K-algebra K[X] of polynomials in the indeterminate X. 7. Let K be a field and let (Ki)i ∈ I be a chain of subfields of K (∀i, j ∈ I, Ki ⊆ Kj or Kj ⊆ Ki). Prove that ∪i ∈ I Ki is a subfield in K. 8. Find a basis of the extension Q ⊆ Q[ 3 ]. Express (1 + 3 ) −1 in this basis. Find Irr(1 + 3 , Q). The same problem for Q ⊆ Q(α) and the element 1 − α, where α is a root of X 3 + X + 1.


191

9. Let K ⊆ K(α) be an algebraic extension. Suppose Irr(α, K) = f is known, where deg f = n. For 0 ≠ β ∈ K(α) given as a K-linear combination of 1, α, , α n−1, describe a method to find β −1 and Irr(β, K). 10. Let K ⊆ L be an extension of degree 2 (quadratic extension). Show that L = K(α), where α ∈ L is a root of a polynomial of the form X 2 − b (if char K ≠ 2), respectively X 2 − b or X 2 − X − b (if char K = 2), for some b ∈ K. Prove that any quadratic extensions of Q is of the form Q( d ), with d ∈ Z squarefree (d ≠ 0, d ≠ 1 and d is not the multiple of any square of a prime). 11. Let m, n ∈ Z be distinct and squarefree. Then Q( nm , ) has de-gree 4 over Q. (Hint. Show that m ∉ Q( n ).) 12. Let K ⊆ L be an extension such that L = K(x, y), with x, y ∈ L. If there exists m, n ∈ N* such that x m ∈ K, y n ∈ K and (m, n) = 1, then K(x, y) = K(xy). 13. Let K and L be fields and let ϕ : K → L be a field isomorphism. Let ψ : K[X] → L[X] be the ring isomorphism that extends ϕ and with ψ(X) = X. Prove that p ∈ K[X] has a root in K ⇔ ψ(p) has a root in L. 14. Let K and L be extensions of Q and let ϕ : K → L be a field homo-morphism. Show that ϕ|Q = id. Prove a similar property for extensions of Fp (where p is a prime). 15. Let d, e ∈ Z, squarefree. Prove that Q( d ) ≅ Q( e ) ⇔ d = e. 16. Let ϕ : R → R be a field homomorphism.

a) Prove that ∀x ∈ R, x > 0 implies ϕ(x) > 0. Deduce ϕ is an increasing map. (Hint. x = a 2, for some a ∈ R.)

b) Prove that ϕ = id. (Hint. Use that ϕ|Q = id and that between any two real numbers there is a rational number.) 17. Let K ⊆ L be an extension, n ∈ N* and let x1, , xn ∈ L, algebraic over K. Then K[x1, , xn] = K(x1, , xn). More generally, if S is a sub-set of L consisting of algebraic elements over K, then K[S] = K(S).


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18. Let E and F be intermediate fields of the extension K ⊆ L. Prove that E(F) ( = the composite EF) is equal to the set xy −1 | x, y linear combinations of elements in F with coefficients in E, y ≠ 0. Under what assumptions is the composite EF equal to the set of all linear combinations of elements in F with coefficients in E? 19. Let L/K be an extension. Show that IF(L/K) = E | E is a subfield in L, K ⊆ E is a complete lattice with respect to inclusion (any family of subfields (Ei)i∈I has sup and inf in IF(L/K)). What is the form of an arbitrary element in supEi | i∈I? 20. Let K ⊆ L be an algebraic extension. Show that any subring of L that includes K (any K-subalgebra of L) is a field. Is the converse true? 21. Let R ⊆ S be an extension of domains (R is a subring of the do-main S and 1 ∈ R). An x ∈ S is called integral over R if x is the root of a monic polynomial in R[X]. Let x ∈ C be integral over Z. Prove that Irr(x, Q) ∈ Z[X]. Deduce that the integral elements in Q ( )d , where d ∈ Z is squarefree, are roots of a monic polynomial of degree 2 with integer coefficients. 22. Let K ⊆ L be a field extension and let x, y ∈ L be algebraic over K. Let p = Irr(x, K), q = Irr(y, K), deg p = m, deg q = n.

a) Prove that deg Irr(x + y, K) ≤ mn and deg Irr(xy, K) ≤ mn. b) deg Irr(x, K(y)) = m ⇔ deg Irr(y, K(x)) = n ⇔ [K(x, y) : K] = mn. c) Formulate a sufficient condition for [K(x, y) : K] = mn.

23. Find Irr(x, Q) and a basis of the extension Q ⊆ Q(x), ∀ x ∈ ( )53 21,31,3 ii −+− . 24. For any n ∈ N*, give an example of extension of degree n of Q. 25. Let K ⊆ L ⊆ E be a tower of extensions and let x ∈ E be algebraic over K. Then [L(x) : L] ≤ [K(x) : K]. 26. Let L and E be intermediate fields of the extension K ⊆ F.

IV.2 Roots of polynomials. Algebraically closed fields

193

a) If K ⊆ L is algebraic, then E ⊆ EL is algebraic. b) If S is a subset of L and K < S > = L (S generates the K-vector

space L), then E < S > = EL. (Hint. Reduce to the case when S is finite.) c) If [L : K] is finite, then [EL : E] ≤ [L : K]. d) If the degrees [L : K] and [E : K] are finite and coprime, then

[EL : E] = [L : K] and [EL : K] = [E : K]·[L : K]. e) If [EL : K] = [E : K]·[L : K], then K = L ∩ E. f) If [E : K] = 2 and K = L ∩ E, then [EL : K] = [E : K]·[L : K]. g) Give an example such that [E : K] = [L : K] = 3, K = L ∩ E, but

[EL : K] < 9. 27. Let K ⊆ L be an extension of degree n and let g ∈ K[X], deg g = p, with p prime, (p, n) = 1. If g is irreducible in K[X], then g is irreducible in L[X]. 28. Show that X 5 − 2 is irreducible in Q(ω)[X], where ω = cos(2π/5) + i·sin(2π/5).


Let R be a domain. Bézout's theorem says that the polynomial f ∈ R[X] has the root a ∈ R if and only if X − a divides f in R[X]. It is thus natural to consider the following definition:

L ∩ E

E L

EL

K


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2.1 Definition. Suppose R is a domain, f ∈ R[X] is a nonzero polynomial, a ∈ R and n ∈ N. We say that a is a multiple root of f with multiplicity n if (X − a)n | f and (X − a)n + 1- f. The natural number n is called the multiplicity of the root a. If n = 1, a is called a simple root of f. If n > 1, a is called a multiple root of f (double if n = 2, triple if n = 3 etc).

When counting the roots of a polynomial, each root counts with its multiplicity (unless otherwise specified).

2.2 Proposition. Let R be a UFD and let f ∈ R[X] be nonzero. If a1, , an ∈ R are distinct roots of f, of multiplicities m1, , mn respec-tively, then ( ) ( ) nm

nnm aXaX −− 1

11 … divides f in R[X]. Proof. Use induction on n. If n = 1, this is the definition. Suppose

the claim is true for n − 1 and let f be as in the statement. By the induction hypothesis, f = ( ) ( ) gaXaX nm

nnm 1−1

1−1−11 −− … , for some g ∈ R[X]. Since the polynomials X − ai, 1 ≤ i ≤ n, are irreducible and pairwise not associated in divisibility, ( ) 1

1maX − , , ( ) nm

naX − are pairwise relatively prime. By Prop. I.5.6, ( ) nm

naX − is coprime with ( ) ( ) 11

11−

−−− nmn

m aXaX … . Since ( ) nm

naX − divides ( ) ( ) gaXaX nmn

m 1111

−−−− … , we obtain

that ( ) nmnaX − divides g. !

2.3 Corollary. Let R be a domain and let f ∈ R[X], deg f = n. Then f has at most n roots in R.

Proof. Of course, each root is counted with its multiplicity. Let K be the field of fractions of R. We see f as a polynomial in K[X] and ap-ply the preceding proposition. !

We now seek a criterion to decide if a polynomial has multiple roots. It is useful to introduce the notion of formal derivative of a polynomial with coefficients in an arbitrary ring.


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2.4 Definition. Let R be a commutative unitary ring and let f = a0 + a1X + + anX

n ∈ R[X]. The (formal) derivative of f is the polynomial

df := a1 + 2a2 X + + nan X n −1.

Other notations: df = f ' or df = f (1).

Direct calculations show that the formal derivative has the usual properties of the derivative of a function as known from calculus:

f + g' = f ' + g', af ' = af ', fg' = f 'g + fg', ∀a ∈ R, ∀f,g ∈ R[X]. Note that d : R[X] → R[X] is the unique R-module homomorphism

with the property that d1 = 0 and dX n = nX n−1, ∀n ∈ N* (apply the universality property of a free module). Evidently, deg df ≤ deg f − 1.

Composing the homomorphism d with itself n times (n ∈ N*) is de-noted by dn; dn : R[X] → R[X]. Thus, dn = ddn−1, ∀n ∈ N*, d0 = id.

dnf is often denoted by f (n), ∀f ∈ R[X].

2.5 Proposition. Let R be a domain, let f ∈ R[X] of degree n > 0 and let α ∈ R.

a) There exist and are unique b0, , bn ∈ R such that ( )∑

≤≤−=

ni

ii Xbf

0α .

b) If α is a root of multiplicity m (m ∈ N*) of f, then f (i)(α) = 0, for any i ∈ 0, , m − 1.

c) If f (α) = f '(α) = 0, then α is a multiple root of f (of multiplicity at least 2).

Proof. a) Use induction on deg f. If f = a0 + a1X, then f = (a0+ a1α) + a1(X − α). If deg f = n > 1, by the integer division with remainder theorem applied to f and X − α (I.6.2), we obtain f = (X − α)g + b0, with b0 ∈ R and g ∈ R[X], deg g = n − 1. Writing g in the form given by the induction hypothesis and replacing in the previous equality, we obtain the result.

The uniqueness part amounts to the R-linear independence of (X − α)i | i ∈ N in R[X], easy to prove.


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b) By a), (X − α)m divides f if and only if b0, b1,, bm−1 are all 0. On the other hand, f (i)(α) = i!bi, ∀i ∈ 0, , n (proof by induction on n). This implies that f (i)(α) = 0, ∀i ∈ 0, , m − 1.

c) Let f be written as in a). We obtain f (α) = b0 = 0 and f '(α) = b1 = 0. So, (X − α)2 | f. !

Suppose K is a field, Ω is an extension of K, α ∈ Ω and f ∈ K[X]. Applying the result above, we obtain that α is a multiple root of f if and only if is simultaneously a root of f and of its derivative: (X − α)| f and (X − α)| f ' in Ω[X]. This implies that the GCD of f and f ' in Ω[X] has degree ≥ 1. But the GCD of two polynomials can be obtained us-ing Euclid's algorithm and does not depend on the field considered: if f, g ∈ K[X], then ( f, g)K[X] = ( f, g)Ω[X]. Thus:

2.6 Proposition. Let K be a field and let f ∈ K[X]. Then f has multi-ple roots (in some extension of K) if and only if f and f ' are not rela-tively prime.

Proof. Let Ω be an extension of K in which f splits (IV.1.25). If f has multiple roots, we saw that f and f ' are not relatively prime. Con-versely, if g = GCD(f, f ') has degree ≥ 1, then g has a root α in Ω (the roots of g are among the roots of f ) and α is a multiple root of f since f(α) = f '(α) = 0. !

Without knowing the roots of a polynomial, the criterion above al-lows to decide if it has multiple roots.

2.7 Proposition. (Viète's relations) 44 Suppose R is a domain, f = a0 + a1X ++ anX n is a polynomial in R[X], an ≠ 0, having the roots x1, , xn ∈ R. Then:

f = an(X − x1)(X − xn) For any k, 1 ≤ k ≤ n, we have: 44 In honor of François Viète, 1540-1603, French mathematician.


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( ) knk

nii iin axxak k −⊆ −=∑ 1,,2,1,,1 1…… … .

In particular, an(x1 + + xn) = an − 1

an(x1 x2 + x1 x3 + + xn −1 xn) = an − 2

an x1 xn = (1)na0. Consequently, any root in R of f divides a0. Proof. The polynomial g = an(X − x1)(X − xn) divides f, by prop.

2.2. The polynomials g and f have the same degree and g | f, so they are associated in divisibility in K[X] (K is the field of fractions of R). Since g and f have the same leading coefficient, g = f. The other equalities result by identifying the coefficients of g and f. !

2.8 Corollary. Let R be a subring of the domain S and let f ∈ R[X] be monic. If deg f = n and f has the roots x1, , xn ∈ S, then g(x1,, xn) ∈ R, for any symmetric polynomial g ∈ R[X1,, Xn].

Proof. By the fundamental theorem of the symmetric polynomials, there exists h ∈ R[X1,, Xn] such that g = h(s1,, sn), where s1,, sn are the fundamental symmetric polynomials in the indeterminates X1,, Xn. The relations between roots and coefficients show that si(x1,, xn) ∈ R, for any 1 ≤ i ≤ n. Thus:

g(x1,, xn) = h(s1(x1,, xn),, sn(x1,, xn)), which is an element of R. !

The fields that have no proper algebraic extensions are highly interesting. They are characterized by the next theorem.

2.9 Theorem. Let K be a field. The following statements are equivalent:

a) There exist no proper algebraic extensions of K. b) There exist no proper finite extensions of K.


198

c) For any extension L of K, the algebraic closure of K in L coin-cides with K. (K is algebraically closed in L).

d) Any polynomial of degree at least 1 with coefficients in K has a root in K.

e) Any polynomial f of degree n ≥ 1 with coefficients in K splits over K ( f has n roots in K).

f) The irreducible polynomials in K[X] are the polynomials of de-gree 1.

Proof. a)⇒b) Clear, since any finite extension is algebraic. b)⇒c) Let x ∈ K'L. Then K ⊆ K(x) is finite, so K = K(x) and x ∈ K. c)⇒d) Let f ∈ K[X], deg f ≥ 1. Prop. 1.23 ensures the existence of

an extension L of K in which f has a root x. Since x is in K'L, x ∈ K. d)⇒e) Suppose by contradiction that there exists f ∈ K[X], deg f

≥ 1, such that f does not split in K[X]. Choose f of minimum degree with this property. By hypothesis, f has a root a ∈ K, so f = (X − a)g, with g ∈ K[X]. But deg g < deg f, so g splits over K. Since f = (X − a)g, f also splits over K, contradiction.

e)⇒f ) Evident. f )⇒a) Let L be an algebraic extension of K and let a ∈ L. Then

Irr(a, K) is an irreducible polynomial in K[X], hence it has degree 1, so a ∈ K. !

2.10 Definition. A field satisfying the equivalent properties above is called an algebraically closed field. An extension L of the field K is called an algebraic closure of K if L is algebraically closed and K ⊆ L is an algebraic extension.

2.11 Examples. a) A finite field is never algebraically closed. In-deed, if F is a field with n elements (n ≥ 2), the polynomial f = ( )∏ ∈ −+ Fa aX1 ∈ F[X] has degree n and f (a) = 1, ∀a ∈ F, so f

has no roots in F.


199

b) The complex field C is algebraically closed. This fact is known as The Fundamental Theorem of Algebra 45 and will be proven later. C is not an algebraic closure of Q, because C is uncountable and any algebraic extension of Q is countable (see next lemma). But Q'C, the algebraic closure of Q in C, is an algebraic closure of Q. More gener-ally, if K ⊆ L is an extension and L is algebraically closed, then K'L is an algebraic closure of K (prove this!).

c) If K and L are isomorphic fields and K is algebraically closed, then L is algebraically closed.

An important problem is the existence (and uniqueness) of an alge-braic closure of a given field K. First, let us prove a technical lemma, also interesting in its own right.

2.12 Lemma. Let K ⊆ L be an algebraic extension. Then |L| ≤ max(|K|, |N|).

Proof. Let P be the set of all irreducible monic polynomials in K[X] and let a ∈ L. Any f ∈ P has a finite number n f of roots in L. We establish an indexing of these, a1, , an f . Define ϕ : L → P × N as follows: if a ∈ L, let ϕ(a) = ( f, r), where f = Irr(a, K) and a = ar.

The function ϕ is injective, so |L| ≤ |P| × |N| = max(|P|, |N|). We have to show |P| ≤ max(|K|, |N|). Note that P is the union of the dis-joint sets (Pn)n ≥1 , where Pn is the set of polynomials in P of degree n, so: |P| = |∪n≥1 Pn| = ∑n≥1 |Pn|. Of course, |Pn | ≤ |K|n, since ψ : Pn → K n, ψ(a0 + a1X + + an − 1X

n −1 + X n) = (a0, , an − 1), is injective. On the other hand, |K|n = |K| if K is infinite, so |P| = ∑n≥1 |Pn| ≤ ∑n≥1 |K| =

45 This name (kept for historic reasons) expresses the concept that Algebra

studies mainly complex numbers. This is no longer true since the 19th century, although complex numbers continue to play an important role in mathematics.


200

|K × N| = |K|. If K is finite, then |K|n is finite and |P| ≤ ∑n≥1 |K|n = |N| (as a countable union of finite sets). !

An immediate consequence is that any algebraic extension of Q is countable; thus, the extension Q ⊆ R is not algebraic, because R is not countable. This means that there exist real transcendental num-bers (even uncountably many!).

2.13 Theorem. Let K be a field. Then an algebraic closure of K ex-ists.

Proof. Th. 2.9.a) suggests looking for the algebraic closure as a maximal element46 in the set of all algebraic extensions of K. The problem is that the algebraic extensions of K do not form a set! Fortu-nately, we can include K in a sufficiently large set M (that contains all algebraic extensions of K) and confine our search to algebraic extensions of K included in M.

Let M be a set such that47 | M | > max(|K|, |N|) and K ⊆ M.48 Let A be the set of algebraic extensions of K included in M. The set A is or-dered: for any F, L ∈ A, F ≤ L if and only if F is a subfield of L. The set A is nonempty (K ∈ A) and is inductively ordered. Indeed, if B is a chain in A, the union of the elements in B is in A (prove!) and is an up-per bound for B. Zorn's Lemma assures the existence of a maximal element F of A.

We prove that F is an algebraic closure of K. Let E be an algebraic extension of F. By the transitivity property of algebraic extensions, E is an algebraic extension of K. The previous lemma says that

46 You guessed, Zorn's Lemma will be used. All proofs of the existence of the

algebraic closure use some form of the Axiom of Choice. 47 This choice for | M | is suggested by the previous lemma. We want to be sure

that M includes a "copy" of any algebraic extension of K. 48 In fact, we need just an injective function α : K → M, but we assume K ⊆ M to

simplify notations.


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| E | < | M |. We can then define ϕ : E → M, injective, such that ϕ|F = idF. By transport of structure, ϕ(E) becomes a field (for instance, the addition in ϕ(E) is defined by ϕ(x) + ϕ(y) := ϕ(x + y), ∀x, y ∈ E). Then ϕ : E → ϕ(E) is an F-isomorphism of fields and thus ϕ (E) is an algebraic extension of K. So, ϕ(E) ∈ A and F ⊆ ϕ(E). Because F is maximal in A, F = ϕ(E), so F = E. This shows that F is algebraically closed. !

For a field K and a polynomial (or a family of polynomials), it is interesting to study the smallest field over which the polynomial (respectively, any polynomial in the family) splits.

2.14 Definition. Let K be a field, let Ω be an algebraic closure of K and let F be a family of polynomials in K[X]. The splitting field of the family F over K is the field S obtained by adjoining to K all the roots in Ω of the polynomials in F,

S := K(x ∈ Ω | ∃f ∈ F such that f(x) = 0). If F = f , S is called the splitting field of f over K. If f ∈ K[X] has

the roots x1, , xn ∈ Ω, then the splitting field of f over K is K(x1, , xn) = K[x1, , xn].

Of course, every polynomial in the family F splits in the splitting field of F over K.

2.15 Remarks. a) At first glance, the splitting field of a family F over K depends on the choice of an algebraic closure Ω of K. We shall prove though that any two splitting fields of F over K are K-isomorphic. This is the reason we say the (and not a) splitting field of F over K.

b) The algebraic closure of K is the splitting field over K of the family of all nonconstant polynomials in K[X].

2.16 Examples. a) The splitting field of X 2 − 2 over Q is Q( 2 ). b) The splitting field of X 2 − 2 over R is R.


202

c) The splitting field of X 3 − 2 over Q is Q( ω,23 ), where ω ∈ C is a root of X 2 + X + 1. Indeed, the roots of X 3 − 2 are 3233 2,2,2 ωω and Q ( )ω,23 = ( )3233 2,2,2 ωωQ .

d) If f ∈ K[X] is a polynomial of degree n, and L is its splitting field over K, then [L : K] ≤ n!.

Indeed, if x1, , xn ∈ L are the roots of f, then [K(x1) : K] = deg(Irr(x1, K)) ≤ deg f = n. Note that L is a splitting field over K(x1) of g := f/(X − x1) ∈ K(x1)[X]. Since deg g = n − 1, apply an induction to obtain that [L : K(x1)] ≤ (n − 1)! and so [L : K] ≤ n!.

Proving the uniqueness (up to a K-isomorphism) of the splitting field of a family of polynomials over K requires some results on the extension of field homomorphisms. These results have also other important applications.

We shall use frequently the following elementary fact: if σ : K → L is a field homomorphism, then σ has a unique extension to a ring homomorphism τ : K[X] → L[X], namely τ(a0 + a1X + + anX

n) = σ(a0) + σ(a1)X + + σ(an)X

n. This is the unique ring homomorph-ism τ : K[X] → L[X] satisfying τ|K = σ and τ(X) = X. The existence and uniqueness of τ are a consequence of the universality property of the polynomial ring K[X]. By a harmless abuse of notation, the extension to K[X] of the homomorphism σ is denoted also by σ.

The following property is very simple, but has deep implications; in particular, it is instrumental in the determination of the Galois group of an extension.

2.17 Proposition. If E and L are extensions of K, α ∈ E is a root of f ∈ K[X], and ϕ : E → L is a K-homomorphism, then ϕ(α) is also a root of f.

Proof. Let f = a0 + a1 X + + an X n ∈ K[X]. Then f(α) = a0 + a1α +

+ anα n = 0. Apply the K-homomorphism ϕ to this equality:


203

0 = ϕ(a0 + a1α + + anα n) = a0 + a1ϕ(α) + + anϕ(α) n = f(ϕ(α)).!

The proposition below uses the isomorphism K[X]/ f ≅ K(α), where α is algebraic over K and f = Irr(α, K). In other words, for an irreducible polynomial f ∈ K[X], regardless of the choice of an exten-sion E in which f has a root and regardless of the choice of a root (of f ) α ∈ E, K(α) is the same (up to a K-isomorphism).

2.18 Proposition. Suppose σ : K → K' is a field isomorphism, f ∈ K[X] is irreducible and α is a root of f in some extension E of K. If α' is a root of f ' := σ f ∈ K[X] in some extension E' of K', then K(α) is isomorphic to K'(α') by an isomorphism σ' that extends σ and σ'(α) = α'.

Proof. We saw that there exists a K-isomorphism η : K[X]/ f → K(α) such that ηX + f = α. Similarly, there exists a K'-isomorphism γ : K'[X]/ f ' → K'(α'), such that γ X + f ' = α'. We also have an isomorphism ϕ : K[X]/ f → K'[X]/ f ', ϕh + f = σ (h) + f ', for any h + f ∈ [ ] ( )fXK . Then γ ϕ η-1 is the isomor-phism σ' we need: ( ) [ ] ( ) [ ] ( ) ( )αα γϕη ′⎯→⎯′′⎯→⎯⎯→⎯

−

KfXKfXKK1

!

The seemingly pointless degree of generality in the previous proposition (taking two isomorphic fields K and K' instead of just K) is in fact useful in the next Theorem.

2.19 Theorem. Let K be a field, let L be an algebraic extension of K, let Ω be an algebraic closure of K and let σ : L → Ω be a K-homomorphism. If E is an algebraic extension of L, then there exists a K-homomorphism τ : E → Ω that extends σ.

Proof. Let E be the set of extensions of the homomorphism σ, E = (F, ϕ) | F is a subfield of E, L ⊆ F, ϕ : F → Ω , ϕ is a

homomorphism, ϕ|L = σ.


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E is ordered by " ≤ ", defined by: (F, ϕ) ≤ (F', ϕ') if and only if F ⊆ F' and ϕ'|F = ϕ. A straightforward proof shows that " ≤ " is indeed an order relation. Moreover, if (Fi, ϕi)i∈I is a chain in E, it is bounded above by ∪i∈I Fi, ϕ, where ϕ : ∪i∈I Fi → Ω is defined by ϕ(x) = ϕi(x) if x ∈ Fi (this definition is independent of the choice of i ∈ I such that x ∈ Fi). Thus, E is inductively ordered and has, by Zorn's Lemma, a maximal element (F, ϕ).

Let us prove that F = E, which finishes the proof. If F ≠ E, pick x ∈ E \ F. The element x is algebraic over F; let f = Irr(x, F). Apply Prop. 2.18 to the following situation: ϕ : F → ϕ(F) field isomorphism, f ∈ F[X], x is a root of f (in E), x' is a root of ϕ f (in Ω). We obtain an isomorphism ϕ' : F(x) → ϕ(F)(x') that extends ϕ. Since ϕ(F)(x') ⊆ Ω, ϕ' is an extension of ϕ to F(x), contradicting the maximality of (F, ϕ).!

2.20 Corollary. Let K be a field. Then: a) Any two algebraic closures of K are K-isomorphic. b) If f ∈ K[X], then any two splitting fields of f over K are

K-isomorphic. c) If F is a family of polynomials in K[X], then any two splitting

fields of the family F over K are K-isomorphic. d) Let Ω be an algebraic closure of K. Then any algebraic exten-

sion of K is K-isomorphic to a subfield of Ω that includes K. Proof. a) Let Ω and Ω' be algebraic closures of K. The canonical

inclusion ι : K → Ω has an extension to a K-homomorphism σ : Ω → Ω' (by the previous theorem). The image of σ is a subfield σ(Ω) of Ω', isomorphic to Ω (by σ), so it is algebraically closed. Since Ω' is algebraic over σ(Ω), σ(Ω) = Ω'. Thus, σ is an isomorphism.

c) Let Ω and Ω' be algebraic closures of K. Let R (respectively R') the set of all roots in Ω (respectively Ω') of the polynomials in F. We have to prove that K(R) and K(R') are K-isomorphic. Let σ : Ω → Ω' be a K-isomorphism, as in a). If f ∈ F, and α ∈ Ω is a root of f, then σ(α) is a root of f in Ω'. This shows that σ (R) ⊆ R'. Considering


205

σ −1 : Ω' → Ω, we obtain analogously that σ −1(R') ⊆ R, so σ estab-lishes a bijection between R and R'. If we remember the form of the elements in K(R), we obtain that σ (K(R)) = K(σ (R)) = K(R'), so the restriction of σ to K(R) is a K-isomorphism between K(R) and K(R').

b) is a particular case of c). d) Let K ⊆ L be algebraic. Then the canonical inclusion ι : K → Ω

extends to a K-homomorphism ϕ : L → Ω by 2.19. Thus, L is K-isomorphic to ϕ(L), a subfield of K. !

Part d) above says that a given algebraic closure Ω of K includes all the algebraic extensions of K.

2.21 Theorem. (the Fundamental Theorem of Algebra)49 The field C is algebraically closed.

Proof. Let us prove first that any polynomial f of degree ≥ 1 with real coefficients has a complex root.

If deg f is odd, then f has a real root. This follows from the well known fact from analysis: the polynomial function ϕ : R → R associ-ated to f has both positive and negative values (since the limits of ϕ at + ∞ and − ∞ are infinite and have opposite signs); the continuity of ϕ implies the existence of c ∈ R with ϕ(c) = f (c) = 0.

Let deg f = n ∈ N and let s f be the exponent of 2 in the prime factor decomposition of n (2 s f | n and 2 s f + 1 - n). We prove the claim by induction on s f . If s f = 0, then deg f is odd and f has a real root.

49 Also known as the dAlembert-Gauss theorem. Jean le Rond dAlembert

proposes an incomplete proof in 1746. C.F. Gauss gives four correct proofs of this theorem, the first one in 1797. Other proofs are due to Jean Argand (1814), Augustin Louis Cauchy (1820). The theorem of Liouville (which is due in fact to Cauchy, 1844) any holomorphic bounded function on C is constant proves the theorem in one line. The present proof belongs to Pierre Samuel and has the advantage (?) of being more algebraic. Note that all proofs use some analysis, because fundamental (topological) properties of R do not possess purely algebraic descriptions. The essential role is played in fact by the order properties of R.


206

Let s ∈ N, s > 0. Suppose that any polynomial g ∈ R[X], with s(g) < s, has a complex root. Let f ∈ R[X], deg f = n = 2sm, s f = s. Let t1, , tn be the roots of f in a splitting field of f over C, K = C(t1, , tn). Let P be the family of subsets of 1, 2, , n having two ele-ments, fix a ∈ R and let i, j ∈ P. Consider the following element in K:

uij(a) := titj + a(ti + tj) = uji(a). Let ga := ( )( ) ∏ ∈ −Pji ij auX, , a polynomial in K[X].

The polynomial ga has degree n(n − 1)/2 and (a priori) coefficients in K. We claim ga has coefficients in R. Indeed, the coefficients of ga are polynomial expressions of t1, , tn with real coefficients. To show that these polynomial expressions are real, let us study how they change under a permutation of the roots t1, , tn. If σ is a permutation of the set 1, 2, , n, the polynomial obtained from ga by the action of σ on the roots (i.e. ti & tσ(i), ∀i) is

( ) ( )( )( ) ∏ ∈−

Pji ji auX, σσ

Since σ(i),σ( j) | i, j ∈ P = P, the polynomial above coin-cides with ga. This means that ga has coefficients symmetric polynomials in t1, , tn with real coefficients. Prop. 2.8 implies that ga ∈ R[X].

We have ( ) ( )1222

1 1 −=− − mmnn ss , so s(ga) = s − 1 and the induc-

tion hypothesis applies to ga ∈ R[X]. Thus, ga has a complex root. On the other hand, the roots of ga are uij(a), so there exists i, j ∈ P, such that uij(a) ∈ C.

The above proves that, for any a ∈ R, there exists i, j ∈ P, with uij(a) ∈ C. Because R is infinite and P is finite, there exist a, b ∈ R, with a ≠ b, such that uij(a) ∈ C and uij(b) ∈ C. It follows immediately that s := ti + tj and p := titj are complex numbers and so ti, tj ∈ C, as roots of X 2 − sX + p ∈ C[X].


207

In the general case of a polynomial g ∈ C[X], let g be the conju-gate of g (the polynomial obtained by the complex conjugation of the coefficients of g). An elementary computation shows that g ⋅ g ∈ R[X] and the first part applies. Therefore, there exists α ∈ C with g (α)⋅g(α) = 0. If g(α) = 0, we are done. If g (α) = 0, then g has the root α . !

Exercises

Throughout the exercises, K is a field and Ω is an algebraic closure of K. 1. Let R be a commutative ring. The following statements are equiva-lent:

a) R is a domain. b) Any polynomial p ∈ R[X] of degree n ≥ 1 has at most n roots in

R (the roots being counted with their multiplicity). c) Any polynomial of degree1 in R[X] has at most one root in R.

2. (The skew field of quaternions) Let

H = ( )⎭⎬⎫

⎩⎨⎧

∈∈⎟⎠

⎞⎜⎝

⎛−

CC vuMuvvu

,2 .

a) Show that H is a skew field50 (division ring) with respect to addi-tion and multiplication of matrices.

50 The skew field of quaternions was discovered (invented?) by W.R. Hamilton.


208

b) Let 1 := ⎟⎠

⎞⎜⎝

⎛1001

, i := ⎟⎠

⎞⎜⎝

⎛− i

i0

0, j := ⎟

⎠

⎞⎜⎝

⎛− 01

10, k := ⎟

⎠

⎞⎜⎝

⎛0

0i

i. Show

that any element of H is written uniquely as a1 + bi + cj + dk, where a, b, c, d ∈ R.

c) Let Q be the multiplicative subgroup of H* generated by i and j. Prove that Q is not commutative and has 8 elements. Write down the multiplication table of Q. (Q is called the quaternion group). Deduce that H is a skew field.

d) Prove that X 2 + 1 ∈ H[X] has an infinity of solutions in H. Does this contradict 2.3?

e) Prove that H is an extension of C. f) Give an example of a countable skew field.

3. Give an example of an extension K ⊆ L, where L is algebraically closed and L is not algebraic over K. 4. Let K ⊆ L ⊆ E be a tower of extensions and let f ∈ K[X]. If E is a splitting field of f over K, then E is a splitting field of f over L. 5. Let K ⊆ L ⊆ E be field extensions and let f ∈ K[X]. Suppose L = K(x1, , xn), where x1, , xn are roots of f (not necessarily all of them). Show that E is a splitting field of f over K if and only if E a splitting field of f over L. 6. Let f ∈ K[X] be a polynomial of degree n and let x1, , xn be the roots of f in some extension L of K. Show that K(x1, , xn −1) is a split-ting field of f over K. 7. If the extension K ⊆ L ⊆ Ω is such that any nonconstant polynomial in K[X] splits over L, then L = Ω. 8. Show that (Q'R)(i) = Q'C. 9. Let char K = 0 and let g ∈ K[X], irreducible, deg g ≥ 2. If α, β ∈ Ω are roots of g, then α − β ∉ K. (Hint. Suppose α − β ∈ K and let β = α + b, b ∈ K. Then K(α) = K(α + b). Because g is irreducible, there exists a K-isomorphism ϕ : K(α) → K(β) = K(α) that takes α to

IV.3 Finite fields

209

β = α + b. Then α, ϕ(α), ϕ(ϕ(α)) = ϕ 2(α),, ϕ n(α), are roots of ϕ, for any n. Since ϕ n(α) = α + nb and char K = 0, these are all distinct.) 10. Let f, g ∈ K[X], nonconstant and let β be a root of f in Ω. Then: f(g(X)) is irreducible in K[X] ⇔ f is irreducible in K[X] and g(X) − β is irreducible in K(β)[X]. (Hint. Let f = a∏(X − βi), where β = β1, , βn ∈ Ω. If α is a root of g(X) − β, then f(g(α)) = 0 and: f(g(X)) is irreducible in K[X] ⇔ [K(α) : K] = deg f(g(X)) = deg f ·deg g. In the tower of extensions K ⊆ K(β) ⊆ K(α), [K(β) : K] = deg Irr(β, K) and [K(α) : K(β)] = deg Irr(α, K(β)).)

IV.3 Finite fields

We now apply the results obtained so far (notably the existence of the algebraic closure of a field) to determine all finite fields. We al-ready know (1.10) that, if F is a finite field, then:

- char F is a prime p > 0; - F is a finite extension of Fp ( = Zp, the field of integers modulo p); - |F| = p n, where n = [F : Fp].

These facts raise the following questions: Given a prime p and n ∈ N*, does there exist a finite field with p n elements? If it exists, is it unique (up to isomorphism)? How can one construct it?

3.1 Definition. Let K be a commutative ring, char K = p > 0, p a prime. Let:

ϕ : K → K, ϕ(x) = x p, ∀x ∈ K.

Of course, ϕ(xy) = ϕ(x)ϕ(y), ∀x, y ∈ K. Also:


210

ϕ(x + y) = (x + y) p = ( )0

p i i

i p

pi x y−

≤ ≤∑ = x p + y p,

The last equality holds because p divides the binomial coefficients

( ) ( ) ( )1 1!

p p p ii

pi

− − += … , for any 0 < i < p.

Thus, ϕ is a ring homomorphism, called the Frobenius endomorph-ism51 of K. The image of ϕ is K p: = x p | x ∈ K, a subring in K. If char K = 0, the Frobenius endomorphism is taken to be the identity function, ϕ = id : K → K. We use this endomorphism especially if K is a field (but also for rings, for example in K[X], where K is a field of characteristic p). The Frobenius endomorphism is the identity in the case of the field Fp (by the little Fermat theorem, x p = x, for any x ∈ Fp; see also the proof below).

3.2 Theorem. Let p be a prime and let Ω be an algebraic closure of the field Fp.

a) Let n ∈ N* and let q := p n. There exists a unique subfield F of Ω having q elements, namely the splitting field of X q − X over Fp. In par-ticular, there exists a finite field with p n elements and any two fields with p n elements are isomorphic.

b) If F is a field with p n elements and K is a subfield of F, then |K| = p m, where m divides n. Conversely, for any divisor m of n there exists a unique subfield of F with p m =: r elements, namely K = x ∈ F| x r = x.

Proof. a) Suppose F ⊆ Ω is a field with q elements. The group (F*, ·) has q − 1 elements. Applying the Lagrange theorem for the or-der of an element in a finite group, we obtain x q −1 = 1, so x q = x, ∀x ∈ F*. The prime subfield of F is Fp. Consequently, any element of F is a root of the polynomial X q − X, considered with coefficients in

51 Ferdinand Georg Frobenius (1849-1917), German mathematician.

IV.3 Finite fields

211

Fp. This polynomial can have at most q distinct roots in Ω; since |F| = q, the elements of F are exactly the roots of X q − X in Ω. This means that F is the splitting field of X q − X over Fp and F is the unique subfield with q elements of Ω.

Let us prove that there exists a field with q elements F ⊆ Ω. The argument above says F must be the splitting field of X q − X over Fp. So, let g = X q − X. Let F be the set of all roots of g in Ω, F = x ∈ Ω | x q = x. We claim F is a subfield of Ω. Indeed, ψ : Ω → Ω, ψ(x) = x q, ∀x ∈ Ω, is a field homomorphism (ψ = ϕ n where ϕ is the Frobenius); then F is the subfield of all elements fixed by this homomorphism. On the other hand, the number of roots of g in Ω is exactly q: g has at most q roots and has no multiple roots, since its derivative is qX q−1 − 1 = −1 (see 2.6). We deduce that F, the split-ting field of g over Fp, has q elements.

b) Let K be a subfield of F (we suppose that F ⊆ Ω). Let s := |K| and let t := [F : K]. We have p n = |F| = s t; since p is a prime, this is possible only if s is of the form p m and p mt = p n, which implies m|n. Conversely, if m divides n, then n = mt for some t. Let Ω be an alge-braic closure of F and let ψ : Ω → Ω, ψ(x) = x r, ∀x ∈ Ω, where r = p m. Let L := x ∈ Ω | x r = x = x ∈ Ω |ψ(x) = x. L is a subfield with r elements of Ω. We show that L ⊆ F. We have (ψ ψ) (x) = ψ(x r) =

2rx and, by induction, ψ t(x) =trx , ∀x ∈ Ω. So,

x ∈ L ⇒ ψ(x) = x ⇒ ψ t(x) = x ⇒ xxtr = . But r t = p n, and F =

x ∈ Ω |npx = x; thus, x ∈ L ⇒ x ∈ F. Any subfield E of F with r

elements is equal to L because all elements of E satisfy the equation x r = x. !

The finite field with p n elements (unique up to an isomorphism) is

denoted by GF(p n) or Fpn . The notation GF(p n), an acronym for

Galois Field, honors Evariste Galois, who determined the structure of the finite fields in 1830.


212

3.3 Proposition. For any n ∈ N*, there exists at least an irreduci-ble polynomial of degree n in Fp[X]; for any such polynomial f, Fp[X]/ f is a field with p n elements.

Proof. Let F be the splitting field of g = X q − X over Fp, where q = p n. The next lemma says that (F*, ⋅) is a cyclic group. Let α be a generator of F* and let f = Irr(α, Fp). We have that F = Fp(α) and deg f = [Fp(α) : Fp] = [F : Fp] = n. So, f is irreducible of degree n in Fp[X]. Also, Fp[X]/ f ≅ Fp(α), a field with q elements. !

3.4 Lemma. If R is a domain and G is a finite subgroup of U(R) (the multiplicative group of units of R), then G is cyclic. In particular, any finite subgroup of K* (where K is a field) is cyclic.

Proof. Let G be a finite group with n elements, G ≤ (U(R), ·). Let m be the exponent of G, (the GCD of the orders of the elements of G). Lagrange's theorem implies that m | n. On the other hand, any element in G is a root of X m − 1, which has at most m roots in R. Thus, n ≤ m and m = n. Since G is Abelian, there exists an element x having order m in G (for an elementary proof of this, see exercise 3.2), so G is cy-clic, generated by x.

Another proof. Use the invariant factors theorem, applied to the fi-nite Abelian group G. Suppose G is not cyclic. Then G ≅ C1××Cm, where m ≥ 2 and C1, , Cm are cyclic, |C1| = d1, , |Cm| = dm and d1 | d2 || dm. Then 1−mdX has n roots in R, where n = d1··dm > dm, contradiction. !

We prove now that any finite division ring is commutative (it is a field). The proof of this result needs some facts on roots of unity and cyclotomic fields, which have also numerous other applications in algebra, number theory etc.

3.5 Definition. Let K be a field and let n be a positive integer. The element ζ ∈ K is called an nth root of unity in K if ζ n = 1. Let Un(K)

IV.3 Finite fields

213

(or simply Un if K cannot be confused) be the set of the nth roots of unity in K, Un(K) := x ∈ K | x n = 1.

Since X n − 1 has at most n roots, Un is a subgroup with at most n elements of the multiplicative group (K*, ·). The element ζ ∈ Un is called a primitive nth root of unity in K if the order of ζ in Un is n. (⇔ ζ n = 1, but ζ m ≠ 1, ∀m < n ⇔ ζ generates Un). We denote by Pn(K) (or simply Pn) the set of all primitive nth roots of unity in K:

Pn(K) := x ∈ Un(K) | ord x = n. If ζ is a root of unity in some extension of K, the extension

K ⊆ K(ζ ) is called a cyclotomic extension.

3.6 Remark. Un(K) is cyclic (as a finite subgroup of K*). Thus, there exists a primitive nth root of unity in K if and only if Un(K) has n elements.

3.7 Examples. a) Un(Q) = −1, 1 if n is even and Un(Q) = 1 if n is odd.

b) Un(C) = e2kπi/n | k ∈ 1, , n, where e2kπi/n = cos(2kπ/n) + isin(2kπ/n). There are ϕ(n) complex primi-tive nth roots of unity, namely e2kπi/n, cu (k, n) = 1, 1 ≤ k < n. Here ϕ(n) is the number of all integers k, (k, n) = 1, 1 ≤ k < n. The function ϕ : N* → N* is called the Euler phi function. We have ϕ(n) = |U(Zn)|, the number of invertible elements in the ring Zn.

c) U4(F5) = F5* = 1, 2, 3, 4; P4(F5) = 2, 3; U4(F7) = 1, 6;

P4(F7) = ∅.

3.8 Proposition. Let Ω be an algebraic closure of the field K and let n ∈ N*.

a) Un(Ω) has n elements if and only if char K does not divide n. b) If char K = p > 0 and n = p tm, with (p, m) = 1, then

Un(K) = Um(K).


214

Proof. a) Un(Ω) has n elements if and only if f = X n − 1 has no multiple roots ⇔ f ', f = 1 ⇔ nX n −1, X n − 1 = 1⇔ n·1 ≠ 0 ⇔ char K does not divide n.

b) In general, m | n implies Um ⊆ Un. Let p t =: q. If x ∈ Un, then x n = ( )qmx = 1. But y & y q, ∀y ∈ K, is a field endomorphism (necessarily injective) of K (it is a power of the Frobenius endomorph-ism), so y q = 1 implies y = 1. Thus, x ∈ Un implies x m = 1, so x ∈ Um.!

According to the above, it is sufficient to study Un(K) when char K does not divide n. In this case, there exists a primitive nth root of unity in some extension of K.

3.9 Definition. Let n ∈ N* and let K be a field containing a primi-tive nth root of unity. The polynomial (in K[X])

Φn := ( )( )∏ ∈ −KPnXζ ζ

is called the nth cyclotomic polynomial over K. The roots of Φn are distinct and they are exactly the primitive nth roots of unity in K, so deg Φn = ϕ(n).

If char K = 0 (one usually takes K = C to ensure a primitive nth root of unity exists), then

Φn := ( )( )∏

=<−

1,,

2

nknk

nikeX π

is called the nth cyclotomic polynomial (dropping any reference to the field).

3.10 Remark. In the complex plane, the complex nth roots of unity are the vertices of a regular polygon with n sides inscribed in the unit circle. This justifies the name cyclotomic, of Greek origin and mean-ing approximately circle dividing.

3.11 Lemma. a) Suppose K ⊆ L is an extension, f, g ∈ K[X] and h ∈ L[X] such that f = gh. Then h ∈ K[X]. In other words, if g divides f in L[X], then g divides f in K[X].

IV.3 Finite fields

215

b) If f ∈ Q[X] is monic and divides (in C[X]) a monic polynomial u ∈ Z[X], then f ∈ Z[X] and u = fg, where g ∈ Z[X] and g is monic.

Proof. a) The division with remainder theorem in K[X] guarantees the existence of q, r ∈ K[X] such that f = gq + r, with deg r < deg g. The division with remainder in L[X] of f to g is f = gh + 0. The unique-ness of the quotient and of the remainder in L[X] implies that r = 0 and h = q ∈ K[X].

b) By a), u = fg, with g ∈ Q[X]. Let a (respectively b) be the LCM of denominators of the coefficients of f (respectively g). Then af, bg ∈ Z[X] and af·bg = abu. Taking the contents of the polynomials, c(af )·c(bg) = ab, since c(u) = 1. So af = c(af )·f1, bg = c(bg )·g1, with f1, g1 ∈ Z[X], primitive. Thus c(af )·f1·c(bg )·g1 = abu. Simplify by ab = c(af )·c(bg) to obtain f1·g1 = u; since u is monic, and f1, g1 ∈ Z[X], f1, g1 have leading coefficients 1 or − 1. Since f (which is monic) is associated in divisibility with f1, we obtain f = ± f1 ∈ Z[X]. Likewise, g ∈ Z[X]. !

3.12 Proposition. Let n ∈ N* and let K be a field containing a primitive nth root of unity. Then:

a) For any d dividing n, there exists a primitive dth root of unity in K.

b) ∏ Φ=− nd dnX 1 holds.

c) The coefficients of Φn are in the prime subfield P of K. d) If char K = 0 (for example K = C), then Φn ∈ Z[X]. Proof. a) If ζ is a primitive nth root of unity and m = n/d, then ζ m

is a primitive dth root of unity. b) The polynomials in the equality are monic and have the same

roots, namely the nth roots of unity in K. c) We prove by induction by m (1 ≤ m ≤ n) the statement: "for any

d with 1 ≤ d ≤ m, d | n implies Φd ∈ P[X]".


216

If m = 1, then Φ1 = X − 1 ∈ P[X]. Let 1 < m ≤ n. We suppose the claim true for any q < m and we prove for m. If m - n, there is nothing to prove. If m | n, then X m − 1 = Φm·g, where

g = ∏<Φmdmd

d,

∈ P[X],

since by the induction hypothesis Φd ∈ P[X], ∀d < m, d | m. Apply now the lemma 3.11.a) to obtain that Φm is also in P[X].

d) Repeat word for word the proof of c) (replace P with Z), to ob-tain that X m − 1 = Φm·g, where g ∈ Z[X], monic. Thus, Φm ∈ Q[X]. Use lemma 3.11.b) to deduce Φm ∈ Z[X]. !

One can use the formula ∏ Φ=− nd dnX 1 to recursively compute

Φn (for this, all Φd are needed, for d | n, d < n). The Möbius inversion formula (see e.g. SPINDLER [1994]) gives an explicit expression for Φn, ∀n ∈ N*, in terms of products of polynomials X d − 1 or their inverses.

3.13 Example. Consider the 7th cyclotomic polynomial Φ7 over an algebraic closure of F2. From Φ1Φ7 = X 7 − 1 we deduce Φ7 = X 6 + X 5 ++ 1. By trial and error (or using a computer) we ob-tain

Φ7 = (X 3 + X 2 + 1)(X 3 + X + 1), where the right hand side is the irreducible factorization of Φ7 in F2[X] (the polynomials have degree 3 and no roots in F2).

So, Φ7 is reducible in F2[X]. If α is a primitive 7th root of unity, then Irr(α, F2) = X 3 + X 2 + 1 or X 3 + X + 1.

F2(α) is the field with 8 elements F8. Any nonzero element of F8 is a primitive 7th root of unity. Why?

In characteristic 0 the behavior above is not present: the cyclotomic polynomials over C are irreducible (in Z[X] and in Q[X]):

IV.3 Finite fields

217

3.14 Theorem. Let n ∈ N*. Then Φn is irreducible in Z[X] and it is the minimal polynomial over Q of any complex primitive nth root of unity.

Proof. Let ζ be a primitive complex nth root of unity and let f = Irr(ζ, Q). We show that f = Φn; more precisely, we show that f and Φn have the same roots.

Since f divides X n − 1 in C[X] and f is monic, 3.11.b) implies that f ∈ Z[X].

Any root of f is a primitive nth root of unity. Indeed, if β is a root of f, then K(ζ) and K(β) are K-isomorphic via an isomorphism that takes ζ to β. In particular, we have, ∀m ∈ N*, ζ m = 1 if and only if β m = 1, so ord ζ = ord β.

This shows that any root of f is a root of Φn, so f | Φn. We need now to prove that any primitive nth root of unity is a root of f. We will show that: If α ∈ Pn and f (α) = 0, then f (α p) = 0, for any prime p not dividing n.(*)

Since f (ζ ) = 0, an inductive argument based on (*) shows that f (ζ m) = 0, for any m relatively prime to n. Since any primitive nth root of unity is of the form ζ m for some m coprime with n, we obtain that all the primitive nth roots of unity are roots of f.

In order to prove (*), let α ∈ Pn and let p be prime, p - n. By lemma 3.11.b), Φn = fg, with g ∈ Z[X], monic. Since α p ∈ Pn, f (α p)g(α p) = Φn(α

p) = 0. Suppose by contradiction that f (α p) ≠ 0, so α p is a root of g. Thus, α is a root for g1 := g(X p) ∈ Z[X]. Because f = Irr(α, Q), f | g1 in Z[X] (by 3.11), so g1 = fh, with h ∈ Z[X], monic.

In what follows, for any q ∈ Z[X], let π(q) denote the polynomial q reduced modulo p (π(q) is the polynomial in Zp[X] whose coefficients are the images in Zp of the coefficients of q; in other words, π : Z[X] → Zp[X] is the unique extension to Z[X] of the canonical homomorphism Z → Zp). Let g = a0 + a1X + + X n. Then:


218

π(g p) = π(g)p = (π(a0) + π(a1)X + + X n) p = π(a0)p + π(a1)

pX p + + X pn= π(a0) + π(a1)X

p + + X pn = π(g1). Since g1 = fh, π(g1) = π(g) p = π f π(h). Thus π f | π(g) p in Zp[X],

so all irreducible divisors of π f divide also π(g). Because deg π f = deg f, there exists an irreducible common factor h of π f and π(g), deg h ≥ 1. But then π(Φn) = π f π(g) is divisible with h2, so it has multiple roots. This is absurd, since Φn | X

n − 1, so π(Φn) divides π(X n − 1), who has no multiple roots: its derivative is nX n −1 ≠ 0 in Zp[X] (p - n implies n invertible in Zp) and π(X n − 1) is coprime to nX n −1. !

We will encounter cyclotomic extensions again in the study of Ga-lois Theory. We need now some elementary facts on the conjugacy classes of a finite group, which will be used here to prove that any fi-nite division ring is commutative.

3.15 Proposition. Let G be a group and let a ∈ G. a) The set C(G) := x ∈ G | xy = yx, ∀y ∈ G (called the center of

G) is a normal Abelian subgroup of G. b) The set C(a) := x ∈ G | xa = ax (called the centralizer of a in

G) is a subgroup of G. c) The conjugacy relation in G, defined by: ∀x, y ∈ G, x ~ y if

and only if exists z ∈ G such that y = z −1xz, is an equivalence relation on G. If x ~ y, we say that x is conjugate with y (or y is a conjugate of x).

d) Let Ca := x ∈ G | x ~ a (the conjugacy class of the element a). Then |Ca| = [G : C(a)] (the index of the subgroup C(a) in G).

e) (conjugacy classes formula) We have: |G| = |C(G)| + ( )[ ]∑ ∈Sa aCG : ,

where the sum runs on a system S of representatives of the conjugacy classes of the elements not contained in the center of G.

Proof. a), b), c) have standard proofs and are left to the reader.

IV.3 Finite fields

219

d) It is clear that Ca = z −1az | z ∈ G. Let G/C(a) := C(a)x | x ∈ G

be the set of right cosets of the subgroup C(a) in G. Define ϕ : G/C(a) → Ca by ϕ(C(a)x) = x −1ax, ∀x ∈ G. ϕ is correctly defined (x −1ax is independent on the representative x of the class C(a)x). In-deed, for any x, y ∈ G: C(a)x = C(a)y ⇔ xy −1 ∈ C(a) ⇔ axy −1 = xy −1a ⇔ x −1ax = y −1ay. The injectivity of ϕ also follows, since x −1ax = y −1ay implies C(a)x = C(a)y. Because ϕ is obviously surjec-tive, |Ca| = |G/C(a)| = [G : C(a)].

e) G is the disjoint union of the conjugacy classes. Clearly, a ∈ C(G) ⇔ C(a) = a ⇔ |Ca| = 1. Let S be a system of representatives as in the statement. Then S ∪ C(G) is a system of representatives for the conjugacy classes, so

G = ∪Ca | a ∈ S ∪C(G) = ∪Ca | a ∈ S ∪ C(G). The unions are disjoint, so, taking cardinals and using

|Ca| = [G : C(a)], we obtain the formula. !

We can prove now the following celebrated result:

3.16 Theorem. (Wedderburn52, 1909) Any finite division ring is commutative.

Proof. Let K be a finite division ring. Let p = char K > 0. Let C be the center of the ring K, C := x ∈ K | xy = yx, ∀y ∈ K. One easily verifies that C is a commutative subring of K, and C is a field (of characteristic p). We have thus the tower of extensions of division rings Zp ⊆ C ⊆ K. Let m = [C : Zp] and let n = [K : C]. Then |C| = p m := q and |K| = q n. It is sufficient to prove that n = 1.

Suppose that n > 1. The center of the group (K*, ·) is C*. For any a ∈ K, consider x ∈ K | xa = ax =: Z(a), which is easily seen to be a subfield of K that includes C. The centralizer of a in the group K* is

52 Joseph Henry Maclagen Wedderburn (1882-1948), Scottish mathematician.


220

Z(a)* := Z(a)\0. Let d(a) = [Z(a) : C], so |Z(a)| = q d(a). By 1.27, the multiplicativity of degrees holds also for extensions of division rings, so Zp ⊆ C ⊆ Z(a) ⊆ K implies da | n. In the group (K*, ·) apply the conjugacy classes formula:

|K*| = |C*| + ( )[ ]∑ ∈Sa aZK ** : ,

S being a system of representatives of the conjugacy classes of ele-ments not contained in C*. Note that a ∈ S implies da ≠ n (otherwise Z(a) = K, so a ∈ C*, contradicting the choice of S). In the formula

above, |K*| = q n − 1, |C*| = q − 1, [K* : Z(a)*] = ( )11

n

d aq

q−−

, so:

q n − 1 = q − 1 + ( )∑∈ −

−

Saad

n

qq

11 . (1)

The cyclotomic polynomial (over C) Φn divides (X n − 1)/(X d(a) − 1) in Z[X], ∀a ∈ S. Indeed, 3.12.b) implies X n − 1 = Φn· ∏

≠Φndnd

d,

; since

da | n, da ≠ n and X d(a) − 1 = ( )∏ Φadd d , the claim follows. So, Φn(q)

divides q n − 1 and also divides (q n − 1)/(q d(a) − 1), ∀a ∈ S; from (1) we obtain that Φn(q) | (q − 1).

On the other hand, |Φn(q)| = ∏ ∈ −nP qζ ζ . We have |q − ζ | > q − 1,

∀n ≥ 2, ∀ζ ∈ Pn (to see this, represent the complex nth roots of unity in a plane). So |Φn(q)| > q − 1, contradiction with Φn(q) | q − 1. !

IV.3 Finite fields

221

Exercises

1. Let K ⊆ L be an extension of degree n of finite fields. Show that the lattice of all intermediate fields of the extension is isomorphic to the lattice of the divisors of n (ordered by divisibility). 2. Let G be a finite group and let exp(G) := LCMord a | a ∈ G (the exponent of G). Prove that:

a) If a, b ∈ G, ab = ba and (ord a, ord b) = 1, then ord ab = (ord a)(ord b).

b) For any a, b ∈ G, ab = ba implies that there exists c ∈ G with ordc = [ord a, ord b].

c) If G is Abelian, there exists an element g of G such that ord(g) = exp(G). 3. Let K be a finite field of characteristic p and let α ∈ K*. If α generates the group K*, then K = Fp(α). Is the converse true? 4. Let K be a finite field. Show that any function ϕ : K → K is polyno-mial (there exists a polynomial g ∈ K[X] such that ϕ(x) = g(x), ∀x ∈ K). More generally, show that any function ϕ : K n → K is polynomial (there is a polynomial g ∈ K[X1,, Xn] such that ϕ(x1,, xn) = g(x1,, xn), ∀(x1,, xn) ∈ K n). What happens if K is infinite? 5. Let K be a finite field with q elements.

a) Show that, ∀n ∈ N*, the product of all irreducible monic polynomials in K[X] with degree dividing n is XX

nq − . (Hint. Show they have the same roots.)

b) Let N(n, q) be the number of irreducible monic polynomials in K[X] having degree n. Then:

∑d|n d·N(d, q) = q n c) Compute N(p, q) for p a prime.


222

6. Let K be a finite field with q elements. Determine the number of equivalence classes in Mn(K) with respect to similarity, for any n ≤ 3. (Hint. Use that two matrices are similar ⇔ they have the same elementary divisors. Make a connection to the N(r, q) for r ≤ n.) 7. Let K ⊆ L be an extension of finite fields, [L : K] = n. Then any irre-ducible polynomial of degree n in K[X] splits in L[X]. 8. Give an example of a finite field K and of an irreducible polynomial g ∈ K[X], deg g ≥ 2, such that, if α, β are roots of g in Ω, then α − β ∈ K (cf. exercise 2.9). (Hint. If K = Fp, g = X p − X − a, with a ∈ Fp, then g(X + c) = g(X), ∀c ∈ Fp.) 9. Find the number of irreducible polynomials of degree 2 in F5[X]. If α is o root of X 2 + 2 ∈ F5[X], then any polynomial of degree 2 in F5[X] has a root in F5[α]. 10. Let p be a prime and let k ∈ N. Then Φpk = Φp(

1−kpX ). 11. Let Ω be an algebraic closure of the field Fp. Then Ω has uncount-able many subfields. (Hint. For any subset S of the set of all prime numbers P, consider the subfield CS of Ω, CS is the composite of the fields Fpn | n is a product of primes from S.)

The following exercises are about the solutions in finite fields of polynomial equations in several unknowns. In the particular case Zp, these yield results on congruences mod p. For some developments, see BOREVICH, SHAFAREVICH [1985]. 12. Let F be a finite field with q elements and let r ∈ N*. Show that:

a) Ur(F) = x ∈ F* | x r = 1 is a cyclic subgroup of F*, with d ele-ments, where d = (q − 1, r).

b) Let Sr = Sr, q = ∑x | x ∈ Ur(F). Then Sr, q = 1 if |Ur(F)| = 1; Sr, q = 0 if |Ur(F)| > 1. (Hint. If α is a generator of Ur(F), then αSr = Sr).

c) Let m ∈ N* and Tm = Tm, q = ∑x m | x ∈ F. Then Tm, q = −1 if (q − 1) | m ; Tm, q = 0 if (q − 1) - m.

d) Let f ∈ F[X1,, Xn], of (total) degree < n(q − 1). Then:

IV.4 Transcendental extensions

223

∑ f (x1, , xn) | (x1, , xn) ∈ F n = 0. (Hint. It is sufficient to suppose f is a monomial. Then the sum

above is a product of n sums of type Tm, and at least one m is < q − 1.) 13. (Warning) Let F be a field with q elements, char F = p and g ∈ F[X1,, Xn], deg g < n. Then the number N of solutions in F n of the equation g(x1, , xn) = 0 is a multiple of p. (Hint. Let f := 1 − g q − 1. Then deg f < n(q − 1) and g(x1, , xn) ≠ 0 ⇔ f (x1, , xn) = 0; also g(x1, , xn) = 0 ⇔ f (x1, , xn) = 1. Calculate ∑f (x1, , xn) | (x1, , xn) ∈ F n and use the previous exercise.) 14. (Chevalley) Suppose F is a field with q elements, char F = p and g ∈ Zp[X1,, Xn] is a homogeneous polynomial, deg g < n. Then the number of solutions in F n of the equation g(x1,, xn) = 0 is a nonzero multiple of p. In particular, the equation has solutions other than (0,, 0).


Recall that the extension K ⊆ L is called transcendental if it is not algebraic: L contains a transcendental element over K. For example, R is a transcendental extension of Q. The field K(X) of rational fractions with coefficients in K is a transcendental extension of K. More gener-ally, if (Xi)i∈I is a family of indeterminates and K(X; I) is the field of rational fractions in the indeterminates (Xi)i∈I with coefficients in K, then K ⊆ K(X; I) is a transcendental extension.

4.1 Definition. a) Let K ⊆ L be a field extension and let x1, , xn be a finite subset of L. We call x1, , xn an algebraically independ-ent set over K (or say that x1, , xn are algebraically independent over


224

K) if, for any polynomial f ∈ K[X1, , Xn], f(x1, , xn) = 0 implies f = 0. An arbitrary subset S of L is called algebraically independent over K if any finite subset of S is algebraically independent over K. The set S is called algebraically dependent over K if S is not algebrai-cally independent over K. This means that there exist x1, , xn ∈ S, distinct and f ∈ K[X1, , Xn], with f ≠ 0 and f(x1, , xn) = 0; we call the equality f(x1, , xn) = 0 an algebraic dependence relation over K for S.

b) The extension K ⊆ L is called a purely transcendental extension if L is generated over K by a set of algebraically independent elements over K: there exists a subset S of L, algebraically independent over K, such that L = K(S).

4.2 Remark. Let (Xs)s∈S be a family of indeterminates indexed by S, a subset of L. The subset S is algebraically independent over K if and only if the unique homomorphism of K-algebras v : K[X; S] → L such that v(Xs) = s, ∀s ∈ S, is injective. The image of this homomorph-ism is K[S], the subring generated by K and S in L (the K-subalgebra of L generated by S). Thus, S is algebraically independent over K if and only if v : K[X; S] → K[S] v(Xs) = s, ∀s ∈ S, is an isomorphism. This isomorphism between the domains K[X; S] and K[S] induces an isomorphism between the fields of fractions K(X; S) and K(S).

In other words, an algebraically independent set over K behaves the same as a set of indeterminates in the field of fractions of a polyno-mial ring with coefficients in K. another way to say this is: the purely transcendental extensions of K are (up to K-isomorphism) the exten-sions K ⊆ K(X; I), where (Xi)i∈I is a family of indeterminates.

4.3 Examples. a) The singleton x is algebraically independent over K if and only if x is transcendental over K. Thus, e is algebrai-cally independent over Q. Likewise, π and e3 are algebraically independent sets over Q. The set e, e5 is not algebraically independ-


225

ent over Q because f (e, e5) = 0, where f = Y 5 − X ∈ Q[X, Y]. It is not known if e, π is algebraically independent over Q.

b) The empty set is algebraically independent over any field. c) Let K be a field and let n ∈ N*. The symmetric elementary

polynomials s1, , sn are algebraically independent over K in K(X1,, Xn). This statement is equivalent to the uniqueness part in the fundamental theorem of symmetric polynomials: Any symmetric polynomial in K[X1,, Xn] is written uniquely as a polynomial with coefficients in K of s1, , sn.

d) In K ⊆ K(X, Y), the elements X 5 and Y are algebraically independent over K. Prove this. Can you generalize? So, K[X, Y] ≅ K[X 5, Y] (K-isomorphism), although K[X 5, Y] ( K[X,Y].

4.4 Proposition. Let K ⊆ L be an extension and let S, T be disjoint subsets of L, with S algebraically independent over K. Then S ∪ T is algebraically independent over K if and only if T is algebraically independent over K(S).

Proof. Suppose S ∪ T is algebraically independent over K. If T is algebraically dependent over K(S), then there exist t1, , tn ∈ T and a polynomial f in n indeterminates with coefficients in K[S] such that f (t1, , tn) = 0. Write down this equality, taking into account the form of the elements in K[S], to obtain an algebraic dependence relation over K for S ∪ T.

Conversely, let T be algebraically independent over K(S) and sup-pose S ∪ T is algebraically dependent over K. Then there exist s1, , sm ∈ S, t1, , tn ∈ T, distinct, and a nonzero polynomial f in m + n indeterminates with coefficients in K such that

f (s1, , sm, t1, , tn) = 0. Group the terms by the monomials in t1, , tn to obtain a relation

of the form g(t1,, tn) = 0, with g nonzero polynomial in n indetermi-nates with coefficients in K(S). !


226

4.5 Definition. Let K ⊆ L be an extension and let S be a subset of L.

a) S is called a set of algebraic generators of L over K (or say that S algebraically generates L over K) if K(S) ⊆ L is an algebraic exten-sion. This is equivalent to the fact that any element α ∈ L satisfies an equation of the form:

( )∑≤≤

=nk

kmk xxf

01 0,, α… ,

for some x1, , xm ∈ S and fk ∈ K[X1, , Xm], ∀k ∈ 0, , n, with fn(x1, , xm) ≠ 0.

b) S is called a transcendence basis for the extension K ⊆ L (or a transcendence basis of L over K) if S is algebraically independent over K and S algebraically generates L over K.

4.6 Examples. a) The extension K ⊆ L is algebraic if and only if the empty set ∅ is a transcendence basis of L over K.

b) If (Xi)i∈I is a set of indeterminates, then (Xi)i∈I is a transcendence basis of the field of rational fractions K(X; I) over K.

c) e is a transcendence basis of Q ⊆ Q(e, 2 ).

4.7 Proposition. Let K ⊆ L be an extension and let S be a subset of L. The following statements are equivalent:

a) S is a transcendence basis of L over K. b) S is a maximal algebraically independent set over K. c) S is a minimal set of algebraic generators of L over K. Proof. a)⇒b) Let α ∈ L \ S. Since S algebraically generates L over

K, there exist x1, , xm ∈ S and fk ∈ K[X1, , Xm] such that α satisfies an equation of the form:

( )∑≤≤

=nk

kmk xxf

01 0,, α… ,

so S ∪α is not algebraically independent over K.


227

b)⇒a) We must prove that K(S) ⊆ L is algebraic. Suppose α ∈ L is not algebraic over K(S). Then the previous result shows that S ∪α is algebraically independent over K, contradicting the maximality of S.

a)⇒c) We must prove that, for any α ∈ S, S \α is not a set of algebraic generators of L over K, i.e. L is a transcendental extension of K(S \α). Indeed, α is transcendental over K(S \α) (otherwise S would be algebraically dependent over K).

c)⇒a) It is sufficient to see that S is algebraically independent over K. If it is not so, then there exist s1, , sn ∈ S and f ∈ K[X1,, Xn], with f ≠ 0 and f (s1, , sn) = 0. Relabeling if necessary, we may sup-pose that there exists a monomial in f that contains a power of s1. This means that s1 is algebraic over K(s2, , sn), so S \s1 algebraically generates L over K, contradicting the minimality of S. !

There is a similarity between the concepts of linear independence (respectively generating set, basis) in vector spaces and algebraic independence (respectively set of algebraic generators, transcendence basis) in field extensions. The same method used at vector spaces (Zorn's Lemma) is used to prove the existence of a transcendence basis of an extension.

4.8 Theorem. Let K ⊆ L be a field extension and let S, T be subsets of L such that S ⊆ T, S is algebraically independent over K and T is a set of algebraic generators of L over K. Then:

a) There exists a transcendence basis B of L over K such that S ⊆ B ⊆ T.

b) The extension K ⊆ L has a transcendence basis. Proof. a) We look for a transcendence basis as a maximal algebrai-

cally independent set B over K, with S ⊆ B ⊆ T. To this end, define B := C | S ⊆ C ⊆ T, C is algebraically independent over K.

B is nonempty, since S ∈ B. Order B by inclusion. The set B is inductively ordered (straightforward proof) and has thus a maximal


228

element B. We claim that B is a transcendence basis. B is algebraically independent over K since B ∈ B. There is left to show that L is alge-braic over K(B). Since L is algebraic over K(S), it is sufficient to prove that any α in S is algebraic over K(B). If α ∈ S is transcendental over K(B), then B ∪α is algebraically independent over K (by 4.4), contradicting the maximality of B.

b) Apply a) with S = ∅ and T = L. !

Thus, any field extension K ⊆ L can be decomposed in a tower of extensions K ⊆ K(S) ⊆ L, with S a transcendence basis of L over K; K ⊆ K(S) is purely transcendental and K(S) ⊆ L is algebraic.

As the bases in a vector space, any two transcendence bases of an extension have the same cardinal. For the proof, we use the following lemma, which says that, for two transcendence bases, any element in the first base can be replaced with some element in the second base, to obtain a transcendence basis.

4.9 Lemma. Let K ⊆ L be a field extension and let S, T be transcendence bases of K ⊆ L. Then, for any s ∈ S, there exists t ∈ T such that (S \s) ∪t is a transcendence basis.

Proof. For any s ∈ S and t ∈ T, let Sst := (S \s) ∪t. Let s ∈ S. Suppose any t ∈ T is algebraic over K(S \s). Then L is algebraic over K(S \s), because L is algebraic over K(T) and the transitivity of alge-braic extension applies. But this contradicts the fact that S is a minimal set of algebraic generators of L over K (see 4.7). Thus, there exists t ∈ T, transcendental over K(S \s). By 4.4, Sst is algebraically independent over K. Let us show that Sst is also a set of algebraic generators of L over K. We remark that s is algebraic over K(Sst). In-deed, if not, s is transcendental over K(Sst), so Sst ∪s = S ∪t is algebraically independent over K, (use 4.4). But this contradicts the fact that S is a maximal algebraically independent set over K. So, any


229

element in S is algebraic over K(Sst). Since L is algebraic over K(S), the transitivity property implies L is algebraic over K(Sst). !

4.10 Theorem. Any two transcendence bases of a field extension K ⊆ L have the same cardinal.

Proof. Suppose first that K ⊆ L has a finite transcendence basis S = s1, , sn. Let T be another transcendence basis of L over K. For s1 ∈ S, the previous lemma yields t1 ∈ T such that S1 := t1, s2, , sn is a transcendence basis. Let 1 ≤ i < n and suppose we have found t1, , ti ∈ T such that Si := t1, t2, , ti, si+1, , sn is a transcendence basis. Apply again the previous lemma for Si, T and si+1 to obtain ti+1 ∈ T such that Si+1 := t1, t2, , ti+1, si+2, , sn is a transcendence basis. Thus, there exist t1,, tn ∈ T such that Sn := t1, , tn is a tran-scendence basis. Since Sn ⊆ T and T is also a transcendence basis, we deduce Sn = T, so |T| = |Sn| = n = |S|.

In the case when any transcendence basis of L over K is infinite, let S and T be two such bases. For any t ∈ T, since t is algebraic over K(S), there exists a finite subset St of S such that t is algebraic over K(St). Let S' = ∪t∈T St. Since T is algebraic over K(S') and L is alge-braic over K(T), L is algebraic over K(S'). So S' = S, since S is a mini-mal set of algebraic generators of L over K. We have:

|S| = |S'| = |∪t∈T St| ≤ |T × N| = |T|.

We used the fact that T is infinite, so |T × N| = |T|. Thus, |S| ≤ |T|. By symmetry, |T| ≤ |S|, so |S| = |T|. !

This theorem shows that the following definition is correct.

4.11 Definition. Let K ⊆ L be a field extension. The cardinal of a transcendence basis of L over K is called the transcendence degree of the extension L/K and is denoted trdeg (L/K).


230

4.12 Example. Let K be a field and let L := K(X1, , Xn). Example 4.3 says that the symmetric fundamental polynomials s1, , sn are algebraically independent over K, so there exists a transcendence basis of K ⊆ L that includes s1, , sn. On the other hand, X1, , Xn is clearly a transcendence basis of K ⊆ L, so trdeg (L/K) = n. This means that s1, , sn is a transcendence basis of K(X1, , Xn) over K.

The transcendence degree is additive (compare with the property 1.26 of the degree of an extension).

4.13 Proposition. Let K ⊆ L ⊆ M be a tower of field extensions. Then: trdeg (M/K) = trdeg (M/L) + trdeg (L/K).

Proof. Let S, T be transcendence bases for K ⊆ L, respectively L ⊆ M. It is clear that S ∩ T = ∅, so it is enough to prove that S ∪ T is a transcendence basis for K ⊆ M.

Since T is algebraically independent over L and K(S) ⊆ L, T is algebraically independent over K(S). By 4.4, S ∪ T is algebraically independent over K.

In order to prove that S ∪ T algebraically generates M over K, re-mark that M is algebraic over L(T). But L is algebraic over K(S), so L(T) is algebraic over K(S)(T). The transitivity of algebraic extensions implies that M is algebraic over K(S)(T) = K(S ∪ T). !

Exercises

1. Any transcendental extension has an infinity of intermediate fields. 2. Prove that an extension K ⊆ L is finitely generated if and only if its transcendence degree is finite and, if B is a transcendence basis, the


231

degree [L : K(B)] is finite. (Hint: If L = K(S), with S finite, there exists a transcendence basis B included in S. Then L is algebraic and finitely generated -by S- over K(B).) 3. Let K ⊆ L be a field extension and let α, β ∈ L. Then there exists a K-isomorphism K(α) ≅ K(β) that takes α in β if and only if α and β are either both transcendental, either both algebraic and have the same minimal polynomial over K. 4. Prove that the following statement is false: If E and F are subfields of the field L such that E ⊆ L and F ⊆ L are algebraic, then E ∩ F ⊆ L is algebraic. (Hint. K(X 2 + X) ∩ K(X 2) = K.) 5. Let K ⊆ L be a purely transcendental extension. Then L is not an algebraically closed field. (Hint. Prove that for any field K, K(X) is not algebraically closed.) 6. Let K ⊆ L be an extension and let α ∈ L be algebraic over K. If t ∈ L is transcendental over K, then Irr(α, K) = Irr(α, K(t)). 7. Let Ω be an algebraically closed field and let K be a subfield in Ω such that trdeg (Ω/K) is infinite. Show that there exists a K-endomorphism of Ω that is not an automorphism. 8. Let Ω be an algebraically closed field and let K be a subfield in Ω such that trdeg (Ω/K) is finite. Show that any K-endomorphism of Ω is an automorphism. 9. Let Ω be an algebraic closure of C(X). Show that there exists a field isomorphism C ≅ Ω. 10. Let K ⊆ C be a field extension such that C is algebraically closed. Let Ω = K'C. Prove that, for any x1, , xn ∈ C, K(x1, , xn) ∩ Ω =: L is a finite extension of K. (Hint. Let B = x1, , xr be a transcendence basis of K ⊆ K(x1, , xn), with r ≤ n. Then B is a transcendence basis of L ⊆ L(x1, , xn) = K(x1, , xn).)

232

V. Galois Theory

The idea behind modern Galois Theory is the following: for a given field extension, one associates to it a group (the Galois group of the extension). Various properties of the extension can then be deduced by investigating its Galois group. The idea of studying a certain structure (in our case, a field extension) by associating to it another structure (in our case, a group) has been very fertile in 20th century mathematics. It can be found in many areas: Algebraic Topology, Class Field Theory (recently generalized in the form of the Langlands Correspondence), Algebraic Geometry, Representation Theory, and the list is far from complete. The recent proof of Fermat's Last Theorem1 uses Galois Theory as a basic tool.

1 Pierre de Fermat (1601-1665), French mathematician. Fermats Last

Theorem claims that if n ≥ 3 is an integer, then the equation x n + y n = z n has no positive integer solutions. All attempts at proving this assertion failed until 1995, although a lot of partial results were proven. In 1995, the English mathematician Andrew Wiles proved the Shimura-Taniyama conjecture, a statement which implies FLT (a fact proven by Gerhard Frey and Ken Ribet in 1986). It appeared soon that some parts of the proof were wrong, but A. Wiles and R. Taylor finally gave a proof which is now accepted as correct. A major part of modern number theory and Algebra owes its existence to the efforts of proving FLT.

V.1 Automorphisms

233

V.1 Automorphisms

Recall that, given a field extension K ⊆ L, a K-automorphism of L is a bijective field homomorphism σ : L → L with σ |K = id; in this case, σ −1 is also a K-automorphism of L.

1.1 Definition. Let K ⊆ L be a field extension. The Galois group of the extension K ⊆ L is the set of K-automorphisms of L (where the group operation is the composition of maps), denoted by Gal(L/K) or G(L/K).

If f is a polynomial in K[X] and F is the splitting field of f over K, G(F/K) is called the Galois group of the polynomial f over K, denoted Gf /K, or Gf, if K is understood.

The transfer of properties from intermediate fields of the extension K ⊆ L to subgroups of the Galois group and vice versa is accom-plished by means of two natural maps, called the Galois Connections, which we now define.

We denote by IF(L/K) the set of intermediate fields of the extension K ⊆ L and by Subg(G(L/K)) the set of subgroups of the group G(L/K).

To any intermediate field E ∈ IF(L/K) we associate the subgroup G(L/E), i.e. the set of those σ ∈ G(L/K) for which σ|E = <id. Checking that G(L/E) is a subgroup in G(L/K) is trivial. The other way around, to any subgroup H of G(L/K) associate the fixed field of H, defined as x ∈ L | σ(x) = x, ∀σ ∈ H and denoted by LH. The proof of the fact that LH is an intermediate field of K ⊆ L is immediate2. Thus, we de-fine:

2 More generally, for any subset S of G(L/K), defining LS = x ∈ L | σ(x) = x,

∀σ ∈ S, one can easily verify that LS is a subfield of L containing K and that LS = L< S >, where < S > is the subgroup generated by S.

V. Galois Theory

234

Φ : IF(L/K) → Subg(G(L/K)), Φ(E) = G(L/E), ∀E ∈ IF(L/K);

Ψ : Subg(G(L/K)) → IF(L/K), Ψ(H) := LH, ∀H ∈ Subg(G(L/K)). A natural question arises: Under what conditions the maps

Φ and Ψ are bijective and inverse to each other? The answer is given, for algebraic extensions, by the Fundamental

Theorem of Galois Theory. First, let us review some simple but far-reaching properties of K-homomorphisms.

1.2 Proposition. Let K ⊆ L, K ⊆ E be field extensions and ϕ : L → E a K-homomorphism.

a) If L = E and K ⊆ L is finite, then ϕ is an automorphism. b) For any f ∈ K[X] and any root α of f in L, ϕ(α) is a root of f in

E. c) If S is system of generators of the extension K ⊆ L (that is,

L = K(S)), and σ : L → E is a K-homomorphism such that σ|S = ϕ|S, then σ = ϕ.

Proof. a) Viewing L as a K-vector space, ϕ is a K-vector space endomorphism of L. Since ϕ is injective (being a field homomorph-ism), we have dim Imϕ = dim L. So Imϕ = L.

b) Let f = a0 + a1X + + anX n. Then a0 + a1α + + anα

n = 0. Applying ϕ, it follows that

a0 + a1ϕ(α) + + anϕ(α) n = ϕ(0) = 0, since ϕ is a K-homomorphism and the ai's are in K. Thus, ϕ(α) is a root of f.

c) Any element of L is of the type xy −1, where y ≠ 0, and x, y ∈ K[S], so they are of the form

( )

1

1

1

1, ,

n

nn

n

ii'i i n

i i

a x x∈

∑ …… '

… , where

1 ni ia K∈… and ∑' indicates the fact that the sum has a finite number of

terms. For such an element, we have:

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235

( ) ( ) ( )∑∑ = nn

nn

in

iii

'in

iii

' xxaxxa ϕϕϕ …… ……1

11

1 11 =

( ) ( )∑ nn

in

iii

' xxa σσ ……1

1 1 = ( )∑ nn

in

iii

' xxa ……1

1 1σ ,

which shows that ϕ = σ. !

1.3 Corollary. Let K ⊆ L be a field extension, let σ ∈ G(L/K) and let f ∈ K[X]. Then:

a) σ permutes the roots of f in L. b) If L = K(S) (S generates L over K), then σ is determined by its ac-

tion on S. c) If x1,, xn are the distinct roots of f in an algebraic closure of K

and L = K(x1, , xn) is the splitting field of f over K, then G(L/K) = Gf is isomorphic to a subgroup of the symmetric group Sn.3

Proof. c) View Sn as the group of permutations of the set R := x1, , xn. The map ϕ : G(L/K) → Sn, σ & σ|R, ∀σ ∈ G(L/K), is a homomorphism (injective, by b)). !

These simple facts are fundamental in determining the Galois group of an extension. It is useful to introduce the following concept:

1.4 Definition. Let K ⊆ L be a field extension and let α, β elements of L, algebraic over K. We say that α and β are conjugate over K if they have the same minimal polynomial over K: Irr(α, K) = Irr(β, K). In other words, α and β are roots of the same irreducible polynomial f with coefficients in K. One also says β is a conjugate of α.

This definition agrees with the terminology conjugate numbers, used to designate the complex numbers α = a + ib and β = a − ib (where a, b ∈ R). Indeed, α and β have the same minimal polynomial

3 This is in fact the original view that Evariste Galois had on the notion of group

associated to a polynomial. We shall exploit this point of view later in order to obtain data on the Galois group of a polynomial.

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over R (find it!). The same remark can be made for the conjugates γ = a + b d and δ = a − b d , where a, b ∈ Q; γ and δ are conjugate over Q, in the sense of the above definition.

An algebraic element α over K can have at most n conjugates, where n is the degree of α over K (n = deg Irr(α, K)). The corollary 3.15.a) says that σ(α) is a conjugate of α, ∀σ ∈ G(L/K).

1.5 Corollary. Let K ⊆ K(α) be a simple field extension. Then its Galois group G has at most n elements, where n = [K(α) : K] is the de-gree of α.

Proof. A K-automorphism σ ∈ G is determined by its value in α. Since σ(α) is a conjugate of α, there are at most n possibilities to choose σ(α) from. !

Thus, the elements of the Galois group can be found by inspecting their action on a generating set. Moreover, one must remember that they transport any element in one of its conjugates. The following examples illustrate this technique.

1.6 Examples. a) Consider the extension R ⊆ C and let G = Gal(C/R). Because C = R[i], where i = 1− , it is sufficient to look for the action of the automorphisms in G on i. Because the mini-mal polynomial of i is X 2 + 1, σ(i) ∈ i, −i, ∀σ ∈ G. If σ (i) = i, then σ = id; if σ (i) = −i, then σ is the complex conjugation: σ (a + bi) = a + bσ (i) = a − bi, ∀a, b ∈ R. So, G(C/R) consists of two elements: the identity map and the conjugation map.

b) Let us find G = ( )( )QQ 3 2G . For any σ ∈ G, ( )3 2σ is a root of X 3 − 2; but 3 2 is the only root of X 3 − 2 in Q 3 2 , so ( )3 2σ = 3 2 . Since 3 2 generates the extension, σ = id. So, G = id, but Q ( Q 3 2 .

c) Let K = F2(X) (the rational function field with coefficients in F2) and let F := ( )2

2 XF , the subfield generated by F2 and X 2. Who is

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237

G(K/F)? Obviously, K = F(X), where X is algebraic over F as a root of h = Y 2 − X 2 ∈ F[Y]. This is even the minimal polynomial of X over F: otherwise, the minimal polynomial would be of degree 1, which means that X ∈ F. But X ∉ F: indeed, if X = f /g, with f, g ∈ F2 [X 2], then Xg = f. In this equality of polynomials in X, the left hand side has odd degree and the right hand side has even degree, contradiction. Note that h can be written (in K[Y]) h = (Y − X)2, so X is a double root of h.

Take now σ ∈ G(K/F); then σ (X) can only be X (the unique root of h in K). The fact that X generates K over F implies σ = id. So G(K/F) = id, but F ( K.

In the cases b) and c) above a bijective correspondence between the subgroups of the Galois group and the intermediate fields cannot ex-ist: the Galois group is trivial (thus it has only one subgroup), but the extensions are not trivial (they have at least two intermediate fields).

Let us look closer at the reasons behind this phenomenon. Note that the given extensions are simple (each is generated by a single element α), so any automorphism σ in the Galois group is perfectly known by its action on α ; moreover, σ (α) must be a conjugate of α. But in both cases α has no conjugates in the extension: in b), the conjugates exist, but they are in a larger extension (for instance, in the splitting field of X 3 − 2 over Q); in c), α has no conjugates distinct from itself because its minimal polynomial has only one root (which is a double root).

Thus, it appears natural to consider the following two properties that an algebraic extension K ⊆ L may have:

Normality: For any α ∈ L, Irr(α, K) has all its roots in L. Separability: For any α ∈ L, Irr(α, K) has no multiple roots.

It turns out that for finite extensions, these are necessary and suffi-cient conditions for the bijectivity of the Galois connections. The

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following two sections are devoted to the study of the conditions above.

V.2 Normal extensions

2.1 Definition. A field extension K ⊆ L is called a normal exten-sion if it is algebraic and for any α ∈ L, Irr(α, K) has all its roots in L. In this case, one says L is normal over K.

Remark that the algebraic extension K ⊆ L is normal iff any irreducible polynomial in K[X], which has a root in L, has all its roots in L. An easy example of normal extension is the extension K ⊆ Ω, where Ω is an algebraic closure of K.

In concrete situations, checking the above definition is rarely practical (for any α, one must find Irr(α, K) and its roots). Before giv-ing other examples, let us give some necessary and sufficient condi-tions for an algebraic extension to be normal.

2.2 Proposition. Let K ⊆ L be an algebraic extension and let Ω be an algebraic closure of L. The following statements are equivalent:

a) K ⊆ L is normal. b) For any K-homomorphism ϕ : Ω → Ω, we have ϕ(L) ⊆ L. c) For any K-homomorphism ϕ : Ω → Ω, we have ϕ(L) = L. d) There exists a family of polynomials F ⊆ K[X] such that L is the

splitting field of F over K. Proof. a)⇒b) Let ϕ : Ω → Ω be a K-homomorphism and take

α ∈ L. If f is Irr(α, K), then ϕ(α) is a root of f, so α ∈ L.


239

b)⇒a) Fix an α ∈ L and denote by f its minimal polynomial over K. All roots of f lie in Ω; let β ∈ Ω be another root of f. There exists a K-isomorphism σ : K(α) → K(β) which takes α to β. Extend σ (see theorem IV.2.19) to a K-homomorphism τ : Ω → Ω. Because τ(L) ⊆ L and τ(α) = σ(α) = β, we obtain β ∈ L. So, all roots of f are in L.

a)⇒c) By b), we already know that ϕ(L) ⊆ L. It remains to show that ϕ is surjective. Take α ∈ L, let f = Irr(α, K) and let S be the set of all roots of f in Ω. We saw that S ⊆ L and that ϕ(S) ⊆ S. But ϕ is injec-tive (it is a field homomorphism) and S is finite, so ϕ|S : S → S is a bijection. Thus there is a β ∈ S with ϕ(β) = α.

a)⇒d) Let F = Irr(x, K) | x ∈ L. Because K ⊆ L is normal, the roots of any polynomial in F are in L, so the splitting field of F over K is included in L. The fact that L is included in the splitting field of F over K is obvious.

d)⇒b) Let S be the set of the roots in Ω of the polynomials in F, S = α ∈ Ω | ∃f ∈ F such that f(α) = 0. By hypothesis, L = K(S). If ϕ : Ω → Ω is a K-homomorphism, then ϕ(S) ⊆ S (by 1.2.b)), so ϕ(L) ⊆ L. !

2.3 Examples. a) The extension ( ) QQ 3 2 is not normal, because Irr( 3 2 , Q) = X 3 − 2 has also complex non real roots.

b) If we want a normal extension of Q that contains 3 2 , we must

adjoin the other roots of X 3 − 2. We obtain the extension ( ) QQ ω,23 , where ω ∈ C is a root of the polynomial X 2 + X + 1. In-

deed, the roots of X 3 − 2 are 2333 2,2,2 ωω and so ( )ω,23Q = ( )2333 2,2,2 ωωQ is the splitting field of X 3 − 2 over Q. c) Any extension K ⊆ L of degree 2 is normal. Indeed, pick

α ∈ L \ K and let f = Irr(α, K). Since f has degree 2 (because K(α) = L) and f is divisible by X − α in L[X], f decomposes in linear factors in L[X].

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240

In order to prove that a given finite extension K ⊆ L is normal, the most frequent approach is to prove that L is a splitting field for some polynomial in K[X].

2.4 Remark. If K ⊆ L is normal, Ω is an algebraic closure of L and ϕ : Ω → Ω is a K-homomorphism, then ϕ|L is a K-automorphism of L (an element of G(L/K)). Since Ω is a normal extension of K, every K-homomorphism ϕ : Ω → Ω is an automorphism. So, in the characterizations above, K-homomorphism can be replaced by K-automorphism.

2.5 Corollary. Let K ⊆ L be a finite extension. Then the extension K ⊆ L is normal iff there exists f ∈ K[X] such that L is the splitting field of f over K.

Proof. Suppose K ⊆ L is normal; let S = x1, , xn ⊆ L be such that L = K(S). If we take f = Irr(x1, K)⋅ ⋅ Irr(xn, K), then L is the split-ting field of f over K. The converse is already proven. !

2.6 Definition. If K ⊆ L is algebraic, Ω is an algebraic closure of L and σ is a K-automorphism of Ω, then σ(L) is called a conjugate extension of L over K in Ω.

The situation in Examples a), b) above can be generalized: if K ⊆ L is not normal, by adjoining to L the roots of the minimal polynomials of a set of generators of L over K, one gets a normal extension.

2.7 Proposition. Let K be a field, Ω an algebraic closure of K and L an extension of K, L ⊆ Ω. Then there exists a unique normal exten-sion N of K such that L ⊆ N ⊆ Ω, which is the smallest with this prop-erty. More precisely, for any normal extension K ⊆ F with L ⊆ F ⊆ Ω, we have N ⊆ F. Also, N is the composite of the conjugates of L over K in Ω:

N = K(∪σ(L) | σ ∈ AutK(Ω))


241

Moreover, if K ⊆ L is finite, then K ⊆ N is finite. Proof. Let S ⊆ L such that L = K(S). We consider I = Irr(x, K) |

x ∈ S; take N to be the splitting field of the family I over K. It is clear that N includes L and N is normal over K. If L is finite over K, we can take S finite, so I is also finite. So, N is a finite extension of K.

If F is a normal extension of K with L ⊆ F ⊆ Ω, then S ⊆ F, so the splitting field of Irr(x, K) over K is included in F, for any x ∈ S. Thus, the splitting field of I over K (that is, N) is included in F.

The uniqueness of N results from the minimal condition proved above: if M is another extension with the same properties, M ⊆ N and N ⊆ M, so M = N.

Let M be the composite of the conjugates of L over K in Ω. If τ is a K-automorphism of Ω, then

σ(L) | σ ∈ AutK(Ω) = τσ(L) | σ ∈ AutK(Ω) So, τ(M) = K(τσ(L) | σ ∈ AutK(Ω)) = K(σ(L) | σ ∈ AutK(Ω)) = M.

This shows that M is a normal extension of K. Obviously, L ⊆ M. On the other hand, any normal extension E of K, that includes L, has the property that σ(E) = E, ∀σ ∈ AutK(Ω), so σ(L) ⊆ σ(E) = E, which means that M ⊆ E. So, M satisfies the same conditions as N. From the uniqueness of N it follows that M = N. !

The extension N constructed above is the smallest (in the sense of inclusion) normal extension of K that includes L and is called the nor-mal closure (in Ω) of the extension L/K (or, of L over K). Since N is the splitting field over K of a family of polynomials, it follows that the normal closure does not depend (up to a K-isomorphism) on the alge-braic closure Ω that we choose.

Normal extensions are not transitive: ( )2⊆, , and ( )2Q ⊆ ( )4 2Q are normal, being of degree 2, but Q ⊆ ( )4 2Q is not

normal (why?). We have however the following:

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242

2.8 Proposition. Let K ⊆ L be a normal extension and E an intermediate field. Then E ⊆ L is normal.

Proof. Let F be a family of polynomials over K such that L is the splitting field of F over K. Then L is the splitting field of F (viewed as included in E[X]) over E. !

2.9 Proposition. Let K ⊆ L be an extension and E, F algebraic extensions of K, included in L. Then:

a) If K ⊆ E is normal, then F ⊆ FE is normal. b) If K ⊆ E and K ⊆ F are normal, then K ⊆ EF and K ⊆ E ∩ F are

normal. Proof. a) E is a splitting field over K for some family I ⊆ K[X], so

E = K(S), where S is the set of all roots in Ω of the polynomials in I. Then FE = F(K(S)) = F(S), so FE is a splitting field over F for I (viewed as a subset of F[X]).

b) Let Ω be an algebraic closure of L and ϕ : Ω → Ω be K-automorphism. Then ϕ(E) ⊆ E, ϕ(F) ⊆ F (as normal extensions), so ϕ(EF) ⊆ EF and ϕ(E ∩ F) ⊆ E ∩ F. So, K ⊆ EF and K ⊆ E ∩ F are normal. !

One can see that the results on normal extensions can be proven us-ing either one of the characterizations given at 2.2 (using the automor-phisms of the algebraic closure or using splitting fields). We propose to the reader to give alternate proofs for the results on normal exten-sions.

Exercises

In the exercises, K is a field and Ω is an algebraic closure of K.


243

1. Let x, y ∈ Ω. Prove that: x and y are conjugate over K ⇔ there ex-ists a K-homomorphism ϕ : Ω → Ω such that ϕ(x) = y ⇔ for any nor-mal extension K ⊆ L such that L ⊆ Ω and x, y ∈ L, there exists a K-homomorphism ϕ : L → L such that ϕ(x) = y. 2. Let K ⊆ L be a normal extension, K ⊆ E ⊆ L an intermediate field and F an algebraically closed field. Then any K-homomorphism ϕ : E → F can be extended to a K-homomorphism ψ : L → F. 3. Let x ∈ Ω and p, q ∈ K[X].

a) If q(x) ≠ 0 then q(x') ≠ 0, for any conjugate x' of x over K. b) Suppose q(x) ≠ 0. Let g ∈ K[X] such that g(p(x)/q(x)) = 0. Then,

for any conjugate x' of x over K, g(p(x')/q(x')) = 0. 4. Let L be the splitting field over Q of the polynomial X 4 − 9. Find the degree and a basis of the extension Q ⊆ L. 5. Find the normal closures for:

Q ⊆ Q( 3 ), Q ⊆ Q( 3 5 ), Q ⊆ Q( 3 5,3 ), Q ⊆ Q( 4 2 ). 6. For any α ∈ Ω, denote by Cα the splitting field over K of the polynomial Irr(α, K). Prove that ∀α, β ∈ Ω, K(α) ⊆ K(β) ⇒ Cα ⊆ Cβ and that K(α) = K(β) ⇒ Cα = Cβ. Give an example to show that Cα = Cβ does not necessarily imply K(α) = K(β). 7. Let L be an extension of K, included in Ω. Show that F := x ∈ Ω | L contains all roots of Irr(x, K) is a subfield of L, and F is a normal extension of K. If K ⊆ E ⊆ L is such that K ⊆ E is normal, then E ⊆ F. 8. Let K ⊆ L be a normal extension and f ∈ K[X] a irreducible monic polynomial. If g, h ∈ L[X] are irreducible divisors of f (in L[X]), then there exists ϕ ∈ AutK(L) such that h = ( )gϕ (where ϕ : L[X] → L[X] is the unique K-algebra homomorphism that extends ϕ and takes X to X). 9. Give an example showing that the assumption that K ⊆ L is normal cannot be omitted in the previous statement.

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10. Show that any extension of finite fields is normal. 11. Give examples of normal extensions of Q of degrees: 4, 6, and 8. 12. For any n ∈ N*, give an example of a normal extension of degree n. 13. Let L, E be two distinct extensions of degree 3 of K, included in Ω. Then [LE : K] ∈ 6, 9. We have [LE : K] = 6 ⇔ L and E are conju-gate over K.

V.3 Separability

In this section, we fix a field K. It is convenient to suppose (without any loss of generality) that all algebraic extensions of the field K that we consider are included in Ω, a fixed algebraic closure of K.

3.1 Definition. Let K ⊆ L be a field extension. a) An element α of L is called separable 4 over K if it is algebraic

over L and Irr(α, K) has no multiple roots (in Ω). b) An irreducible polynomial f ∈ K[X] is called a separable polyno-

mial if f has no multiple roots (in Ω). c) An arbitrary polynomial in K[X] is called separable if all its

irreducible factors are separable. Note that f ∈ K[X] is separable if and only if every root of f (in Ω) is separable over K.

d) An element of L is called inseparable over K if it is algebraic and not separable over K.

4 This terminology, introduced by B. L. van der Waerden, expresses the idea that

the roots of f are "separated" (distinct).

V.3 Separability

245

e) An irreducible polynomial f ∈ K[X] is called an inseparable polynomial if it has repeated roots

f) The extension K ⊆ L is called a separable extension if it is alge-braic and every element of L is separable over K (one also says L is separable over K).

g) If K ⊆ L is algebraic and not separable (there exists at least one element of L, inseparable over K), K ⊆ L is called an inseparable extension (or we say that L is inseparable over K).

Note that the irreducible polynomial f ∈ K[X] is separable iff the number of distinct roots of f in Ω equals its degree.

3.2 Example. a) The polynomial X 2 + 1 ∈ Q[X] is separable over Q; its roots i and − i are thus separable over Q.

The following simple result will be often used. We remark that the converse is also true (to be proven in 3.14).

3.3 Proposition. Let K ⊆ L ⊆ M be algebraic extensions. If K ⊆ M is separable, then K ⊆ L and L ⊆ M are separable.

Proof. Clearly, K ⊆ L is separable. Let us show that any α in M is separable over L. We know that Irr(α, K) has no multiple roots. On the other hand, Irr(α, L) divides Irr(α, K) in L[X], so Irr(α, L) cannot have multiple roots. !

The next result characterizes the irreducible separable polynomi-als. The characterization of the polynomials having repeated roots us-ing the formal derivative (IV.2.6) is essential in the proof.

3.4 Proposition. Let f = a0 + a1 X + + an X n, with an ≠ 0, be an irreducible polynomial with coefficients in K.

a) f has repeated roots (is inseparable) iff f ' = 0 ( f' is the formal derivative of f ).

b) If char K = 0, then f is separable (f has no repeated roots).

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246

c) If char K = p > 0, then f is inseparable iff f ∈ K[X p]. Further-more, there exists e ∈ N* and an irreducible separable polynomial

g ∈ K[X], such that f = ( )epg X .

Proof. a) Suppose f is inseparable. Let d = GCD f, f '. If f ' ≠ 0, d is a divisor of f ' and deg d ≤ deg f ' < deg f. Since f is irreducible, we must have d = 1, so f has no repeated roots. Conversely, if f ' = 0, then d = f, so deg d ≥ 1 and f has multiple roots.

b) If char K ≠ 0, f ' = a1 + + nan X n − 1≠ 0 because nan ≠ 0. c) Let char K = p > 0 and f ' = 0. The coefficients of f ' are

iai = (i·1)·ai, i ∈ 1, , n. Since iai = 0, we have i·1 = 0 or ai = 0. If p - i, then i·1 ≠ 0, so ai = 0. Thus ai = 0, ∀i ∈ 1, , n with p - i. In other words, f ∈ K[X p].

Let e be the greatest integer such that f ∈ K[ epX ]. Let g ∈ K[X] with f = ( )epXg ; then g is irreducible (otherwise f would be reducible!)

and g ∉ K[X p] (if g ∈ K[X p], then f ∈ K[ 1+epX ], contradicting the maximality of e). So, g is separable. !

The fields that pose no problems of separability are called perfect:

3.5 Definition. A field K is called a perfect field if any algebraic extension of K is separable.

Rephrasing, a field K is perfect iff any irreducible polynomial in K[X] is separable. Proposition 3.4.b) says that any field of characteris-tic 0 is perfect. So, the study of separability is meaningful only in characteristic p > 0.

3.6 Proposition. Every algebraic extension of a perfect field is a perfect field.

Proof. Suppose K ⊆ L is algebraic and K is perfect. If α is an alge-braic element over L, then α is algebraic over K (by transitivity) and

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247

Irr(α, L) divides Irr(α, K) in L[X]. Irr(α, K) has only simple roots, so Irr(α, L) has the same property. Therefore, α is separable over L. !

3.7 Proposition. The field K is perfect iff the Frobenius endomorphism is an automorphism (if char K = p > 0, this means Kp = K, where K p = x p | x ∈ K).

Proof. If char K = 0, the Frobenius is the identity map and all is evident. Let char K = p > 0. Suppose that the Frobenius endomorph-ism (ϕ : K → K, ϕ(x) = x p, ∀x ∈ K) is an automorphism. If, on the contrary, an irreducible inseparable polynomial f ∈ K[X] would exist, then f ∈ K[X p]. Let

f = ∑=

n

i

pii Xa

0, ai ∈ K.

Since ϕ is surjective, there exist bi ∈ K such that ai = pib , ∀i ∈ 0,

, n, so

f = pn

i

ii

n

i

pipi XbXb ⎟

⎠⎞

⎜⎝⎛= ∑∑

== 00, bi ∈ K.

(we used the fact that g & g p is an endomorphism of the ring K[X], of characteristic p). But this shows that f is reducible!

Conversely, let K be a perfect field. It will be enough to show that, if ϕ is not surjective, then there exists an inseparable algebraic ele-ment over K. Let a ∈ K \ K p. Consider the polynomial f = X p − a ∈ K[X] and a root α of f in Ω. We have α ∉ K (because other-wise a ∈ K p), so deg Irr(α, K) > 1. If β is another root of f, then β p = α p = a. The injectivity of the Frobenius endomorphism of Ω im-plies that β = α. So, f has only one root, with multiplicity p. The mini-mal polynomial of α over K divides f and has the root α, with multiplicity equal to its degree, so α is inseparable over K. !

3.8 Corollary. The finite fields, the fields of characteristic 0 and the algebraically closed fields are perfect. !

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248

3.9 Example. The corollary above shows that, in order to find a field that is not perfect, we must look among infinite fields of characteristic p > 0. A natural example of such a field is Fp(X) =: K. The choice is correct. Indeed, consider h = Y p − X ∈ K[Y]. We have that: h is irreducible (by Eisenstein's criterion applied for the prime element X ∈ Fp[X]); h is not separable, because h ∈ K[Y p]. Denoting by α a root of h, the extension K ⊆ K(α) is inseparable.

A K-homomorphism of extensions of K carries a root of a polyno-mial f ∈ K[X] in another root of f. This leads to the idea of using K-homomorphisms as a tool in the study of separability.

3.10 Definition. Let K ⊆ L be an algebraic extension. The cardinal of the set HomK(L, Ω) (the K-homomorphisms defined on L with val-ues in Ω) is denoted by [L : K]s and is called the separable degree of the extension K ⊆ L. In other words, [L : K]s is the cardinal of the set

σ : L → Ω | σ homomorphism, σ|K = ι, where ι : K → Ω is the canonic inclusion. For simple extensions, an-other way to characterize the separable degree is given at c) below.

3.11 Proposition. a) Let K ⊆ L be an algebraic extension, Ω' an-other algebraic closure of L and ϕ : K → Ω' a field homomorphism5. Then HomK(L, Ω) is in a bijective correspondence with the set of homomorphisms from L to Ω' that extend ϕ,

P(L/K, ϕ, Ω') := η : L → Ω' | η|K = ϕ, η homomorphism. Consequently, |P(L/K, ϕ, Ω')| = |HomK(L, Ω)| = [L : K]s does not

depend on ϕ and Ω'. b) The separable degree is multiplicative: if K ⊆ L ⊆ M are alge-

braic extensions, then [M : K]s = [M : L]s [L : K]s.

5 Recall that Ω is a fixed algebraic closure of K that contains L.

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249

c) If L = K(α) is a simple algebraic extension of K, then [K(α) : K]s is equal to the number of conjugates of α in Ω.

d) If α is algebraic over K, then [K(α) : K]s ≤ [K(α) : K]. Proof. a) We must find a bijection between P(L/K, ϕ, Ω') and

P(L/K, ι, Ω), where ι : K → Ω is the canonic inclusion. There exists an isomorphism ψ : Ω → Ω' which extends ϕ (by IV.2.19). To every σ ∈ P(L/K, ϕ, Ω') we associate ψ −1σ, which belongs to P(L/K, ι, Ω): if x ∈ K, then ψ −1(σ(x)) = ψ −1(ϕ(x)) = x = ι(x). The other way round, to every η ∈ P(L/K, ι, Ω) we associate ψ η ∈ P(L/K, ϕ, Ω'). It is easy to see that these maps are inverse one to each other.

b) For the sake of simplifying the notation, denote P(L/K, ϕ, Ω) by P(L/K, ϕ). Thus, [M : K]s = |P(M/K, ι)|. For any η ∈ P(M/K, ι), η|L ∈ P(L/K, ι); this shows that P(M/K, ι) is the union of the sets P(M/L, σ) when σ runs over P(L/K, ι). These sets are mutually disjoint and |P(M/L, σ)| = [M : L]s, ∀σ ∈ P(L/K, ι), as shown before. There-fore, we can write:

[M : K]s = |P(M/K, ι)| = |P(L/K, ι)|⋅[M : L]s = [L : K]s⋅[M : L]s. c) Let f = Irr(α, K) and let R be the set of all conjugates of α (the

roots of f in Ω). If ϕ : K(α) → Ω is a K-homomorphism, then ϕ(α) is a root of f; on the other hand, any K-homomorphism defined on K(α) is uniquely determined by its action on α. In other words, the map v : HomK(K(α), Ω) → R, v(ϕ) = ϕ(α) is injective. This map is also surjective: for any β ∈ R, there exists a K-isomorphism between K(α) and K(β).

d) Irr(α, K) has at most deg Irr(α, K) roots; but deg Irr(α, K) = [K(α) : K]. The previous part yields the desired inequality. !

3.12 Proposition. If L = K(α) is a simple algebraic extension of K, then the following conditions are equivalent:

a) [K(α) : K]s = [K(α) : K]. b) α is separable over K.

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c) K(α) is a separable extension of K. Proof. a) ⇔ b) is clear, if we look at d) above, and c) ⇒ b) is obvi-

ous. a)⇒c) Let β ∈ K(α). We must show that β is separable over K,

which amounts to [K(β) : K]s = [K(β) : K]. The multiplicativity of the separable degree shows that [K(α) : K]s = [K(α) : K(β)]s[K(β) : K]s. (*)

α is separable over K, so it is also over K(β), which means that [K(α) : K(β)] = [K(α) : K(β)]s and [K(α) : K]s = [K(α) : K]. Rewriting (*), we get

[K(α) : K] = [K(α) : K(β)][K(β) : K]s. But [K(α) : K] = [K(α) : K(β)][K(β) : K]; comparing the last two

relations, it follows that [K(β) : K]s = [K(β) : K]. !

The characterization of separable simple extensions using the separable degree, given above, holds in the general case of finite extensions:

3.13 Proposition. Let K ⊆ L be a finite extension. Then: a) [L : K]s ≤ [L : K]. b) K ⊆ L is separable iff [L : K]s = [L : K]. Proof. a) If K ⊆ L is simple, the inequality it is already proven.

Suppose that the inequality holds for any finite extension that has a set of generators with at most n − 1 elements and let us prove it for the extension K ⊆ K(x1, , xn). We have:

[K(x1) : K]s ≤ [K(x1) : K] and [K(x1)(x2, , xn) : K(x1)]s ≤ [K(x1)(x2, , xn) : K(x1)].

So: [K(x1)(x2, , xn) : K(x1)]s[K(x1) : K]s ≤ [K(x1)(x2, , xn) : K(x1)][K(x1) : K].

Since the (separable) degree is multiplicative, we get the conclu-sion.

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b) Suppose K ⊆ L is separable. We prove by induction on n the next Proposition: For any n ∈ N*, and any separable extension K ⊆ L of degree n, [L : K]s = [L : K]. The case n = 1 is trivial. If n > 1, let α ∈ L \ K. If L = K(α), α is separable and the previous proposition ap-plies. If Κ(α) ≠ L, then the extensions K ⊆ K(α) and K(α) ⊆ L are separable, with degrees strictly smaller than [L : K] = n. By induction, we get [L : K(α)] = [L : K(α)]s and [K(α) : K] = [K(α) : K]s. Multiply-ing these equalities gets us to the conclusion.

Suppose now [L : K]s = [L : K] and yet there exists α ∈ L, insepara-ble over K. Then [K(α) : K]s < [K(α) : K]. Since [L : K(α)]s ≤ [L : K(α)], multiplying these relations yields [L : K]s < [L : K], contradiction. !

3.14 Theorem (transitivity of separable extensions) Let K ⊆ L ⊆ M be algebraic extensions. If K ⊆ L and L ⊆ M are separable, then K ⊆ M is separable.

Proof. We shall use the characterization of separability with the separable degree in the finite case. Take α ∈ M and let us show that α is separable over K. Let a0, , an ∈ L be the coefficients of Irr(α, L) and let L' = K(a0, , an). Evidently, Irr(α, L) = Irr(α, L'), so α is separable over L'. Thus [L'(α) : L']s = [L'(α) : L']. The finite extension K ⊆ L' is separable, because K ⊆ L is separable; so [L' : K]s = [L' : K]. The multiplication of the last two equalities yields [L'(α) : K]s = [L'(α) : K], which means that K ⊆ L'(α) is separable. So, α is separable over K. !

Notice the resemblance with the proof of the transitivity of alge-braic extensions.

3.15 Corollary. Let K ⊆ L be any extension and let A be a set of elements in L, separable over K. Then K ⊆ K(A) is a separable exten-sion.

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Proof. Every element in K(A) is a polynomial expression with coefficients in K in a finite set of elements of A. Thus, we can suppose that A is finite. Suppose A = x1, x2. Since x1 is separable over K, K ⊆ K(x1) is separable (by 3.12). As x2 is separable over K, it is separa-ble over K(x1), so K(x1) ⊆ K(x1)(x2) is separable. By transitivity, K ⊆ K(x1, x2) is separable. The case A = x1, x2, , xn, n > 2, reduces to the already proven case, by induction. !

A remarkable fact, very important in Galois Theory, is that any fi-nite separable extension is simple. The following lemma is the essen-tial step in proving this.

3.16 Lemma. Let K be an infinite field, L an extension of K and α, β ∈ L, algebraic over K, where β is separable over K. Then K ⊆ K(α, β) is a simple extension (it has a primitive element).

Proof. Let K(α, β) = E, f = Irr(α, K), m = deg f, g = Irr(β, K), n = deg g, α = α1, , αm the roots of f in Ω, β = β1, , βn the roots of g in Ω. Since β is separable, β1, , βn are distinct. We claim that there exists c ∈ K* such that: ∀i ∈ 1, , m, ∀j ∈ 1, , n, α + cβ = αi + cβj ⇔ i = 1 and j = 1.(i)

Indeed, the condition on c amounts to c ∉ (αi − α)(β − βj)−1 |

1 ≤ i ≤ m, 2 ≤ j ≤ n; since K is infinite, and the set above is finite, there exists c ∈ K satisfying (i). With c chosen this way, let γ = α + cβ. We show that γ is a primitive element, i.e. K(α, β) = K(γ). Because K(γ) ⊆ K(α, β) is evident, it is enough to show that β ∈ K(γ) (this will imply also α = γ − cβ ∈ K(γ)).

Let h(X) = f (γ − cX), a polynomial with coefficients in K(γ). The

idea is to show that the GCD of h and g is X − β ; the fact that d = X − β ∈ K(γ)[X] implies then β ∈ K(γ).

We have h(β) = f (γ − cβ) = f (α) = 0, which means that X − β di-vides h (and g, too) in Ω[X]. The polynomials h and g have no other

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common roots in Ω but β: if h(βj) = f (γ − cβj) = 0, then γ − cβj is a root of f, which means it is among α1, , αm. But then condition (i) en-sures that j = 1, so βj = β1 = β. Thus, in Ω[X], GCD(h, g) = X − β.

But h, g ∈ K(γ)[X], so GCD(h, g) = X − β ∈ K(γ)[X], that is, β ∈ K(γ). !

3.17 Theorem. (Primitive element theorem) Any separable finite extension K ⊆ L is simple (it has a primitive element).

Proof. If K is a finite field, then L is also finite. The multiplicative group L* is cyclic, and any generator of L* can be taken as a primitive element.

Suppose now that K is infinite. If there exist x1, x2 ∈ L such that L = K(x1, x2), the preceding lemma applies. The general case follows by an easy induction argument on n ∈ N* with the property that exist x1, , xn ∈ L such that K(x1,, xn) = L. !

3.18 Remark. For the extensions of the type K ⊆ K(α, β), with K infinite, lemma 3.16 gives also a practical procedure to find a primi-tive element (or, at least, a class of good candidates for it, if the condi-tion (i) is hard to verify).

The remaining results (until the end of the section) are a somewhat deeper study of (in)separability. The reader interested primarily in the Fundamental Theorem of Galois theory may skip directly to the next section.

Recall that every algebraic extension of a field of characteristic 0 is separable; thus, all that follows is relevant only in characteristic p > 0.

3.19 Definition. If K ⊆ L is a field extension, the separable closure of K in L is the set Ks

L := x ∈ L | x separable over K. By 3.15, the separable closure of K in L is a subfield of L and a separable exten-sion of K. Clearly, Ks

L is the largest separable extension of K in-

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cluded in L, (KsL includes any separable extension K ⊆ E with E ⊆ L).

The separable closure of K in Ω (an algebraic closure of K) is called the separable closure of K, denoted Ks. Notice that Ks is the splitting field over K of the family of all separable polynomials in K[X] (thus Ks is unique up to a K-isomorphism).

The following concept is, in a certain sense, the opposite of separability:

3.20 Definition. An algebraic element α of the extension K ⊆ L is called purely inseparable over K if Irr(α, K) has only one root in Ω (α itself, which has order of multiplicity deg Irr(α, K)). In other words, Irr(α, K) is of the form (X − α)n. Evidently, any purely inseparable ele-ment is inseparable. An extension K ⊆ L is called purely inseparable if any element in L is purely inseparable over K.

Directly from the definitions, it follows that: an element is simultaneously purely inseparable and separable over K iff it belongs to K.

The following result gives a clearer picture of purely inseparable elements over a field K with char K = p: roughly put, they are roots of order pe of some element in K.

3.21 Proposition. Let char K = p > 0 and let α ∈ Ω. The following statements are equivalent:

a) α is purely inseparable over K. b) The only conjugate of α in Ω is α. c) There exists e ∈ N such that

epα ∈ K. d) The minimal polynomial of α over K is of the form

( )eee ppp XX αα −=− .

e) The separable degree of the extension K ⊆ K(α) is 1. If α is purely inseparable over K, then Irr(α, K) = ( )

epX α− , where e is minimal with

epα ∈ K.

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Proof. a)⇔b) b) just rephrase the definition of pure inseparability. a)⇒c) and a)⇒d). Let f := Irr(α, K). Since α is inseparable over K,

f ∈ K[X p]. Let n ∈ N* be such that f = (X − α)n. Then there exist e, m ∈ N with n = pem and GCD(p, m) = 1. We have

f = ( )( ) ( )mppmp eee

XX αα −=− ,

so the coefficient of ( ) 1−mpeX is ( )

epm α1⋅ . Since m·1 ≠ 0 in K

(because p - m), we get epα ∈ K. The polynomial f is irreducible in

K[X], so m = 1; thus f = ( )epX α− . If

dpα ∈ K, for some d ∈ N, then

g := ( ) ddd ppp XX αα −=− ∈ K[X] and g(α) = 0. So, g is divisible by f = Irr(α, K). Then deg g ≥ deg f, that is, d ≥ e. c)⇒d), d)⇒ a) Evident. b)⇔e) It follows from 3.11.c): [K(α) : K]s is the number of conju-

gates of α in Ω. !

3.22 Example. There exist algebraic elements that are inseparable, but not purely inseparable. Take K = F2(X) and g = Y 6 − X ∈ K[Y], irreducible (by Eisenstein: X is prime in the UFD F2[X]). A root α of g is inseparable over K, because g' = 0. But α is not purely inseparable over K, since g = Y 6 − α 6 and 6 is not a power of 2 = char K.

3.23 Definition. Let K ⊆ L be a field extension. The set Ki

L := x ∈ L | x is purely inseparable over K is called the purely inseparable closure of K in L.

3.24 Proposition. Let K ⊆ L be an algebraic extension, char K = p > 0.

a) K ⊆ L is purely inseparable iff [L : K]s = 1. In this case, K ⊆ L is normal and Gal(L/K) = id.

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b) Let K ⊆ L ⊆ M be a tower of extensions. Then: K ⊆ M is purely inseparable iff K ⊆ L and L ⊆ M are purely inseparable.

c) If K ⊆ L is purely inseparable and finite, then [L : K] is a power of p.

d) The purely inseparable closure KiL of K in L is a subfield of L. If

E is a purely inseparable extension of K, E ⊆ L, then E ⊆ KiL. More-

over, KiL ∩ K

sL = K.

Proof. a) Let K ⊆ L be purely inseparable. For any α ∈ L, Irr(α, K) has all its roots (namely α itself) in L, so K ⊆ L is normal. If σ : L → Ω is a K-homomorphism, then σ(α) is a root of Irr(α, K), so σ(α) = α, ∀α ∈ L. Thus, the canonic inclusion is the only element in HomK(L, Ω), so [L : K]s = 1 and G(L/K) = id. Conversely, if [L : K]s = 1 and α ∈ L, then [K(α) : K]s ≤ [L : K]s = 1, so α is purely inseparable over K.

b) Let K ⊆ M be purely inseparable. Then, clearly, K ⊆ L is purely inseparable. Any α ∈ M has the property that

epα belongs to K (and to L) for some e, that is, α is purely inseparable over L. Conversely, let K ⊆ L and L ⊆ M be purely inseparable and let α ∈ M. Then there ex-ists e ∈ N with β =

epα ∈ L. Since β is purely inseparable over K, dpβ ∈ K for some d. So

dep +α ∈ K.

c) If L = K(α) is a simple extension, then [L : K] = deg Irr(α, K), which is a power of p. In the general case, L = K(α1,, αn) for some finite set of purely inseparable elements α1, , αn in L. Let Ki = K(α1, , αi), ∀i ≤ n; we have a tower of simple extensions (which are purely inseparable, by b)) K = K0 ⊆ K1 ⊆ ⊆ Kn = L. For every i, the degree of Ki ⊆ Ki + 1 is a power of p, so [L : K] is still a power of p, be-ing the product of these degrees.

d) Let α, β be purely inseparable over K, with β ≠ 0. We must show that α − β, αβ, β −1 are purely inseparable. Let e, d ∈ N with

epα ∈ K,dpβ ∈ K. Then:

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257

( ) ( ) ( ) ed

dede ppppp βαβα −=−

+

∈ K. A similar argument works for αβ and β −1. Since Ki

L contains all purely inseparable elements over K in L, Ki

L includes any purely inseparable extension K ⊆ E with E ⊆ L. Finally, if α is separable and purely inseparable over K, then α ∈ K, so Ks

L ∩ KiL = K. !

3.25 Proposition. Let K ⊆ L be an algebraic extension. Then: a) Ks

L ⊆ L is purely inseparable. b) [L : K]s = [Ks

L : K]. c) Ki

L ⊆ L is separable iff L = KsL Ki

L.

Proof. a) Let α ∈ L and let g ∈ K[X] be the irreducible separable polynomial such that Irr(α, K)(X) = g

epX (see 3.4.c)). Let a =epα . It

follows that g(a) = 0, so g = Irr(a, K); thus, a is separable over K. So epα = a ∈ Ks

L, that is, α is purely inseparable over KsL.

b) By 3.11.b), [L : K]s = [L : KsL]s·[K

sL : K]s. Since Ks

L ⊆ L is purely inseparable, [L : Ks

L]s = 1, so [L : K]s = [KsL : K]s = [Ks

L : K] (because K ⊆ Ks

L is separable). c) Let C be the composite Ks

L KiL. Suppose L is separable over Ki

L. Then L is separable over C; L is also purely inseparable over Ks

L, so it is purely inseparable over C. This shows that L = C.

Suppose now that L = KiL K

sL = Ki

L(KsL). This means that L is gener-

ated over KiL by separable elements over K (which are also separable

over KiL). It follows that L is separable over Ki

L. !

K

KiL Ks

L

Lseparable ⇔ L = Ks

LKiL

purely inseparable

purely inseparable

separable

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The proposition above shows that, for every algebraic extension K ⊆ L, the separable closure Ks

L is important in the sense that we can decompose K ⊆ L in a tower of extensions: the separable extension K ⊆ Ks

L followed by the purely inseparable extension KsL ⊆ L. The

situation is not symmetric for the purely inseparable closure KiL, since

K ⊆ KiL is purely inseparable, but Ki

L ⊆ L is not always separable (see exercise 13).

3.26 Definition. For an algebraic extension K ⊆ L, define the inseparable degree of L over K to be [L : K]i := [L : Ks

L].

3.27 Remark. If K ⊆ L is finite and char K = p > 0, then [L : K]i is a power of p, as Ks

L ⊆ L is purely inseparable. But [L : K]i is not necessarily equal to the greatest power of p that divides [L : K] and neither equal to [Ki

L : K]. The results proven until now show that the following properties

hold: a) K ⊆ L is finite ⇒ [L : K] = [L : K]s·[L : K]i. b) K ⊆ L is separable ⇔ L = Ks

L ⇔ [L : K]i = 1. c) K ⊆ L is purely inseparable ⇔ L = Ki

L ⇔ [L : K]s = 1. Exercise 1.14 shows that if K ⊆ L is normal, then Ki

L ⊆ L is separa-ble (so L = Ks

LKiL).

3.28 Proposition. Let K have characteristic p > 0 and let a ∈ K. If a ∉ K p, then the polynomial f = aX

mp − is irreducible in K[X], ∀m ≥ 1. Conversely, if aX

mp − is irreducible in K[X] for some m ≥ 1, then a ∉ K p.

Proof. Let q := pm and take α, β to be roots of f in Ω. Then α q = a = β q, so α = β (by the injectivity of the field homomorphism x & x q). Consequently, α is the only root of f in Ω : f = (X − α)q. Let g ∈ K[X] be a monic irreducible factor of f. If h ∈ K[X] is irreducible monic and h| f, then h = g. Indeed, g and h have α as a common root

V.3 Separability

259

and hence are not mutually prime; being irreducible and monic, they are equal. So f = g t, where t ∈ N*; t is a power of p, because t·grad g = pm. Let b := g(0) ∈ K. We have b t = f(0) = −a. If t > 1, −a ∈ K p. Since K p is a subfield, a ∈ K p, contradiction. It follows that t = 1, so f = g, which is irreducible.

Conversely, if a = b p for some b ∈ K, then f = ( )pp bXm

−−1

, contradicting that f is irreducible. !

This proposition gives a method to exhibit purely inseparable ele-ments. In addition, any purely inseparable element can be constructed by this method (cf. 3.21).

Exercises

In the exercises, K denotes a field and Ω an algebraic closure of K. 1. Let g = X 3 + 3X + 1 ∈ K[X] and α a root of g in Ω. Under what conditions is K ⊆ K(α) separable? 2. Let L/K be a finite extension. We want to prove that L/K is simple iff it has a finite number of intermediate fields. Denote by S the set of all intermediate fields of L/K.

a) Suppose L = K(a), with a ∈ L. Let f = Irr(a, K). Show that, for any E ∈ S, Irr(a, E) is a monic divisor (in Ω[X]) of f and that E = K(C), where C is the set of coefficients of Irr(a, E). Deduce that E & Irr(a, E) is injective and that S is finite.

b) Suppose S is finite and L = K(a, b). If K is infinite, show that L is simple considering the set of intermediate fields K(a + bc), c ∈ K. If L = K(a1, , an), use an induction on n to prove that L/K is simple.

c) Is the statement still true if L/K is infinite?

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260

3. Let K ⊆ L be a simple extension of degree n. Then K ⊆ L has at most 2 n − 1 intermediate fields. 4. Let char K = p > 0 and f ∈ K[X]. Suppose f has a repeated root (in Ω). Prove that, if p -deg f, then f is reducible in K[X]. 5. Let α be a separable element over K and g = Irr(α, K). Prove that, if α is a multiple root (having multiplicity n) of some f ∈ K[X], then g n | f. (Hint: Using formal derivatives, show that any conjugate of α is a root of f, of multiplicity ≥ n.) 6. Suppose K is perfect and f ∈ K[X] has a repeated root α (in Ω), of multiplicity > (deg f )/2. Show that α ∈ K. 7. Let K ⊆ L be a normal extension. Then K ⊆ Ks

L is normal. 8. Let n ∈ N and let K ⊆ L be an algebraic separable extension such that ∀x ∈ L, [K(x) : K] ≤ n. Then [L : K] ≤ n. 9. Let K ⊆ L be a normal extension. Then every irreducible polyno-mial in K[X] decomposes in L[X] in a product of irreducible polynomials that have the same degree. 10. Let K ⊆ L be algebraic and let x, x' ∈ L be conjugate over K. Let y ∈ K(x) and g ∈ K[X] such that y = g(x). Then y = g(x') ⇔ x' is conju-gate to x over K(y). 11. Suppose K ⊆ L is an algebraic extension and S is an intermediate field with K ⊆ S separable and S ⊆ L purely inseparable. Then S = Ks

L. 12. Suppose char K = p > 0 and K ⊆ L is an algebraic extension. Prove that Ki

L = K is equivalent to: ∀x ∈ L, if x p ∈ K, then x ∈ K. 13. Suppose F is a field of characteristic 2 and put K = F(X, Y ). Let u be a root of the polynomial T 2 + T + X ∈ K[T], S = K(u) and v = uY (v is a root of T 2 − uY ∈ S[T]). Let L = S(v). Prove that:

a) K ⊆ S is separable of degree 2 and S ⊆ L is purely inseparable of degree 2. Deduce that Ks

L = S. b) If t ∈ L and t 2 ∈ K, then t ∈ K. (Hint. Find a base in L/K and use

the form of the elements in L).

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261

c) KiL = K.

14. Let K ⊆ L ⊆ E be a tower of algebraic extensions, with L ⊆ E nor-mal and K ⊆ L purely inseparable. Then K ⊆ E is normal. 15. Suppose K ⊆ L is a finite extension of characteristic p > 0, such that L p ⊆ K.

a) Prove that K ⊆ L is purely inseparable. b) Suppose x1, , xn ⊆ L is such that

K ( K(x1) ( K(x1, x2) ( ( K(x1,, xn) = L. Show that [L : K] = p n (a set x1, , xn with these properties is

called a p-basis of K ⊆ L). c) Any two p-bases of K ⊆ L have the same cardinal (called the

p-dimension of the extension K ⊆ L). d) The extension K ⊆ L has a p-basis. e) x1, , xn is a p-basis ⇔ 1,,0,,11

1 −∈ piixx nin

i n ……… is a linearly independent set over K ⇔ ∀i ∈ 1, , n, xi ∉ K(x1, , xn \ xi).

16. Let char K = p > 0 and consider L = K(X, Y), the field of rational functions in two indeterminates X and Y over K. Let F = K(X p, Y p). Show that:

a) F ⊆ L is purely inseparable. b) [L : F] = p 2. c) For any α ∈ L, α p ∈ F. d) F ⊆ L is not simple. e) Deduce that F ⊆ L has an infinity of intermediate fields. Prove

that ∀β, γ ∈ F, F(X + βY) = F(X + γY) iff β = γ. f) X, Y is a p-basis of F ⊆ L. Can an extension of degree p have an infinity of intermediate

fields? 17. Take p = 2 and K = F2 in the previous problem.

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262

a) Show that 1, X, Y, XY is a basis of the extension F ⊆ L and write the general form of an element of L.

b) Let F ( E ( L be an intermediate field. Show that there exists a unique polynomial PE ∈ F2[X, Y ], of the form

PE = Xu + Yv + XYw, with u, v, w ∈ F2[X

2, Y 2], (u, v, w) = 1, such that E = F(PE). Show that E & PE establishes a bijection between the set E | F ( E ( L is an intermediate field and the set of the polynomials PE of the form above. 18. Give an example of a normal extension K ⊆ L such that L is a splitting field for a polynomial f ∈ K[X], but such that there is no g ∈ K[X], irreducible, with L the splitting field of g over K.

V.4 The Fundamental Theorem of Galois Theory

We gathered sufficient data on normal extensions and on separable extensions to deal with the problem of bijectivity of the Galois connections we stated at the beginning of the chapter. First, we generalize Corollary 1.5, concerning the order of the Galois group of a finite extension. An essential step in the proof is the Dedekind lemma6. Notice the linear algebra methods.

4.1 Proposition. a) (Dedekind's Lemma) Suppose (G, ·) is a semi-group, K is a field, n ∈ N* and σi : G → (K*, ·), ∀i ∈ 1, , n are distinct semigroup homomorphisms. If α1,, αn ∈ K, such that

6 Julius Wihelm Richard Dedekind (1831-1916), German mathematician, one of

the creators of algebraic number theory.


263

α1σ1(x) + + αnσn(x) = 0 for any x ∈ G, then α1 = = αn = 0.

b) If K ⊆ L is a finite extension, then |G(L/K)| ≤ [L : K]. Proof. a) The statement can be rephrased as follows: in the

K-vector space K G of functions defined on G with values in K, σ1, , σn are linearly independent.

Suppose the statement is false. Relabelling if necessary, there exists m ≤ n and α1, , αm ∈ K, all nonzero, such that α1σ1(x) + + αmσm(x) = 0, ∀x ∈ K. (1)

We may even suppose that m is the smallest having this property, in the sense that any linear dependence relation between σ1, , σn has at least m terms. Since σ1 ≠ σ2, there exists y ∈ G such that σ1(y) ≠ σ2(y). Replacing x with xy in (1), we have: α1σ1(xy) + + αmσm(xy) = α1σ1(x)σ1(y) + + αmσm(x)σm(y) = 0 (2) α1σ1(x)σ1(y) + + αmσm(x)σ1(y) = 0, (3) (3) is obtained by multiplying (1) with σ1(y). Subtracting (2) from (3) we get

α2(σ2(y) − σ1(y))σ2(x) + + αm(σm(y) − σ1(y))σm(x) = 0, ∀x ∈ K. In this equality, α2(σ2(y) − σ1(y)) is nonzero, so we obtained a linear

dependence relation with less than m terms, contradiction with the minimality of m.

b) First, we notice that G(L/K) is finite. Indeed, let [L : K] = n and take x1, , xn a K-basis of L. Then any σ ∈ G(L/K) is determined by its values in x1, , xn. But σ(x) is a conjugate of x, ∀x ∈ L, and the number of conjugates of x is finite. So let G(L/K) = σ1, , σm and suppose, by contradiction, that m > n. Consider the matrix:

A =

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

nmmm

n

n

xxx

xxxxxx

σσσ

σσσσσσ

+!*!!

++

21

22212

12111

∈ Mm, n(L).

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The rank of A is at most n, so its rows are linearly dependent. Let α1, , αm ∈ L, not all zero, such that α1σ1(xi) + + αmσm(xi) = 0, ∀i ∈ 1, , n. (4)

Since any x ∈ L is a K-linear combination of x1, , xn, and σi are K-homomorphisms, (4) holds for any x ∈ L.

Viewing σi as homomorphisms from L* to L*, Dedekind's lemma applies. We get that α1 = = αm = 0, contradiction. !

4.2 Corollary. Let K ⊆ L be a finite, normal and separable exten-sion. Then |G(L/K)| = [L : K].

Proof. It is enough to show that |G(L/K)| ≥ [L : K] =: n. The exten-sion is finite and separable, so it has a primitive element α. So deg Irr(α, K) = n. The extension is normal, so Irr(α, K) has n roots in L. For any root β of Irr(α, K), [K(β) : K] = n, thus K(β) = L. But K(α) and K(β) are K-isomorphic by an isomorphism σβ that transports α to β. Summarizing, for every conjugate β of α, we exhibited a K-isomorphism σβ : L → L, which is evidently an element of G(L/K).!

The next proposition is an important step in proving the fundamen-tal theorem and shows that the converse of the preceding result also holds.

4.3 Proposition. (E. Artin7) Suppose L is a field, H is a finite sub-group of the group Aut(L) of the automorphisms of L and LH is the fixed field of H. Then the extension LH ⊆ L is finite, normal, separable, [L : LH] = |H| and G(L/LH) = H.

7 Emil Artin (1898-1962), Austrian mathematician (he lived most in Germany

and the USA).


265

Proof. Let x ∈ L. The set Cx = σ(x) | σ ∈ H is finite and |Cx| ≤ |H|. If σ ∈ H, then σ(Cx) = Cx. Consider the polynomial

fx := ( )∏ ∈ −xCy yX .

For any automorphism σ ∈ H, the isomorphism σ': L[X] → L[X] that extends σ leaves fx invariant, so the coefficients of fx are in LH. But fx(x) = 0, hence every element in L is algebraic over LH. Moreover, Irr(x, LH) = fx. Indeed, if g ∈ LH[X] is such that g(x) = 0, then g(σ(x)) = 0, ∀σ ∈ H, so any root of fx is a root of g. Thus, fx | g. Note that fx has all roots in L, and these are distinct. Summarizing, LH ⊆ L is algebraic, normal and separable. Let us show that it is finite. The previous arguments show that, for any x ∈ L, the simple extension LH ⊆ LH(x) is finite and its degree is at most |H|. If LH ⊆ E is a finite extension (where E ⊆ L), it is simple by the primitive element theo-rem, so [E : LH] ≤ |H|. Suppose that LH ⊆ L is infinite. Then, ∀x ∈ L, LH(x) ≠ L; take then some y ∈ L \ LH(x). This implies that there exists a sequence (xn)n≥1 of elements of L such that LH(x1,, xn) ( LH(x1,, xn+1), ∀n ∈ N*. This in turn means that the sequence [LH(x1,, xn) : L

H] is strictly increasing. Pick n ∈ N* such that [LH(x1, , xn) : L

H] > |H|. We have reached a contradiction, since the finite extension LH ⊆ LH(x1, , xn) is of degree at most |H|, as we saw. In conclusion, LH ⊆ L is finite and its degree is at most |H|. As H ⊆ G(L/LH) (and |G(L/LH)| = [L : LH] from the preceding Corollary), |G(L/LH)| = [L : LH] = |H| and H = G(L/LH). !

4.4 Corollary. Let K ⊆ L be a finite extension with |G(L/K)| ≥ [L : K]. Then K ⊆ L is normal and separable and |G(L/K)| = [L : K].

Proof. Let G = G(L/K). We have |G| = [L : K] from 4.1.b). Let K0 = LG. The proposition above ensures that K0 ⊆ L is normal, separa-ble and [L : K0] = |G| = [L : K]. Because K ⊆ K0 ⊆ L, we have K = K0.!

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266

Recall that the Galois connections are defined as follows: for an extension K ⊆ L, whose Galois group is G, denote by IF(L/K) the set of its intermediate fields and by Subg(G) the set of subgroups of G. Then we define:

Φ : IF(L/K) → Subg(G), Φ(E) = G(L/E), ∀E ∈ IF(L/K); Ψ : Subg(G) → IF(L/K), Ψ(H) = x ∈ L|σ(x) = x, ∀σ ∈ H:= LH,

∀H ∈ Subg(G).

The result that follows holds for any extension and collects some general properties of Galois connections. Its proof is a mere applica-tion of the definitions:

4.5 Proposition. Let K ⊆ L be an extension. Then: a) For any E ∈ IF(L/K), LG(L/E) ⊇ E. b) For any H ∈ Subg(G(L/K)), G(L/LH) ⊇ H. c) Φ is inclusion reversing: ∀E1, E2 ∈ IF(L/K), E1 ⊆ E2 ⇒

G(L/E1) ⊇ G(L/E2). d) Ψ is inclusion reversing: ∀H1, H2 ∈ Subg(G(L/K)), H1 ⊆ H2 ⇒

21 HH LL ⊇ . Proof. a) LG(L/E) = x ∈ L|σ(x) = x, ∀σ ∈ G(L/E) ⊇ E. b), c), d) are proposed as exercises. !

4.6 Theorem. (The fundamental theorem of Galois Theory) Let K ⊆ L be a finite, normal and separable extension of fields. Then the Galois connections are inclusion-reversing maps that are bijective and inverse to each other. Via these maps, the intermediate fields E with K ⊆ E normal correspond to normal subgroups of G(L/K). More precisely:

a) For any intermediate field K ⊆ E ⊆ L, LG(L/E) = E and [L : E] = |G(L/E)|.

b) For any subgroup H ≤ G(L/K), G(L/LH) = H.


267

c) For any intermediate field K ⊆ E ⊆ L with K ⊆ E normal, G(L/E) is a normal subgroup in G(L/K). Besides, G(E/K) is canoni-cally isomorphic to the factor group G(L/K)/G(L/E).

d) If H is a normal subgroup in G(L/K), then LH is a normal exten-sion of K, and G(LH/K) is canonically isomorphic to the factor group G(L/K)/H.

Proof. a) Let E be an intermediate field of K ⊆ L. Then E ⊆ L is fi-nite, normal and separable, so [L : E] = |G(L/E)| by Cor. 4.2. On the other hand, Prop. 4.3 ensures that [L : LG(L/E)] =|G(L/E)|. As LG(L/E) ⊇ E (by 4.5), we have E = LG(L/E).

b) Proved at 4.3. c) Consider the restriction homomorphism

res : G(L/K) → G(E/K), res(σ) = σ|E, for any σ ∈ G(L/K). We have σ|E ∈ G(E/K) because K ⊆ E is normal. The homomorphism res is surjective, since every τ ∈ G(E/K) extends to a K-homomorphism τ' : L → Ω (where Ω is an algebraic closure of L), by IV.2.19. But τ'(L) = L, since K ⊆ L is normal, so τ' ∈ G(L/K) and res(τ') = τ. The kernel of res is a normal subgroup in G(L/K),

Ker res = σ ∈ G(L/K) | σ|E = id = G(L/E). Applying the isomorphism theorem, we get G(L/K)/G(L/E) ≅

G(E/K). d) Let Ω be an algebraic closure of L and η a K-automorphism of

Ω. We must prove that η(LH) ⊆ LH. This amounts to show that, ∀x ∈ LH, we have τη(x) = η(x), ∀τ ∈ H. First, observe that η(L) = L (because K ⊆ L is normal), so we can consider η as belonging to G(L/K). We have τη(x) = η(η −1τη)(x) = ησ(x) = η(x), where we de-noted η −1τη by σ ∈ H and used the fact that σ(x) = x, ∀x ∈ LH. The isomorphism follows by c), keeping in mind that H = G(L/LH). !

A normal and separable extension is called a Galois extension. The fundamental theorem of Galois Theory says that, for a finite Galois extension, the Galois connections are bijective. A natural question

V. Galois Theory

268

arises: Which are all field extensions for which the Galois connec-tions are bijective? The answer to this problem was given in 1951 by the Romanian mathematician Dan Barbilian8 (in the paper Soluţia exhaustivă a problemei lui Steinitz (The exhaustive solution of the Steinitz problem), Acad. R.P.R., Stud. Cerc. Mat. 2 (1951), 195-259). His result states that: Any field extension for which the Galois connec-tions are bijective and inverse to each other is a finite Galois exten-sion. Notice that, if we suppose that the extension is finite, the result is a consequence of 4.3.

If K ⊆ L is finite Galois, IF(L/K) and Subg(G(L/K)) are anti-isomor-phic as ordered sets, since the Galois connections are inclusion-reversing bijections. Moreover, they are anti-isomorphic as lattices: sup(E, F) = EF in IF(L/K) corresponds to inf(Φ(E), Φ(F)) = Φ(E)∩Φ(F) ∈ Subg(L/K), that is: G(L/EF) = G(L/E)∩G(L/F). A similar statement holds for inf(E, F) = E∩F (see problem 7).

We remark that, if K ⊆ L is Galois of degree n and x ∈ L, the conjugates of x over K (the roots of Irr(x, K)) are exactly σx | σ ∈ G(L/K) =: x1, , xm (where necessarily m | n. Why? We have m = n iff L = K(x)). So, Irr(x, K) = (X − x1)(X − xm).

4.7 Example. The extension Q ⊆ ( )3,2Q =: L is Galois: separability is automatic (the characteristic is 0), and normality fol-lows from the fact that L is a splitting field of (X 2 − 2)(X 2 − 3) over Q. The degree is 4 (see example IV.1.28.b), so the Galois group G has 4 elements, by the fundamental theorem. In order to find this group, we look on the action of the automorphisms in G on the generators 2 and 3 . If σ ∈ G, then ( ) 2,22 −∈σ (these are the roots of

8 Also known as a poet, under the pen name Ion Barbu (1895-1964).


269

Irr( 2 , Q) = X 2 − 2); likewise, ( ) 3,33 −∈σ . Since G has 4 ele-ments, and the number of all possible choices is also 4, the automor-phisms in G are determined by their action on generators, according to the following table:

id σ τ η

2 2 − 2 2 − 2 3 3 3 − 3 − 3

For instance, τ( 2 ) = 2 and τ( 3 ) = − 3 . Based on the table above one can compile the multiplication table of G.

We remark that G ≅ Z2×Z2 (the 4 Klein group), any element of G being of order 2. The subgroups of G are id, <σ > = id, σ, <τ > = id, τ, <η > = id, η and G. So, the extension has 3 proper intermediate fields, corresponding to the proper subgroups <σ >, <τ >, <η >. On the other hand, one sees immediately that

( )2Q , ( )3Q , ( )6Q are proper distinct intermediate fields, so these are all intermediate fields. We have σ( 3 ) = 3 , so L< σ > ⊇ ( )3Q ; the equality holds since the degrees match:

( )[ ]QQ :3 = [L< σ > : Q]. But [L< σ > : Q] = [L : Q]/[L : L< σ >] = 4/2 = 2 = ( )[ ]QQ :3 . The connections between the remaining subgroups and the remaining intermediate fields are established similarly.

Here is a sample of the applications of Galois Theory. The follow-ing result is often used in arguments on field extensions:

4.8 Proposition. Let K ⊆ L be finite Galois and let K ⊆ M be an extension (suppose L and M are subfields of a larger field F). Then M ⊆ ML is finite Galois and G(ML/M) is isomorphic to a subgroup of G(L/K) (namely, with G(L/M ∩ L)).

V. Galois Theory

270

L ∩ M

M L

ML

KProof. L is the splitting field over K of a separable polynomial

f ∈ K[X], so ML is the splitting field over M of f (considered in M[X]). So, M ⊆ ML is normal, finite and separable. Let σ be an automor-phism in G := G(ML/M). Then σ|L ∈ G(L/K). Consider the group homomorphism res : G → G(L/K), res(σ) = σ|L, ∀σ ∈ L. It is injective: ∀σ ∈ G with σ|L = id, we have σ = id (since σ and id agree on M and on L, they agree on ML). Thus G ≅ I, where I is the image of the res homomorphism; I is a subgroup of G(L/K). Let us show that I = G(L/M ∩ L). By the fundamental theorem, this is tantamount to the fact that the fixed fields of I and G(L/M∩L) are equal. We have LI = x ∈ L|σ(x) = x, ∀σ ∈ G(ML/M). But x ∈ ML| σ(x) = x, ∀σ ∈ G(ML/M) = M, so LI = M ∩ L. !

The classical Galois theory that we presented here using a lin-earized approach, due to Dedekind and Artin, has generalizations and counterparts in multiple directions:

Infinite Galois Theory treats the case of an extension that is alge-braic, normal and separable, but not necessarily finite. The idea is to make the Galois group of the extension a topological group (by means of the Krull 9 topology) ; the fundamental theorem reads in this case:

9 Wolfgang Adolf Ludwig Helmuth Krull (1899-1971), German mathematician.


271

The Galois connections establish bijections from the set of all intermediate fields of the extension to the set of all closed subgroups of the Galois group.

A Galois theory for commutative rings was developed by CHASE, HARRISON, ROSENBERG, in a paper published in 1965.

Differential Galois theory looks into the problem of explicit solutions of differential equations. For an introduction and references, see for instance GOZARD [1997].

Co-Galois theories. An example of such a theory is the Kummer theory that we present in the next chapter. The name reflects the fact that these theories establish inclusion preserving bijections between the lattice of intermediate fields and the lattice of subgroups of a cer-tain group associated to the extension (as opposed to Galois theory, where they are order reversing). Recently, the Romanian mathemati-cians T. Albu and F. Nicolae devised a theory of this type that generalizes, among others, Kummer theory (T. ALBU, Cogalois The-ory, Marcel Dekker, New York, 2003).

Exercises

1. Let L be the splitting field over Q of X 3 − 2. Find G(L/Q) and all the subfields of L. (Hint: Look for the action of the automorphisms on the generators 3 2 and ω, where ω 2 + ω + 1 = 0). 2. The same problem, for the polynomial X 4 − 2 ∈ Q[X]. 3. Let G = G(L/Q), where L = Q( 8 2 ). Find G and the fixed field of G. 4. Let K be a field and let K(X) be the rational function field over K.

V. Galois Theory

272

a) Let ψ : K(X) → K(X) be the unique K-homomorphism with ψ(X) = X + 1. Determine the subgroup H generated by ψ ; find the fixed field of H.

b) Prove that ϕ ∈ AutK(K(X)) = Gal(K(X)/K) iff there exist a, b, c,

d ∈ K, with ad − bc ≠ 0 such that ϕ(X) = dcXbaX

++ . In particular,

H ≠ Gal(K(X)/K), but these subgroups have the same fixed field. (Hint. Use IV.1.28c))

c) Show that Φ : GL(2, K) → AutK(K(X)), ⎟⎠

⎞⎜⎝

⎛Φdcba

= ϕ, with

ϕ(X) = dcXbaX

++ , is a surjective group homomorphism, whose kernel is

the subgroup S of scalar matrices (matrices of the type aI, with a ∈ K and I is the identity matrix). The group GL(2, K)/S is called the projective linear group of degree 2 over K, denoted PGL(2, K). So AutK(K(X)) ≅ PGL(2, K).

5. Suppose L/K is a normal finite extension and S is the separable clo-sure of K in L. Show that:

a) K ⊆ S is normal and res : G(L/K) → G(S/K), res(σ) = σ |S is an isomorphism.

b) The fixed field of G(L/K) is I, the purely inseparable closure of K in L (so I ⊆ L is Galois).

c) L = SI. 6. Let K ⊆ L, K ⊆ E be Galois extensions such that L ∩ E = K. Then:

G(LE/K) ≅ G(L/K) × G(E/K). 7. Let L/K be a Galois extension and let E, F be intermediate fields. Show that:

a) G(L/EF) = G(L/E) ∩ G(L/F). b) G(L/E∩F) = < G(L/E) ∪ G(L/F) >, the subgroup generated by

G(L/E) and G(L/F).

VI.1 Ruler and compass constructions

273

c) If A, B ≤ G(L/K), which are the fixed fields of A ∩ B, respec-tively of < A ∪ B > ? 8. This exercise shows how to obtain a normal extension of degree 3 of Q using Galois theory10. The idea is to take an Abelian extension of degree 6 of Q; taking a subgroup H of order 2 of the Galois group, the fixed field of H is normal and has degree 3 over Q. Consider the cyclotomic extension Q ⊆ Q(ω), where ω is a primitive root of unity of order 9. Prove that the Galois group of the extension is isomorphic to U(Z9) and the fixed field of the complex conjugation automorphism is Q(cos(2π/9)), a normal extension of degree 3 of Q. Determine all subfields of Q(ω).

VI. Applications of Galois Theory


Geometric ruler and compass constructions have a substantial historical interest. Such problems were one of the focal points of the antique Greek geometry. A number of celebrated problems stayed un-

10 A more general way to construct such extensions uses discriminants.


274

solved for millennia, despite the efforts of some of the best mathematicians. The solution (often in the negative) was given only in the 19th century, at the dawn of the theory of field extensions. It is remarkable that the proof of impossibility of famous constructions re-quires only elementary concepts of field extensions theory, giving yet another demonstration of the power of the algebraic methods. The main difficulties seem to have been the transfer of the problem from Geometry to Algebra and a precise formulation of the concept of constructible by ruler and compass.

In this section, construction means exclusively ruler and com-pass construction.

We want to obtain general constructibility criteria, which will an-swer the following famous problems:

- Angle trisection: Construct an angle whose measure is 1/3 of the measure of a given angle.

- Cube duplication: Construct a cube whose volume is the double of the volume of a given cube.

- Square the circle: Construct a square whose area is equal to the area of a given circle.

- Construct a regular heptagon. More generally, for what natural numbers n is it possible to construct a regular polygon with n sides?

The rules of construction are simple and well-known: - A set of initial points is given (usually two points); - For any two points (given, or already constructed as intersections

of lines or circles), with the ruler one can draw the straight line pass-ing through these points.

- The compass can draw the circle centered in a constructed (or given) point and passing through another constructed (or given) point.


275

These are the only constructions allowed. Note that drawing a line (a circle) does not mean that all points belonging to it are constructed. A point is constructed only if it is identified as an intersection between lines (or circles, or lines and circles).

Analyzing the various problems of construction, one realizes that all are equivalent to a problem of the following type:

For a given set S of points in the plane, construct a set T of points (satisfying some property).

Indeed, the construction of a line reduces to the construction of two distinct points on that line; an angle is determined by three points (the vertex of the angle and a point on each side) etc. The reader is invited to formulate in this form the problems in the list above.

1.1 Definitions and notations. In what follows, P denotes an Euclidian plane. If A and B are distinct points in P, AB denotes the line determined by the points A and B, [AB] is the (closed) segment deter-mined by A and B, and |AB| is the length of the segment [AB]. Let S be a set of points in P (called initially constructible points). Let DS be the set of the lines determined by two distinct points in S and let CS be the set of circles centered in a point in S and passing through a point in S.11

The point M ∈ P is called constructible12 in one step from S if it satisfies one of the following conditions:

- P belongs to the intersection of two distinct lines in DS.

11 Sometimes CS is defined as the set of circles centered in a point in S with

radius equal to the distance between two arbitrary points in S. The definition we have adopted for CS corresponds to a collapsible compass: one cannot transport with the compass the distance between two points. The two definitions are in fact equivalent, in the sense that they lead to the same set of constructible points C(S). Prove this!

12 We omit saying by the ruler and compass in what follows.


276

- P belongs to the intersection of two distinct circles in CS. - P belongs to the intersection of a circle in CS with a line in DS. Let C1(S) denote the set of points that are constructible in one step

from S. Let C0(S) := S and define by recurrence Cn(S) := C1(Cn − 1(S)), ∀n ∈ N*.

A point in P is called constructible from S if the point is in C(S) := ∪n∈N Cn(S).

A line (a circle, an angle,) is said to be constructible from S if the line (the circle, the angle,) is determined by points in C(S).

If the set S is understood, we say constructible instead of construct-ible from S.

1.2 Remark. a) |S| = 1 ⇔ C1(S) = S. If one takes only one initially constructible point, nothing else can be constructed from it.

b) S ⊆ C1(S). Indeed, if |S| ≥ 2, let A, B ∈ S. The point A is the intersection of the line AB with the circle of center B passing through A. So, Cn(S) ⊆ Cn + 1(S), ∀n ∈ N.

1.3 Lemma. Let S ⊆ P with |S| ≥ 2 and let A, B, C ∈ C(S) be dis-tinct.

a) The symmetric of B with respect to A is constructible from S. b) The midpoint D of the segment [AB] and the perpendicular on

[AB] in D are constructible. c) The perpendicular from C on AB is constructible. d) If C ∉ AB, then the parallel through C to AB is constructible. Proof. Exercise in elementary geometry. !

Assumptions and notations. We suppose from now on that S is a subset of P with at least two points. Fix two distinct points O and I in S. There exists a unique system of Cartesian coordinates in the plane P such that O is the origin, of coordinates (0,0), and I has coordinates (1,0). The segment OI has thus length 1. Let κ : P → R × R the map-ping that associates to a point in P its coordinates in the system above.


277

We often identify a point M ∈ P with the couple (x, y) of its coordi-nates, and say the point (x, y). Thus, S can be seen as a subset of R × R. If T is a set of points in P, define TR as the set of real numbers that occur as coordinates for the points in T. Therefore,

TR := x ∈ R | ∃ y ∈ R such that (x, y) ∈ T or (y, x) ∈ T. In the hypotheses above, SR contains at least 0 and 1.

As a consequence of the constructions in 1.3, the coordinate axes Ox and Oy are constructible from O and I.

1.4 Definition. The real number x is called constructible from S if the point of coordinates (x, 0) is constructible from S. If the set S is clear from the context, we say simply x is constructible.

In our hypotheses, 0 and 1 are always constructible.

1.5 Proposition. Let x ∈ R. The following assertions are equiva-lent:

a) x is constructible from S. b) The point (0, x) is constructible. c) There exists y ∈ R such that (x, y) is constructible. d) There exists y ∈ R such that (y, x) is constructible. Proof. a)⇒b) Let M be the (constructible) point (x, 0). By 1.3, con-

struct the symmetric M' of M with respect to O, then the midperpendicular of the segment [MM'] (i.e. the Oy axis). The point (0, x) belongs to the intersection of the circle centered in O of radius [OM] with Oy.

The other implications are equally easy and are left to the reader. !

The concept of constructible complex number is also natural (and useful, as we will see shortly):


278

1.6 Definition. The complex number z = x + iy (where x, y ∈ R) is called constructible from S if the real numbers x and y are construct-ible from S.

The set C is in one-to-one correspondence with the plane P (composing the bijection κ above with (x, y) & x + iy from R × R to C). Let SC be the set of complex numbers that correspond to the points in S (i.e., the affixes of the points in S):

SC = x + iy ∈ C | (x, y) ∈ S. We have 0, 1 ⊆ SC . It is natural to show that the following result

holds:

1.7 Proposition. Let a, b ∈ R. The following statements are equivalent:

a) The complex number a + ib is constructible. b) The point (a, b) is constructible. c) The real numbers a and b are constructible. Proof. a)⇒b) If a + ib is constructible, then the points A = (a, 0)

and B = (0, b) are constructible (use prop. 1.5). Thus, (a, b) is the intersection of the perpendicular in A on Ox with the perpendicular in B on Oy.

b)⇒c) Let (a, b) be a constructible point. Then the perpendiculars from (a, b) on the axes Ox and Oy are constructible, so (a, 0) and (0, b) are constructible. !

1.8 Remark. Let K be the set of constructible reals13. The previous proposition says that the point (x, y) is constructible if and only if (x, y) ∈ K × K. In other words, C(S) = K × K. The set of constructible complex numbers is K[i] = x + iy | x, y ∈ K.

13 Of course, we mean the real numbers constructible from S. Omitting S from

the notation K simplifies the notation.


279

1.9 Proposition. a) K is a subfield of R, closed under square roots: for any x ∈ K with x > 0, x ∈ K. Moreover, K is the smallest sub-field of R that includes SR, the coordinates of the points in S, and is closed under square roots.

b) K[i] is a subfield of C, closed under square roots and under complex conjugation: for any z ∈ K, z and z ∈ K. Moreover, K[i] is the smallest subfield of C that includes the set SC of the affixes of points in S and is closed under square roots and conjugation.

Proof. a) 0 and 1 belong to K, since O(0, 0) and I(0, 1) are in S. Let a, b ∈ K, a, b ≠ 0. It is enough to prove that − a, a + b, ab −1 are con-structible. We use the constructions from 1.3.

The point (a, 0) is constructible, so (− a, 0), its symmetric with re-spect to O, is constructible, whence − a ∈ K.

The constructible points (0, a) and (− a, 0) determine a line d. The perpendicular on Ox in (b, 0) is constructible and intersects d in (b, a + b). So, a + b ∈ K.

For the constructibility of ab −1, consider the line determined by (b,0) and (0, 1). The parallel to this line through (a, 0) intersects Oy in (0, ab −1).

Suppose now that a > 0. Since K is a field, ((a + 1)/2, 0) is constructible. The circle centered in ((a + 1)/2, 0) passing through (0, 0) intersects the perpendicular on Ox in (a, 0) in two points, of coordinates (a, a ) and (a, − a ). So, a ∈ K.

It is clear that K contains the coordinates of any point in S. The next lemma implies that, if L is a subfield of R, L contains SR and L is closed under square roots, then K ⊆ L.

b) Since K is a subfield in R, K[i] is a subfield in C. If a + bi, with a, b ∈ K, is an element in K[i], then its conjugate a − bi ∈ K[i]. For proving that K[i] is closed under square roots, we use the trigonomet-ric form of complex numbers. Let z = r(cosα + isinα), with r > 0, r·cosα, r·sinα ∈ K. Then ( ) ( )( )2sin2cos αα irz +±= . We have


280

r 2cos2α + r 2sin2α = r 2, so r 2 ∈ K, r and r ∈ K (which is closed un-der square roots). We get also cosα, sinα ∈ K. We have

( ) ( ) 2cos12cos αα +±= ∈ K. Likewise, sin(α/2) ∈ K, so z ∈ K[i]. The next lemma shows that any subfield of C containing

the affixes of points in S and is closed under conjugation and square roots must include K[i]. !

1.10 Lemma. a) Let L be a subfield of R and let (x, y) ∈ R × R. If (x, y) is constructible in one step from L × L, then x, y ∈ L or there ex-ists u ∈ L, u > 0, such that x, y ∈ ( )uL .

b) If L is a subfield of R, closed under square roots, then C(L × L) = L × L: the set of points constructible from L × L coincides with L × L. In particular, if SR ⊆ L, then K ⊆ L.

c) Let E be a subfield of C such that E is closed under conjugation and square roots and E ⊇ SC. Then K[i] ⊆ E.

Proof. a) It is enough to prove that any point M(x, y), constructible in one step from L × L, is in L × L or in ( ) ( )uLuL × for some u ∈ L, u > 0. Use the notations in Definition 1.1.

Case I. M(x, y) is the intersection of two lines in DL × L. Let Ai(xi, yi) ∈ L × L, ∀i ∈ 1, 2, 3, 4, such that A1A2 and A3A4 are not parallel and M is their intersection. The equation of the line A1A2 is (x − x1)(y2 − y1) = (y − y1)(x2 − x1). Since L is a field, this equation can be written as ax + by = c, with a, b, c ∈ L. In the same way, the equa-tion of A3A4 is dx + ey = f, for some d, e, f ∈ L. Then M(x, y) is the solution of the system

⎩⎨⎧

=+=+

feydxcbyax

The determinant of the system is nonzero, (otherwise the lines would be parallel); Cramer's formulas show that the solution (x, y) of the system is in L × L.


281

Case II. M belongs to the intersection of a line in DL × L with a cir-cle in CL × L. The equation of a circle in CL × L is (x − a)2 + (y − b)2 = r 2, for some a, b, r ∈ L. We saw that the equation of a line in DL × L is dx + ey = f, for some d, e, f ∈ L. Then M(x, y) is the solution of the sys-tem:

( ) ( )⎩⎨⎧

=+=−+−

feydxrbyax 222

Suppose d ≠ 0 (if d = 0, then e ≠ 0 and swap d with e). So, x = − d −1ey − d −1f. The first equation becomes py 2 + qy + s = 0, for some p, q, s ∈ L. This equations must have real solutions (the circle and the line intersect), y = ( )spqqp 22 21 −±−⋅− . So, x, y ∈ ( )uL , where u = q2 − 2sp ∈ L.

Case III. M belongs to the intersection of two circles in CL × L. We must show that the solutions (if any) of a system of the form

( ) ( )( ) ( )⎪⎩

⎪⎨⎧

=−+−

=−+−222

222

tdycx

rbyax,

where a, b, r, c, d, t ∈ L, are in ( ) ( )uLuL × for some u ∈ L, u > 0. Subtracting the equations, the quadratic terms cancel out and we ob-tain linear equation in x and y; together with one of the initial equa-tions, we obtain a system of the type studied at case II.

b) The subfield L being closed under square roots, the result fol-lows from a).

If SR ⊆ L (i.e. S ⊆ L × L), then C(S) ⊆ C(L × L). By a), C(L × L) = L × L. Thus, K ⊆ L.

c) Let L = E ∩ R. If x ∈ L, x > 0, then x ∈ E (E is closed under square roots) and x ∈ R, so x ∈ L. Thus, L is a subfield of R, closed under square roots. On the other hand, if x, y ∈ R and x + iy ∈ E, then x − iy ∈ E (E is closed under conjugation), so x, y ∈ E. Since x, y ∈ R, x, y ∈ L. Conversely, if x, y ∈ L, then x + iy ∈ E, be-cause i = 1− ∈ E. So, E = L[i] = x + iy | x, y ∈ L.


282

The hypothesis SC ⊆ E is thus equivalent to SR ⊆ L; by a), we have K ⊆ L, so K[i] ⊆ L[i] = E. !

We have proven that the geometric statement the point M(x, y) is constructible from the set of points S has a purely algebraic form: x and y belong to the smallest subfield of R that includes Q(SR) and is closed under square roots. This algebraic translation allows us to say about some real number x: x is constructible from U (where U is a set of real numbers), meaning that x belongs to the smallest subfield of R that includes Q(U) and is closed under square roots.

Note also that any rational number is constructible (from any U ⊆ R, |U| ≥ 2).

1.11Theorem.14 A real number x is constructible from S if and only if there exists a tower K0 ⊆ K1 ⊆ ⊆ Kn of subfields of R, such that x ∈ Kn, K0 = Q(SR) and [Ki : Ki −1] ≤ 2 for any i, 1 ≤ i ≤ n.

Proof. ⇒ It is enough to prove by induction on m that, for any fi-nite set of points T, if T ⊆ Cm(S), then there exists a tower of exten-sions K0 ⊆ K1 ⊆ ⊆ Kn such that TR ⊆ Kn and [Ki : Ki − 1] ≤ 2, ∀i ∈ 1, , n. If m = 0, T ⊆ S, so TR ⊆ SR ⊆ Q(SR) = K0. Let m > 0 and let T = A1, , Ar be a finite subset of Cm(S). This means that there exists a finite set U ⊆ Cm−1(S) such that any point in T is constructible in one step from U. From the induction hypothesis, there is a tower K0 ⊆ K1 ⊆ ⊆ Kn such that UR ⊆ Kn and [Ki : Ki − 1] ≤ 2, ∀i ∈ 1, , n. For any point As ∈ T (s ∈ 1, , r), lemma 1.10.a) says that As has its coordinates in Kn (then set us = 0) or in a quadratic extension of Kn of the form ( )sn uK , for some us ∈ Kn, us > 0. So, TR ⊆ ( )rn uuK ,,1 … and we have a tower of extensions

14 The result belongs to M.L. Wantzel, who proved it and published it in 1837

(when he was elève-ingénieur des Ponts-et-Chaussées). It seems though that Gauss knew as early as 1796 this criterion of constructibility.


283

K0 ⊆ ⊆ Kn ⊆ ( ) ( )rnn uuKuK ,,11 ……⊆⊆ , satisfying the condition that each extension has degree at most 2.

⇐ We prove by induction on n that, for any tower Q(SR) = K0 ⊆ ⊆ Kn ⊆ R, with [Ki : Ki−1] ≤ 2, 1 ≤ i ≤ n , we have Kn ⊆ K (recall K is the field of the real numbers constructible from S). If n = 0, then K0 = Q(SR). Using 1.9.a), we deduce that K includes the subfield of R generated by SR, which is exactly Q(SR). If n > 0, [Kn : Kn−1] ≤ 2 implies either that Kn = Kn−1 (and the induction hypothesis shows that we are done) or that Kn is an extension of de-gree 2 of Kn−1. In this case there exists u ∈ Kn−1, u > 0, such that Kn = Kn−1( u ). By induction, we know that Kn−1 ⊆ K; since K is closed under square roots, we deduce Kn ⊆ K. !

1.12 Corollary. (Necessary condition for constructibility) Let x be a real number constructible over S and let K0 = Q(SR). Then x is alge-braic over K0 and its degree over K0 is a power of 2 (there exists e ∈ N such that [K0(x) : K0] = 2e ).

Proof. The extension Kn in the previous theorem is such that x ∈ Kn and [Kn : K0] is a power of 2. Thus, x is algebraic over K0 and K0 ⊆ K0(x) ⊆ Kn; so, [K0(x) : K0] divides [Kn : K0], which means that [K0(x) : K0] is a power of 2. !

We can easily prove now that some classic ruler and compass constructions problems are impossible:

1.13 Proposition. a) The trisection of an angle of measure θ is equivalent to the construction of the point (cos(θ/3), sin(θ/3)) from O,


284

I, (cosθ, sinθ). A 60° angle cannot be trisected by ruler and compass (a 20° angle is not constructible).15

b) Cube duplication is impossible (it is impossible to construct a cube whose volume is the double of the volume of a given cube).

c) Squaring the circle is impossible (it is impossible to construct a square whose area is equal to the area of a given circle).

Proof. a) Constructing an angle of measure α is equivalent to con-structing the point (cos α, sin α). Indeed, if the angle has its vertex in O one of the sides is Ox, the intersection of the other side with the unit circle is (cos α, sin α) or (cos α, − sin α). Conversely, if P(cos α, sin α) is given, then the angle between OP and Ox has meas-ure α.

On the other hand, if cos α is constructible, then sin α = α2cos1−± is constructible. Thus, trisecting the angle of measure θ is tantamount to the constructibility of cos(θ/3) from cosθ. The formula cos(3x) = 4cos3x − 3cosx implies that u := cos(θ/3) satisfies the equation:

4u 3 − 3u = cosθ.

We are led to the study of the polynomial 4X 3 − 3X − cosθ , with coefficients in Q(cosθ). For θ = 60°, we obtain that u := cos20° is a root of g := 8X 3 − 6X − 1 ∈ Q[X]. We have Irr(u, Q) = g (since g has no rational roots), so [Q(u) : Q] = 3. This shows that u is not construct-ible, since [Q(u) : Q] is not a power of 2.

b) Choose the unit length to be the length of the side of the cube. Thus, the initially constructible points are O and I. The cube having double the volume has the side 3 2 . The real number 3 2 is not con-structible, since [Q( 3 2 ) : Q] = 3 is not a power of 2.

15 This result does not say that no angle can be trisected (for instance, a 90° angle

can be trisected), but that some angles (the 60° angle) cannot be trisected. Thus, there exists no ruler and compass construction of the trisection of an arbitrary angle.


285

c) Choosing the unit length to be the radius of the circle, the ini-tially constructible points are O and I. The area of the circle of radius 1 is π, so the side of the square with area π is π . But π is transcendental over Q and π is also transcendental, so it is not con-structible. !

For the formulation of a necessary and sufficient criterion of constructibility, the following complex version of Theorem 1.11 is useful:

1.14 Theorem. a) The complex number z is constructible from S (i.e. z ∈ K(i)) if and only if there exists a chain of subfields of C, L0 ⊆ L1 ⊆ ⊆ Ln, such that z ∈ Ln, L0 = Q(SR)(i) and [Lt : Lt −1] ≤ 2, for any t ∈ 1, , n.

b) If z is a complex number constructible over S, then z is algebraic over L0 = Q(SR)(i) and its degree over L0 is a power of 2 (there exists e ∈ N such that [L0(x) : L0] = 2e ).

Proof. a) ⇒ Let z = x + yi, with x, y ∈ K. By 1.11, there exists a chain K0 ⊆ K1 ⊆ ⊆ Kn of subfields of R, with K0 = Q(SR), x, y ∈ Kn and [Kt : Kt − 1] ≤ 2, for any t ∈ 1, , n. Let Lt := Kt(i). The chain of subfields L0 ⊆ L1 ⊆ ⊆ Ln satisfies the required conditions.

⇐ The same technique as in Theorem 1.11 is used. By induction on n, we show that for any chain Q(SR)(i) = L0 ⊆ L1 ⊆ ⊆ Ln of sub-fields of C, with [Lt : Lt−1] ≤ 2, ∀t ∈ 1, , n, we have Ln ⊆ K(i). If n = 0, apply 1.9. For n > 0, [Ln : Ln−1] ≤ 2 implies Ln = Ln−1 or Ln is a quadratic extension of Ln−1. Since K(i) is closed under square roots, and Ln−1 ⊆ K(i) by hypothesis, Ln ⊆ K(i).

b) Exercise. !

1.15 Theorem. (Characterization of constructible real num-bers) Suppose U ⊆ R, x ∈ R is algebraic over Q(U) and N is the nor-mal closure of Q(U) ⊆ Q(U)(x). Then x is constructible from U if and only if [N : Q(U)] is a power of 2.


286

Proof. Let g := Irr(x, Q(U)) ∈ Q(U)[X]. Then N is the splitting field of g over Q(U) := K0. We must show that x ∈ K (the numbers constructible from U) if and only if [N : K0] is a power of 2.

⇒ Theorem 1.11 shows there exists a tower K0 ⊆ K1 ⊆ ⊆ Kn of subfields of R, such that [Kt : Kt − 1] ≤ 2, ∀t ∈ 1, , n and x ∈ Kn. Let E be the normal closure of K0 ⊆ Kn. Since x ∈ E and E is normal over K0, g splits over E, so N ⊆ E. On the other hand, E = K0∪σ(Kn)|σ ∈ H, where H is the set of the K0-homomorphisms from Kn to C (see 2.7). For any σ ∈ H, the chain of extensions K0 = σ(K0) ⊆ σ(K1) ⊆ ⊆ σ(Kn) has the property that [σ(Kt) : σ(Kt−1)] ≤ 2, ∀t ∈ 1, , n. The proof of theorem 1.14 im-plies that σ(Kn) ⊆ K(i). So, E ⊆ K(i) and thus N ⊆ K(i). Let α be a primitive element of the separable finite extension K0 ⊆ N. Because α ∈ K(i), theorem 1.14.b) guarantees that [K0(α) : K] = [N : K0] is a power of 2.

⇐ The Galois extension K0 ⊆ N has the degree 2e for some e, so its Galois group G has 2e elements. From (Appendix 6, Proposition 11) there exists a chain of subgroups of G, G = G0 ⊇ G1 ⊇ ⊇ Ge = id, with [Gt − 1 : Gt] = 2, ∀t ∈ 1, , e (so |Gt| = 2 e − t). Denoting by Kt the fixed field of Gt, ∀t ∈ 1, , e, we have K0 ⊆ K1 ⊆ ⊆ Ke = N (we applied the Galois correspondence). We have [Kt : Kt−1] = [Gt−1 : Gt] = 2, ∀t ∈ 1, , n. Theorem 1.14. implies that N ⊆ K(i). Therefore x ∈ R ∩ K(i) = K. !

1.16 Remark. Keeping the same notations, we also have: x is con-structible from U if and only if there exists a normal extension L of Q(U), with [L : Q(U)] a power of 2 and x ∈ L. Indeed, if x belongs to such an extension L, then the normal closure of Q(U) ⊆ Q(U)(x) is in-cluded in L and has its degree over Q(U) a power of 2.

We now solve the classical problem of the constructibility of regu-lar polygons.


287

1.17 Theorem. (Gauss, 1801) Let n ≥ 3. The regular n-gon is con-structible by ruler and compass if and only if the prime factor decomposition of n is n = 2e·p1··pt, where e, t ∈ N and p1, , pt are distinct primes of the form 122 +

m, for some m ∈ N.16

Proof. Here, constructible means constructible from two points (O and I ). To construct a regular n-gon is equivalent to construct the angle17 2π/n. Let a, b ∈ N*, with (a, b) = 1. We claim that the angle 2π/ab is constructible if and only if the angles 2π/a and 2π/b are con-structible. The direct implication is simple: for a given angle (for in-stance 2π/ab), any multiple of it can be constructed (so the angles 2π/a, 2π/b are constructible). Conversely, let u, v ∈ Z such that ua + vb = 1. Multiplying by 2π/ab, we get u·2π/b + v·2π/a = 2π/ab and we apply the fact that if the angles α and β are constructible, then, for any u, v ∈ Z, the angle uα + vβ is constructible.

Thus, if kkppn αα …11= , with p1, , pk distinct primes, the construc-

tion of the angle 2π/n is equivalent to the construction of every angle i

ipαπ2 , for any i ∈ 1, , k. This allows us to suppose from now on that n is a power of a prime.

If n = 2e, e > 1, then the angle 2π/n is constructible (induction on e: the angle 2π/2e is obtained by bisecting the angle 2π/2e − 1).

At 1.13. we saw that the angle α is constructible if and only if cosα is constructible.

16 Such a prime number is called a Fermat prime. For m = 0, 1, 2, 3, 4, one

obtains 3, 5, 17, 257, 65537, which are indeed prime. No other Fermat primes have been found yet. Using computers, it has been proven that the above are all the Fermat primes less than 1040000.

17 We say the angle u instead of the angle of measure u.


288

Suppose from now on that n = pα, p a prime, p ≠ 2, α ∈ N*. We prove that cos(2π/pα ) is constructible ⇔ α = 1 and p is a Fermat prime.

Suppose cos(2π/n) is constructible. Then, using 1.12, [Q(cos(2π/n )) : Q] is a power of 2. Let ζ := cos(2π/n) + isin(2π/n). The next lemma says that [Q(ζ ) : Q] = 2·[Q(cos(2π/n)) : Q]. On the other hand, the degree of the cyclotomic extension Q ⊆ Q(ζ ) is ϕ(n) = ϕ(pα ) = pα − pα − 1, so pα − pα − 1 is a power of 2. Let e ∈ N with pα − 1(p − 1) = 2e. If α ≥ 2, then p | 2

e, absurd. So α = 1 and p − 1 = 2e, whence p = 2e + 1. Suppose e is not a power of 2. Then e = a ·2b, for some a, b ∈ N, a ≥ 3, a odd. So, p = ab22 + 1. Let d :=

b22 , so p = ad + 1. Since a is odd , ad + 1 is divisible by d + 1, so p is not a prime, contradiction. Hence e must be a power of 2, which means that p is a Fermat prime.

Conversely, let p be a Fermat prime. Let us prove that cos(2π/p) is constructible. The extension Q ⊆ Q(ζ ) is normal, of degree ϕ(p) = p − 1, which is a power of 2. Since ζ = cos(2π/p) + isin(2π/p) and ζ − 1 = cos(2π/p) − isin(2π/p) are in Q(ζ ), cos(2π/p) belongs to Q(ζ ). The remark following Theorem 1.15. applies and we obtain the constructibility of cos(2π/p). !

1.18 Lemma. Let n ∈ N, n ≥ 2 and let ζ be a primitive complex nth root of unity. Then

[Q(ζ) : Q(cos(2π/n ))] = 2. Proof. We may suppose that ζ = cos(2π/n) + i·sin(2π/n). Clearly,

ζ ∉ Q(cos(2π/n)), so [Q(ζ ) : Q(cos(2π/n))] ≥ 2. But i·sin(2π/n) is a root of (cos(2π/n)2 − X 2 = 1, so [Q(ζ ) : Q(cos(2π/n))] ≤ 2. !

1.19 Example. A regular heptagon is impossible to construct by ruler and compass, as is a regular 9-gon. In fact, for n ≤ 20, a regular n-gon is constructible ⇔ n ∈ 3, 4, 5, 6, 8, 10, 12, 15, 16, 17.

VI.2 Trace and norm

289

VI.2 Trace and norm

If K ⊆ L is a finite extension and x ∈ L, the conjugates of x over K are the roots of the minimal polynomial of x over K. The sum (and the product) of these conjugates, each occurring with its multiplicity, are elements in K, as the Viète relations show. For instance, if x = a + b d , with d a square free integer and a, b ∈ Q, then a − b d is the only conjugate of x over Q. The sum of all conjugates of a + b d is thus 2a, and the product of all its conjugates is a2 − db2. This is what we called the norm of a + b d and appeared in the study or arithmetic properties of the ring Z[ d ]. These are particular cases of concepts that we study now.

Although we use the notions of trace and norm mainly in the case of a finite extension of fields, the definitions and the proofs are the same as in the more general case of an extension of commutative rings R ⊆ S such that S is a free R-module of finite rank. The reader less familiar with these concepts may assume in what follows that R ⊆ S is a finite extension of fields (so, S is an R-vector space of finite dimen-sion). In this case, free module translates into vector space, mod-ule homomorphism means linear mapping etc.

Let R ⊆ S be an extension of commutative rings with identity18, such that S, with its canonical R-module structure, is free and has rank n. We recall some facts on free modules from II.4. We denote EndR(S) := ϕ : S → S | ϕ is an R-module homomorphism. (EndR(S), +, °) is a ring with identity; "+" denotes homomorphism addition and "°" the usual map composition. Let e = (e1,,en) be a basis of the free R-module S. If ϕ ∈ EndR(S), the matrix of ϕ in the basis e is the n×n matrix Me(ϕ) = (aij)1 ≤ i, j ≤ n, where aij ∈ R are defined by the relations

18 In other words, S is a commutative ring with identity and R is a subring of S

containing the identity.


290

ϕ(ei) = ∑=

n

jjijea

1, ∀i ∈ 1, , n.

We obtain a function Me : EndR(S) → Mn(R), ϕ & Me(ϕ), which is a ring anti-isomorphism. If f = (f1,, fn) is an n-uple of elements in S, then there exists a unique matrix U = (uij)1 ≤ i, j ≤ n ∈ Mn(R) such that:

fi = ∑=

n

jjijeu

1.

f = (f1, , fn) is a basis in R S ⇔ U is invertible in Mn(R) ⇔ det U ∈ U(R) (the group of units of R); in this case,

Mf(ϕ) = U·Me(ϕ)·U −1.

The characteristic polynomial of a vector space endomorphism in-spires the following definition (cf. III.4.18.):

2.1 Definition. Let I ∈ Mn(R) the identity matrix and let A = (aij) ∈ Mn(R). The matrix:

XI − A :=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−−−−

nnnn

n

n

aXaa

aaXaaaaX

……

……

21

22221

11211

∈ Mn(R[X])

is called the characteristic matrix of A. Let ϕ ∈ EndR(S) and let e be an arbitrary basis in R S. Define

P(ϕ) := det(XI − Me(ϕ)) ∈ R[X], called the characteristic polynomial of ϕ.

The characteristic polynomial of ϕ is correctly defined (it is independent on the choice of a basis): if f is another basis, there exists U ∈ U(Mn(R)) such that Mf(ϕ) = U·Me(ϕ)·U −1, so XI − Mf(ϕ) = U·(XI − Me(ϕ))·U −1. Thus,

det(XI − Mf(ϕ)) = detU·det(XI − Me(ϕ))·det(U −1) = det(XI − Me(ϕ)).

VI.2 Trace and norm

291

Let P(ϕ) := X n + an − 1X n − 1 + + a1 X + a0; the coefficients ai ∈ R

depend only on ϕ (not on the choice of the basis), so the following definitions are correct:

Tr(ϕ) := − an − 1 (called the trace of ϕ) det(ϕ) := (−1)na0 (called the determinant of ϕ).

If A = (aij) ∈ Mn(R) is the matrix of ϕ in some basis, then Tr(ϕ) = Tr(A) = a11 + a22 + + ann; det(ϕ) = det(A).

For any x ∈ S, the function θx : S → S, θx(y) = xy, ∀y ∈ S,

is an R-module homomorphism. We apply the definitions above to θx ∈ EndR(S). The following notations and terms are used:

P(θx) =: P(x, S/R) ∈ R[X] is called the characteristic polyno-mial of x;

Tr(θx) =: TrS/R(x) ∈ R is called the trace of x; det(θx) =: NS/R(x) ∈ R is called the norm of x. It is clear that these notions depend not only on x, but on the exten-

sion S of R. The notations we use take this into account, but not the terminology, so some caution is recommended. We defined therefore the mappings trace TrS/R : S → R, and norm NS/R : S → R.

2.2 Remark. θ : (S, +, ·) → (EndR(S), +, ), x & θx, is an R-module homomorphism and also a ring homomorphism (in other words, θ is an R-algebra homomorphism). Prove this!

Using this remark, the next properties are easy to prove:

2.3 Proposition. In the conditions above, we have, ∀x, y ∈ S, ∀a ∈ R: TrS/R(x + y) = TrS/R(x) + TrS/R(y); NS/R(xy) = NS/R(x)·NS/R(y);

TrS/R(ax) = aTrS/R(x); NS/R(ax) = a nNS/R(x); TrS/R(a) = na; NS/R(a) = a n. !

So, TrS/R : S → R is an R-module homomorphism.


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The case we are interested in is when R and S are fields. Fix a finite extension of fields K ⊆ L. We suppose that all algebraic extensions of K are subfields of Ω, a fixed algebraic closure of K.

2.4 Theorem. Let K ⊆ L be a finite extension of fields, let x ∈ L and let P(x, L/K) be its characteristic polynomial. Let H := HomK(L, Ω) be the set of K-homomorphisms from L to Ω. Then: P(x, L/K) = Irr(x, K)[L : K(x)] (1) Irr(x, K) = P(x, K(x)/K) and P(x, L/K) = ∏σ ∈H (X − σx)d, (2) where d = [L : K]i is the inseparability degree of L over K. In particu-lar: TrL/K(x) = [L : K(x)]·TrK(x)/K(x) = d·∑σ ∈H σx (3)

NL/K(x) = (NK(x)/K(x))[L : K(x)] = ∏σ ∈H σxd. (4) Proof. Note that, for any g ∈ K[X], g(θx) = θg(x) (use 2.2 for the

proof). Let P := P(x, L/K) ∈ K[X]. The Cayley-Hamilton theorem for θx ∈ EndK(L) says that P(θx) = 0. So, θP(x) = 0, which means P(x) = θP(x)(1) = 0.

Let f := Irr(x, K). We have f | P, since P(x) = 0. Also, f (θx) = θf(x) = θ0 = 0; if g ∈ K[X] is such that g(θx) = 0, then θg(x) = 0, so g(x) = 0, which means f |g. This shows that f is the minimal polyno-mial of the endomorphism θx. The Frobenius theorem implies that P and f have the same irreducible factors. Since f is irreducible, f is the only irreducible factor of P. Thus, P = f t, for some t ∈ N*. Taking the degrees, we obtain n := deg P = [L : K] = t·deg f = t·[K(x) : K], so t = [L : K(x)]. This proves (1).

Let Q := P(x, K(x)/K). As for P(x, L/K), we have Q(x) = 0, so f |Q; since deg Q = [K(x) : K] = deg f, we obtain f = Q. Identifying the

coefficients of X n−1 in (1), written as P(x, L/K) = P(x, K(x)/K)[L : K(x)],

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we get TrL/K(x) = [L : K(x)]·TrK(x)/K(x), which is the first equality in (3). Setting X = 0 in (1), we obtain the first equality in (4.)

Formula (1) says that Irr(x, K) = P(x, K(x)/K) and reduces the computation of P(x, L/K) to the case L = K(x). Assume for the moment that L = K(x). Let n := |H| = [L : K]s (the separable degree of L over K). The polynomial f has n distinct roots in Ω, namely σx |σ ∈ H. If m is the multiplicity of the root x of f and σ ∈ H, then the root σx ∈ Ω has also multiplicity m. Indeed, for any q ∈ N, (X − x)q| f in Ω[X] ⇔ ∃g ∈ Ω[X] with f = (X − x)qg. Apply σ' (σ : L → Ω extends to a K-isomorphism σ0 : Ω → Ω, and σ0 extends to a unique K-algebra isomorphism σ' : Ω[X] → Ω[X] with σ'(X) = X) and obtain f = (X − σx)qσ'(g). This implies that: for any q ∈ N, (X − x)q| f ⇔ (X − σx)q| f, which proves the claim. In conclusion,

f =∏σ ∈ H (X − σx)m. Comparing the degrees, we obtain [L : K] = mn = m[L : K]s. So,

m = [L : K]i = [K(x) : K]i, the inseparability degree of L over K. In the general case, each σ ∈ HomK(K(x), Ω) has r := [L : K(x)]s

extensions to L (see V.3.11.). So, ∏σ ∈Hom(L, Ω) (X − σx) = ∏τ ∈ Hom(K(x), Ω) (X − τx)r.

By (1), we have: P(x, L/K) = f [L : K(x)] = ∏τ ∈Hom(K(x), Ω) (X −τx)m[L : K(x)]

= ∏σ ∈Hom(L, Ω) (X −σx)m[L : K(x)]/r, where m[L : K(x)]/r = [K(x) : K]i·[L : K(x)]/[L : K(x)]s = [K(x) : K]i·[L : K(x)]i = [L : K]i.

This proves (2). The second part of (3) and (4) follows from (2). !

2.5 Corollary. Let K ⊆ L be a Galois extension and let G be its Galois group. Then, for any x ∈ L, TrL/K(x) = ∑σ ∈G σx NL/K(x) = ∏σ ∈G σx.


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In particular, for any σ ∈ G, NL/K(σx) = NL/K(x) and TrL/K(σx) = TrL/K(x). !

In the case of separable extensions, d = 1 in (2), (3), (4). In fact the inseparable extensions are exactly the extensions for which Tr = 0:

2.6 Proposition. Let K ⊆ L be a finite extension. Then TrL/K : L → K is the 0 mapping if and only if K ⊆ L is inseparable.

Proof. If K ⊆ L is inseparable, then char K = p > 0 and the inse-parability degree d = [L : K]i is a power of p (3.27). Formula (3) im-plies TrL/K = 0.

Conversely, suppose K ⊆ L is separable. Then HomK(L, Ω) has n = [L : K] elements σ1, , σn and TrL/K(x) = σ1x ++ σnx. If TrL/K = 0, then σ1 ++ σn = 0, so σ1, , σn are linearly dependent, contradicting Dedekind's lemma. !

2.7 Example. Let K be a field and let d ∈ K. If g = X 2 − d is irreducible in K[X] (⇔ d is not a square in K) and e = d is a root of g in Ω, the extension K ⊆ K(e) =: L has degree 2 and it is normal. So, HomK(L, Ω) = HomK(L, L) = G(L/K). If char K = 2, then g is insepara-ble and G(L/K) = id; if char K ≠ 2, g is separable and G(L/K) = id, σ, where σ(e) = −e. A basis is 1, e. From (3) and (4) we have, for any a, b ∈ K:

- if char K = 2: NL/K(a + be) = (a + be)2 = a2 + b2d, TrL/K(a + be) = 2(a + be) = 0;

- if char K ≠ 2: NL/K(a + be) = (a + be)(a − be) = a2 − b2d, TrL/K(a + be) = 2a.

An important property of the trace and the norm is the transitivity property: If R ⊆ S ⊆ T are extensions of commutative rings such that S

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is a free R-module and T is a free S-module (of finite ranks), then T is a free R-module of finite rank19 and, for any x ∈ T:

NT/R(x) = NS/R(NT/S(x)), TrT/R(x) = TrS/R(TrT/S(x)). We shall prove this for field extensions; a proof for the general case

uses linear algebra methods.

2.8 Theorem. (The transitivity of the trace and norm) Let K ⊆ L ⊆ E be finite extensions of fields. Then: TrE/K = TrL/KTrE/L and NE/K = NL/KNE/L.

Proof. We use 2.4.(3), 2.4.(4). There exists a bijection

Ψ : HomK(L, Ω)×HomL(E, Ω) → HomK(E, Ω), defined as follows: for any σ ∈ HomK(L, Ω), fix a K-automorphism σ : Ω → Ω with σ |L = σ. For any (σ, τ) ∈ HomK(L, Ω)×HomL(E, Ω), define Ψ(σ, τ) := σ τ (which is a K-homomorphism from E to Ω since σ and τ are K-homomorphisms). We show that Ψ is surjective. If η : E → Ω is a K-homomorphism, let σ := η|L ∈ HomK(L, Ω). Let τ := 1−σ η : E → Ω; τ is an L-homomorphism : ∀x ∈ L, τ(x) = ( )( ) ( )( )xx σσησ 11 −− = = x. Moreover, η = Ψ(σ, τ), since, ∀y ∈ E, ( ) ( )yy ησστσ 1−= = η(y). Since HomK(L, Ω)×HomL(E, Ω) and HomK(E, Ω) are finite and have the same cardinal (see V.3.11), Ψ is also injective.

Let H := HomK(L, Ω), G := HomL(E, Ω), J := HomK(E, Ω). We prove for the trace Tr, the proof for the norm being similar. We have, ∀x ∈ E:

TrL/K(TrE/L(x)) = [L : K]i·∑σ ∈H σ [E : L]i·∑τ ∈Gτx

= [E : L]i[L : K]i·∑(σ,τ)∈H×G xτσ = [E : K]i·∑η∈Jηx = TrE/K(x). !

19 This fact is proven exactly as the theorem of transitivity of finite extensions.


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Exercises

1. Let K ⊆ L be an extension of finite fields, | K | = q, | L | = q n. a) Write explicit formulas for NL/K(x) and TrL/K(x), ∀x ∈ L. b) Prove that Tr : L → K is surjective. Can the hypothesis L is a fi-

nite field be weakened? c) Prove that the norm NL/K is a homomorphism of multiplicative

groups N : L* → K*. d) If α is a generator of the cyclic group L*, then N(α) is a genera-

tor of K*. Deduce that the norm N : L → K is surjective. e) Calculate the cardinal of the set x ∈ L | NL/K(x) = 1.

2. Let n > 1 and let ω be a primitive complex nth root of unity. Show that NL/Q(ω) = 1, where L = Q(ω). 3. Write down formulas for the trace and the norm of an arbitrary ele-ment of a quadratic extension Q ⊆ Q ( )d , with d ∈ Z squarefree. The same problem for Q ⊆ Q ( )3 2 . 4. Let K ⊆ L be a finite extension and let x ∈ L such that L = K(x) and

Irr(x, K) = X n + an − 1 X n − 1 + + a0.

Then NL/K(x) = (−1) na0 and TrL/K(x) = − an − 1.

5. Let p > 2 be a prime and let ω be a primitive complex pth root of unity. Show that NL/Q(1 − ω) = p, where L = Q(ω). Compute NL/Q(a − ω), where a ∈ Q. (Hint. Find Irr(1 − ω, Q)).

6. Let L/K be finite Galois. If K ⊆ F ⊆ L is an intermediate field, then F = K(TrL/F(x) | x ∈ L). Is it true in general that F = K (NL/F(x) | x ∈ L)? 7. Let K be a field and let K(X) be the field of fractions of K[X]. Let ϕ : K(X) → K(X) be the unique K-homomorphism with ϕ(X) = 1/(1 − X). Prove that ϕ ∈ AutK(K(X)). Determine the fixed field of ϕ. (Ind. ϕ 3 = id. Use the trace to find an element in the fixed field.)

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8. The same problem, for ϕ(X) = 1/X, respectively ψ(X) = 1 − X. Determine the subfield of K(X) fixed by ϕ and ψ.


We fix a field K, an algebraic closure Ω of K and we suppose that all the algebraic extensions of K are subfields of Ω. We study first the splitting field of the polynomial X n − 1 over K.

Let L be an extension of K. Recall some notations and facts from IV.3: Un(L) := x ∈ L | x n = 1 is a finite subgroup (necessarily cyclic) of (L*, ·); if |Un(L)| = n, then any generator of Un(L) is called a primi-tive nth root of unity. A primitive nth root of unity exists in Ω ⇔ |Un(L)| = n ⇔ char K - n.

Pn(L) := ω ∈ Un(L) | ordω = n is the set of primitive nth roots of unity in L. If L contains a primitive nth root of unity ω, then Pn(L) = ω i | 1 ≤ i < n, (i, n) = 1, so |Pn(L)| = ϕ(n). The splitting field of X n − 1 over K is also called the n-th cyclotomic field over K.

3.1 Proposition. Suppose n ∈ N*, K is a field with char K - n and L is a splitting field of X n − 1 over K. Let ω be a primitive nth root of unity in L. Then L = K(ω), K ⊆ L is a Galois extension and its Galois group G is isomorphic to a subgroup of (U(Z/nZ), ·), the group of invertible elements of Z/nZ. In particular, G is Abelian and [L : K] is a divisor of ϕ(n).

Proof. K ⊆ L is a separable extension, since (X n − 1)' = nX n−1 ≠ 0 is relatively prime to X n − 1. The extension K ⊆ L is also normal, being a splitting field. If ω generates Un(L), then Un(L) = ω i | 0 ≤ i < n and clearly L = K(ω). For any σ ∈ G, σω ∈ Un(L) (σ permutes the roots of X n − 1). For any k ∈ N, ω k = 1 ⇔ σ(ω k) = 1 ⇔ (σω )k = 1. Thus, the


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order of σω in Un(L) is equal to ordω = n, whence σω ∈ Pn(L). But Pn(L) = ω i | 0 ≤ i < n, (i, n) = 1. So, ∀σ ∈ G, ∃! i < n, (i, n) = 1, such that σω = ω i. Define then ψ(σ ) = i + nZ. We obtain a function ψ : G → U(Z/nZ). An easy check shows that ψ is a homomorphism; ψ is injective, since ψ(σ ) = 1 + nZ implies σω = ω ⇒ σ = id. Since U(Z/nZ) = i + nZ | 0 ≤ i < n, (i, n) = 1 is Abelian and has ϕ(n) ele-ments, the other claims are evident. !

3.2 Remark. a) If K already contains some nth roots of unity, and ω is a primitive nth root of unity, then we can have [K(ω) : K] < ϕ(n). For instance, if n = 8 and K = Q(i) (i 2 = −1), then (ω 2)4 = 1, so ω 2 = ±i. Thus Irr(ω, K) = X 2 − i. So, [K(ω) : K] = 2 < ϕ(8) = 4.

b) If K = Q, then G(Q(ω)/Q) ≅ U(Z/nZ). Indeed, Irr(ω, K) = Φn (the nth cyclotomic polynomial) whose degree is ϕ(n). So, |G(Q(ω)/Q)| = [Q(ω) : Q] = ϕ(n) and ψ in the proof must be an isomorphism.

The extensions of the form K ⊆ K(a), where a has the property that a n = b ∈ K for some n ∈ N, are very important in the problem of solvability by radicals of polynomial equations. A suggestive writing for this situation is a = n b or a = b1/n and an extension as above is written K ⊆ K( n b ). These extensions have a simple description if K contains a primitive nth root of unity.

3.3 Proposition. Let K be a field containing a primitive nth root of unity ω. Let b ∈ K* and let L = K(a), where a is a root of X n − b (in Ω). Then:

a) K ⊆ L is a Galois extension and G(L/K) is cyclic of order m, with m | n .

b) [L : K] = m is the order of bK* n in the group (K*/K* n, ·). (K* n := x n | x ∈ K* is a subgroup in (K*, ·))

c) Irr(a, K) = X m − c, for some c ∈ K.


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Proof. a) The roots of X n − b are aω i | 0 ≤ i < n, thus K ⊆ L is normal. It is also separable (because char K does not divide n), so it is Galois. If σ ∈ G := G(L/K), then σ(a) ∈ aω i | 0 ≤ i < n. For any σ ∈ G, there exists a unique i < n such that σ(a) = aω i; define ϕ : G → Z/nZ, ϕ(σ ) = i + nZ. The function ϕ is a injective homomorphism, since ϕ(σ ) = 0 + nZ implies σa = a ⇒ σ = id. Thus, G is isomorphic to Imϕ, a subgroup of Z/nZ, so it is cyclic. |G| di-vides n, by the theorem of Lagrange.

b) Let η be a generator of G. Then m = |G| = ordη. There exists s < n such that η(a) = aω s. We have

NL/K(a) = ∏σ ∈G σa = ∏0 ≤ i < m aω si = a mω t ∈ K, with t ∈ N, so a m ∈ K. Therefore, b m = (a n)m = (a m)n ∈ K* n. Let q de-note the order of bK* n in grupul (K*/K* n, ·). We have q | m. Let us show that m | q. Since b q ∈ K* n, b q = c n, for some c ∈ K. So, (a q)n = b q = c n , whence a q = cω i, with i < n. Thus, a q ∈ K, so a q is fixed by η: η(a q) = η(a)q = a qω sq = a q. We obtain ω sq = 1 ⇒ η q = id. This shows that ordη = m divides q. From m | q and q | m we deduce that m = q.

c) deg Irr(a, K) = [L : K] = m, and a is a root of X m − a m ∈ K[X]. !

3.4 Definition. A Galois extension (normal and separable) whose Galois group is cyclic (respectively Abelian) is called a cyclic exten-sion (respectively Abelian extension).

The previous proposition says: if K contains a primitive nth root of unity and b ∈ K*, then K ⊆ K( n b ) is cyclic.

We want to prove that, conversely, any finite cyclic extension of K is of the form K ⊆ K( n b ) (if K contains a primitive nth root of unity). This step is crucial in the proof of the characterization of solvability by radicals of an equation. We use the following result (very important by itself):


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3.5 Lemma. (Hilbert Satz 90)20 Let K ⊆ L be a cyclic extension of degree n, let σ be a generator of G(L/K) and let x ∈ L*. Then:

NL/K(x) = 1 ⇔ there exists y ∈ L* such that x = σ(y)/y. Proof. One implication is easy: if x = σ(y)/y, using NL/K(y) =

NL/K(σy) we obtain that NL/K(x) = NL/K(σy)/NL/K(y) = 1. Suppose now that NL/K(x) = xσ(x)σ n−1(x) = 1. Let η := σ −1 and let

x0, , xn−1 ∈ L, defined as follows: x0 := x, x1 := xη(x0) = xη(x), , xn−1 := xη(xn−2) = xη(x)η n−1(x).

Let u := x0·id + x1·η + + xn−1η n−1 ( the Lagrange-Hilbert resol-vent ).

Consider u : L → L as an endomorphism of the K-vector space L. The homomorphisms id, η, , η n−1 from L* to L* are distinct. Dede-kind's lemma says they are linearly independent in the L-vector space LL*; since x0 ≠ 0, we have u ≠ 0, so there exists z ∈ L such that u(z) =: t ≠ 0. Thus, t = u(z) = x0z + x1η(z) + + xn−1η n−1(z), so

η(t) = η(x0)η(z ) + η(x1)η 2(z) + + η(xn−1)η n(z)

= x1η(z ) + x2η 2(z) + + z (we used that xn−1 = NL/K(x) = 1 and

η n = id) We remark that η(t) = t/x0, so x = x0 = t/η(t). Setting y := σ (t) fin-

ishes the proof. !

3.6 Proposition. Let K be a field containing a primitive nth root of unity ω and let K ⊆ L be a cyclic extension of degree n. Then there ex-ists a ∈ L*, with a n ∈ K, such that L = K(a).

Proof. Let G := G(L/K) and let σ be a generator of G. Since ω ∈ K, NL/K(ω) = ω n = 1. By Hilbert's Satz 90, there exists a ∈ L with σ(a) = ω a, so σ i(a) = ω ia, 0 ≤ i ≤ n − 1. Thus, ω ia | 0 ≤ i ≤ n − 1 are n distinct conjugates of a over K, so Irr(a, K) has degree ≥ n. Because

20 This result is included as "Satz 90" in the Hilbert's monumental Bericht über

die Theorie der algebraischen Zahlkörper (Zahlbericht for short, published in 1897).


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σ(a n) = ω na n = a n, a n is fixed by the subgroup <σ > = G, so a n ∈ K. The polynomial X n − a n ∈ K[X] has the root a and its degree is n, so it is equal to Irr(a, K). Thus, [K(a) : K] = n and K(a) = L. !

The propositions 3.3 and 3.6 determine all cyclic extensions of de-gree n of a field K of characteristic exponent p, such that p - n and a K contains a primitive nth root of unity. The cyclic extensions of degree p of a field of characteristic p > 0 are described in what follows. This part is not used in the theory of solvability by radicals and may be skipped in a first reading.

3.7 Lemma. (Hilbert Satz 90, additive version) Let K ⊆ L be a cy-clic extension of degree n, let σ be a generator of G(L/K) and let x ∈ L*. Then:

TrL/K(x) = 0 ⇔ there exists y ∈ L* such that x = σ(y) − y. Proof. The argument is essentially the same as in the multiplicative

version (3.5). If x = σ(y) − y, then TrL/K(x) = TrL/K(σy) − TrL/K(y) = 0, since TrL/K(y) = TrL/K(σy).

Suppose TrL/K(x) = x +σ(x) + + σ n−1(x) = 0. Let x0, , xn−1 ∈ L, defined as follows: x0 := x, x1 := x + σ(x0) = x + σ(x), , xn−1 := x + σ(xn−2) = x + σ(x) ++ σ n−1(x) ( = TrL/K(x) = 0)

Let u = x0σ + x1σ 2 ++ xn−2σ n−1∈ EndK(L). We have:

σu = σ(x0)·σ 2 + σ(x1)·σ

3 + + σ(xn−2)·σ n = (x1 − x)·σ 2 + (x2 − x)·σ 3 + + (xn−1 − x)·σ n = x1·σ

2 + x2·σ 3 + + xn−1·σ n + x·σ − x·(σ +σ 2 + +σ n−1 + σ n)

= u − x·TrL/K. We have TrL/K = id + σ ++ σ n−1 ≠ 0 since, by Dedekind's

Lemma, id, σ, , σ n−1 are L-linearly independent. Let z ∈ L* such that TrL/K(z) ≠ 0. If t := z/TrL/K(z), then TrL/K(t) = 1 and thus σ(u(t)) = u(t) − x. Thus, x = σ(y) − y, where y = −u(t). !


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3.8 Theorem. (Artin-Schreier21) Let K be a field of characteristic p > 0 and let K ⊆ L be a cyclic extension of degree p. Then there exists a ∈ L, with a p − a =: b ∈ K, such that L = K(a).

Conversely, let b ∈ K such that f = X p − X − b does not split over K. Then f is irreducible in K[X] and, for any root a ∈ Ω of f, the exten-sion K ⊆ K(a) is cyclic of degree p.

Proof. Note that it is enough to suppose that K ⊆ L is Galois; its Galois group must then be cyclic, having prime order.

Let σ be a generator of G := G(L/K). We have TrL/K(1) = [L : K]·1 = p·1 = 0. Applying the additive Hilbert Satz 90, there exists a ∈ L such that 1 = σa − a, so σa = a + 1. Then σ 2a = a + 2 and, by induction, σ na = a + n·1, for any n ∈ N. Thus, a has p distinct conjugates in L, namely a, σa = a + 1, , σ p−1a = a + (p − 1), so Irr(a, K) has degree > p. Since K(a) ⊆ L and [L : K] = p, we have K(a) = L. Because

σ(a p − a) = (σa) p − σa = (a + 1) p − a − 1 = a p − a, a p − a is fixed by σ, whence a p − a ∈ L < σ > = K.

Conversely, let b ∈ K, f = X p − X − b and let a ∈ Ω \ K be a root of f. Then a + 1 is also a root : (a + 1) p − (a + 1) − b = a p − a − b = 0. Then there exists a K-isomorphism σ : K(a) → K(a + 1), with σ(a) = a + 1. Therefore σ ∈ G(K(a)/K) and ord σ = p, since σ na = a + n·1, for any n ∈ N. Because |G(K(a)/K)| = p ≥ [K(a) : K], K ⊆ K(a) is Galois and p = [K(a) : K] (use V.4.4). Since deg Irr(a, K) = p = deg f, we deduce that f = Irr(a, K), so f is irreduci-ble. G(K(a)/K) has order p, so it is cyclic. !

3.9 Remark. The proof also shows that, for any b ∈ K, where char p > 0, X p − X − b is either irreducible in K[X] or splits over K.

21 Otto Schreier (1901-1929), German mathematician.


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The theorems that characterize cyclic extensions, combined with the fact that any Abelian finite group is a product of cyclic groups, lead to structure theorems of finite Abelian extensions of the field K. The theory we developed so far allows us to obtain such theorems in two situations (described by the hypotheses in prop. 3.6, respectively 3.8):

- for Abelian extensions of exponent n, if K contains a primitive nth root of unity;

- for Abelian extensions of exponent p, if char K = p. This type of theorems are known as Kummer Theory.22 If K does not contain a primitive nth root of unity, the theory is

considerably more complicated. In some cases (when K is a local field or when K is a finite extension of Q), a theory that describes all Abelian extensions of K using only the internal structure of K is the Class Field Theory, one of the major accomplishments of 20th century Algebra. (see e.g. NEUKIRCH [1986]). We describe now Kummer the-ory.

3.10 Definition. If G is a group such that ordx | x ∈ G is a finite set, then the LCM of ordx | x ∈ G is called the exponent of G, de-noted exp G. If G is finite, then its exponent is defined and exp G di-vides |G|, by Lagrange's theorem. If K ⊆ L is a finite Galois extension, the exponent of the extension is the exponent of its Galois group.

Let n ∈ N* and let K be a field containing a primitive nth root of unity. An n-Kummer extension is a finite Galois extension K ⊆ L, such that G(L/K) is an Abelian group whose exponent divides n. An exten-sion is called a Kummer extension if it is n-Kummer for some n ∈ N*.

22 Ernst Eduard Kummer (1810-1893),German mathematician.


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If K contains a primitive nth root of unity and b ∈ K*, 3.6 says that K( n b )/K is a cyclic extension and its degree divides n, so it is an n-Kummer extension.

3.11 Lemma. Let L/K be an Abelian finite extension of exponent n. Then L is the composite of a finite set of cyclic intermediate fields K ⊆ Li, 1 ≤ i ≤ m, and each degree [Li :K] divides n.

Proof. Let G := G(L/K). The finite Abelian group G is a direct product of cyclic subgroups: G = C1××Cm; since exp G = n = |C1|··|Cm|, it follows that |Ci| divides n, 1 ≤ i ≤ m. Take the subgroup Hi := ∏ j≠i Cj and let Li be the fixed subfield of Hi. The extension Li/K is Galois and G(Li/K) ≅ G/Hi ≅ Ci, so Li/K is cyclic and [Li :K] divides n. Let E be the composite L1Lm. We have G(L/E) = H1∩∩Hm = id, so L = E. !

For any extension K ⊆ L and any subset S of K, let n S := x ∈ L| x n ∈ S.

3.12 Proposition. Let n ∈ N* and let K be a field containing a primitive nth root of unity ω. Then a finite extension K ⊆ L is n-Kummer if and only if there exist a1, , am ∈ K* \ K* n such that L = K( n

mn aa ,,1 … ). For such an extension, L = K( n S ), where S :=

L*n ∩ K*is a subgroup in (K*, ·). Proof. a) Let L = K(b1, , bm), with bi ∈ L and i

ni ab = , 1 ≤ i ≤ m.

Since ω ∈ K, X n − ai has n distinct roots in L: bi, biω, , biω n −1. So,

L is a splitting field over K of the separable polynomial ∏1≤ i ≤ m(X n − ai), which implies that K ⊆ L is Galois. If σ ∈ G(L/K), σ is determined by its values on the generators b1, , bm. But σ(bi) = biω

j, with j ∈ N depending on σ and i. Let τ ∈ G(L/K), with τ(bi) = biω

k. Then: (σ τ)(bi) = σ(biω

k) = σ(bi)σ(ω k) = biω jω k = biω

j + k = (τ σ)(bi), ∀i,


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so σ τ = τ σ. Thus, G(L/K) is Abelian. Also, σ n(bi) = biω jn = bi, so

σ n = id and the exponent of G(L/K) divides n. Suppose now that K ⊆ L is n-Kummer. By 3.11, L = L1Lm, where

each Li is a cyclic extension of K, included in L; G(Li/K) has order di, with di|n. For the extension K ⊆ Li we can apply prop. 3.6, since idnω is a primitive di-th root of unity in K. So, Li = K(bi), with bi ∈ L* and

idib ∈ K*. Then L = K(b1, , bm) and clearly n

ib =: ai ∈ K*. We have di > 1 (otherwise K = Li), so ai ∉ K*n (if ai = c n, for some c ∈ K, then, for some j, bi = cω j ∈ K and Li = K(bi) = K, absurd).

The last statement is justified as follows: S is a subgroup in (K*, ·) and K( n S ) ⊆ L. But L = K(b1,, bm), with bi ∈ L* and i

ni ab = ∈ K*,

1 ≤ i ≤ m, so ai ∈ S. So, L ⊆ K( n S ). !

This result generalizes 3.3 and 3.6. In order to investigate the struc-ture of the Galois group of an n-Kummer extension, we need some background.

3.13 Definition. Let (H, ·) be an Abelian group with the operation denoted multiplicatively, let 1 be its identity element and let n ∈ N*. Let tn(H) := x ∈ H | x n = 1.

An easy check shows that tn(H) is a subgroup in H (it is the kernel of the homomorphism x & x n) and exp tn(H) divides n. We use this notion for an extension K ⊆ L, for the factor group L*/K* (L* is a group with respect to multiplication; K* is a subgroup). Then:

tn(L*/K*) = xK* | x ∈ L*, x n ∈ K*.

3.14 Proposition. Let K ⊆ L be an n-Kummer extension. Then the canonical group homomorphism

ϕ : tn(L*/K*) → K*/K* n,ϕ(xK*) = x nK* n, ∀x ∈ L* with x n ∈ K*, is injective. The homomorphism ϕ induces an isomorphism tn(L*/K*) ≅ Imϕ = (L* n ∩ K*)/K* n, which is a subgroup in K*/K* n.


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Proof. ϕ is well defined: if xK*∈ tn(L*/K*), then x n ∈ K*; if x, y ∈ L* such that xK* = yK* ∈ tn(L*/K*), then xy −1 ∈ K*, so (xy −1) n ∈ K* n, i.e. x nK* n = y nK* n. It is immediate that ϕ preserves multiplication.

Let us show Kerϕ = 1K*. Let x nK* n = 1K* n for x ∈ L* with x n ∈ K. Hence x n ∈ K* n, so there exists a ∈ K* such that x n = a n. Then x = aζ, where ζ is an nth root of unity (ζ ∈ K by hypothesis). So, x ∈ K* and x K* = 1K*.

Imϕ = x nK* n| x ∈ L*, x n ∈ K* = x nK* n| x n ∈ L* n ∩ K* = (L* n ∩ K*)/K* n. The inverse of ϕ is

ψ : (L* n ∩ K*)/K* n → tn(L*/K*), ψ(yK* n) = n y K*.

The definition of ψ is correct: ∀y ∈ L* n ∩ K*, ∃x := n y ∈ L* such

that x n = y. The class xK* is independent on the choice of the root x of X n − y, since any other root is of the form xα, with α ∈ Un(K), so xαK* = xK*. !

The main theorem says that the n-Kummer extensions of K are in one-to-one correspondence with the finite subgroups of K*/K* n. If K ⊆ L is n-Kummer, then

G(L/K) ≅ tn(L/K) ≅ (L* n ∩ K*)/K* n. We need some group theoretical facts, stated in the following

lemma.

3.15 Lemma. Let (G, ·), (H, ·) be finite Abelian groups and let (C, ·) be a cyclic group. Denote by 1 the neutral elements.

a) Hom(G, C*) ≅ G. (C* is the multiplicative group of nonzero com-plex numbers). If exp G divides |C|, then

Hom(G, C*) ≅ Hom(G, C) ≅ G.23

23 Hom(G, C*) is called the dual of the group G.


307

b) Let p : G × H → C be a bilinear function: for any a, b ∈ G, ∀x, y ∈ H, we have: p(ab, x) = p(a, x)p(b, x), (*) p(a, xy) = p(a, x)p(a, y). (**)

Then, for any x ∈ G, px : G → C, px(a) = p(a, x), ∀a ∈ G, is a group homomorphism. The function η : H → Hom(G, C), η(x) = px, ∀x ∈ H is a group homomorphism.

c) Suppose that the bilinear function p is nondegenerate: x ∈ H | p(a, x) = 1, ∀a ∈ G = 1 and

a ∈ G | p(a, x) = 1, ∀x ∈ H = 1. Then exp G divides |C| and η : H → Hom(G, C) is an isomorphism,

so: H ≅ Hom(G, C) ≅ G.

Proof. Note first that, for any Abelian group (A,·), Hom(G, A) = ϕ : G → A |ϕ is a homomorphism is an Abelian group with respect to the law (αβ)(x) = α(x)β(x), ∀α, β ∈ Hom(G, A), ∀x ∈ G.

a) Let G' := Hom(G, C*). Let exp G = n. Then α(x)n = α(x n ) = 1, ∀α ∈ G', ∀x ∈ G. So, Imα ⊆ Un, the group of complex nth roots of unity, cyclic of order n. Thus, G' = Hom(G, Un). If n divides |C|, then C includes a unique subgroup with n elements Cn (Cn is cyclic) and, as above, Hom(G, C) = Hom(G, Cn).

Since Cn ≅ Un, we have Hom(G, Cn) ≅ Hom(G, Un) = G'. Let us prove that G' ≅ G. Suppose first that G is cyclic of order n

and let g ∈ G be a generator. Let α ∈ G' and let ω generate Un. Then α is determined by α(g), and α(g) = ω s, for some s ∈ N. Define ϕ : Z → G', ϕ(s) = αs, ∀s ∈ Z, where αs(g) = ω s. A quick check shows that ϕ is a surjective group homomorphism and Kerϕ = nZ. So, Z/nZ ≅ G', which means that G' is cyclic of order n, isomorphic to G.

If G is an arbitrary Abelian finite group, the invariant factors theo-rem says that G = G1××Gm, with Gi cyclic groups. But

Hom(G1××Gm, C) ≅ Hom(G1, C)××Hom(Gm, C)


308

(see II.3.20) and each Hom(Gi, C) ≅ Gi, as shown. Thus Hom(G, C) ≅ G1××Gm = G.

b) Condition (*) means that px is a homomorphism. Let x, y ∈ H and let a ∈ G. We have

η(xy)(a) = pxy(a) = p(a, xy) = p(a, x)p(a, y) = η(x)(a)·η(y)(a), so η(xy) = η(x)η(y).

c) Ker η = x ∈ H | η(x) = 1 = x ∈ H | p(a, x) = 1, ∀a ∈ G = 1 (p is nondegenerate). So η is injective. Let m = |C|. For any x ∈ G, p(x m, y) = p(x, y) m = 1, ∀y ∈ H. Because p is nondegenerate, x m = 1, ∀x ∈ G. But exp G | m. So, G ≅ Hom(G, C) by a).

Therefore, H ≅ Imη ≤ Hom(G, C) ≅ G, so H is isomorphic to a sub-group of G, in particular |H| ≤ |G|. The situation is symmetric in G and H, so G is isomorphic to a subgroup of H and |G| ≤ |H|. But G and H are finite, so |G| = |H| and G ≅ H. !

3.16 Theorem. (multiplicative Kummer Theory) Let K be a field containing a primitive nth root of unity ω.

a) If K ⊆ L is an n-Kummer extension, then there is a canonical isomorphism:

tn(L*/K*) ≅ Hom(G(L/K), Un) aK* & χa, where χa(σ) = σ(a)/a,

for any aK* ∈ tn(L*/K*), ∀σ ∈ G(L/K). Also, there is a canonical isomorphism

(L* n ∩ K*)/K* n ≅ Hom(G(L/K), Un), bK* n & χb, where χb(σ) = ( ) nn bbσ ,

∀ bK* n ∈ (L* n ∩ K*)/K* n, ∀σ ∈ G(L/K). Therefore, G(L/K) is isomorphic to the finite subgroup

(L* n ∩ K*)/K* n of K*/K* n. b) There exists an order preserving bijection between the set of all

n-Kummer extensions of K included in Ω and the finite subgroups of K*/K* n:


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the extension K ⊆ L corresponds to the subgroup (L*n ∩ K*)/K* n;

the subgroup S/K*n, with K*n≤ S ≤ K*, corresponds to the extension K ⊆ K( n S ).

The degree of the extension is equal to the order of its correspond-ing subgroup.

Proof. a) Let Un = Un(K) be the group of the nth roots of unity in K (we know that Un = Un(Ω) is a cyclic group of order n). Define:

p : G(L/K)×tn(L*/K*) → Un, p(σ, xK*) := σ(x)/x, for any σ ∈ G(L/K) and any x ∈ L* with xK*∈ tn(L*/K*).

The definition is correct: ∀x ∈ L* with xK*∈ tn(L*/K*), we have x n = a ∈ K, so σ(x) is a root of X n − a, so (σ(x)/x) n = 1; thus, σ(x)/x ∈ Un. It is easily seen that p(σ, xK*) is independent on the choice of the representative of the class xK*.

p is bilinear. Indeed, ∀σ, τ ∈ G(L/K) and ∀xK*, yK*∈ tn(L*/K*), we have:

p(στ, xK*) = ( ) ( )( )

( ) ( ) ( ) ( ) ( )xx

xx

xx

xx

xx

xx

xx τστσττ

τστστ

⋅=⋅⎟⎠⎞

⎜⎝⎛=⋅= =

p(σ, xK*)p(τ, xK*) We used first that στ = τσ (G(L/K) is Abelian), then that

σ(x)/x ∈ K, so it is fixed by τ. Also, p(σ,xK*·yK*) = p(σ,xyK*) = σ(xy)/xy = (σ(x)/x)·(σ(y)/y) = p(σ, xK*)p(σ, yK*).

p is nondegenerate: if σ ∈ G(L/K) and p(σ, xK*) = 1, ∀xK* ∈ tn(L*/K*), then σ(x) = x, ∀x ∈ L with x n ∈ K. But L = K(x ∈ L*| x n ∈ K) (see 3.12), so σ = id. If xK* ∈ tn(L*/K*) with p(σ, xK*) = 1, ∀σ ∈ G(L/K), then x is fixed by any σ ∈ G(L/K), so x ∈ K. Thus, xK* = 1K*.

We have tn(L*/K*) ≅ Hom(G(L/K), Un), via the isomorphism in Lemma 3.15.c). Using ψ : (L* n ∩ K*)/K* n → tn(L*/K*), the isomor-phism in 3.14, ψ(yK* n) = n y K*, we obtain the other isomorphism. Fi-nally, 3.15.a) implies


310

tn(L*/K*) ≅ (L* n ∩ K*)/K* n ≅ G(L/K). b) Any finite subgroup of K*/K* n can be written uniquely as S/K* n,

with K* n ≤ S ≤ K* and [S : K* n] finite. If K ⊆ L is n-Kummer, (L* n ∩ K*)/K* n ≅ G(L/K), a finite group.

Thus, the subgroup corresponding to the extension K ⊆ L is finite. Conversely, if K* n ≤ S ≤ K*, S/K* n is finite, and a1, , am ∈ K* are the representatives of the classes in S/K* n, then

K ⊆ K n S = K nm

n aa ,,1 … is an n-Kummer extension (see 3.12). Clearly, these correspondences are inclusion preserving. They are also inverse to each other, as we now prove.

Let K ⊆ L be n-Kummer. Its associated subgroup is S = L* n ∩ K*. To S/K* n corresponds K( n S ), equal to L (by 3.12).

Conversely, let K* n ≤ S ≤ K*, where ∆ := S/K* n is a finite group. Its associated n-Kummer extension is K ⊆ K( n S ) =: L. Let (L* n ∩ K*)/K* n =: ∆'. We have to prove that ∆ = ∆' (⇔ S = L* n ∩ K*). Evidently, S ⊆ L* n ∩ K*, so ∆ ⊆ ∆'. For the other inclusion, define

p : G(L/K) × ∆ → Un, p(σ, xK* n) := σ( n x )/ n x , ∀σ ∈ G(L/K), ∀xK* n ∈ ∆.

As in a), one shows that p is correctly defined and bilinear. It is nondegenerate: if σ ∈ G(L/K) and p(σ, xK* n) = 1, ∀xK* n ∈ ∆, then σ( n x ) = n x , ∀x ∈ S. But L = K( n S ), so σ = id. If xK* n ∈ S/K* n with p(σ, xK*) = 1, ∀σ ∈ G(L/K), then n x is fixed by any σ ∈ G(L/K), so n x ∈ K and x ∈ K* n. Consequently, xK* n = 1K* n.

From 3.15.c), ∆ ≅ Hom(G(L/K), Un) via the isomorphism bK* n & χb, χb(σ) = ( ) nn bbσ , ∀bK* n ∈ S/K* n , ∀σ ∈ G(L/K). But ∆' ≅ Hom(G(L/K), Un), by a). Since ∆ ⊆ ∆' and the groups are finite, they are equal. !

Using the same methods, an analogous result can be obtained for Abelian extensions of exponent p of a field K having characteristic p > 0. For any extension K ⊆ L, we use the following notations:


311

L+ is the additive group (L, +); P : Ω+ → Ω+ is the group homomorphism given by P(x) = x p − x,

∀x ∈ Ω; tP(L

+/K +) := x + K + ∈ L+/K + | P(x) ∈ K +. The additive homomorphism P plays the role of the multiplicative

homomorphism x & x n in the multiplicative theory. For any a ∈ K +, P

−1(a) denotes any root in Ω of X p − X − a (3.8 ensures that, if b is a root, the roots of X p − X − a are b, b + 1, , b + (p − 1)).

The following results are the additive versions of 3.12, 3.14, 3.16. The proofs are proposed as an exercise.

3.17 Proposition. Let K be a field of characteristic p > 0 and let K ⊆ L be a finite extension. Then K ⊆ L is Abelian of exponent p if and only if there exist a1, , am ∈ K + \ P(K +) such that L = K(P −1(a1), , P −1(am)). For such an extension, L = K(P −1(S)), where S := P(L+) ∩ K

+, a subgroup in K +. !

3.18 Proposition. Let K be a field of characteristic p > 0 and let K ⊆ L be a finite Abelian extension of exponent p. Then the canonical group homomorphism

ψ : tP(L+/K +) → K +/P(K +), ψ(x + K +) = P(x) + P(K +), ∀x ∈ L+ cu

P(x) ∈ K +, is injective. The homomorphism ψ induces an isomorphism

tP(L+/K +) ≅ Imψ = (P(L +) ∩ K

+))/P(K +), a subgroup in K +/P(K +). !

3.19 Theorem. (Additive Kummer Theory) Let K be a field of characteristic p > 0 and let Fp be its prime subfield. Fp

+ is a cyclic group of order p.

a) If K ⊆ L is Abelian of exponent p, then there exists a canonical isomorphism

tP(L+/K +) ≅ Hom(G(L/K), Fp

+),


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a + K + & χa, where χa(σ) = σ(a) − a, ∀ a + K + ∈ tP(L

+/K +), ∀σ ∈ G(L/K). We also have a canonical isomorphism

(P(L+) ∩ K +)/P(K +) ≅ Hom(G(L/K), Fp+),

b + P(K +) & χb, where χb(σ) = σ(P −1(b)) − P −1(b), ∀ b + P(K +) ∈ (P(L+) ∩ K +)/P(K +), ∀σ ∈ G(L/K).

Therefore G(L/K) is isomorphic to the finite subgroup (P(L+) ∩ K +)/P(K +) of K +/P(K +).

b) The set of Abelian extensions of exponent p of K included in Ω is in a bijective inclusion preserving correspondence with the finite sub-groups of K +/P(K +): the extension K ⊆ L corresponds to the subgroup (P(L+) ∩ K +)/P(K +);

the subgroup S/K +, with P(K +) ≤ S ≤ K +, corresponds to the extension K ⊆ K(P −1(S)).

The degree of the extension is equal to the order of the correspond-ing subgroup. !

Exercises

1. Let ω be a primitive 12th root of unity. Determine the Galois group of Q ⊆ Q(ω, 12 2 ) and all its intermediate fields. 2. For any n ∈ N*, let ζn be a primitive nth root of unity. Prove that, if m, n ∈ N, (m, n) = 1, then Q(ζm) ∩ Q(ζn) = Q. 3. Let p1, , pn be distinct primes. Then Q ⊆ Q npp ,,1 … has de-gree 2 n. Find its Galois group. 4. Determine the degree of Q ⊆ Q d | 1 ≤ d ≤ 20. Generalization.

VI.4 Solvability by radicals

313

5. Let K = C(X, Y) (the fraction field of the polynomial ring C[X, Y ]). Let L = K 4 34 , YXY , where, ∀t ∈ K, n t denotes a root of T n − t (in some extension of K). Prove that K ⊆ L is a 4-Kummer extension and determine its degree and all its intermediate fields. 6. Construct a 6-Kummer extension of degree 18 of C(X, Y).


The classical problem of solving by radicals an algebraic equa-tion requires expressing the roots of a given polynomial as a function of the coefficients of the polynomial, using only the four arithmetic operations and radicals (of any order). A typical example is the for-mula for the roots of the equation of second degree (known to the Babylonians, about 1900-1600 B.C.). In the 16th century, the formulas for the roots of any polynomial equation of degree 3 or 4 were discov-ered. These successes led many mathematicians of that time to think that such formulas exist for algebraic equations of any degree. For in-stance, Euler24, around 1749, believed that the expressions for the roots contain no other operations than radicals, except for the four vul-gar operations, and one cannot sustain that transcendental operations might be involved. These ideas were proven to be false at the begin-ning of the 19th century by Paolo Ruffini, who proved the impossibil-ity of solving the general equation of degree 5 by radicals. Independently from Ruffini (whose intricate proof failed to convince many fellow mathematicians), the Norwegian mathematician Niels

24 Leonhard Euler (1707-1783), famous Swiss mathematician.


314

Henrik Abel gave in 1824 (at the age of 22) a clear and rigorous proof of this impossibility. In 1829, two months before his death, Abel pub-lished a memoir in which he describes a class of polynomials solvable by radicals (namely the polynomials whose Galois group is commuta-tive). The commutative groups are called today Abelian, in his honor. In 1830, Evariste Galois, unaware of Abel's results, creates the notion of group (of permutations) and formulates a general criterion of solvability by radicals of a polynomial equation, using what we call today the Galois group of the polynomial. The ideas and results of Galois had a decisive contribution to the development of Algebra.

4.1 Definition. A field extension K ⊆ L is called a radical exten-sion if there exists a finite tower of intermediate fields K = K0 ⊆ K1⊆ ⊆ Km = L, such that, for any i, 0 ≤ i < m, there exists ai ∈ Ki+1 and ni ∈ N*, cu Ki+1 = Ki(ai) and in

ia ∈ Ki. If n1 = = nm = n, K ⊆ L is called an n-radical extension.

Let f ∈ K[X] and let E be the splitting field of f over K. The Galois group G(E/K) is called the Galois group of the polynomial f over K. We say that the polynomial f is solvable by radicals over K (or that the equation f = 0 is solvable by radicals over K) if there exists a radical extension L that includes the splitting field of f over K.

These definitions articulate rigorously the notion of formula using radicals and the four operations. For instance, if a, b ∈ Q,

Q ⊆ ( ) ( )( )3 3⊆ −a a b aQ Q = L,

is a 6-radical extension. Any element of L is expressed in a basis of L over Q (for instance 1, α, β, β 2, αβ, αβ 2, where α = a ,

β = 3 3b a− ) as an expression using radicals and the four opera-

tions.


315

4.2 Examples. a) The equation aX 2 + bX + c = 0, where a, b, c ∈ C, a ≠ 0, is solvable by radicals over Q(a,b,c). Indeed, the formula

for the roots, 2

1,24

2− ± −

=b b acx

a shows that the roots xi are in the

2-radical extension Q(a,b,c) ⊆ ( )( )2, , 4a b c b ac−Q . Note that, if a,

b, c ∉ Q, then the equation may not be solvable over Q. b) Any polynomial f ∈ R[X] is solvable by radicals over R. Indeed

the extension R ⊆ C is radical (why?) and C is algebraically closed, so it includes the splitting field of f over R.

Historically, solvability by radicals means solvability by radicals over Q of polynomials in Q[X].

We want to prove the following result (Galois' criterion of solvabil-ity by radicals): If char K = 0, a polynomial is solvable by radicals over K if and only if its Galois group over K is solvable.

4.3 Remarks. a) If K ⊆ L is a radical extension, as in the definition above, then K ⊆ L is n-radical, where n = LCM(n1, , nm).

b) The (n-)radical extensions are transitive: If K ⊆ L and L ⊆ M are (n-)radical, then K ⊆ M is (n-)radical. The converse is false: if K ⊆ M is n-radical and K ⊆ L ⊆ M, then L ⊆ M is n-radical, but K ⊆ L is not necessarily radical (see example 5.6).

c) If K ⊆ L is such that L = K(x1, , xn), where, for each i, xim ∈ K

for some m ∈ N*, then K ⊆ L is m-radical.

4.4 Proposition. The normal closure of an n-radical extension is an n-radical extension.

Proof. Let K ⊆ L be n-radical and let N be its normal closure. There exists a tower K = K0 ⊆ K1⊆ ⊆ Km = L and ai ∈ Ki+1, 0 ≤ i < m, with n

ia ∈ Ki and Ki+1 = Ki(ai).


316

We prove by induction on m that K ⊆ N is radical. If m = 1, then L = K(a), with a n = b ∈ K, and N = K(x1, , xr), where x1, , xr are the roots of g := Irr(a, K). Since g | X n − b, we have n

ix = b, 1 ≤ i ≤ r, so K ⊆ N is n-radical.

If m > 1, let M be the normal closure of K ⊆ Km −1. By induction, K ⊆ M is n-radical. It is enough to prove that M ⊆ N is n-radical (and then apply the transitivity of n-radical extensions). Note that M is the splitting field over K of the family of polynomials Irr(ai, K)| 1 ≤ i ≤ m − 1, and N is the splitting field over K of the family of polynomials Irr(ai, K)| 1 ≤ i ≤ m. So, N = M(c1, , ct), where c1, , ct are the roots of h := Irr(am, K). Let b ∈ Km −1 such that n

ma = b. For a fixed i, 1 ≤ i ≤ t , K(am) and K(ci) are K-isomorphic via an isomorphism that takes am in ci. Extend this homomorphism to a K-automorphism σi : Ω → Ω such that σi(am) = ci (see IV.2.18 and IV.2.19). Thus, )( n

mini ac σ= = σi(b). Since K ⊆ M is normal, σi(b) ∈ M.

So, nic ∈ M, ∀i, 1 ≤ i ≤ t, and therefore M ⊆ M(c1, , ct) = N is

n-radical. !

Thus, if an extension K ⊆ L is contained in a radical extension K ⊆ E, we may assume (taking the normal closure) that K ⊆ E is radi-cal and normal.

The following proof of Galois' characterization of solvability by radicals uses basic facts on solvable groups (any subgroup and any factor group of a solvable group are solvable; any finite solvable group has a normal series with the factors cyclic groups), which can be found in the Appendices.

4.5 Theorem. (Galois) Let K be a field of characteristic 0 and let f ∈ K[X]. The polynomial f is solvable by radicals over K if and only if the Galois group of f over K is solvable.

Proof. The theorem follows by setting L = the splitting field of f over K in the statement below:


317

If char K = 0 and K ⊆ L is a normal extension, there exists a radi-cal extension K ⊆ M with L ⊆ M if and only if G(L/K) is a solvable group.

Let us prove this claim. Let K ⊆ L ⊆ M with L/K normal and M/K n-radical (for some n ∈ N). Let us show that G(L/K) is solvable. By 4.4 we may suppose that M/K is normal; since char K = 0, it is separable (thus Galois) and n-radical. By Galois Theory, G(M/L) . G(M/K) and G(L/K) ≅ G(M/K)/G(M/L). If we show that G(M/K) is solvable, G(L/K) will be solvable, as a factor of a solvable group.

Let us show that G(M/K) has a solvable series (a normal series with the factors Abelian groups). We use the characterization of the exten-sions of the form K ⊆ K( n a ) (prop. 3.3), extensions that come up in the definition of a radical extension. In order to apply this proposition, we need that ω ∈ K (ω is a primitive nth root of unity in Ω; ω exists, since char K = 0). Consider the extension K ⊆ M(ω). Since K ⊆ M and M ⊆ M(ω) are n-radical, K ⊆ M(ω) is n-radical. In the tower of exten-sions

K ⊆ K(ω) ⊆ M(ω), K ⊆ K(ω) is an Abelian extension (by 3.1) and K(ω) ⊆ M(ω) is n-radical. Thus, there exists a sequence

K(ω) = K0 ⊆ K1⊆ ⊆ Km = M(ω) and ai ∈ Ki+1, 0 ≤ i < m, with n

ia ∈ Ki and Ki+1 = Ki(ai). Because ω ∈ Ki, ∀i, 3.3 implies that Ki ⊆ Ki+1 is cyclic. Let Hi = G(M(ω)/Ki). Thus, the group G := G(M(ω)/K) has the following series

1 = Hm ⊆ Hm−1 ⊆ ⊆ H1 ⊆ H0 ⊆ G. We claim that this is a solvable series. Ki −1 ⊆ Ki is normal, ∀i,

1 ≤ i ≤ m, so Hi . Hi −1 and Hi −1/Hi ≅ G(Ki/Ki −1) is cyclic; also, K ⊆ K(ω) = K0 is normal, so H0 . G and G/H0 ≅ G(K(ω)/K) is Abe-lian. Therefore G is solvable.


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Conversely, suppose G(L/K) is a solvable group. There exists a se-ries of subgroups

1 = Hm ⊆ Hm−1 ⊆ ⊆ H1 ⊆ H0 ⊆ G, with Hi . Hi −1 and Hi −1/Hi cyclic, ∀i, 1 ≤ i ≤ m. Let Ki be the fixed field of Hi. The extension K ⊆ L is Galois, so Ki −1 ⊆ Ki is Galois, ∀i, 1 ≤ i ≤ m, and Hi −1/Hi ≅ G(Ki/Ki −1) is cyclic. Let ω be a primitive nth root of unity in Ω. The extension Ki −1(ω) ⊆ Ki(ω) is Galois and G(Ki(ω)/Ki −1(ω)) is isomorphic to a subgroup of G(Ki/Ki −1) (we used V.4.8), so it is cyclic. By 3.3, there exists ai ∈ Ki+1(ω), with n

ia ∈ Ki(ω) and Ki+1(ω) = Ki(ω)(ai), ∀i, 0 ≤ i < m. So, K(ω) ⊆ L(ω) is radical; since K ⊆ K(ω) is radical, K ⊆ L(ω) is radical (and clearly L ⊆ K(ω)). !

4.6 Example (The general equation of degree n). Let K be a field, let n ∈ N* and let F be the field K(X1,, Xn) of rational fractions in n indeterminates. Consider S := K(s1,, sn), the subfield of symmetric rational fractions in F (where s1,, sn are the fundamental symmetric polynomials in X1,, Xn). The following polynomial in the indetermi-nate X

P := (X − X1)(X − Xn) = X n − s1 X n − 1 + s2 X

n − 2 − + (−1)nsn has coefficients in S and its roots are X1,, Xn ∈ F. The polynomial P is called the generic polynomial of degree n; the equation P(x) = 0 is called the general equation of degree n. The name comes from the fact that s1, , sn are algebraically independent over K (see IV.4.3), so they behave like indeterminates.

Since F = S(X1,, Xn), F is the splitting field of P over S. The roots of P are distinct, so the extension S ⊆ F is separable and normal. The Galois group G := G(F/S) is isomorphic to Sn, the symmetric group on n elements. To prove this, we remark that any σ ∈ Sn induces a K-algebra homomorphism σ * : K[X1,, Xn] → K[X1,, Xn] such that σ *(Xi) = Xσ(i). It is immediate that (στ)* = σ *τ * and id* = id, ∀σ, τ ∈ Sn. Thus, σ * is an isomorphism and its inverse is (σ −1)*. Pass-ing to the fraction field K(X1,, Xn) = F, we obtain an isomorphism


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(denoted also by σ *) σ * : F → F. Since σ *(si) = si, 1 ≤ i ≤ n, σ * fixes the subfield S = K(s1,, sn). In other words, σ * ∈ G(F/S). Thus, σ & σ * defines a homomorphism (clearly injective) from Sn to G. On the other hand, G has at most n! elements, as the Galois group of the splitting field of f a polynomial of degree n (see Example IV.2.16). Sn has n! elements, so Sn ≅ G. Also note that P is irreducible over S (the root X1 of P has n = deg P conjugates X1, , Xn).

In conclusion, given a field K, we constructed a (transcendental) extension S of K such that there exists a Galois extension S ⊆ F whose Galois group is Sn. More generally, given a field K and a finite group G, does there exist a Galois extension K ⊆ L whose Galois group is G? This problem is extremely difficult in the general case. For example, a theorem of I.R. Shafarevitch (proven around 1954 in several papers totaling more than 100 pages) has as a consequence that, for any solv-able finite group G, there exists a Galois extension of Q, whose Galois group is G. The results of this type are known as Constructive Galois Theory.

Back to our example, recalling that Sn is solvable for any n ≤ 4 im-plies the solvability of the extension S ⊆ F (if char K = 0). In other words, there exists formulas that express by radicals the roots of P if n ≤ 4. If a1, , an ∈ K, there exists a unique K-algebra homomorph-ism ϕ : K[s1,, sn] → K such that ϕ(si) = ai, 1 ≤ i ≤ n (K[s1,, sn] is isomorphic to the K-algebra of polynomials in n indeterminates and use the universality property of this K-algebra). So, replacing the indeterminates s1,, sn with a1,, an, we obtain formulas that solve by radicals any polynomial equation of degree n ≤ 4 with coefficients in K.

For n ≥ 5, Sn is not solvable, so the general equation of degree n is not solvable by radicals over K (this is the Abel-Ruffini Theorem). But there exist particular equations of degree ≥ 5, solvable by radicals (like X 5 = 0 ).


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VI.5 Discriminants, resultants

The notion of discriminant of a polynomial of degree 2 is well-known. If g = aX 2 + bX + c ∈ R[X], g has a double root if and only if the discriminant D = b 2 − 4ac is zero. If x1, x2 are the roots of g, then D = a2(x1 − x2)

2 (easy exercise). This suggests the following general definition:

5.1 Definition. a) Let K be a field, let g ∈ K[X] be a polynomial of degree n and let x1, , xn be the roots of g (in some algebraic closure Ω of K). The discriminant of g is

D(g) := a 2n − 2∏i < j (xj − xi)2,

where a is the leading coefficient of g. Note that D(g) is a symmetric polynomial in the roots x1, , xn of g.

This implies that D(g) ∈ K and that D(g) is independent on the label-ing of the roots of g. Moreover, g has a double root if and only if D(g) = 0.

It is useful to define also ∆ := ∏i< j (xj − xi) ∈ Ω. In general, ∆ ∉ K and relabeling the roots x1, , xn may change ∆ in −∆. Thus, ∆ is de-fined up to a sign.

b) If K ⊆ L is a field extension and x ∈ L is algebraic over K, then the discriminant of the element x over K is DK(x) := D(Irr(x, K)).

We use the discriminant to obtain data on the Galois group G of the polynomial g, assuming g has only simple roots (it is separable). To this end, we see the elements of the Galois group as permutations of the roots.

5.2 Definition. Let L be the splitting field of g over K and let x1, , xn be its (distinct) roots in L. For any σ ∈ G = G(L/K), let ϕ(σ) ∈ Sn be the permutation given by ϕ(σ)(i) = j ⇔ σ(xi) = xj, ∀i, j ∈ 1, , n. The mapping ϕ : G → Sn is an injective group homo-


321

morphism (see IV.3.15.c)) and identifies G with a subgroup of Sn. If we reorder the roots (say x'1, , x'n, where x'i = xτ(i) for some τ ∈ Sn) then ϕ is replaced by ϕ' : G → Sn. We have ϕ'(σ)(i) = j ⇔ σ(x'i) = x'j ⇔ ϕ(σ)(τ(i)) = τ( j); thus ϕ'(σ) = τ −1ϕ(σ)τ.

This shows that a reordering of the roots leads to replacing ϕ(σ) with a conjugate permutation τ −1ϕ(σ)τ. Thus, if a permutation (a sub-group of permutations) has a property that is invariant under any conjugation25, then this property can be transferred to the correspond-ing element in G(L/K) (respectively to a subgroup in G(L/K)). Thus, we define:

Let σ ∈ G(L/K). Define the signature of σ, ε(σ) := ε(ϕ(σ)), where ε(ϕ(σ)) denotes the signature of the permutation ϕ(σ) ∈ Sn. We say that σ is even if ε(σ) = 1 ⇔ ϕ(σ) is an even permutation in Sn. We say that σ is odd if ε(σ) = − 1.

If 1 ≤ k ≤ n, we say that σ is a cycle of length k if ϕ(σ) is a cycle of length k.

These definitions are independent on the labeling of the roots of g. Indeed, ε(τ −1ϕ(σ)τ) = ε(ϕ(σ)) (see above); also, the conjugate of a cy-cle of length k is also a cycle of length k.

5.3 Proposition. Keep the previous notations and suppose char K ≠ 2. Let g ∈ K[X] be a polynomial without multiple roots and let σ ∈ G(L/K). Then:

a) σ(∆ ) = ε(σ)∆. In other words: σ ∈ G(L/K) is even ⇔ σ(∆ ) = ∆. Also, σ ∈ G(L/K) is odd ⇔ σ(∆ ) = −∆.

b) All σ ∈ G(L/K) are even ⇔ ∆ ∈ K ⇔ D(g) ∈ K 2.

25 If τ ∈ Sn, the conjugation by τ is the automorphism κτ : Sn → Sn defined by

κτ(η) = τ −1ητ, ∀η ∈ Sn.


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c) Gal(L/K(∆)) = σ ∈ G(L/K) | σ is even. In other words, the Ga-lois correspondences take the intermediate field K(∆) to the subgroup G(L/K) ∩ ϕ −1(An) of even elements in G(L/K).

Proof. a) In our hypotheses, ∆ ≠ 0, ∆ ≠ −∆ and K ⊆ L is a Galois extension. If τ ∈ Sn, let τ * : K[X1,, Xn] → K[X1,, Xn] denote the unique K-algebra homomorphism such that τ *(Xi) = Xτi. Let d := ∏i< j (Xj − Xi) ∈ K[X1,, Xn]. We show now that τ *(d) = ε(τ )d.

For any 1 ≤ i < j ≤ n, define:

ε(i, j) := ⎩⎪⎨⎪⎧ 1 if τi < τj−1 if τi > τj

Note that (Xj − Xi) | 1 ≤ i < j ≤ n = (Xτi − Xτj)·ε(i, j) | 1 ≤ i < j ≤ n,

so τ *(d) = ∏i< j (Xτj − Xτi) = ∏i< j ε(i, j)(Xj − Xi) = ε(τ )d.

We have ∆ = d(x1, , xn). Let σ ∈ G(L/K). Then: σ(∆) = ϕ(σ)*(d)(x1, , xn) = ε(ϕ(σ))d(x1, , xn) = ε(σ)∆.

b) If G(L/K) ⊆ An (i.e. any σ ∈ G(L/K) is even), then σ(∆) = ∆, for any σ ∈ G(L/K). Since K ⊆ L is Galois, this implies ∆ ∈ K, so D(g) = ∆ 2 ∈ K 2.

If D(g) ∈ K 2, then D(g) = a2, for some a ∈ K. But D(g) = ∆ 2, so ∆ 2 = a 2 ⇒ ∆ = a or ∆ = −a. Anyway, ∆ ∈ K, thus, ∀σ ∈ G(L/K), σ(∆) = ∆, which means that σ is even.

c) G(L/K(∆)) = σ ∈ G(L/K) | σ(∆) = ∆ = σ ∈ G(L/K) | σ is even = G(L/K) ∩ ϕ −1(An). !

We need a method to compute the discriminant of a polynomial without knowing its roots.

5.4 Proposition. Let g = a0 + a1 X ++ an −1 X n −1 + X n ∈ K[X] and let x1, , xn be the roots of g in Ω. For any m ∈ N, let tm := x m

1 + + x mn ∈ K. Then


323

t0 t1 tn − 1

t1 t2 tnD(g) = det ! ! * !

tn − 1 tn t2n−2

The sums tm can be computed by recurrence using the following relations (Newton's identities): t0 = n; t1 = − an − 1; −tm = an − 1 tm − 1 + an − 2 tm − 2 + + an − m + 1 t1 + an − m m, if 2 ≤ m ≤ n; −tm = an − 1 tm − 1 + an − 2 tm − 2 + + a0 tm − n, if m > n.

Proof. ∆ = ∏i< j (xj − xi) = det A, where A is the Vandermonde ma-trix:

1 x1 x n-11

1 x2 x n-12

! ! * ! A =

1 xn x n-1n

The matrix tA·A is the matrix in the statement, and det(tA·A) = det(A)2 = ∆ 2 = D(g).

Viète's relations coupled with Newton's identities (see Appendix) yield the formulas for the tm. !

The discriminant can also be computed using the resultant, defined below.

5.5 Example. Let us compute the discriminant of a polynomial of degree 3, X 3 + pX + q. We have n = 3, a0 = q, a1 = p, a2 = 0. We have t0 = 3; t1 = 0. Newton's identities yield: t2 = −2p; t3 = − 3q; t4 = 2p 2. Thus

D(X 3 + pX + q) = 2232

320203

pqpqpp

−−−−−

= −4p 3 − 27q 2.


324

5.6 Example. We want to give an example of polynomial g ∈ Q[X] solvable by radicals over Q, but such L, the splitting field of g over Q, is not a radical extension of Q. We claim that any irreducible polyno-mial g ∈ Q[X], of degree 3, such that the splitting field of g over Q has degree 3, has this property. Indeed, a normal extension Q ⊆ L of degree 3 cannot be radical. Assuming Q ⊆ L is radical, the definition implies L = Q(b), where b ∈ L is a root of h = X 3 − a, for some a ∈ Q. Since the extension is normal, all roots of h are in L, so bω and bω 2 ∈ L, where ω is a primitive third root of unity. Thus, ω ∈ L, so Q ⊆ Q(ω) ⊆ L. It follows that 2 = [Q(ω) : Q] divides 3 = [L : Q], ab-surd. On the other hand, g is solvable by radicals over Q because its Galois group G(L/Q) is a subgroup of S3, which is solvable.

An example of such polynomial is the minimal polynomial of cos(2π/9) (see exercise V.4.8). More generally, if g = X 3 + pX + q ∈ Q[X] is irreducible and has the discriminant D(g) = −4p 3 − 27q 2 a square of a rational number, then (by Proposi-tion 5.3) the Galois group of g is A3 (the alternating group on 3 ele-ments, which is cyclic of degree 3) and g is solvable by radicals. For instance, g = X 3 − 3X + 1 has D(g) = 81 ∈ Q2 and is irreducible (g has no roots in Q and has degree 3).

The resultant of two polynomials g and h ∈ K[X] appears in the problem of finding conditions under which g and h have a common nonconstant factor (equivalently, they have a common root in the alge-braic closure Ω of K).

5.7 Definition. Let K be a field and let g, h ∈ K[X] be nonzero. As-sume that the roots of g are x1,, xm ∈ Ω and the roots of h are y1,, yn ∈ Ω. Then, in Ω[X],

g = a(X − x1)(X − xm), h = b(X − y1)(X − yn), where a, b ∈ K* and m, n ∈ N. The resultant of g and h is the product

Res(g, h) := a nb m· ∏1≤ i≤ m,1≤ j≤ n

(xi − yj).


325

By convention, Res(0, g) = Res(g, 0) = 0, ∀g ∈ K[X].

The next proposition shows that Res(g, h) ∈ K and provides an algorithm that computes the resultant using divisions with remainder.

5.8 Proposition. Let g, h ∈ K[X] be nonzero. Then: a) Res(g, h) = 0 ⇔ g and h have a common root in Ω ⇔ GCD(g, h)

is a nonconstant polynomial in K[X]. b) Let g = a(X − x1)(X − xm), h = b(X − y1)(X − yn), with a,

b ∈ K* and m, n ∈ N. The following formulas hold: Res(g, h) = a deg(h) ( )∏

≤≤ miixh

1 (R1)

Res(h, g) = (−1) mnRes(g, h) (R2)

Res(g, h) = a n − deg(r)Res(g, r) (R3) (r is the remainder of the division of h by g: h = gq + r, where q, r ∈ K[X], deg r < deg g or r = 0) Res(g, b) = b m, ∀b ∈ K a constant polynomial. (R4)

c) Res(g, h) ∈ K. Proof. a) Clear, keeping in mind that the GCD of g and h in Ω[X]

is the same as their GCD in K[X]. b) We have h(xi) = b(xi − y1)(xi − yn), for any i, 1 ≤ i ≤ m; R1 fol-

lows by multiplying these relations. R2 is clear from the definition. Let h = gq + r, where r = 0 or deg r < m = deg h. From R1,

Res(g, h) = a n ∏i h(xi) = a n ∏i (g(xi)q(xi) + r(xi)) = a n ∏i r(xi), since g(xi) = 0, 1 ≤ i ≤ m. Bur Res(g, r) = a deg(r)∏i r(xi) and R3 follows.

Finally, R1 implies R4. c) By induction on min(deg g, deg h), using R3 and R4. !

A polynomial g has multiple roots if and only if g and its formal derivative g' have common roots. Thus D(g) = 0 ⇔ Res(g, g') = 0. In fact, D(g) and Res(g, g') differ only by ± the dominant coefficient of g.


326

5.9 Proposition. Let g ∈ K[X] have degree m ∈ N and let a ∈ K* be its dominant coefficient. Then D(g) = a −1(−1)m(m−1)/2Res(g, g').

Proof. Let g = a(X − x1)(X − xm), where xi ∈ Ω. Then g' = a∑i∏j≠i (X − xj). By R1,

Res(g, g') = a m−1∏i g'(xi) = a m−1∏i a∏j≠i (xi − xj). For each couple (i, j), 1 ≤ i < j ≤ m, the product above contains

(xi − xj) and (xj − xi). There are m(m − 1)/2 such couples, so Res(g, g') = a 2m−1(−1)m(m−1)/2∏i<j (xi − xj)

2 = a(−1)m(m−1)/2D(g). !

The resultant of two polynomials can be also computed by means of a determinant formed with their coefficients (see the exercises). If x is a root of f, and y is a root of g, with the help of resultants one can find a polynomial that has the root x + y (see Exercise 9).

Exercises

1. Let K ⊆ L be an extension of prime degree p. If the extension is radical, then there exists b ∈ L such that L = K(b) and b p ∈ K. Deduce that a normal extension of degree p of Q cannot be radical. 2. Let K be a field of characteristic 0 and let f ∈ K[X] be a polynomial solvable by radicals over K, deg f = n. Prove that, if K contains a primitive n!-th root of unity, then the splitting field of f over K is a radical extension of K. 3. Let K be a field and let p = X n + an − 1 X

n − 1 + + a1 X + a0 ∈ K[X], such that the characteristic exponent of K does not divide n. Prove that solving the equation p(X) = 0 is equivalent to solving an equation in Y of the form Y n + bn − 2 X

n − 2 + + b1 Y + b0, by setting Y = X + a/n.


327

4. (Solving the cubic equation)26 Let K be a field whose characteristic is not 2 or 3 and let the equation

f(x) = x 3 + px + q = 0, where p, q ∈ K. Prove the following assertions:

a) Setting x = u + v, one obtains the equation u 3 + v 3 + q + (3uv + p)(u + v) = 0.

b) Requiring u 3 + v 3 + q = 0 and 3uv + p = 0, an equation of degree

2 in u 3 is obtained, whose solutions are 2274 32

3 pqqu +±−= . Let

D = − 4p 3 − 27q 2 be the discriminant of the equation f = 0. Then u 3 = − q/2 ± γ , where γ = − D/108.

c) Let A = − q/2 + γ , B = − q/2 − γ and let ω be a primitive 3rd root of unity. The condition uv = − p/3 implies that the roots of f are:

33232333 ,, BABABA ωωωω +++ .

5. (Solving the quartic equation)27 Let K be a field whose characteris-tic is not 2 or 3 and let the equation

f(x) = x 4 + px 2 + qx + r = 0, where p, q, r ∈ K. Verify the details in the following steps, leading to a solution of this equation:

a) Solve the equation if q = 0. b) Let q ≠ 0. Then f(x) = 0 is rewritten:

42

222 prqxpx +−−=⎟

⎠⎞⎜

⎝⎛ +

c) Let u ∈ K. Then any solution x of the equation satisfies

26 The formula was obtained around 1515, by Scipione del Ferro (? - 1526), but

was not published. Niccolò Fontana Tartaglia (1500-1557) rediscovered the formula in 1535 and communicated it to Girolamo Cardano (1501-1576), who published it in his book Ars Magna, sive de regulis algebraicis.

27 The formula was found by Lodovico Ferrari (1522-1565), a student of Cardano.


328

puuxuprqxupx ++++−−=⎟⎠⎞⎜

⎝⎛ ++ 22

222 242

The right hand side is a perfect square of some polynomial in x if and only if

222

2

22224 ⎟

⎠⎞⎜

⎝⎛ −=++++−− u

quxpuuxuprqx

d) The last equality is equivalent to ( ) 08288 2223 =−−++ qurppuu .

(called the cubic resolvent of Ferrari). e) Let u be a solution of the cubic resolvent. Then the 4 solutions of

f = 0 are:

uqpuux22222

εε −−−±= , where ε ∈ − 1, 1.

6. Let g = X 3 + pX + q ∈ Q[X] be irreducible, and let α ∈ R be a root of g. Prove that, if ∆ = − 4p 3 − 27q 2 is the discriminant of g, then Q(α, ∆ ) is the splitting field of g over Q. Deduce that Q ⊆ Q(α) is normal if and only if ∆ is the square of a rational number. 7. Let g = X 3 + X + 3 ∈ Q[X]. Show that g is irreducible. Let α be the real root of g and let K be the splitting field of g over Q. Show that K ∩ Q(α, d ) = Q(α), for any d ∈ Z a positive squarefree integer. 8. (The resultant as a determinant) Let R be a domain, let a, b ∈ R and let m, n ∈ N*. Consider the polynomials in X, with coefficients in R[X1,, Xm, Y1,, Yn]:

f := a ∏1 ≤ i ≤ m (X − Xi) = ∑0 ≤ j ≤ m ajX j

g := b ∏1 ≤ i ≤ n (X − Yi) = ∑0 ≤ j ≤ n bjX j.

Let D the square matrix in Mm + n(R[X1,, Xm, Y1,, Yn]),


329

D =

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

−−

−−

00

10

010

0

11

11

ba

babbaa

bababbaa

ba

n

nm

m

nm

nnmm

nm

!!!!!!!!!!!!!!

!!!

There are n columns of a's and m columns of b's (the empty places contain 0's). The point is to prove that Res f, g = det D.

a) Let M be the Vandermonde matrix of dimension m + n,

M =

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−+

−+

−+

−+

1

11

1

1

11

1

1

11

1

mnm

m

nmn

nmn

nm

XX

XXYY

YY

………………………………

Computing det(MD) in two ways, show that det D = Res f, g b) Prove that Res f, g ∈ R[a0, , am, b0, , bn].

9. Let K be a field and let x, y be algebraic elements over K; let f = Irr(x, K) and let g = Irr(y, K). Let Z be an indeterminate and let gZ := g(Z + X) ∈ K[Z][X]. View f and gZ as polynomials in X (with coefficients in K[Z]), and let h := Res f, gZ ∈ K[Z]. Prove that x + y is a root of h. More precisely, the roots of h are xi + yj, 1 ≤ i ≤ n, 1 ≤ j ≤ m, where x1, , xn are the roots of f and y1, , ym are the roots of g. 10. Let C be the curve in R 2 given by the parametric representation

( )( )⎩

⎨⎧

+−==+==

12

2

tttyytttxx , t ∈ R. This means that C = (x, y) ∈ R 2 | ∃t ∈ R


330

such that x = x(t), y = y(t). Using resultants, find an implicit representation of the curve (i.e., a polynomial g ∈ R[X, Y] such that: ∀t ∈ R, g(x(t)), y(t)) = 0; and, conversely, g(x, y) = 0 implies that ∃t ∈ R such that x = x(t), y = y(t)). (Hint. ∀(x, y) ∈ R 2, (x, y) ∈ C ⇔ the polynomials t 2 + t − x and t 2 − t + 1 − y have a common root.)

331

Appendices

The following appendices contain a part of the background re-quired for reading the book. They include themes less likely to be in-cluded in a first course on Algebra and themes that are necessary, but auxiliary to the main ideas in the book.

The reader should feel free to consult the appendices as needed, when reading the main text.

1. Prime ideals and maximal ideals

All rings considered are assumed commutative rings with identity. If I is an ideal of the ring R (i.e. ∀x, y ∈ I and ∀r ∈ R, we have x + y ∈ I and rx ∈ R), we denote this by I ≤ R. An ideal I is called a proper ideal if I ≠ R. If I and J are ideals in R with I ⊆ J, we write this also I ≤ J. The set of all ideals of a ring is a lattice with respect to inclusion: for any two ideals I and J of R, inf(I, J) = I ∩ J, sup(I, J) = I + J, where I + J denotes the sum of the ideals I and J, I + J := i + j | i ∈ I, j ∈ J (the ideal generated by I ∪ J).

Throughout this section, R denotes a commutative ring with iden-tity.

Appendices

332

1.1 Definition. An ideal P of the ring R is called a prime ideal if P ≠ R and, for any x, y ∈ P, xy ∈ P implies x ∈ P or y ∈ P.

An ideal M of R is called a maximal ideal if M ≠ R and there exist no proper ideals of R that strictly include M. In other words, for any J ≤ R, M ≤ J implies M = J or J = R.

1.2 Examples. a) If p ∈ Z is a prime number, then the ideal gener-ated by p in Z, denoted pZ, is a prime ideal in Z. Conversely, if pZ is a prime ideal, then p is a prime number.

b) An ideal I is maximal in the ring R iff I is a maximal element of the set (ordered by inclusion) of all proper ideals of R. In the ring Z, any ideal is of the form nZ, for some n ∈ Z. This implies that nZ is maximal if and only if n is a prime number. Indeed, let nZ be a maxi-mal ideal. Then, ∀m ∈ Z, nZ ⊆ mZ implies nZ = mZ or mZ = Z; in other words, m|n implies m ∼ n or m = 1. Since nZ ≠ Z, this means that n is irreducible, hence it is a prime. The converse is left to the reader.

c) The ring R is a domain (has no zero divisors) if and only if (0) is a prime ideal.

d) If K is a field, (0) is its only proper ideal; (0) is also a maximal ideal and a prime ideal.

Here is a useful characterization of maximal (respectively prime) ideals, by means of the factor ring.

1.3 Theorem. Let I be a proper ideal of R. a) I is a prime ideal if and only if the factor ring R/I is a domain. b) I is a maximal ideal if and only if the factor ring R/I is a field. Proof. a) Let I be a prime ideal. Let α = a + I, β = b + I (where a,

b ∈ R) be elements in R/I. If αβ = 0 ∈ R/I, then (a + I)(b + I) = 0 + I, so ab ∈ I. Since I is prime, a ∈ I or b ∈ I, hence a + I = α = 0 + I or b + I = β = 0 + I. Therefore, R/I has no zero divisors. Conversely, as-sume that R/I is a domain and let a, b ∈ R with ab ∈ I. This implies

2. Algebras. Polynomial and monoid algebras

333

that (a + I)(b + I) = 0 + I, so a + I = 0 + I or b + I = 0 + I. Thus, a ∈ I or b ∈ I.

b) Suppose I is a maximal ideal in R. We want to show that any nonzero element of R/I is invertible. Let α = a + I ≠ 0 + I, hence a ∉ I. Then the ideal generated by I and a, I + Ra, includes strictly I; since I is maximal, I + Ra = R. In particular, 1 ∈ R is written i + ra, for some i ∈ I and r ∈ R. Thus, 1 + I = (ra + i) + I = ra + I = (r + I)(a + I), so a + I is invertible.

If R/I is a field and J is an ideal strictly including I, there exists x ∈ J, x ∉ I. Hence, x + I ≠ 0 + I, so x + I is invertible in R/I. For some r ∈ R, 1 + I = (r + I)(x + I), whence there exists i ∈ I such that 1 = rx + i. It follows that 1 ∈ J ⇔ J = R. !

1.4 Corollary. Any maximal ideal is a prime ideal. !

The converse is false: the ideal (X) of the ring Z[X] is prime and not maximal, because the factor ring [ ] ( )XXZ ≅ Z is a domain, but not a field.

If R is a principal ideal domain and not a field, the prime nonzero ideals are exactly the maximal ideals and they are the ideals generated by irreducible elements.

Krull's Lemma (II.1.20) says that any proper ideal is included in some maximal ideal.


We fix (R, +, ·), a commutative ring with identity.

Appendices

334

2.1 Definition. An R-algebra is a ring (A, +, ·) (not necessarily associative or having an identity), which is simultaneously an R-module, such that, ∀r ∈ R, ∀a, b ∈ A:

r(ab) = (ra)b = a(rb). The R-algebra A is called associative (respectively with identity,

commutative) if the ring A has the corresponding property. Recall that the center of the ring A, Cen(A), is the set of all ele-

ments in A that commute with any element: Cen(A) := a ∈ A | ab = ba, ∀b ∈ A

Cen(A) is a subring of A (easy exercise).

We are interested in R-algebras that are associative and have an identity. For this type of algebras there is the following characteriza-tion (often taken as the definition):

2.2 Proposition. a) Let A be an associative R-algebra with identity element e. Then the function α : R → A defined by

α(r) := re, ∀r ∈ R is an identity preserving ring homomorphism and α(r)a = aα(r), ∀r ∈ R, ∀a ∈ A (in other words, α(R) ⊆ Cen(A)).

b) Conversely, if A is an associative ring with identity, and α : R → A is an identity preserving ring homomorphism such that α(R) ⊆ Cen(A), then A is an R-algebra if we define the R-module multiplication by

ra := α(r)a, ∀r ∈ R, ∀a ∈ A. Proof. a) If r, s ∈ R, then, by definition:

α(r + s) = (r + s)e = re + se = α(r) + α(s) α(r)α(s) = (re)(se) = r(e(se)) = r(se) = (rs)e = α(rs).

Also, α(1) = 1e = e (since A is an R-module). Thus, α is an identity preserving ring homomorphism. If r ∈ R, a ∈ A, α(r)a = (re)a = r(ea) = ra = r(ae) = a(re) = aα(r).

b) Exercise. !


335

2.3 Remark. The homomorphism α : R → A in the proposition above us called the structural homomorphism of the associative R-algebra with identity A. Naturally, a ring A can have several differ-ent R-algebra structures (corresponding to different structural homomorphisms).

2.4 Examples. a) The ring Mn(R) of all square matrices of type n×n with entries in R is an associative R-algebra with identity (noncommutative if n ≥ 2). The structural homomorphism takes r ∈ R to the matrix having r on the diagonal and 0 elsewhere.

b) The polynomial ring R[X] is a commutative R-algebra. If K ⊆ L is a field extension, L is a K-algebra. Which are the structural homomorphisms (equivalently, which is the module structure) for these examples?

2.5 Definition. Let A and B be two R-algebras. A ring homomorph-ism ϕ : A → B that is simultaneously an R-module homomorphism is called an R-algebra homomorphism.

If A and B are associative and have identities, and α, respectively β, are the structural homomorphisms, then a ring homomorphism ϕ : A → B is an R-algebra homomorphism if and only if ϕ α = β.

In what follows, we consider only associative algebras with iden-tity.

A subset C of the R-algebra A is called an R-subalgebra of A if C is a subring in A and also a submodule in R A: ∀r ∈ R, ∀a ∈ C ⇒ ra ∈ C. It is immediate that the intersection of a family of subalgebras of A is still a subalgebra of A (for the proof, see the corresponding property for rings or for modules). This allows to define, given a sub-set S of A, the subalgebra generated by S as the intersection of all subalgebras of A that include S. For commutative R-algebras, the subalgebra generated by S is denoted by R[S] and it is the set of all

Appendices

336

polynomial expressions in the elements of S, with coefficients in R (cf. prop. IV.1.14).

If ϕ : A → B is an R-algebra homomorphism, then ϕ(A) is a subalgebra of B. An ideal I of the ring A is also called an ideal of the R-algebra A. If I is an ideal of the R-algebra A, then the factor ring A/I is canonically an R-algebra, its structural homomorphism being π α, where π : A → A/I is the canonical surjection. This algebra is called the factor algebra of A relative to the ideal I.

We fix a commutative ring with identity R and a monoid (G, ·). We shall now describe the construction of the monoid algebra80 R[G]. Recall that (G, ·) is a monoid (or a semigroup with identity) if the operation · is associative and has an identity element e.

In particular, we obtain the construction of polynomial algebras (in a finite or infinite number of indeterminates). The idea is the follow-ing: we define on R (G) (the free R-module on the set G) a multiplica-tion law that is associative and distributive, and that coincides with the multiplication in G for the elements of G. More precisely, any ele-ment in R (G) is written uniquely as a finite sum

∑∈Gg

g ga

,

where ag ∈ R, for any g ∈ G and supp(ag)G is finite. The product between g and h ∈ G (seen as elements in the basis of

R (G)) is gh (element in the basis of R (G)); this product extends by linearity to any two elements of R (G):

( ) , g h g h

g G h G g h G G

a g b h a b gh∈ ∈ ∈ ×

⎛ ⎞ ⎛ ⎞⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

∑ ∑ ∑

The rigorous construction is described in the sequel.

80 Called group algebra if G is a group.


337

The support of a function ϕ : G → R is the set supp(ϕ) := g ∈ G | ϕ(g) ≠ 0. Define

R[G] := ϕ : G → R | supp(ϕ) is finite. A function in R[G] is called a function of finite support. We define

on R[G] the following operations: ∀ϕ, ψ ∈ R[G], ∀g ∈ G, (ϕ +ψ)(g) := ϕ(g) +ψ (g)

(ϕ ·ψ)(g) := ( ) ( )( )

∑=

×∈guv

GGvuvu

,ψϕ .

The first equality clearly defines a function ϕ +ψ : G → R. It is not as obvious that ϕ ·ψ is correctly defined: we must show that the sum defining ϕ ·ψ has a finite number of nonzero terms. Indeed, the set of all (u,v) ∈ G×G such that ϕ(u)ψ(v) ≠ 0 is included in supp(ϕ)×supp(ψ), which is finite.

We must also show that ϕ +ψ and ϕ ·ψ have finite support. We have:

supp(ϕ +ψ) ⊆ supp(ϕ) ∪ supp(ψ), a finite set. For ϕ ·ψ: if g ∈ G \ uv |u ∈ supp(ϕ) and v ∈ supp(ψ), then

(ϕ ·ψ)(g) = 0, since all terms in the sum in the definition are zero. So, supp(ϕψ) is included in uv | u ∈ supp(ϕ) and v ∈ supp(ψ), which is finite.

Therefore, "+" and " · " are correctly defined and are composition laws on R[G]. We define an external operation "·" : R × R[G] → R[G],

(r ·ϕ)(g) := rϕ(g), ∀r ∈ R, ∀ϕ ∈ R[G], ∀g ∈ G. Endowed with these operations, R[G] is an R-module, namely the

free R-module of basis G (if we disregard the multiplication in R[G]).

2.1 Remark. The construction we described generalizes (and is in-spired from) the usual construction of the polynomial ring in one indeterminate over R, R[X]. The reader is encouraged to verify that if -

Appendices

338

(G, ·) is (N, +), the additive monoid of the natural numbers, then R[G] is exactly the ring R[X].

2.2 Proposition. (R[G], + , ·) is an associative ring with identity. Proof. We check the associativity of multiplication. Let ϕ, ψ, η ∈

R[G] and let g ∈ G. ( )( )( ) ( )( ) ( )

( )∑

=∈

=

guvGvu

vug2,

ηϕψηϕψ =

( ) ( )( )

( )( )

( ) ( ) ( )( )

∑∑ ∑=∈

=∈

=∈

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

gstvGvts

guvGvu

ustGts

vtsvts32 2 ,,, ,

ηψϕηψϕ .

Computing (ϕ(ψη))(g), we obtain the same thing, so (ϕψ)η = ϕ(ψη).

The existence of neutral elements for addition and multiplication is proven below. The other ring axioms are left to the reader. !

It is necessary to show that this construction satisfies the demands stated at the beginning of this section and. To this end, we define the following elements in R[G]:

∀g ∈ G, define ηg : G → R by ( )0, if 1, if g

h gh

h gη

≠⎧= ⎨ =⎩

, ∀h ∈ G;

∀r ∈ R, define ψr : G → R by ( )0, if , if r

h eh

r h eψ

≠⎧= ⎨ =⎩

, ∀h ∈ G.

It is clear that ηg, ψr ∈ R[G], ∀ g ∈ G, ∀r ∈ R. We also have:

2.3 Proposition. a) The function i : R → R[G], given by i(r) = ψr, ∀r ∈ R, is an injective ring homomorphism. Furthermore, Im i is in-cluded in the center of R[G] (so, R[G] is an R-algebra of structural homomorphism i). That entitles us to write r instead of ψr (identifying r ∈ R with its image ψr ∈ R[G]).


339

b) The function j : G → (R[G], ·), j(g) = ηg, ∀g ∈ G, is an injective monoid homomorphism. We write g instead of ηg (identifying g ∈ G with its image ηg ∈ R[G]).

c) For any g, h ∈ G and any r ∈ R, (ψr·ηg)(h) = 0, if , if

h gr h g

≠⎧⎨ =⎩

.

d) Any element ϕ ∈ R[G] is written as a finite sum: ϕ = ( )

( ) ( )∑∑

∈∈=

ϕϕϕ ηψ

supp supp gg

ggg ga ,

In the second sum, ϕ(g) is denoted ag, ηg is identified with g and ψϕ(g) is identified with ag = ϕ(g), for any g ∈ G (thus (ag)g ∈ G has finite support).

The writing of ϕ is unique: if ∑∑∈∈

=Gg

gGg

g gbga

, where (ag)g ∈ G and

(bg)g ∈ G have finite support, then ag = bg, ∀g ∈ G. e) The zero of the ring R[G] (the neutral element for addition) is

ψ0 (written as a sum of the type ∑∈Gg

g ga

as the sum with one term 0e).

The identity of the ring R[G] (the neutral element for multiplication) is

ηe = 1e. Proof. a) It is evident that ψr + s = ψr + ψs, ∀r, s ∈ R. Computing

ψr·ψs, we obtain ψr·ψs(g) = ( ) ( )∑=guv

sr vu ψψ . If g ≠ e, then, for any cou-

ple (u, v) ∈ G×G such that uv = g, we have u ≠ e or v ≠ e, so ( ) ( )vu sr ψψ = 0. Therefore, if g ≠ e, then (ψr·ψs)(g) = 0. Also,

(ψr·ψs)(e) = ψr(e)·ψs(e) = rs. In conclusion, ψr·ψs = ψrs. The injectivity

is clear. b) The injectivity is easy. We prove that ηgηh =ηgh, ∀g, h ∈ G. For

any x ∈ G, x ≠ gh, (ηgηh)(x) = ( ) ( )∑=xuv

hg vu ηη = 0, since uv = x ≠ gh im-

plies u ≠ g or v ≠ h. Also, (ηgηh)(gh) = 1.

Appendices

340

c) Exercise. d) For any h ∈ G, we have

( ) ( )( )( )( )

( )( ) ( )

( ) supp supp

0, if supp , if suppg gg

g g

ha g h h h

h hϕϕ ϕ

ϕψ η ϕ

ϕ ϕ∈ ∈

∉⎧⎛ ⎞= = =⎨⎜ ⎟ ∈⎝ ⎠ ⎩

∑ ∑

.

We used in the last equality that ( )( ) ( )( )

0, if , if gg

h gh

g h gϕψ ηϕ

≠⎧= ⎨ =⎩

, as

seen at c).

The uniqueness ensues from ( ) hGg

g ahga =⎟⎟⎠

⎞⎜⎜⎝

⎛∑∈

, ∀h ∈ G.

e) We show that ηe is the identity of the ring R[G]. For any g ∈ G, ηgηe = ηge = ηg = ηeηg, by c). The general case is proven using d) and the distributivity. !

2.4 Remarks. a) By construction, R[G] is isomorphic to the free R-module of basis G. The elements of R[G] can be seen as formal finite sums of the form ∑g∈G agg, where (ag)g∈G is a family of elements

of R having finite support. We can identify a ∈ R with the sum with one term a·e; also, we can identify g ∈ G with 1·g. The addition obeys the rule ∑g∈G agg + ∑g∈G bgg = ∑g∈G (ag + bg)g; the multiplication is (left and right) distributive relative to addition and obeys the rule (1·g)·(1·h) = 1·(gh). Therefore:

( )∑ ∑∑∑∈ =∈∈

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛

Gg guvvu

Ggg

Ggg gbagbga

.

Thus, R[G] meets the conditions stated in the introduction. Any ele-ment of R[G] is written uniquely ∑g∈G agg, the sums being always fi-nite. In particular, ∑g∈G agg = 0 ⇔ ag = 0, ∀g ∈ G.


341

b) If G is a commutative monoid, then R[G] is also a commutative ring. If G is not commutative, then R[G] is not commutative, as part b) of the proposition shows.

The classical polynomial algebras.

I. For (G, ·) = (N, +), one obtains the usual construction of the

polynomial algebra in one indeterminate with coefficients in R. In-deed, an arbitrary element in R[N] is a function ϕ : N → R with finite support (a sequence of elements of R with finite support). Setting ϕ(i) =: ai, ∀i ∈ N, the general form of an element f in R[N] is f = ∑

∈N iiiaη . Since ηiηj = ηi + j, for any i, j ∈ N, ηi = (η1)

i, ∀i ∈ N.

Denoting η1 by X, one obtains the usual form:

f = ∑i∈N ai X i = a0 + a1 X + + an X

n

where n = max i ∈ N | ai ≠ 0 is the degree of f.

II. Consider the commutative monoid (N n, +) (for a fixed n ∈ N*), where the addition is defined component-wise:

(i1, , in) + (j1, , jn) := (i1 + j1, , in + jn), ∀(i1, , in), (j1, , jn) ∈ N n.

The R-algebra R[N n] is called the polynomial algebra in n indeterminates81 with coefficients in R. An element of R[N n] is called a polynomial (in n indeterminates).

We make the connection with the usual form of a polynomial. For each i ∈ 1, , n, let

ei := (0,,1,,0) ∈ N n (1 on the i-th place, 0 elsewhere).

81 Often encountered terminologies are unknown or variable instead of

indeterminate.

Appendices

342

Any element in N n is written uniquely - up to a order of the terms - as a sum of ei's (in other words, e1, , en generate the monoid N n). The element

ieη ∈ R[N n] is denoted by Xi and is called an indetermi-

nate. A product of indeterminates (of the form nin

i XX …11 ) is called a term. Any polynomial g ∈ R[N n] is then written uniquely as a finite

sum: g =

( )

1

1

1

1, ,

n

nn

n

iii i n

i i

a X X∈

∑ …… '

… ,

where ( )( )1 1 , , nn ni i i i

a∈… … '

is a family of elements in R, having finite sup-

port. Thus, g is a linear combination of terms, with coefficients in R. The polynomial g is thus a sum of products between a nonzero ele-ment of R and a term, of the form 1

1 1n

n

iii i na X X… … . If g is like above

and 1 ni ia … ≠ 0, then 1

1 1n

n

iii i na X X… … is called a monomial of g.

R[N n] is denoted usually by R[X1,, Xn].

III. The previous construction can be generalized to the polynomial algebra in S indeterminates, where S is an arbitrary nonempty set. To this end, we replace the monoid (N n, +) with (N(S), +),

where N(S) := f : S → N | the support of f is finite. Continuing the analogy with N n, we see the elements of N(S) as multi-indices and use notations like i, j, . The addition in the monoid N(S) is defined naturally as follows: for any i, j ∈ N(S), (i + j)(s) = i(s) + j(s), ∀s ∈ S.

The axioms of a commutative monoid are readily checked. The R-algebra R[N(S)] is called the polynomial algebra in S indeterminates with coefficients in R. For any s ∈ S, define es ∈ N(S),

( )0, if 1, if s

s tt

s t=⎧

= ⎨ ≠⎩e , ∀t ∈ S and let Xs be the element

seη ∈ R[N(S)].


343

Any element i ∈ N(S) is written uniquely as i = ∑∈Ss

ssm e , where (ms)s∈S

is a family of natural numbers82 indexed by S, having finite support. Therefore,

( )∏

∈=

ii

supps

ms

sXη . An arbitrary polynomial f ∈ R[N(S)] has a

unique writing as f = ∑∈F

ai

iiη , where F is a finite subset of N(S); if

∪i∈F supp(i) =: s1, , sn (a finite subset of S), then

f =( )

∑∈ n

n

nnn

mm

ms

msmm XXa

N,,1

111

…… … ,

where the sum is finite (the family of elements of R ( )

( ) nnn mmmma

N∈,,11 …… has finite support).

In other words, any polynomial in S indeterminates is a polynomial in a finite set of indeterminates in S. The R-algebra R[N(S)] is usually denoted as R[(Xs)s∈S] or R[Xs]s∈S or R[X; S].

2.5 Theorem. (The universality property of the monoid algebra) Let (G, ·) be a monoid and let i : R → R[G], j : G → R[G] be the canonical mappings defined at 2.3. The triple (R[G], i, j) has the following universality property: for any R-algebra T having the struc-tural homomorphism α : R → T and any homomorphism of monoids β : G → (T, ·), there exists a unique homomorphism of R-algebras ϕ : R[G] → T such that ϕ i = α and ϕ j = β:

82 For any m ∈ N and i ∈ R[N(S)], mi is defined as (mi)(s) := m·i(s), ∀s ∈ S.

β

[ ]GRjG ⎯→⎯

T

ϕ

Appendices

344

Proof. Suppose ϕ is as stated. Thus, ϕ(r) = α(r), ∀r ∈ R and ϕ(g) = β(g), ∀g ∈ G. If ∑g∈G agg ∈ R[G], then

( ) ( )∑∑∈∈

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Ggg

Ggg gaga

ϕϕϕ = ( ) ( )∑

∈Ggg ga

βα ,

which shows that ϕ is uniquely determined by α and β. A direct check shows that ϕ given by the above is a ring homomorphism and satisfies the stated requirements. !

The universality property of the monoid algebra determines this algebra up to an unique isomorphism: if (A, γ, δ) (where A is an R-algebra, γ : R → A is its structural homomorphism and δ : G → (A, ·) is a homomorphism of monoids) satisfies the same universality property as (R[G], i, j), then there exists a unique R-algebra isomorphism ϕ : R[G] → A such that ϕi = γ and ϕj = δ (cf. 5.7).

For the classical polynomial algebras, using the remarkable fact that the monoid (N, +) is generated by one element (respectively (N n, +) is generated by n elements), this theorem reads:

2.6 Theorem. Let A be an R-algebra. a) (The universality property of the polynomial algebra R[X]) For

any a ∈ A there exists a unique R-algebra homomorphism eva : R[X] → A such that eva(X) = a. b) (The universality property of the polynomial algebra R[X1,,

Xn]) Let n ∈ N*. For any n-uple a = (a1,,an) ∈ A n there exists a unique R-algebra homomorphism eva : R[X1,, Xn] → A such that eva(Xi) = ai, ∀i ∈ 1,, n.

c) (The universality property of the polynomial algebra R[X; S]) Let S be a nonempty set. For any mapping γ : S → A there exists a unique R-algebra homomorphism evγ : R[X; S] → A such that evγ(Xs) = γ(s), ∀s ∈ S.


345

Proof. Of course, a) follows from b), which is a particular case of c). Nevertheless, we sketch a proof for every case.

a) For any f = ∑=

n

i

ii Xb

0∈ R[X], let eva( f ) = ∑

=

n

i

iiab

0 (usually called

the value of f at a and denoted f (a)). The fact that eva is a homomorph-

ism amounts to that, ∀f, g ∈ R[X], (f + g)(a) = f(a) + g(a),

(f·g)(a) = f(a)·g(a),

A standard check proves these familiar properties and the rest of the claims.

b) For any f =( )

∑∈ n

n

nn

ii

in

iii XXb

N,,1

1

11

…… … ∈ R[X1,, Xn], and any

a = (a1, , an) ∈ A n, let eva( f ) = ( )

( )∑

∈

=n

n

nn

ii

in

iiin aabaaf

N,,11

1

11

,,…

… …… ,

the value of f at a = (a1, , an).

c) The mapping γ induces a homomorphism of monoids β : N(S) → (A, ·), β(i) = ( )

( )

( )s

ss i

i∏

∈suppγ , ∀i ∈ N(S)

Applying the universality property of the monoid algebra R[N(S)] = R[X; S], there exists a unique R-algebra homomorphism evγ : R[X;S] → A such that vγ j = β, where j : N(S)→ R[X; S] is the canonical mapping; in our case, j(es) = Xs, ∀s ∈ S. Thus vγ (Xs) = β(es) = γ(s), ∀s ∈ S.

Let us prove the uniqueness of evα. If v : R[N(S)] → A is an R-algebra homomorphism with v(Xs) = γ(s), then vj = β, where β is the

homomorphism defined above. The uniqueness part of the universality property implies v = evα. !

Appendices

346

The homomorphism eva (respectively eva) above is called the evaluation homomorphism.

The previous theorem formalizes and gives a precise meaning to the procedure of assigning the values a1, , an to the indeterminates X1, , Xn.

A useful property of the R-algebras R[X1,, Xn], sometimes used to define them by recurrence on n, is the following:

2.7 Theorem. Let n ≥ 1. Then there exists a canonical isomorphism of R-algebras:

R[X1,, Xn] ≅ R[X1,, Xn−1][Xn]. Proof. Use 2.6.b): there exists a unique R-algebra homomorphism

ϕ : R[X1,, Xn] → R[X1,, Xn−1][Xn], with ϕ(Xi) = Xi, 1 ≤ i ≤ n. Let A := R[X1,, Xn−1]. Conversely, 2.6.b) applied to R[X1,, Xn−1], yields a unique

R-algebra homomorphism α : R[X1,, Xn−1] → R[X1,, Xn], with α(Xi) = Xi, 1 ≤ i ≤ n − 1. Thus, R[X1,, Xn] is an A-algebra with α its structural homomorphism. The universality property of the A-algebra A[Xn] shows that there exists a unique A-algebra homomorphism β : A[Xn] → R[X1,, Xn], such that β(Xn) = Xn. Of course, β is also an R-algebra homomorphism.

We have βϕ = id. Indeed, βϕ : R[X1,, Xn] → R[X1,, Xn] is an R-algebra homomorphism such that βϕ(Xi) = Xi, 1 ≤ i ≤ n, and id : R[X1,, Xn] → R[X1,, Xn] has the same properties. The unique-ness part in 2.6.b) implies that βϕ = id. Similarly, ϕβ = id, so ϕ is an isomorphism. !

In what follows we discuss the important notion of degree of a polynomial.

If aX n is a monomial in R[X] (with a ≠ 0), n is called the degree of aX n.

For an arbitrary polynomial g ∈ R[X],


347

g = a0 + a1X + + anX n, where an ≠ 0,

the natural number n is called the degree of g, denoted deg g. Thus, the degree of g is the greatest degree of the monomials of g. We de-fine by convention deg 0 = −∞. Sometimes deg 0 is not defined.

The elements a0, , an ∈ R are called the coefficients of the polynomial g; among these, an is called the leading coefficient of g, a0 is called the constant term.

If the leading coefficient an is 1, then the polynomial g is called monic.

If R is a domain, then the degree is additive: ∀g, h ∈ R[X1,, Xn], deg (gh) = deg g + deg h.

Also: deg (g + h) ≤ max(deg g, deg h).

If nin

i XaX …11 is a monomial in R[X1,, Xn] (where a ≠ 0), and

1 ≤ k ≤ n, then its degree in Xk is ( )11deg ,nii

n kaX X X… := ik (the expo-

nent of Xk in the monomial), also denoted ( )11deg n

k

iiX naX X… .

For any g ∈ R[X1,, Xn], deg (g, Xk) is the greatest degree in Xk of

the monomials of g. If R is a domain, then the degree in Xk satisfies the same relations as above: ∀g, h ∈ R[X1,, Xn],

deg (gh, Xk) = deg (g, Xk) + deg (h, Xk). deg (g + h, Xk) ≤ max(deg (g, Xk), deg (h, Xk)).

The total degree of the monomial nin

i XaX …11 is i1 + + in; the to-tal degree of an arbitrary polynomial g ∈ R[X1,, Xn] is the largest to-tal degree of its monomials. Usually, the degree of a polynomial in several indeterminates is its total degree, unless otherwise specified. A polynomial whose monomials have all the same degree is called a homogeneous polynomial. The total degree satisfies relations similar to the above, if R is a domain.

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348

3. Symmetric polynomials

Let R be a fixed commutative ring with identity. Let n ∈ N* and let σ ∈ Sn, the symmetric group of all permutations of n objects. Then there exists a unique homomorphism of R-algebras ϕσ : R[X1,, Xn] → R[X1,, Xn] such that ϕσ(Xi) = Xσ(i), ∀i = 1,, n (see 2.6, the universality property of the polynomial R-algebra R[X1,, Xn]). If g ∈ R[X1,, Xn], then

ϕσ(g) = g(Xσ(1),, Xσ(n)). If R is a domain and K is its field of fractions, consider K(X1,, Xn)

(the field of fractions of the domain R[X1,, Xn], called the field of ra-tional fractions in the indeterminates X1,, Xn with coefficients in K). Then ϕσ extends to a unique field homomorphism (denoted also ϕσ) ϕσ : K(X1,, Xn) → K(X1,, Xn). For any g, h ∈ R[X1,, Xn], h ≠ 0, ϕσ(g/h) = ϕσ(g)/ϕσ(h).

3.1 Definition. Let g ∈ R[X1,, Xn]. We call g a symmetric polynomial in R[X1,, Xn] if, for any σ ∈ Sn, we have ϕσ(g) = g.

If R is a domain and K is its field of fractions, a rational fraction g/h ∈ K(X1,, Xn) is called symmetric if, for any σ ∈ Sn, ϕσ(g/h) = g/h.

3.2 Example. In R[X1, X2, X3], the following polynomials are symmetric:

X1 + X2 + X3, X1 X2 X3, 2

231

233

221

223

212

21 XXXXXXXXXXXX +++++ .

The polynomial X1 + X2 is not symmetric in R[X1, X2, X3] (but it is symmetric in R[X1, X2]).

3.3 Remarks. a) Let S = g ∈ R[X1,, Xn] | ϕσ(g) = g, ∀σ ∈ Sn be the set of symmetric polynomials. Then S is an R-subalgebra of R[X1,, Xn]. For instance, if g, h ∈ S, then


349

ϕσ(g + h) = ϕσ(g) + ϕσ(h) = g + h, ∀σ ∈ Sn. The other conditions are checked similarly. It is easy to see that the

symmetric rational fractions form a subfield in K(X1,, Xn). b) If ni

ni XaX …11 is a monomial of the symmetric polynomial g,

then ( ) ( )ni

ni XaX σσ …1

1 is also a monomial of g, for any σ ∈ Sn.

3.4 Definition. Let n ∈ N* and let 0 ≤ k ≤ n. The polynomial sk := ∑∏i∈I Xi | I ⊆ 1, , n, |I| = k.

is called the fundamental (or elementary) symmetric polynomial of de-gree k in R[X1,, Xn] .

In other words, sk is the sum of all products of k distinct indetermi-

nates chosen among X1,, Xn; thus sk has ( )nk monomials. By

convention, s0 = 1 and sk = 0 if k > n. The polynomial sk is homogene-ous of degree k (indeed, all its monomials have degree k). Since sk

obviously depends on the number of indeterminates, the notation sk(X1,, Xn) is sometimes used to avoid confusions. For example, the

fundamental symmetric polynomials in 4 indeterminates are: s0 = 1

s1 = X1 + X2 + X3 + X4 s2 = X1 X2 + X1 X3 + X1 X4 + X2 X3 + X2 X4 + X3 X4

s3 = X1 X2 X3 + X1 X2 X4 + X1 X3 X4 + X2 X3 X4 s4 = X1 X2 X3 X4

The fundamental symmetric polynomials appear in the relations be-tween the coefficients of a polynomial and its roots (Viète's relations).

3.5 Theorem. a) Let n ∈ N* and let sk = sk(X1,,Xn). In R[X1,, Xn][X] the following relation holds:

(X − X1)(X − Xn) = X n − s1 X n−1 + s2 X

n−2 − + (−1)nsn.

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350

b) If R is a subring of the domain S and g = a0 + a1X + + anX

n ∈ R[X] has the roots x1, , xn ∈ S, then ansk(x1, , xn) = (1) kan − k.

Proof. a) Induction on n (exercise). b) There exists a unique R-algebra homomorphism

ϕ : R[X1,, Xn][X] → S[X] such that ϕ(Xi) = xi and ϕ(X) = X. We have, by a):

ϕ(an(X − X1)(X − Xn)) = an(X − x1)(X − xn) = an(X

n − s1 X n−1 + s2 X

n−2 − + (−1)nsn). But an(X − x1)(X − xn) = g (in the field of fractions K of S, these

polynomials have the same roots and the same leading coefficient). The relations follow by identifying the coefficients. !

3.6 Theorem. (The fundamental theorem of symmetric polynomi-als) Let g be a symmetric polynomial in R[X1,, Xn]. Then g is a polynomial of the fundamental symmetric polynomials: there exists a unique polynomial h ∈ R[X1,, Xn] such that g = h(s1, , sn).

In other words, denoting by S the R-subalgebra of symmetric polynomials in R[X1,, Xn], the unique R-algebra homomorphism ψ : R[X1,, Xn] → S with ψ(Xi) = si (for 1 ≤ i ≤ n) is an isomorphism.

Proof. Let T := ( ) nn

in

i iiXX n N∈,,111 …… be the set of all terms in

R[X1,, Xn]. Define a total order relation on T (the lexicographic or-der) by: nn k

nki

ni XXXX …… 11

11 ≤ ⇔ ∃r, 1 ≤ r ≤ n, such that it = kt, ∀t < r and ir ≤ kr. For instance, we have X1 > X2 > > Xn and

7 2 23 2 3 1 1 21 X X X X X X< < < < . This order relation is total and it is com-

patible with term multiplication: ∀λ, µ, ν ∈ T, µ ≤ ν implies λµ ≤ λν (one can prove that it is the unique total order on T, compatible with multiplication, such that X1 > X2 > > Xn). Moreover, T is well or-dered by the lexicographic order (any nonempty subset of T has a smallest element), as the following lemma shows. Therefore, one can make induction proofs on this ordered set (as is this proof).


351

The lexicographic order induces a preorder relation83, denoted also " ≤ ", on the set aλ | λ ∈ T, a ∈ R, a ≠ 0 of all monomials in R[X1,, Xn], by aλ ≤ bµ ⇔ λ ≤ µ. The proof of this is straightfor-ward. If p ∈ R[X1,, Xn], there exists a unique greatest monomial of p (with respect to the lexicographic preorder), called the leading mono-mial of p and denoted by lm(p). The following property holds:

If p, q ∈ R[X1,, Xn], such that lm(p) = aλ, lm(q) = bµ, where λ, µ ∈ T, a, b ∈ R and ab ≠ 0, then lm(pq) = lm(p)lm(q) = abλµ.

Indeed, any monomial of pq is a sum of monomials of the form forma rα·sβ, where rα is a monomial of p and sβ is a monomial of q. But α ≤ λ and β ≤ µ, so αβ ≤ λβ ≤ λµ. Thus, abλµ = lm(pq).

We proceed to the proof of the theorem. Let g be a symmetric polynomial and let lm(g) = ni

ni XaX …11 . Then i1 ≥ i2 ≥ ≥ in (if not,

then there exists a k such that ik < ik+1, and nkk in

ik

ik

i XXXaX …… 1111

++ is

also a monomial in g, strictly greater than lm(g), contradiction). We want to find a polynomial p of the form nj

nj sas …11 , such that

lm(p) = lm(g). Using the above property, ( ) ( ) ( )21 1

1 1 1 2 1nn j jjj j

n nlm as s aX X X X X=… … … This monomial equals lm(g) if and only if j1 + + jn = i1,

j2 + + jn = i2, , jn = in. Thus jn = in, jk = ik − ik + 1, for 1 ≤ k < n. The polynomial

g1 := g − njn

j sas …11 is symmetric and lm(g1) < lm(g). If lm(g1) = 0, then g1 = 0 and we are finished. If lm(g1) ≠ 0, replace g by g1 and apply the procedure above. The algorithm terminates in a finite number of steps because any strictly decreasing sequence of terms must be finite, as the next lemma shows. This concludes the existence part of the proof.

Let us prove the uniqueness (or, equivalently, Kerψ = 0). Suppose there exists a nonzero polynomial p ∈ R[X1,, Xn] such that

83 A relation that is reflexive and transitive, but not necessarily antisymmetric.

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352

ψ(p) = p(s1, , sn) = 0. We claim that there exists a unique nonzero monomial λ of p such that lm(ψ(p)) = lm(λ(s1, , sn)). Indeed, if α = ni

ni XX …11 and β = nj

nj XX …11 are distinct terms, then:

lm(α(s1, , sn)) = nnnn jn

jjin

ii XXXX …… …… ++++ ≠ 1111 = lm(β(s1, , sn)).

Thus, there exists a unique nonzero monomial λ of p such that lm(λ(s1, , sn)) = max lm(α(s1, , sn)) | α is a monomial of p. Since p(s1, , sn) = ∑α(s1, , sn) | α is a monomial of p, we

have lm(p(s1, , sn)) = lm(λ(s1, , sn)) ≠ 0, contradicting that p(s1, , sn) = 0. !

3.7 Lemma. a) Let (A, ≤) and (B, ≤) be well ordered sets. Then A×B is well ordered by the (lexicographic) order defined as (a, b) ≤ (a', b') if and only if a < a' or (a = a' and b ≤ b').

b) In a well ordered set (A, ≤) there exist no infinite strictly decreasing sequences.

c) For n ∈ N, the set Tn of the terms in R[X1,, Xn] is well ordered by the lexicographic order (thus any strictly decreasing sequence of terms must be finite).

Proof. a) Recall that the ordered set (A, ≤) is called well ordered if for any nonempty subset S of A, ∃α ∈ S such that α ≤ a, ∀a ∈ S (α is unique with this property and is called the smallest element of S. Thus, A is well ordered if any nonempty subset has a smallest element). Let ∅ ≠ S ⊆ A×B. Since S1 := a ∈ A | ∃b ∈ B cu (a, b) ∈ S ≠ ∅, and A is well ordered, there exists its smallest element α ∈ S1 (so, ∀(a, b) ∈ S, α ≤ a). Let S2 := b ∈ B| (α, b) ∈ S. There exists the smallest element β of S2. Then (α, β) is the smallest element of S: ∀(a, b) ∈ S, we have α < a (thus (α, β) < (a, b)) or α = a, in which case b ∈ S2, so β ≤ b.

b) Let (an)n ≥ 1 be a decreasing sequence of elements in A. Then the set an | n ≥ 1 has a smallest element ak. For any n ≥ k, we must have then ak ≤ an; since an ≤ ak (the sequence is decreasing), an = ak and the sequence is not strictly decreasing.


353

c) Induction on n. If n = 1, T1 = X n | n ∈ N is isomorphic as an ordered set to (N, ≤), which is well ordered. If n > 1, then Tn, ordered lexicographically, is isomorphic to Tn−1 × T1 with the order defined as in a). By induction, Tn−1 is well ordered and, by a), Tn−1 × T1 is well ordered.

The theorem 3.6 extends easily to symmetric rational fractions.

3.8 Corollary. (The fundamental theorem of the symmetric rational fractions) Let R be a domain and let K be its field of fractions. If p, q ∈ R[X1,, Xn], q ≠ 0, are such that p/q is a symmetric rational frac-

tions, then there exist the polynomials f, g ∈ R[X1,, Xn] such that ( )( )n

n

ssgssf

qp

,,,,

1

1

……

= . In other words, the subfield of the rational symmet-

ric fractions of the field K(X1,, Xn) is K(s1,, sn). Proof. If q is a symmetric polynomial, then p is symmetric, being

the product q·(p/q) in the subfield of the symmetric rational fractions. From 3.6, it follows that p, q ∈ R[s1,, sn]. If q is not symmetric, let s = ∏σ ∈Snϕσ(q). Then s is symmetric and

( )s

qpqp ∏ ≠= idσ σϕ

,

and we are in the conditions of the first case. !

The symmetric polynomial tm := X m1 ++ X m

n ∈ R[X1,, Xn] (m ∈ N) is expressible using the fundamental symmetric polynomials s1, , sn. The following identities allow a recursive computation of tm as a polynomial of s1, , sn.

3.9 Proposition. (Newton's identities) Let m ∈ N. In R[X1,, Xn] the following relation holds:

tm = s1 tm − 1 − s2 tm − 2 + + (−1)m − 2sm − 1 t1 + (−1)m − 1msm.

Appendices

354

Proof. If m > n, the convention sk = 0 for k > n truncates the for-mula above (there are only n terms).

Let r ≤ n and let (a1, , ar) be an r-uple of natural numbers with a1 ≥ a2 ≥ ≥ ar. Let s(a1, , ar) be the unique symmetric polynomial in R[X1,, Xn] having the leading monomial ra

raa XXX …2121 .

For instance, s(m, 0, , 0) = X m1 ++ X m

n = tm, s(1, 1, 0, , 0) = X1 X2 + X1 X3 + = s2.

To simplify notations, let 1i := (1, , 1) (1 appears i times) and (a, 1i) := (a, 1, , 1) (1 appears i times); also, we omit writing a se-quence of 0's: s(m, 0, , 0) = s(m), s(1, 1, 0, , 0) = s(1, 1) = s2, s(1i, 0, , 0) = s(1i) = si. The following relations can be verified eas-ily:

s1 tm − 1 = tm + s(m − 1, 1) s2 tm − 2 = s(m − 1, 1) + s(m − 2, 1, 1) s3 tm − 3 = s(m − 2, 1, 1) + s(m − 3, 1, 1, 1)

More generally, for any i ≤ minm − 1, n,

si tm − i = s(m − i + 1,1i) + s(m − i,1i). If m ≤ n and i = m − 1, then

sm − 1t1 = s(2,1m − 2) + msm. If m > n = i, then

sn tm − n = s(m − n + 1,1n − 1). Newton's identities follow by using the relations above in the sum

∑1≤ i< m (−1)i−1si tm−i. !

4. Rings and modules of fractions

The method of construction of the field Q from the domain Z generalizes naturally to any commutative ring R (although in general


355

the result will not be a field). In Q, all nonzero elements in Z become invertible. In many cases, there is no need that all nonzero elements in a ring R become invertible in some extension of R. It is thus natural to define a concept corresponding to the notion of set of denomina-tors. In what follows, all rings are commutative with identity and R denotes such a ring.

4.1 Definition. A subset S of R is called a multiplicatively closed set (or a multiplicative set) if:

a) 1 ∈ S. b) 0 ∉ S. b) ∀s, t ∈ S ⇒ st ∈ S. These conditions are natural for a set of denominators: 1 must be

a denominator, 0 cannot be one, and the product of two denominators is also a denominator.

For example, Z \ 0, Z \ 2Z, 2 n | n ∈ N are multiplicatively closed sets in Z.

For a given ring R and a multiplicatively closed set S ⊆ R, we con-struct a ring T and a ring homomorphism ϕ : R → T, such that the im-ages by ϕ of all elements in S are invertible in T (T is thus an R-algebra and ϕ is its structural homomorphism).

4.2 Definition. On the set R × S = (a, s) | a ∈ R, s ∈ S we define the relation:

∀(a, s), (b, t) ∈ R × S, define: (a, s) ∼ (b, t) ⇔ ∃u ∈ S such that u(ta − sb) = 0.

4.3 Remark. In the classic case of the construction of Q we have: R = Z, S = Z* and the following relation is used: ∀(a, s), (b, t) ∈ R × S, (a, s) ∼ (b,t) ⇔ ta − sb = 0. In this case the relation obtained coincides with the one in definition 4.2 (prove this!). The reason for adopting the

Appendices

356

definition 4.2 is to handle the more general case when the multiplica-tively closed set S possibly contains zero divisors.

4.4 Proposition. The relation " ∼ " is an equivalence relation on R × S.

Proof. We prove only the transitivity. Let (a, s), (b, t), (c, u) ∈ R × S such that (a, s) ∼ (b,t) and (b, t) ∼ (c,u). Then ∃v, w ∈ S such that v(ta − sb) = 0 and w(ub − tc) = 0. Multiply the first equality by uw and the second by sv and add. We obtain

uwvta − uwvsb + svwub − svwtc = 0 ⇔ vwt(ua − sc) = 0. Since S is multiplicatively closed, vwt ∈ S, so (a, s) ∼ (c,u). !

4.5 Definition. Let (a, s) ∈ R × S. The equivalence class of (a, s)

with respect to "∼" is denoted by sa or a/s and is called a fraction or a

quotient (of denominator s and numerator a). Thus:

sa = a/s = (b, t) ∈ R × S | (b, t) ∼ (a, s).

The factor set R × S/∼ (the set of all equivalence classes) is denoted

by S −1R: S −1R := a/s | a ∈ R, s ∈ S.

It easy to see that tsta

sa = , ∀s,t ∈ S, ∀a ∈ R.

We define on S −1R two operations, having in mind the usual rules of addition and multiplication of fractions. For any (a, s), (b, t) ∈ R × S, define:

stsbta

tb

sa +=+ :

stab

tb

sa =⋅ :


357

4.6 Proposition. The operations defined above on S −1R are cor-rectly defined and S −1R becomes a commutative ring with identity. The elements 0 and 1 in S −1R are:

s0

100 == , ∀s ∈ S;

ss== 1

11 , ∀s ∈ S.

The mapping ϕ : R → S −1R, ϕ(a) = a/1, ∀a ∈ R, is a ring homomorphism, called the canonical homomorphism (thus S −1R is an R-algebra).

Proof. We check that the addition is correctly defined. Let (a, s), (b, t), (a', s'), (b', t') ∈ R × S, such that (a, s) ∼ (a', s') and (b, t) ∼ (b', t'). We must show that (ta + sb, st) ∼ (t'a' + s'b', s't'). Let u, v ∈ S such that u(s'a − sa') = 0 and v(t'b − tb') = 0. Multiply the first of these equalities by tt'v and the second by ss'u and add them. We obtain vu((ta + sb)s't' − (t'a' + s'b')st) = 0.

The rest of the proof (the multiplication is correctly defined; check-ing the axioms for the ring S −1R) is left to the reader. !

Note that any s ∈ S is taken by ϕ into an invertible element in S −1R: ϕ(s) = s/1 has the inverse 1/s ∈ S −1R.

Moreover, the homomorphism ϕ is injective ⇔ S contains no zero divisors. Indeed, a/1 = 0/1 ⇔ ∃u ∈ S such that ua = 0.

If 0 ∈ S, then S −1R is the zero ring (with only one element, 0/1 = a/s, ∀a ∈ R, ∀s ∈ S); for this reason the condition 0 ∉ S is im-posed in the definition of a multiplicatively closed set.

The ring S −1R is called the ring of fractions (or the ring of quo-tients) of R with respect to the multiplicatively closed set S.

In the important case when R is a domain and S = R \ 0, S −1R is a field, called the field of fractions (or field of quotients) of R, and de-noted Q(K).

Appendices

358

4.7 Example. a) The field of quotients of Z is Q.

b) For any field K, the field of quotients of the polynomial ring K[X] is denoted by K(X) and is called the field of rational fractions

with coefficients in K. Its elements are fractions of the form fg ,

where f, g ∈ K[X], with g ≠ 0. Similarly, the field of quotients of the polynomial ring K[X1, , Xn] is denoted by K(X1, , Xn) and is called

the field of rational fractions in n indeterminates with coefficients in K.

c) The field of quotients of Z[X] is (isomorphic to) Q(X) (prove this!).

For any domain R, Q(R) is the smallest field that includes R. More generally, S −1R is the smallest ring that includes R (if S con-tains no zero divisors) such that all elements in S are invertible in S −1R. This fact is stated rigorously as follows:

4.8 Theorem. (The universality property of the ring of fractions) Let S be a multiplicatively closed subset of R. Then S −1R is a commutative ring with identity and ϕ : R → S −1R is a ring homomorphism such that ϕ(s) is invertible in S −1R, ∀s ∈ S. Moreover, the pair (ϕ, S −1R) is universal relative to this property, namely:

For any pair (γ, T) where T is a commutative ring with identity and γ : R → T is a ring homomorphism, such that γ(s) is invertible in T, ∀s ∈ S, there exists a unique ring homomorphism g : S −1R → T such that γ = gϕ.

Proof. Define g(a/s) = γ(a)(γ(s)) −1, ∀a ∈ R, ∀s ∈ S. The reader can easily verify that g is correctly defined, that it is a ring homomorphism and it is the only one with γ = gϕ. !


359

The complete structure of S −1R is that of R-algebra, the canoni-cal homomorphism ϕ being the structural homomorphism. In this set-ting, the above property reads:

For any commutative R-algebra (γ, Τ), where γ : R → T is the structural homomorphism, such that γ(s) is invertible in T for any s ∈ S, there exists a unique R-algebra homomorphism g : S −1R → T.

As expected, the universality property of the ring of fractions deter-mines the ring of fractions up to a (unique) isomorphism:

4.9 Theorem. Let S be a multiplicatively closed subset of R. As-sume B is a commutative ring with identity and β : R → B is a homo-morphism satisfying the property:

For any commutative ring with identity T and any ring homomorphism γ : R → T such that γ(s) is invertible in T, ∀s ∈ S, there exists a unique ring homomorphism g : B → T such that γ = gβ.

Then there exists a unique ring isomorphism h : S −1R → B such that hϕ = β. !

The above construction can be applied to an R-module M, with mi-nor modifications. Given a multiplicatively closed subset S ⊆ R and an R-module M, we define on M × S the equivalence relation: ∀(a, s), (b, t) ∈ M × S, (a, s) ∼ (b, t) ⇔ ∃u ∈ S such that u(ta − sb) = 0 (cf. 4.2). The following result is proven exactly like in the case of S −1R:

4.10 Proposition. Let M be an R-module and let S be a multiplica-tively closed subset of R. Then the relation " ∼ " defined above is an

equivalence relation on M × S. Denoting by sx the equivalence class

of (x, s) ∈ M × S and S −1M := ⎭⎬⎫

⎩⎨⎧ ∈∈ SsMxs

x , , S −1M becomes an

Abelian group with the addition: ∀x, y ∈ M, ∀s, t ∈ S,

Appendices

360

stsytx

ty

sx +=+ : .

Moreover, S −1M is an S −1R-module with the multiplication defined by: ∀a ∈ R,∀x ∈ M, ∀s, t ∈ S,

stax

tx

sa =⋅ : . !

The S −1R-module S −1M is called the module of fractions (or quo-tients) of M relative to the multiplicatively closed subset S. The homo-morphism ϕM : M → S −1M, ϕM(x) = x/1, ∀x ∈ M, is called the canoni-cal homomorphism.

The connection between the ideals of R and the ideals of the ring of fractions is very close. An immediate property is:

4.11 Proposition. Let I be an ideal in the ring R. Then S −1I := a/s | a ∈ I, s ∈ S is an ideal in S −1R. Moreover, any ideal in S −1R is of the form S −1I, for some ideal I in R.

Proof. It is immediate that S −1I ≤ S −1R if I ≤ R. If now J ≤ S −1R, let I := ϕ −1(J) (an ideal in R). We have a/1 ∈ J ⇔ ∃s ∈ S such that a/s ∈ J. Thus ϕ −1(J) = a ∈ R | ∃s ∈ S such that a/s ∈ J. Then S −1I = a/s | a ∈ I, s ∈ S = J. !

A similar connection exists between the submodules of M and the submodules of S −1M (can you state and prove it?).

4.12 Definition. A multiplicatively closed subset S ⊆ R is called saturated if all the divisors of the elements in S are also in S: ∀s ∈ S, ∀d, r ∈ R, dr = s implies d ∈ S and r ∈ S. If S is an arbitrary multiplicatively closed set, let

S' := d ∈ R | ∃r ∈ R, ∃s ∈ S such that dr = s. S' is called the saturate of the multiplicative set S. Evidently, S is

saturated ⇔ S = S'.


361

The following property says that the any ring of fractions can be constructed using a saturated multiplicative set.

4.13 Proposition. Let S be multiplicative set in the ring R. Then: a) S' is a saturated multiplicative set of R. b) There exists a canonical isomorphism S −1R ≅ S' −1R. Proof. a) Check the definition. b) We denote the equivalence relation on S' × R (defined as in 4.2)

with ≈ ; the equivalence class of (a, s) in R × S' is denoted by a//s ∈ S' −1R (in order to distinguish from the fraction a/s which may be in S −1R). Define the canonical mapping ϕ : S −1R → S' −1R, ϕ(a/s) = a//s, ∀a/s ∈ S −1R. The definition is independent on the choice of the representatives a and s: if (a, s) ∼ (b, t), then (a, s) ≈ (b, t). Clearly, ϕ is a ring homomorphism. We have Ker ϕ = a/s ∈ S −1R | a//s = 0//1. But a//s = 0//1 ⇔ ∃u ∈ S' such that ua = 0. Thus, ∃r ∈ R such that ur ∈ S and ura = 0, i.e. a/s = 0/1. Therefore, Ker ϕ = 0/1. Also, ϕ is surjective: if a//d ∈ S' −1R, with a ∈ R, d ∈ S', then there exists r ∈ R such that dr ∈ S. It is clear then that r ∈ S', so a//d = ar//dr = ϕ(ar/dr). !

An important example of a multiplicative set and its corresponding ring of fractions is the following:

4.14 Proposition. Let P be a prime ideal in the ring R. Then S := R \ P is a multiplicative subset in R and the ring of fractions S −1R has a unique maximal ideal (it is a local ring).

Proof. The definition of a prime ideal can be stated as follows: for any a, b ∉ P, then ab ∉ P, which means that S is multiplicatively closed. If I ≤ R, with I ∩ S ≠ ∅, then S −1I = S −1R. Indeed, if s ∈ I ∩ S, then s/1 ∈ S −1I and it is invertible, so S −1I = R. Thus, the proper ideals of S −1R are of the form J = S −1I, where I ∩ S = ∅ (⇔ I ⊆ P), so J ⊆ S −1P. But S −1P is a proper ideal: if 1/1 = p/s, with p ∈ P, s ∈ S,

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362

then ∃u ∈ S such that u(s − p) = 0 ⇒ us ∈ P ⇒ u ∈ P or s ∈ P, contra-dicting S = R \ P. Thus, S −1P is the unique maximal ideal in S −1R. !

If u : M → N is an R-module homomorphism, then we define the mapping

S −1u : S −1M → S −1N, (S −1u)(x/s) := u(x)/s, ∀x ∈ M, ∀s ∈ S. S −1u is easily seen to be an S −1R-module homomorphism and it is

the unique S −1R-module homomorphism with the property that (S −1u)ϕM = ϕNu, where ϕM : M → S −1M and ϕN : N → S −1N are the canonical homomorphisms.

Thus, for a fixed multiplicatively closed subset S, we defined a functor:

S −1− : R-Mod → S −1R-Mod. Moreover, S −1− is an additive functor: for any R-module homomor-

phisms u1, u2 : M → N, S −1(u1 + u2) = S −1u1 + S −1u2.

4.15 Proposition. The functor S −1− : R-Mod → S −1R- Mod is ex-act: if the sequence

CBA vu ⎯→⎯⎯→⎯ is exact in R-Mod, then the sequence

CSBAS vSuS 11 11 −− ⎯⎯ →⎯⎯⎯ →⎯−−

is exact in S −1R- Mod. !

The unproven statements above are proposed as exercises.

5. Categories, functors

The category language is nowadays all-pervading throughout mathematics. Introduced in 1942 by MacLane and Eilenberg, the con-


363

cept of category marks a new step of abstraction in mathematics. For instance, from the abstract notion of integer one passes to the notion of set of integers Z (as a new object of study, endowed with a certain structure). Generalizing key properties of Z yields the concept of ring. This leads to the study of structures given by axioms (such as the structures of group, ring, field, topological space etc). For a certain type of structure, usually a natural notion of homomorphism arises (for example, the familiar notion of group homomorphism, or ring homo-morphism).

The philosophy in the category theory is to study the class of all structures of a certain type (for instance the class of all rings) using the homomorphisms between these structures, and ignoring the elements of these structures. One advantage of this approach is given by generality: a result that holds in any category is valid in the category of groups, and also in the category of topological spaces etc. Besides many results and clarifications brought in almost all areas of mathematics, the category theory simplifies, unifies (to a certain ex-tent) and standardizes the language of mathematics.

We present here some basic concepts on categories that are useful for a better understanding of several topics. A detailed presentation of category theory is to be found for instance in HERRLICH, STRECKER

[1979].

Before proceeding to the definition of a category, we briefly de-scribe the concept of class.

The notion of class is introduced in the axiomatic set theory (in or-der to avoid the paradoxes generated by considering very large sets). In this theory, the notion of set and the relation "∈" (belongs to) are primary notions (are undefined); also, all objects are sets. Any ele-ment of a set is thus also a set.

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364

Formally, (in the Zermelo-Fraenkel theory84) a class is an expres-sion of the formal language of the set theory85 that contains exactly one free variable (in other words, a predicate with one variable). Of course, if P(x) is a predicate, the class P defined by P(x) is intuitively the collection of all objects (i.e. sets) x for which P(x) is true. Instead of writing "P(x) is true" one writes "x ∈ P", by analogy with the set language. Any set A is a class (corresponding to the predicate "x ∈ A"), but there exist classes that are not sets (for instance, the class of all sets, defined by the predicate "x = x").

If P and Q are classes, defined by the predicates P(x) and Q(x), their union P ∪ Q is the class P(x) ∨ Q(x); their intersection P ∩ Q is the class P(x) ∧ Q(x). We say that P ⊆ Q if the proposition ∀x(P(x) → Q(x)) is true. Similarly, one can define the analogue for classes of the usual set operations. For details, see

5.1 Definition. A category C consists of the following data: - a class Ob C. The elements of Ob C are called the objects of the

category C. - for every couple (A, B), where A, B ∈ Ob C, a (possibly empty) set

HomC(A, B) is given. The elements of HomC(A, B) are called mor-phisms (or arrows) from A to B. The fact that u ∈ HomC(A, B) is also written u : A → B or BA u⎯→⎯ ; A is the domain (or source) and B is

84 In the Gödel-Bernays-von Neumann theory, the notion of class is a primary

notion. The sets are exactly the classes that are elements of some class. 85 We do not define here this formal language. Roughly speaking, it consists of

expressions (strings of symbols) formed from the atomic expressions (of the type x = y or x ∈ y) by using the logical operators ∨, ¬, ∧ and the quantifiers ∀ and ∃. For instance "(∀x(x ∈ y))∧ (∃z(y = z)∨¬(z = a))" is an expression, in which x and z are bound variables and y is a free variable. If a is assumed to be a constant, then this expression is a predicate (it has one free variable, namely y). An expression with no free variables is a proposition.


365

the codomain (or sink) of u. The class ∪HomC(A, B)| A, B ∈ Ob C is denoted Hom C and is called the class of the morphisms of the cate-gory C.86

- for any triple (A, B, C) of objects of C there exists a function de-fined on HomC(B, C)×HomC(A, B) with values in HomC(A, C). The im-age of the couple (v,u) is denoted by vu (or simply vu) and is called the composition of the morphisms v and u.

In any category C the following axioms must be satisfied: 1) Any morphism has a unique domain and a unique codomain: for

any A, B, C, D ∈ Ob C, (A, B) ≠ (C, D) implies HomC(A, B)∩HomC(C, D) = ∅.

2) The composition of morphisms is associative: ∀A, B, C, D ∈ Ob C and ∀u : A → B, v : B → C, w : C → D, we have w(uv) = (wu)v (denoted usually by wuv).

3) ∀A ∈ Ob C, there exists a morphism 1A : A → A (called the iden-tity morphism of A) such that, ∀B ∈ Ob C and ∀u : A → B, v : B → A, we have u1A = u and 1Av = v.

5.2 Remark. a) The identity morphism of an object A is unique: if j: A → A is an identity morphism of A, then j = j1A = 1A.

b) In the definition of a category, the morphisms are essential. One can identify an object A ∈ Ob C with its identity morphism 1A. The notion of category can be defined using only the concept of morphism.

5.3 Examples. a) The category Set of all sets. Its objects are sets. If A and B are sets, HomSet(A, B) is the set of all functions ϕ : A → B. The composition of morphisms in Set is the usual function composi-tion. The identity morphism of A is the identity function of A.

86 Formally, it is the class defined by the predicate H(u) = "∃A∃B(A ∈ Ob C ∧

B ∈ Ob C ∧ u ∈ Hom(A, B))".

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366

b) The category Gr of groups. Ob Gr is the class of groups and HomGr(G, H) is the set of group homomorphisms from G to H, ∀G, H ∈ Ob Gr. As in Set, the composition is the usual function composi-tion.

c) Let R be a ring with identity. The category R-Mod has as objects left R-modules, and the morphisms are R-module homomorphisms, with the usual composition. In the same way one defines the category Mod-R of right R-modules.

d) One can define similarly the following categories: - Ring: the rings with the ring homomorphisms. - Ringu: the rings with identity, the morphisms are ring homo-

morphism preserving the identity. - Ab: the Abelian groups, with the group homomorphisms - Poset: the (partially) ordered sets, the morphisms being the order

preserving mappings. d) Let (A, ≤) be a set equipped with a preorder relation (a transitive

and reflexive relation). Define a category A, with Ob A := A. For any a, b ∈ Ob A, set

HomA(a, b) = ⎩⎨⎧ (a, b), if a ≤ b ∅, else

The reader is invited to define the composition of morphisms and check the axioms 1) − 3).

e) Let (G, ·) be a monoid (the operation is associative and has a neu-tral element). Define a category G as follows: Ob G is a set with one element (for instance Ob G = G), and HomG(G, G) = G (the mor-phisms are the elements of G); the composition of the morphisms a, b ∈ G is a·b (where · is the operation on G). The identity morphism is the identity element.

The reader can easily produce other examples of categories, based on hers/his mathematical background (the category of semigroups, the category of finite sets, the category of fields, the category of topologi-


367

cal spaces, the continuous mapping being the morphisms etc.). In each situation it is necessary to state exactly the class of the objects of the category, the set of morphisms between two arbitrary objects, the composition of morphisms and check axioms 1)-3).

Often the writing A ∈ C replaces A ∈ Ob C, if no confusion arises.

5.4 Definition. A category C is called a subcategory of a category D if Ob C ⊆ Ob D and, ∀A, B ∈ Ob C, HomC(A, B) ⊆ HomD(A, B); moreover, the composition of two morphisms in C is their composi-tion in D. We call C a full subcategory of D if ∀A, B ∈ Ob C, HomC(A, B) = HomD(A, B).

For instance, Ab is a full subcategory of Gr; Gr is a subcategory of Set, but not a full subcategory.

5.5 Definition. We define now some remarkable objects and mor-phisms in a category C that generalize familiar concepts.

a) An object I ∈ C is called an initial object if ∀A ∈ C, |HomC(I, A)| = 1 (there exists a unique morphism I → A). An object F is called a final object if ∀A ∈ C, |HomC(A, I)| = 1. An object that is simultaneously initial and final is called a zero object.

b) A morphism u : A → B is called: - a monomorphism if ∀C ∈ C, ∀v, w ∈ HomC(B, A), uv = uw im-

plies v = w. - an epimorphism if ∀C ∈ C, ∀v, w ∈ HomC(B, C), vu = wv im-

plies v = w. - a bimorphism if it is both a monomorphism and an epimor-

phism. - an isomorphism if there exists v : B → A such that uv = 1B and

vu = 1A (v is called then the inverse of u). The notation A ~→ B is used to denote an isomorphism.

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368

Two objects A, B ∈ C are called isomorphic (written A ≅ B) if there exists an isomorphism A → B. The relation of isomorphism on the class Ob C is an equivalence relation.

5.6 Examples. a) In Gr there exist initial objects, namely the groups having one element (necessarily the neutral element of that group). These are also final objects (thus they are zero objects) in Gr. The same remark holds for Ab and R-Mod.

b) In Set the empty set ∅ is the only initial object87; any set with one element is a final object. Set has no zero objects.

c) The monomorphisms in Set (as in Gr, Ab, R-Mod) are the mor-phisms that are injective functions. Which are the epimorphisms? In these categories the isomorphisms coincide with the bimorphisms.

d) In the category Ann of rings with identity, the inclusion Z → Q is a monomorphism and an epimorphism, and is not a surjective func-tion or an isomorphism.

5.7 Proposition. Let C be a category and let a A, B be initial ob-jects in C. Then there exists a unique isomorphism A ~→ B.

Proof. Since A is an initial object, there exists a unique morphism ϕ : A → B. But B is an initial object, thus there exists a unique mor-phism ψ : B → A. The morphism ψϕ : A → A is equal to 1A (because there exists a unique morphism A → A). Likewise, ϕψ = 1B. So, ϕ and ψ are isomorphisms, inverse to each other. !

It is important to point out that various universality properties that some objects satisfy are simply a restatement of the fact that those objects are initial (or final) objects in certain categories. In this situa-tion, the propositions of the type the universality property of determines up to a unique isomorphism merely translate for a spe-

87 For any set A, there exists a unique function ∅ → A, namely the function ∅.


369

cific category the assertion between any two initial (final) objects in a category there exists a unique isomorphism.

5.8 Examples. a) The direct sum. Let (Mi)i∈I be a family of objects in R-Mod. Let S be the category whose objects are couples of the form (S, (σi)i∈I), where S ∈ R-Mod and σi : Mi → S are morphisms in R-Mod, ∀i ∈ I. If (S, (σi)i∈I), (T, (τi)i∈I) ∈ S, define the morphisms in S between these objects as the R-module homomorphisms ϕ : S → T such that ϕσi = τi, ∀i ∈ I. Check the axioms for a category and the fol-lowing assertion:

(S, (σi)i∈I) is a direct sum of the family (Mi)i∈I, of canonical injec-tions (σi)i∈I, is tantamount to saying that (S, (σi)i∈I) is an initial object in S.

b) The direct product. Let (Mi)i∈I be a family of objects in R-Mod. Then: (P, (πi)i∈I) is a direct product of the family (Mi)i∈I ⇔ (P, (πi)i∈I) is a final object in a certain category (describe it!).

An important principle, often invoked, is the principle of duality.

5.9 Definition. If C is a category, the dual category C° is defined as follows: Ob C° := Ob C; if A ∈ Ob C, let A° be the object A seen in C°. For any A, B ∈ Ob C, let HomC°(B°, A°) := HomC(A, B). A morphism u : A → B in C is denoted u° : B° → A° in C°. The Composition in C° of the morphisms u° : B° → A° and v° : C° → B° is defined by u°v° := (vu)°, where vu is the composition of u : A → B and v : B → C in C. Evidently, there exists 1A° = (1A)°.

Intuitively, the dual of the category C is obtained by reversing the arrows in C (and reversing the order of composing the arrows).

Let P be a statement formulated in terms of objects and morphisms. For each category C, one obtains a proposition, denoted P(C). Let P°

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370

be the dual statement (obtained from P by reversing the arrows and the order in composing the arrows)88. The principle of duality is the following: If P is valid in any category, then P° is valid in any cate-gory.

Similarly, any notion (definition) in a category has a dual notion, obtained by reversing the arrows and the order in composing the ar-rows. A notion that coincides with its dual is called autodual.

5.10 Example. a) The dual of the notion of initial object is the no-tion of final object.

b) The dual of the notion of monomorphism is the notion of epimorphism.

c) The notion of isomorphism is autodual. d) We saw that: for any category C, any two initial objects in C

(if any) are isomorphic. By dualization, one obtains (no new proof needed): for any category C, any two final objects in C (if any) are isomorphic.

The intuitive concept of morphism of categories is the notion of functor.

5.11 Definition. Let C and D be categories. A covariant functor F from C to D, denoted F : C → D, is a couple F = (F', F"), where F' : Ob C → ObD, F": Hom C → Hom D, such that:

(F1) ∀A, B ∈ Ob C, F"(HomC(A, B)) ⊆ HomD(F'(A), F'(B)); in other words, if u : A → B, then F"(u) : F'(A) → F'(B).

(F2) F preserves the composition of morphisms: ∀A, B, C ∈ Ob C and ∀u : A → B, v : B → C, then F"(vu) = F"(v)F"(u).

(F3) F preserves the identity morphisms: ∀A ∈ C, F"(1A) = 1F"(A).

88 in other words, P°(C) is the same thing as P(C°) interpreted in C.


371

A contravariant functor F : C → D satisfies F3 and the duals of F1 and F2:

(F1*) F"(HomC(A, B)) ⊆ HomD(F'(B), F'(A)), ∀A, B ∈ Ob C. (F2*) ∀A, B, C ∈ Ob C and ∀u : A → B, v : B → C, then

F"(vu) = F"(u)F"(v). A contravariant functor reverses the arrows. A contravariant

functor from C to D is the same as a covariant functor from C to D° (or from C° to D). This is the reason why the results on covariant functors transfer to contravariant functors. In what follows, functor means covariant functor.

The distinction between the components F' and F" of the functor F is usually dropped, denoting F(A) instead of F'(A) and F(u) instead of F"(u). Besides, F is perfectly determined by F", its action on mor-phisms, cf. 5.2).

A functor F : C → D is called: - faithful if ∀A, B ∈ C, FA,B : HomC(A, B) → HomD(FA, FB) is an

injective function. - full if ∀A, B ∈ C, FA,B : HomC(A, B) → HomD(FA, FB) is a

surjective function. - fully faithful if it is full and faithful.

We are interested in categories in which the morphisms are func-tions between sets:

5.12 Definition. The category C is called concrete if there exists a covariant faithful functor F : C → Set.

The category Gr is concrete: define the functor U : Gr → Set, by associating to any group its underlying set89 and sending any group

89 A group G is, formally, a couple (G, ·), where G is the underlying set of the

group and · : G×G → G is the group operation. Thus, a function from a group (G, ·)

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372

homomorphism u to u, seen as a function between the underlying sets. Then U is a faithful functor (which is not full). U is called a forgetful functor: it forgets the group structure. Similarly, Ab, Ring, R-mod are concrete categories (why?).

In concrete categories there is a notion of free object on a set X (compare the following definition with the universality property of the free R-module on a set X ).

5.13 Definition. Let C be a concrete category and let F : C → Set be a covariant faithful functor. The object L in C is called free (relative to F) over the set X ⊆ F(L) if, for any A ∈ C and any function γ : X → F(A), there exists a unique morphism in C, g : L → A, such that F(g)|X = γ.

If C is one of the categories Gr, Ab, R-mod, Ring, and F is the forgetful functor, the definition reads: L ∈ C is free over the set X ⊆ L if, for any object A ∈ C and any function γ : X → A, there exists a unique morphism in C, g : L → A, such that g|X = γ.

The following notion allows comparing two functors and is the analogue of the concept of homomorphism of algebraic structures.

5.14 Definition. Let F, G : C → D be covariant functors. A natural transformation (or a functor morphism) α : F → G is given if for any object A ∈ C, there exists a morphism in D, αA : F(A) → G(A), such that, for any u : A → B morphism in C, αBF(u) = G(u)αA, i.e. the following diagram (of morphisms in D) is commutative:

to a group (H, *) is not the same thing as a function between their underlying sets G and H. For this reason, Gr is not a subcategory of Set.

6. Solvable groups

373

If , for any A ∈ C, αA : F(A) → G(A) is an isomorphism in D, then α : F → G is called a natural isomorphism, in which case the functors F and G are called naturally isomorphic.

Many canonical isomorphisms in module theory (for instance) express the fact that there exists a natural isomorphism between some functors.

6. Solvable groups

The notion of solvable group is closely connected with the solvabil-ity by radicals of a polynomial.

In what follows, (G, ·) is a group and its neutral element is denoted by 1. The trivial subgroup 1 is also denoted 1. The notation H ≤ G signifies H is a subgroup of G, and H . G means H is a normal subgroup of G.

6.1 Definition. The group G is called a solvable group if there ex-ists a finite chain of subgroups of G: 1 = G0 ≤ G1 ≤ ≤ Gn = G, (S) such that:

i) Gi−1 . Gi, ∀i, 1 ≤ i ≤ n; ii) The factor groups Gi/Gi−1 are Abelian, ∀i, 1 ≤ i ≤ n.

( ) ( ) ( )BFuFAF ⎯⎯⎯⎯ →⎯

αB αA

( ) ( ) ( )BGuGAG ⎯⎯⎯⎯ →⎯

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374

A chain of subgroups (S) that satisfies (i) is called a normal series of G.90 A chain (S) satisfying i) and ii) is called a solvable series of G. The number n is called the length of the series (S). The groups Gi/Gi−1 are called the factors of the normal series (S).

6.2 Examples. a) Any Abelian group is solvable. b) Let n ∈ N* and let Sn be the group of all permutations on n ob-

jects (also called the symmetric group on n objects). Let An be the alternating group on n objects, namely the subgroup of Sn formed by the even permutations. An is a subgroup of index 2 in Sn (it is the ker-nel of the sign homomorphism ε : Sn → − 1, 1), thus it is normal in Sn.

The group S3 is solvable (and not Abelian!), a solvable series being 1 ≤ A3 ≤ S3. Indeed, A3 has 3 elements, so it is Abelian; A3 is normal in S3, and the factor S3/A3 is Abelian, since it has 2 elements.

c) S4 is solvable. A solvable series is 1 ≤ V ≤ A4 ≤ S4, where V = 1, (12)(34), (13)(24), (14)(23)91. Check the details!

d) Any nonabelian simple group (i.e. having no proper normal sub-groups) is not solvable.

Before studying solvable groups, let us recall some elementary re-sults in group theory.

6.3 Theorem. (The fundamental isomorphism theorem) Let ϕ : G → G' be a group homomorphism. Then there exists a canonical isomorphism G/Kerϕ ≅ Imϕ xKerϕ & ϕ(x) !

90 Some authors use in this case the term subnormal and call a normal series

a chain (S) of normal subgroups of G. 91 V is also called the Klein group (the Viergruppe).

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375

6.4 Theorem. (Second isomorphism theorem) Let G be a group, let H and N be subgroups of G, such that N is a normal subgroup in the group (H, N) generated by H ∪ N in G. Then (H, N) = HN = hn | h ∈ H, n ∈ N = NH, N ∩ H . H and there is a canonical isomorphism HN/N ≅ H/(N ∩ H). !

If H, N are subgroups of G satisfying the conditions in the previous theorem, HN or NH denotes the subgroup generated by H ∪ N.

6.5 Theorem. (Third isomorphism theorem) Let G be a group, let A and B be normal subgroups of G such that A ≤ B. Then B/A . G/A and there is a canonical isomorphism

BG

ABAG

≅ . !

6.6 Proposition. (The modularity property) Let G be a group and let A, B, N be subgroups of G, such that N is normal and B ≤ A. Then A ∩(BN) = B ∩(AN). !

6.7 Proposition. Let G be a group. a) If G is solvable, then each subgroup of G is solvable. b) If G is solvable, then each factor group of G is solvable. c) If H . G and H and G/H are solvable, then G is solvable. In

other words, if 1 → H → G → F → 1, is an exact sequence92 of groups and group homomorphisms, then G is solvable if and only if H and F are solvable.

92 The notion of exact sequence of groups and group homomorphisms is defined

exactly as in the case of modules.

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376

Proof. a) Let 1 = G0 ≤ G1 ≤ ≤ Gn = G be a solvable series of G, let H ≤ G and let Hi := Gi ∩ H. We claim that 1 = H0 ≤ H1 ≤ ≤ Hn = H is solvable series of H.

Fix i, 1 ≤ i ≤ n. The canonical homomorphism ϕ : Hi → Gi/Gi−1 (ϕ(x) = xGi−1, ∀x ∈ Hi) has kernel Hi−1. Thus Hi−1 . Hi and Hi/Hi−1 ≅ Imϕ ≤ Gi/Gi−1, which is Abelian.

b) Let F be a factor group of G. There exists a surjective homomor-phism ϕ : G → F. If 1 = G0 ≤ G1 ≤ ≤ Gn = G is a solvable series of G, let Fi := ϕ(Gi). We claim that 1 = F0 ≤ F1 ≤ ≤ Fn = F is a solvable series of F. Since Gi−1 . Gi and ϕ is surjective, Fi−1 . Fi. Let

H = Kerϕ. Then Ker ( )iGϕ = H ∩ Gi, so

Fi ≅ Gi/Ker ( )iGϕ = Gi/(H ∩ Gi) ≅ (GiH)/H.

We used the fundamental and the second isomorphism theorems. Therefore:

( )( ) ( )ii

i

ii

i

i

ii

i

i

i

i

i

i

GHGG

GHGG

HGHGG

HGHG

HHGHHG

FF

∩∩ 111

1

111 −−−

−

−−−

=≅=≅≅

We used successively: the third, the second isomorphism theorems and the modularity property for H . G and Gi−1 ≤ Gi. But:

( ) ( ) 11

1

1 −−

−

−

≅iii

ii

ii

i

GGHGGG

GHGG

∩∩,

The last group is Abelian (it is a factor group of Gi/Gi−1). c) Any subgroup of G/H is of the form G'/H, where G' ≤ G, with

H ≤ G. Therefore, a solvable series for G/H is of the form 1 = G0/H ≤ G1/H ≤ ≤ Gn/H = G/H, where

H = G0 ≤ G1 ≤ ≤ Gn = G. If 1 = H0 ≤ H1 ≤ ≤ Hm = H is a solvable series for H, then

1 = H0 ≤ H1 ≤ ≤ Hm = G0 ≤ G1 ≤ ≤ Gn = G is a solvable series for G. !

6. Solvable groups

377

6.8 Proposition. Let G be a group. If G has a normal series whose factors are solvable groups, then G is solvable.

Proof. If the series has length 2, there exists H . G such that H and G/H are solvable and the previous result applies. Continue by induc-tion on the length of the series. !

The finite solvable groups have the following property, which is essential in the theory of solvability by radicals.

6.9 Proposition. A finite group is solvable if and only if it has a normal sequence whose factors are cyclic groups of prime orders.

Proof. Of course, a group with the property above is solvable. Conversely, let G be finite solvable. If |G| ≤ 3, all is clear. We sup-

pose the statement is true for any solvable group whose order is less than |G| and we prove for G.

If G is Abelian, then use the following lemma. Assume now that G is not Abelian. The definition of solvability im-

plies the existence of a proper normal subgroup H of G such that H is solvable and G/H is Abelian. Since H and G/H have orders less than |G|, the induction hypothesis says that H and G/H have each a normal series with cyclic factors of prime orders. As in the proof of prop. 6.7.c), gluing these series yields a solvable series of G, with cyclic factors of prime orders. !

6.10 Lemma. A finite Abelian group has a normal sequence whose factors are cyclic groups of prime orders.

Proof. This is an easy consequence of the structure theorem of the Abelian finite groups. We give a proof that does not use this result.

Assume first that (G, ·) is cyclic of order n, generated by x ∈ G. If n is prime, we are finished; if not, for any d dividing n, there exists a subgroup of order d (generated by x n/d ). Thus, taking a prime divisor p of n, there exists a subgroup H of order p. The factor group G/H and H have orders less than |G|. Apply now an induction, as above.

Appendices

378

If G is not cyclic, let x ∈ G, x ≠ 1. Thus, the cyclic subgroup C generated by x is not equal to G. Apply an induction for C and G/C, of orders less than |G|. !

6.11 Proposition. Let G be a finite group, whose order is a power of a prime p (such a group is called a p-group). Then G is solvable. In particular, there exists a normal series of G whose factors are cyclic groups of order p.

Proof. Let |G| = p n. If n = 1, then G is cyclic, and solvable. If n > 1, then the center of G, C(G) ≠ 1. Indeed, in the conjugacy classes formula (IV.3.15.e)),

|G| = |C(G)| + ( )[ ]∑ ∈Sa aCG : ,

(where S is a system of representatives of the conjugacy classes of G with at least 2 elements), |G| and [G : C(a)] are powers of p, so |C(G)| is a multiple of p (and is nonzero, since 1∈ C(G)).

Consequently, 1 ≠ C(G) . G and the factor group G/C(G) is still a p-group, whose order is less than |G|. Using an induction argument, we obtain that G/C(G) is solvable. Since C(G) is Abelian, thus solv-able, we obtain that G is solvable. !

The next proposition says that among the symmetric groups Sn, the solvable groups are exactly those in Example 6.2. The proof is not in-cluded, being often encountered in introductory Algebra texts (see for instance HUNGERFORD [1974])

6.12 Proposition. a) If n ≤ 4, then Sn is solvable. b) If n ≥ 5, then Sn is not solvable. More precisely, the alternating

subgroup An is simple (has no proper normal subgroups) and noncommutative, so it is not solvable. !

379

Index

A

adjoining elements to a field....... 175 algebra ........................................ 334 algebra homomorphism .............. 335 algebraic (element) ..................... 177 algebraic closure (absolute) ........ 198 algebraic closure (relative) ......... 189 algebraic generators .................... 226 algebraic integer ........................... 27 algebraic number ........................ 178 algebrically independent (set)..... 223 alternating group......................... 374 annihilator........................... 104, 133 arithmetically equivalent matrices121 arrow........................................... 364 Artin-Schreier theorem ............... 302 ascending chain condition............. 34 associated in divisibility ............... 12 autodual ...................................... 370

B

basis change matrix..................... 113

basis of a module.........................106 bilinear function ..........................307

nondegenerate.........................307 bimorphism .................................367

C

canonical homomorphism .............75 canonical injections.......................86 canonical projections.....................81 canonical surjection.......................75 category.......................................364

concrete ..................................371 dual.........................................369

Cayley-Hamilton theorem...........163 center of a ring ............................334 characteristic exponent................173 characteristic matrix ....................290 characteristic of a ring.................172 characteristic polynomial ............290

of a matrix ..............................161 of an endomorphism...............161

Chevalley theorem ......................223 Chinese remainder theorem.........141

380

class ............................................ 363 codomain .................................... 364 coefficient of a polynomial......... 347 cokernel ........................................ 96 comaximal ideals........................ 141 complement .................................. 90 complex of modules ..................... 96 composite of fields ..................... 175 composition of morphisms ......... 365 congruent modulo n...................... 26 conjugate elements ..................... 235 conjugate extension .................... 240 constant term .............................. 347 constructibile complex number .. 278 constructibile point ..................... 276 constructibile real number .......... 277 content of a polynomial ................ 41 coordinates of an element in a basis

............................................... 106 coprime......................................... 14 coproduct...................................... 86 cyclotomic extension. ................. 213 cyclotomic polynomial ............... 214

D

Dedekind's Lemma..................... 262 degree

of an algebraic element.......... 181 of an extension....................... 172

degree of a polynomial ....... 346, 347 denominator................................ 356 determinant................................. 291

determinant of an endomorphism162 diagonal matrix ...........................121 direct product

of homomorphisms ..................92 of modules................................81 of modules................................83

direct sum of modules................................85 of submodules ..........................90

direct summand.............................90 discriminant

of a polynomial ......................320 of an element..........................320

divisibility .....................................10 division with remainder theorem ..22 divisor ...........................................11 domain ..................................11, 364 dual of a group ............................306 duality .........................................369

E

eigenvalue ...................................165 eigenvector..................................165 Eisenstein criterion .......................47 element

integral .....................................52 elementary divisors .....................147 elementary divisors (of a matrix) 154 elementary divisors (of an

endomorphism) ......................154 elementary transformations.........123 endomorphism ..............................68

381

cyclic...................................... 153 endomorphism ring....................... 61 epimorphism ......................... 76, 367 Euclidian Algorithm ..................... 23 Euclidian domain.......................... 22 Euler phi function ....................... 213 evaluation homomorphism ......... 346 exponent (of a group).................. 303 exponent of a group .................... 221 extension..................................... 170

Abelian................................... 299 algebraic................................. 187 cyclic...................................... 299 finite....................................... 172 finitely generated ................... 175 Galois..................................... 267 Kummer ................................. 303 n-radical ................................. 314 purely transcendental ............. 224 radical .................................... 314 simple..................................... 175 transcendental ........................ 188

extension of a module................... 97 external direct sum of modules..... 87

F

factor algebra .............................. 336 factorization algorithm.................. 48 field............................................. 169

algebraically closed................ 198 of rational fractions ................ 348

field extension............................. 170

field generated by a set................174 fixed field ....................................233 fixed subfield...............................172 formal derivative .........................195 Fp 171 fraction ........................................356 free resolution .............................109 free subset of a module ...............105 Frobenius endomorphism............210 Frobenius theorem.......................163 functor

contravariant...........................371 covariant .................................370 fidel ........................................371 forgetful..................................372 fully faithful............................371 plin..........................................371

functor morphism........................372 fundamental isomorphism theorem77 Fundamental Theorem of Algebra205 fundamental theorem of Galois

Theory ....................................266 fundamental theorem of symmetric

polynomials ............................350

G

Galois connections ......................233 Galois Field .................................211 Galois group

of a field extension .................233 of a polynomial.......................233

Galois group of a polynomial......314

382

Gauss integers .............................. 12 GCD ............................................. 14 GCD-domain ................................ 15 general equation of degree n....... 318 generic polynomial of degree n .. 318 GL(n, R) ...................................... 114 Gr 366 greatest common divisor............... 14 group

solvable.................................. 373 group of units ............................... 13

H

Hilbert Satz 90............................ 300 additive version...................... 301

homomorphism of module extensions ............... 97

I

ideal ...................................... 20, 331 maximal ................................. 332 prime...................................... 332 proper..................................... 331

idempotent .................................... 91 identity morphism....................... 365 image ............................................ 71 independent family of submodules90 indeterminate .............................. 342 initially constructibile point........ 275 inseparable

element .................................. 244 extension................................ 245

inseparable degree.......................258 integer division theorem ...............44 integers modulo n..........................26 integral domain .............................11 intermediate field ........................171

proper .....................................171 internal direct sum ........................90 invariant factors ..........................137 invariant factors (of a matrix) .....154 invariant factors (of an

endomorphism) ......................154 Invariant factors theorem ............131 invariant subspace.......................152 inverse (of a morphism) ..............367 Irr(x, K) .......................................179 irreducible .....................................18 isomorphism................................367

J

Jordan canonical form of a matrix164 Jordan canonical matrix ..............157 Jordan cell ...................................156

K

kernel ............................................71 K-homomorphism .......................171 Kummer Theory..........................303

additive...................................311 multiplicative .........................308

L

Lagrange interpolation polynomial54

383

lattice .......................................... 176 LCM ............................................. 14 leading coefficient ...................... 347 leading monomial ....................... 351 least common multiple.................. 14 lexicographic order ..................... 350 linear combination ........................ 63 linearly dependent............... 104, 105 linearly independent............ 104, 105 Lüroths theorem ........................ 187

M

matrix canonically diagonal .............. 121 elementary.............................. 122 in Smith normal form............. 121

matrix companion of a polynomial............................................... 156

matrix of a homomorphism ........ 112 minimal polynomial.................... 179 minimal polynomial (of a matrix)154 minimal polynomial (of an

endomorphism) ...................... 154 Mod-R ......................................... 366 module .......................................... 59

decomposable......................... 140 factor ........................................ 75 finitely generated ..................... 65 free ......................................... 106 indecomposable ..................... 140 left ............................................ 59 of fractions ............................. 360

right ..........................................59 simple .......................................80

module automorphism...................71 module homomorphism ................68 module isomorphism.....................71 monic polynomial........................347 monoid ........................................336 monomial ....................................342 monomorphism .....................76, 367 morphism ....................................364 multiple .........................................11 multiple root..................................45 multiplicative set .........................355

saturated .................................360 multiplicatively closed set...........355 multiplicity of a root..............45, 194 mutually prime ..............................14

N

natural isomorphism....................373 natural transformation .................372 Newton's identities ......................353 Noetherian ring .............................35 norm ......................................27, 291 normal closure.............................241 normal extension .........................238 normal series ...............................374 numerator ....................................356

O

object final ........................................367

384

free......................................... 372 initial...................................... 367 zero ........................................ 367

object (in a category) .................. 364 opposite of a ring.......................... 60 order ........................................... 130

P

p-basis......................................... 261 p-dimension................................ 261 perfect field ................................ 246 p-group ....................................... 378 PID ............................................... 31 polynomial

elementary symmetric............ 349 fundamental symmetric ......... 349 homogeneous......................... 347 monic ....................................... 27 symmetric .............................. 348

polynomial algebras ................... 341 polynomial function ..................... 54 prime ............................................ 18 prime field .................................. 172 primitive element........................ 175 primitive polynomial .................... 41 principal ideal ............................... 20 principal ideal domain .................. 31 projective linear group................ 272 proper divisor ............................... 13 p-submodule ............................... 133 purely inseparable....................... 254 purely inseparable closure .......... 255

Q

quadratic integer ...........................27 quaternion group .........................208 quaternions, skew field of ...........207 quotient .................................22, 356

R

rank of a free module ..................111 rational integers.............................27 remainder ......................................22 resultant.......................................324 ring

integrally closed .......................52 of quadratic integers.................28

R-Mod .........................................366 root

multiple ..................................194 simple.....................................194

root of a polynomial....................177 root of unity ................................212

primitive.................................213

S

separable element ...................................244 extension ................................245 polynomial .............................244

separable closure.........................253 separable degree..........................248 sequence........................................97 sequence of modules .....................96

exact .........................................96

385

semiexact ................................. 96 Set ............................................... 365 similar endomorphisms............... 152 similar matrices........................... 152 simple root .................................... 45 sink ............................................. 364 smallest element ......................... 352 solvable by radicals..................... 314 solvable series............................. 374 source.......................................... 364 split epimorphism ....................... 100 split monomorphism ................... 100 split short exact sequence ........... 100 splits............................................ 183 splitting field............................... 201 squarefree ..................................... 11 structural homomorphism (of an

algebra) .................................. 335 subalgebra................................... 335

generated by a set................... 335 subcategory................................. 367

full.......................................... 367 subextension ............................... 171 subfield ....................................... 170

proper..................................... 170 submodule..................................... 62

cyclic........................................ 65 generated by a set..................... 64 maximal ................................... 67 minimal .................................... 67 proper....................................... 63

subring generated by a set ...........175 sum of ideals .........................20, 331 support...................................86, 337 symmetric group .........................374 system of generators......................64

T

term .............................................342 torsion module.............................133 torsion submodule .......................133 torsion-free module .....................133 total degree..................................347 trace.......................................27, 291 trace of a matrix ..........................161 trace of an endomorphism...........162 transcendence basis .....................226 transcendence degree ..................229 transcendental (element) .............177 transcendental number ................178 trigonable matrix .........................167

U

UFD...............................................36 u-indecomposable space..............152 u-invariant subspace....................152 unique factorization domain..........36 unit ................................................13 universality property

of the direct product ...........82, 83 of the direct sum.......................85 of the free module...................107 of the monoid algebra.............343

386

of the polynomial algebra ...... 344 of the ring of fractions ........... 358

U(R) .............................................. 13

W

Warning Theorem.......................223 Wedderburn's Theorem...............219 well ordered ................................352

387

Bibliography

1. ALBU, T., ION, I.D. [1984] Capitole de teoria algebrică a numerelor, Ed. Academiei R.S.R., Bucureşti.

2. ALBU, T., MANOLACHE, N. [1987] 19 Lecţii de teoria grupurilor, Ed. Universităţii Bucureşti, Bucureşti.

3. ALBU, T., RAIANU, Ş. [1984] Lecţii de algebră comutativă, Ed. Universităţii Bucureşti, Bucureşti.

4. ANDERSON, F.W., FULLER, K.R. [1974] Rings and categories of modules, Springer-Verlag, New York.

5. AYAD, M. [1997] Théorie de Galois. 122 exercices corrigés, Ellipses, Paris. 6. BOREVICI, Z.I, ŞAFAREVICI, I.R. [1985], Teoria numerelor, Ed. Ştiinţifică şi

Enciclopedică, Bucureşti. 7. BOURBAKI, N. [1958] Eléments de mathématique, Fasc. VII, Livre II: Algè-

bre, Chapitre 3, Algèbre multilinéaire, Hermann, Paris. 8. BOURBAKI, N. [1967] Eléments de mathématique, Fasc. VI, Livre II: Algèbre,

Chapitre 2, Algèbre linéaire, Hermann, Paris. 9. BOURBAKI, N. [1981] Algèbre, Chapitres 4 à 7, Masson, Paris. 10. BOURBAKI, N. [1985] Eléments de mathématique: Algèbre commutative,

Chapitres 1 à 4, Masson, Paris. 11. ESCOFIER, J.P. [1997] Théorie de Galois, Masson, Paris. 12. FRIED, M., JARDEN, M. [1986] Field Arithmetic, Springer Verlag, Berlin. 13. GALBURĂ, GH. [1961] Corpuri de funcţii algebrice şi varietăţi algebrice, Ed.

Academiei R.P.R., Bucureşti. 14. GALBURĂ, GH. [1972] Algebră, Ed. Didactică şi pedagogică, Bucureşti. 15. GOZARD, I. [1997] Théorie de Galois, Ellipses, Paris. 16. HALL, M. [1959] The Theory of Groups, Macmillan, New York. 17. HERRLICH, H., STRECKER G.E. [1979], Category Theory, second edition,

Heldermann Verlag, Berlin. 18. HUNGERFORD, T.W. [1974], Algebra, Springer-Verlag, New York. 19. ION, I.D., NĂSTĂSESCU, C., NIŢĂ, C. [1984] Complemente de algebră, Ed.

Ştiinţifică and enciclopedică, Bucureşti. 20. ION, I.D., RADU, N. [1981a] Algebra, Ed. Didactică and pedagogică, Bucur-

eşti.

Bibliography

388

21. ION, I.D., RADU, N., NITA, C., POPESCU, D. [1981b] Probleme de algebră, Ed. Didactică and pedagogică, Bucureşti.

22. JACOBSON, N. [1964], Lectures in Abstract Algebra III. Theory of Fields and Galois Theory, Springer-Verlag, New York.

23. JACOBSON, N. [1974], Basic Algebra I, W.H. Freeman and Co., San Fran-cisco.

24. KAPLANSKY, I. [1973], Fields and Rings, The University of Chicago Press, Chicago.

25. LAFON, J.P. [1977] Algèbre commutative. Langages géometrique et algébri-que, Hermann, Paris.

26. LANG, S. [1964], Algebraic numbers, Addison-Wesley, Reading Massachus-setts.

27. MACCARTHY, P.J. [1966], Algebraic Extensions of Fields, Blaisdell Publish-ing, Waltham, Massachusets.

28. MORANDI, P. [1996] Field and Galois Theory, Springer-Verlag, New York. 29. NĂSTĂSESCU, C. [1974] Introducere în teoria mulţimilor, Ed. Didactică şi

pedagogică, Bucureşti. 30. NĂSTĂSESCU, C. [1976] Inele. Module. Categorii, Ed. Academiei R.S.R., Bu-

cureşti. 31. NĂSTĂSESCU, C., NIŢĂ, C. [1979] Teoria calitativă a ecuaţiilor algebrice, Ed.

Tehnică, Bucureşti. 32. NĂSTĂSESCU, C., NIŢĂ, C., VRACIU, C. [1986] Bazele Algebrei, vol. I, Ed.

Academiei R.S.R., Bucureşti. 33. NEUKIRCH, J. [1986] Class Field Theory, Springer-Verlag, Berlin. 34. NITA, C., SPIRCU, T. [1974] Probleme de structuri algebrice, Ed. Tehnică,

Bucureşti. 35. PARENT, D.P. [1978] Exercices en théorie des nombres, Gauthier-Villars, Pa-

ris. 36. POPESCU, N. [1971] Categorii abeliene, Ed. Academiei R.S.R., Bucureşti. 37. PURDEA, I. [1982] Tratat de algebră modernă, vol II, Ed. Academiei R.S.R.,

Bucureşti. 38. RADU, GH. [1988] Algebra categoriilor şi functorilor, Ed. Junimea, Iaşi. 39. RADU, GH., TOFAN, I., GONTINEAC, V. M. [2000] Introducere în algebra

omologică, Ed Universităţii Al. I. Cuza, Iaşi. 40. RADU, N. [1968] Inele locale, vol. I, Ed. Academiei R.S.R., Bucureşti. 41. REGHIŞ, M. [1981] Elemente de teoria mulţimilor and logică matematică, Ed.

Facla, Timişoara. 42. SAMUEL, P. [1963] Anneaux factoriels, Sociedade de Matemática de São

Paulo. 43. SAMUEL, P. [1968] Théorie algébrique des nombres, Hermann, Paris. 44. SCORPAN, A. [1996] Introducere în teoria axiomatică a mulţimilor, Ed.

Universităţii Bucureşti, Bucureşti. 45. SPINDLER, K. [1994] Abstract Algebra with Applications, vol. II, M. Dekker,

New York.

6. Solvable groups

389

46. ŞTEFĂNESCU, M., [1993] Introducere în teoria grupurilor, Ed. Universităţii Al. I. Cuza, Iaşi.

47. TIGNOL, J.-P. [1987] Galois' Theory of Algebraic Equations, Longman Scien-tifical and Technical.

48. TOFAN, I. [2000] Capitole speciale de structuri algebrice, Ed Universităţii Al. I. Cuza, Iaşi.

49. VAN DER WAERDEN, B.L. [1967], Algebra II (Fünfte auflage der Modernen Algebra) Springer-Verlag, Berlin.

50. VAN DER WAERDEN, B.L. [1971], Algebra I (Achte auflage der Modernen Algebra) Springer-Verlag, Berlin.

51. VAN DERWAERDEN, B.L. [1985], A History of Algebra, Springer-Verlag, Ber-lin.

52. WALKER, R.J. [1950] Algebraic Curves, Dover Publications, New York. 53. WINKLER, F. [1996] Polynomial Algorithms in Computer Algebra, Springer

Wien New York. 54. ZARISKI, O., SAMUEL, P. [1958] Commutative Algebra, vol. I, Princeton. 55. ZARISKI, O., SAMUEL, P. [1960] Commutative Algebra, vol. II, Princeton.

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