Ring Theory Course notes for MAT 3143 (Winter 2013)

Ring TheoryMAT 3143

Winter 2018

Alistair Savage

Department of Mathematics and Statistics

University of Ottawa

This work is licensed under aCreative Commons Attribution-ShareAlike 4.0 International License

http://alistairsavage.ca/mat5145

http://alistairsavage.ca

http://creativecommons.org/licenses/by-sa/4.0/

Contents

Preface 3

1 Rings 41.1 Examples and basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Integral domains and fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3 Ideals and quotient rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2 Polynomials 322.1 Polynomial rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2 Factorization of polynomials over a field . . . . . . . . . . . . . . . . . . . . 412.3 Quotient rings of polynomials over a field . . . . . . . . . . . . . . . . . . . . 51

3 Integral domains 543.1 Unique factorization domains . . . . . . . . . . . . . . . . . . . . . . . . . . 543.2 Principal ideal domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.3 Euclidean domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4 Fields 674.1 Brief review of vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.2 Field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.3 Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.4 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Index 85

2

Preface

These notes are aimed at students in the course Ring Theory (MAT 3143) at the Universityof Ottawa. This is a first course in ring theory (except that students may have seen somebasic ring theory near the end of MAT 2143/2543). In this course, we study the generaldefinition of a ring and the types of maps that we allow between them, before turning ourattention to the important example of polynomials rings. We then discuss classes of ringsthat have some additional nice properties (e.g. euclidean domains, principal ideal domainsand unique factorization domains). We also spend some time studying fields in more depththan we’ve seen in previous courses. For example, we examine the ideas of field extensionsand splitting fields.

Notation: In this course N = Z≥0 = {0, 1, 2, 3, . . . } denotes the set of nonnegative integers.

Acknowledgement : Portions of these notes are based on handwritten notes of Erhard Ne-her and Hadi Salmasian. Other portions follow the text Abstract Algebra: Theory andApplications by Tom Judson, which is the recommended text for the course.

Alistair Savage

Course website: http://alistairsavage.ca/mat3143

3

http://abstract.ups.edu/index.html


http://alistairsavage.ca

http://alistairsavage.ca/mat3143

Chapter 1

Rings

In this chapter we introduce the main object of this course. We start with the basic defi-nition of a ring, give several important examples, and deduce some important properties.We then turn our attention to integral domains and fields, two important types of rings.Next, we discuss the important concept of an ideal and the related notion of quotient rings.Finally, we conclude with a discussion of ring homomorphisms and state the important FirstIsomorphism Theorem. A reference for the material in this chapter is [Jud12, Ch. 16].

1.1 Examples and basic properties

Definition 1.1.1 (Ring). A ring is a nonempty set R with two binary operations, usuallyusually written as (and called) addition and multiplication satisfying the following axioms.

(R1) a+ b = b+ a for all a, b ∈ R.

(R2) (a+ b) + c = a+ (b+ c) for all a, b, c ∈ R.

(R3) There exists an element 0 ∈ R such that a+ 0 = a for all a ∈ R.

(R4) For every a ∈ R, there exists an element −a ∈ R such that a+ (−a) = 0.

(R5) (ab)c = a(bc) for all a, b, c ∈ R.

(R6) There exists an element 1 ∈ R such that 1a = a1 = a for all a ∈ R.

(R7) a(b+ c) = ab+ ac and (a+ b)c = ac+ bc for all a, b, c ∈ R.

A ring R is said to be commutative if, in addition,

(R8) ab = ba for all a, b ∈ R.

When we wish to specify the ring, we sometimes write 0R and 1R for the elements 0 and 1.

Sometimes condition (R6) is omitted from the definition of a ring and one refers to a ringwith unity (or identity) to specify that condition (R6) also holds. However, in this course,we will always assume that our rings have a unity and use the term general ring for objectsthat satisfy all the axioms of a ring other than (R6). Axioms (R1)–(R4) are equivalent to(R,+) being an abelian group. Axioms (R5)–(R6) imply that (R, ·) is a monoid. Thus, theelement 1, called the unity , or identity element , of R, is unique. The zero element 0 is alsounique. See Exercise 1.1.1.

4

1.1. Examples and basic properties 5

Remark 1.1.2. Nonassociative rings are also an important area of study. These are objects forwhich we do not require (R5) to hold. (A better name might be “not necessarily associativerings”.) Lie algebras are a well studied class of nonassociative rings.

Example 1.1.3. Each of Z, R, Q, and C is a commutative ring.

Example 1.1.4. Let n be a positive integer and let Zn = {0, 1, . . . , n− 1} with addition andmultiplication performed modulo n. Then Zn is a commutative ring.

Example 1.1.5 (Matrices). The set Mn(Q) of all n × n matrices with rational entries is aring under matrix addition and multiplication. If n ≥ 2, this ring is noncommutative. Moregenerally, if R is a ring, then Mn(R) is also a ring (with the usual rules for matrix additionand multiplication).

Example 1.1.6 (Polynomial rings). For any ring R, we have the ring

R[x] = {a0 + a1x+ a2x2 + · · ·+ amx

m | a0, . . . , am ∈ R},

called the ring of polynomials with coefficients in R. Here x is an indeterminate (and additionand multiplication of polynomials is “formal”). We will discuss polynomial rings in furtherdetail in the next chapter.

Example 1.1.7 (Function rings). If X is a nonempty set, then the set F(X,R) of real valuedfunctions f : X → R is a commutative ring under pointwise addition and multiplication.

Example 1.1.8 (Direct product of rings). If R1, . . . , Rn are rings, then their direct product isthe cartesian product R1 × · · · ×Rn with the operations

(a1, . . . , an) + (b1, . . . , bn) = (a1 + b1, . . . , an + bn),

(a1, . . . , an) · (b1, . . . , bn) = (a1b1, . . . , anbn).

Example 1.1.9 (The zero ring). The smallest ring is the zero ring R = {0}. A ring R isthe zero ring (i.e. has only one element, its zero element) if and only if 1R = 0R. SeeExercise 1.1.2. Since the zero ring is not very interesting, we usually assume that rings arenonzero.

Example 1.1.10. The set 2Z of even integers, with the usual addition and multiplication, isa general ring that is not a ring. Another such example is the set of all 3× 3 real matriceswhose bottom row is zero, with usual addition and multiplication of matrices.

Definition 1.1.11 (Unit, multiplicative inverse). Let R be a ring. An element a ∈ R iscalled a unit if there exists an element b ∈ R such that ab = ba = 1. The element b is calledthe multiplicative inverse of a. The set of units of R is denoted R×. (In some references, thegroup of units is denoted R∗. We use the notation R× to avoid confusion with the dual of avector space.)

Proposition 1.1.12 (Uniqueness of multiplicative inverses). If a ∈ R has a multiplicativeinverse, then this inverse is unique.

6 Chapter 1. Rings

Proof. If b and c are both multiplicative inverses of a, then

b = b1 = b(ac) = (ba)c = 1c = c.

Proposition 1.1.13. For every ring R, the set R× is a group under multiplication.

Proof. The proof of this proposition is left as Exercise 1.1.5.

From now on, we will call R× the group of units of R.Suppose R is a ring. If m ∈ Z and a ∈ R, then we define

ma :=

a+ a+ · · ·+ a︸︷︷︸

m summands

if m > 0,

0 if m = 0,

−a+ (−a) + · · ·+ (−a)︸︷︷︸m summands

if m < 0.

If m ∈ N, then we defineam := a · a · · · a︸︷︷︸

m factors

.

Here we interpret a0 = 1. If a is a unit, then am is defined for all m ∈ Z. For negative m,we define

am := a−1 · a−1 · · · a−1︸︷︷︸|m| factors

.

It is easy to show (see Exercise 1.1.6) that

(m+ n)a = ma+ na, m(na) = (mn)a for all m,n ∈ Z, a ∈ R, (1.1)

am+n = aman, (am)n = amn for all m,n ∈ N, a ∈ R. (1.2)

If a is a unit, then (1.2) holds for all m,n ∈ Z.

Theorem 1.1.14. Let R be a ring and r, s ∈ R. Then

(a) r0 = 0 = 0r,

(b) (−r)s = r(−s) = −(rs),

(c) (−r)(−s) = rs, and

(d) (mr)(ns) = (mn)(rs) for all m,n ∈ Z.

(Here 0 = 0R, as opposed to the integer zero.)

Proof. For r ∈ R, we have r0 = r(0 + 0) = r0 + r0. Adding −r0 to both sides gives r0 = 0.The proof of the relation 0r = 0 is similar. The remainder of the relations are left as anexercise (or see [Jud12, Prop. 16.1]).

Definition 1.1.15 (Subtraction). If r and s are elements of a ring R, their difference isdefined to be

r − s := r + (−s).In this way, we can define subtraction in a ring.


Definition 1.1.16 (Characteristic). If R is any ring, the characteristic of R, denoted charR,is defined to be the order of 1R in (R,+) if this order is finite and zero if this order is infinite.

Example 1.1.17. We have

charZ = 0, charR = 0, charC = 0, charZn = n, char(zero ring) = 1.

Remark 1.1.18. Note that if the ring R is finite (i.e. |R| <∞), then charR > 0.

Lemma 1.1.19. The characteristic of a ring R is the exponent of the ring’s additive group.That is, it is the smallest positive integer n such that na = 0 for all a ∈ R.

Proof. It suffices to show that n1 = 0 if and only if, for all a ∈ R, we have na = 0. Clearlythe latter condition implies the former (just take a = 1). Now, if n1 = 0, then, for all a ∈ R,we have na = n(1a) = (n1)a = 0a = 0.

Definition 1.1.20 (Idempotent, nilpotent). An element a ∈ R is called an idempotent ifa2 = a. It is called nilpotent if an = 0 for some positive integer n.

Examples 1.1.21. In any ring R, the elements 0 and 1 are idempotents and 0 is nilpotent. InM2(R) we have other idempotents:[

1 00 0

],

[1/2 1/21/2 1/2

], . . . .

In M2(R) we also have many nilpotent elements:[0 10 0

],

[0 0−2 0

], . . . .

Definition 1.1.22 (Subring). A subset S ⊆ R of a ring R is called a subring of R if it isitself a ring with the same operations (and the same unity) as R.

Proposition 1.1.23 (Subring Test). A subset S ⊆ R of a ring R is a subring of R if

(SR1) 0 ∈ S and 1 ∈ S,

(SR2) If s, t ∈ S, then s+ t, st and −s are all in S.

Alternatively, S is a subring of R if

(SR1 ′) 1 ∈ S,

(SR2 ′) rs ∈ S for all r, s ∈ S, and

(SR3 ′) r − s ∈ S for all r, s ∈ S.

Proof. The proof of this proposition is left as Exercise 1.1.7.

Examples 1.1.24. (a) Z is a subring of R.

(b) M2(Z) is a subring of M2(Q).

8 Chapter 1. Rings

(c)

[R R0 R

]:=

{[x y0 z

] ∣∣∣∣ x, y, z ∈ R}

is a subring of M2(R).

(d) 2Z is not a subring of Z.

(e) The gaussian integers Z(i) := {a+ bi | a, b ∈ Z} form a subring of C.

(f) If a, b ∈ R with a < b, then the continuous real valued functions on the interval [a, b]form a subring of F([a, b],R).

Example 1.1.25. Let us find Z(i)×. We have

(a+bi)(c+di) = 1 =⇒ (a−bi)(c−di) = (a+ bi)(c+ di) = 1 = 1 =⇒ (a2+b2)(c2+d2) = 1.

Since a2 + b2, c2 + d2 ∈ N, this implies that a2 + b2 = 1. Thus, either a = 0, b = ±1 ora = ±1, b = 0. Therefore Z(i)× = {±1,±i}.

Definition 1.1.26 (Center). The center of a ring R is defined to be

Z(R) := {r ∈ R | rs = sr for all s ∈ R}.

Example 1.1.27. A ring R is commutative if and only if Z(R) = R.

Lemma 1.1.28. The center Z(R) of a ring R is a subring of R.

Proof. The proof of this lemma is left as Exercise 1.1.10.

Later, in Section 1.4, we will discuss the notion of a ring homomorphism. However,it is useful to give here the definition of the special case of a ring isomorphism (see alsoDefinition 1.4.10).

Definition 1.1.29 (Isomorphic rings). Two rings R, S are said to be isomorphic, and wewrite R ∼= S, if there exists a map σ : R→ S such that

(a) σ is bijective,

(b) σ(a+ b) = σ(a) + σ(b) for all a, b ∈ R, and

(c) σ(ab) = σ(a)σ(b) for all a, b ∈ R.

The map σ is called an isomorphism.

Remark 1.1.30. Note that if σ : R→ S is an isomorphism of rings, then we have the following:

(a) σ is an isomorphism of the corresponding additive groups. In particular, σ(0R) = 0S.

(b) σ(1R) = 1S. This can be seen as follows. For any s ∈ S, there exists r ∈ R such thatσ(r) = s. Thus sσ(1R) = σ(r)σ(1S) = σ(r1S) = σ(r) = s. Similarly σ(1R)s = s. Sinces ∈ S was arbitrary, this implies that σ(1R) is the unity of S.


Example 1.1.31. Let

R1 =

[R R0 R

]=

{[a b0 c

] ∣∣∣∣ a, b, c ∈ R}

and R2 =

[R 0R R

]=

{[a 0b c

] ∣∣∣∣ a, b, c ∈ R}.

Consider the map σ : R1 → R2 given by

σ

([a b0 c

])=

[0 11 0

] [a b0 c

] [0 11 0

]−1

=

[c 0b a

]

It is easy to see that σ2 = id. Thus σ is invertible and hence bijective. If M =

[0 11 0

], we

also have

σ(A+B) = M(A+B)M−1 = MAM−1 +MBM−1 = σ(A) + σ(B),

σ(AB) = (MAM−1)(MBM−1) = MABM−1 = σ(AB).

Thus σ is an isomorphism and so R1∼= R2.

Note that the map A 7→ AT is not a ring isomorphism since (AB)T 6= ATBT in general.

Exercises.

1.1.1. Prove that the identity element in any monoid (hence in any group) is unique.

1.1.2. Show that a ring R is the zero ring (i.e. R = {0}) if and only if 0R = 1R.

1.1.3. Explain why each of the following is not a ring.

(a) N = {0, 1, 2, 3, . . . , } under usual addition and multiplication.

(b) 2Z.

(c) The set of all mappings f : R → R under pointwise addition, but with multiplicationgiven by composition.

1.1.4. Suppose R is a ring. The opposite ring Rop of R is the same set as R, with the sameaddition, but with multiplication given by r · s = sr. (Here · denotes the multiplication inRop and juxtaposition denotes the multiplication in R.) Prove that Rop is a ring.

1.1.5. Prove Proposition 1.1.13.

1.1.6. Prove (1.1) and (1.2).


1.1.8 ([Nic12, Ex. 3.1.3]). For each of the following, show that S is a subring of R.

10 Chapter 1. Rings

(a) S =

{[a bc d

] ∣∣∣∣ a, b, c, d ∈ R, a+ c = b+ d

}, R = M2(R).

(b) S =

{[a b0 a

] ∣∣∣∣ a, b ∈ R}

, R = M2(R).

(c) S =

a 0 b

0 c d0 0 a

∣∣∣∣∣∣ a, b, c, d ∈ R

, R = M3(R).

(d) S =

{[a 2bb a

] ∣∣∣∣ a, b ∈ R}

, R = M2(R).

1.1.9 ([Nic12, Ex. 3.1.4]). If S and T are subrings of R, show that S ∩ T is a subring of R.Is it necessarily true that S + T = {s+ t | s ∈ S, t ∈ T} is a subring of R?

1.1.10. Prove Lemma 1.1.28.

1.1.11. Find the center of M2(R).

1.1.12. Show that if R and S are rings, then (R× S)× = R× × S×.

1.1.13 ([Nic12, Ex. 3.1.5]). Suppose X is a nonempty subset of a ring R. The centralizer ofX in R is defined to be

C(X) = {c ∈ R | cx = xc for all x ∈ X}.

Prove that C(X) is a subring of R.

1.1.14 ([Nic12, Ex. 3.1.10]). Suppose a and b are elements of a ring.

(a) If ab+ ba = 1 and a3 = a, prove that a2 = 1.

(b) If ab = a and ba = b, prove that a2 = a and b2 = b.

1.1.15 ([Nic12, Ex. 3.1.11]). Show that 0 is the only nilpotent in a ring R if and only if a2 = 0implies a = 0.

1.1.16 ([Nic12, Ex. 3.1.14]). Given r and s in a ring R, show that 1 + rs is a unit if and onlyif 1 + sr is a unit. Hint : s(1 + rs) = (1 + sr)s.

1.1.17. Find all the rings of characteristic one.

1.1.18 ([Nic12, Ex. 3.1.18]). Find the characteristics of the following rings.

(a) Zn × Zm.

(b) M2(Zn).

(c) Z× Zn.

1.1.19 ([Nic12, Ex. 3.1.20]). If ua = au, where u is a unit and a is a nilpotent, show thatu+ a is a unit.

1.2. Integral domains and fields 11

1.1.20 ([Nic12, Ex. 3.1.28]). For each of the following rings, find all the units, nilpotents, andidempotents:

(a) Z(b) Z24

(c) M2(Z2)

(d)

[R R0 R

]1.1.21 ([Nic12, Ex. 3.1.36]). For each of the following pairs of rings, show that the two ringsare not isomorphic.

(a) R and C.

(b) Q and R.

(c) Z and Q.

(d) Z8 and Z4 × Z2.

1.2 Integral domains and fields

We now consider certain special types of rings that we will examine in further detail later inthe course.

Definition 1.2.1 (Zero divisor, integral domain, division ring, field). Suppose R is a ring.A nonzero element r ∈ R is said to be a zero divisor if there exists a nonzero s ∈ R suchthat rs = 0 or sr = 0. A nonzero ring is said to be a domain if it has no zero divisors.A commutative domain is called an integral domain. If every nonzero element in a (notnecessarily commutative) nonzero ring is a unit, then R is called a division ring (or skewfield). A commutative division ring is called a field .

Domains

Division

Rings

Commutative

Rings

Integral

Domains

Rings

Fields

Figure 1.1: Types of rings

12 Chapter 1. Rings

Example 1.2.2. The rings R,C,Q are fields. The ring Z is an integral domain, but not afield.

Example 1.2.3. The ring Z(i) of gaussian integers is an integral domain (Exercise 1.2.1) butnot a field, since Z(i)× = {±1,±i} (see Example 1.1.25).

Example 1.2.4. The ring M2(R) is not a domain since, for example,[1 00 0

] [0 00 1

]=

[0 00 0

].

Proposition 1.2.5. The ring Zm is a field if and only if m is a prime number.

Proof. Let m be a prime. For every a ∈ Zm, a 6= 0, we have gcd(a,m) = 1 and so there existx, y ∈ Z such that ax + my = 1. Thus ax = 1 in Zm. Thus every nonzero element of Zm isa unit.

Now suppose m is not prime. Then m = m1m2 for some 1 < m1,m2 < m. In particular,m1, m2 6= 0, but m1m2 = m = 0. Therefore m1 and m2 are not units of Zm.

Proposition 1.2.6. Let R be a ring such that |R| = p, where p is a prime number. Then Ris isomorphic to Zp. In particular, R is a field.

Proof. By Lagrange’s Theorem, the additive group 〈1〉 generated by the unity is equal to allof R and this must be isomorphic to the group Zp (since this is the only group of order pwhen p is a prime number). Define σ : Zp → R by σ(m) = m1R. Then we have the following:

• σ is injective since

σ(a) = σ(b) =⇒ a1 = b1 =⇒ (a− b)1 = 0 =⇒ p|(a− b) =⇒ a = b in Zp,

where the third implication follows from the fact that (a − b)1 = 0 in the additivegroup (R,+) ∼= Zp (isomorphism of groups) if and only if p|(a− b).• σ is bijective because σ is injective and |R| = |Zp|.• We have

σ(a+ b) = σ(a+ b

)= (a+ b)1 = a1 + b1 = σ(a) + σ(b).

• σ(ab) = σ(a)σ(b). The proof of this fact is similar to the above and is left as anexercise.

Proposition 1.2.7. Let R be a ring. Then the following statements are equivalent:

(a) R is a domain.

(b) If ab = ac in R and a 6= 0, then b = c.

(c) If ba = ca in R and a 6= 0, then b = c.

Proof. (a) =⇒ (b). Suppose R is a domain and ab = ac in R. Then a(b− c) = ab− ac = 0.Since a 6= 0 and R is a domain, we must have b− c = 0, hence b = c.

(b) =⇒ (a). Suppose R satisfies (b) and that ab = 0 in R. Then we have ab = a0.Thus, if a 6= 0, we must have b = 0 by (b). Hence R is a domain.

We have proven the equivalence of (a) and (b). The proof that (a) and (c) are equivalentis similar.


Examples 1.2.8. (a) Every division ring is a domain.

(b) Every field is an integral domain.

(c) If R is an (integral) domain and S is a subring of R, then S is an (integral) domain.

(d) Z(i) is an integral domain (since it is a subring of C).

Example 1.2.9. The subring Q(√

2) := {a+ b√

2 | a, b ∈ Q} of the field of complex numbersis a field. Being a subring of C, it is an integral domain. Thus, it remains to show that everynonzero element has a multiplicative inverse. Suppose x ∈ Q(

√2) \ {0}. Then x = a+ b

√2

with a 6= 0 or b 6= 0. Then

x−1 =1

a+ b√

2=a− b

√2

a2 − 2b2=

a

a2 − 2b2− b

a2 − 2b2

√2.

Note that a2 − 2b2 6= 0 since if a2 − 2b2 = 0, then a2

b2= 2, which implies that

√2 = ±a

b∈ Q,

which contradicts the fact that√

2 is an irrational number.

Definition 1.2.10 (Quaternions). The quaternions are the ring

H = {a+ bi+ cj + dk | a, b, c, d ∈ R}

with multiplication determined by the rules

ij = k = −ji, jk = i = −kj, ki = j = −ik, i2 = j2 = k2 = −1.

For w = a+ bi+ cj+ dk ∈ H, we define w∗ := a− bi− cj− dk and N(w) := a2 + b2 + c2 + d2.

For example, we have

(3− 4j)(2i+ k) = 6i+ 3k − 8ji− 4jk = 6i+ 3k + 8k − 4i = 2i+ 11k.

Lemma 1.2.11. If w ∈ H \ {0}, then w−1 = 1N(w)

w∗.

Proof. See [Jud12, Example 6].

By Lemma 1.2.11, the ring of quaternions is a division ring. However, it is not a fieldsince it is not commutative. Thus, the quaternions are an example of a noncommutativedivision ring.

Proposition 1.2.12. The characteristic of an integral domain is either zero or a primenumber.

Proof. Let R be an integral domain and suppose that the characteristic of R is n 6= 0. If nis not prime, then n = ab, with 1 < a < n and 1 < b < n. Then 0 = n1 = (ab)1 = (a1)(b1).Since R has no zero divisors, either a1 = 0 or b1 = 0. Thus the characteristic of R must beless than n, which is a contradiction. Therefore, n must be prime.

Proposition 1.2.13. Every finite integral domain is a field.

14 Chapter 1. Rings

Proof. Let R be an integral domain with |R| = n < ∞. Suppose r ∈ R with r 6= 0. Then,by Proposition 1.2.7, the function f : R → R, f(x) = rx is injective. Since R is finite, thisimplies that f is bijective. Thus f(s) = rs = 1 for some s ∈ R. Hence r is a unit. Since rwas an arbitrary nonzero element of R, this implies that R is a field.

A proof of the following theorem can be found in [Jud12, Th. 16.16]. (We will not usethis theorem.)

Theorem 1.2.14 (Wedderburn’s Theorem). Every finite division ring is a field.

We conclude this section by showing that every integral domain “embeds” into a field.We do this by mimicking the construction of the rational numbers from the integers. Weassume for the rest of this section that R is an integral domain.

LetX = {(r, u) | r, u ∈ R, u 6= 0}

and define a relation on X by

(r, u) ≡ (s, v) ⇐⇒ rv = su.

This is an equivalence relation on X. It is easy to see that this relation is reflexive andsymmetric. We therefore must show that it is transitive. Suppose that

(r, u) ≡ (s, v) and (s, v) ≡ (t, w).

Then, by definition, we have rv = su and sw = tv. Therefore,

(rw)v = (rv)w = (su)w = u(sw) = utv (since R is commutative).

Since (s, v) ∈ X, we have v 6= 0 and so we can cancel v in the above by Proposition 1.2.7.This gives rw = tu and so (r, u) ≡ (t, w). Thus ≡ is an equivalence relation on X. We writers

for the equivalence class of (r, s). Thus rs

= sv

if and only if (r, s) ≡ (s, v).We now define

Q :={ ru

∣∣∣ r, u ∈ R, u 6= 0}. (1.3)

We define addition and multiplication on Q by

r

u+s

v=rv + su

uvand

r

u· sv

=rs

uv.

Note that uv 6= 0 since u, v 6= 0 and R is a domain. Since the quotients above are equivalenceclasses, we must show that these operations are well defined. We show this for multiplicationand refer the reader to [Jud12, Lem. 18.2] for addition. If r

u= r′

u′and s

v= s′

v′, we must show

that rsuv

= r′s′

u′v′. This follows from the fact that

(rs)(u′v′) = (ru′)(sv′) = (r′u)(s′v) = (r′s′)(uv).

We leave it as a Exercise 1.2.14 to show that Q is a field with zero 01

and unity 11. The

negative of an element ru

is −ru

. Furthermore, if ru

in nonzero in Q, then r 6= 0. Thus ur∈ Q

and we have(ru

)−1= u

r.


Define

R′ :={r

1| r ∈ R

}. (1.4)

It is easy (Exercise 1.2.15) to verify that R′ is a subring of R. Furthermore, it is not hardto check that the map

σ : R→ R′, σ(r) =r

1, r ∈ R.

is an isomorphism of rings (see [Jud12, Th. 18.4]) and so R ∼= R′. One usually identifies Rwith R′ via this isomorphism. That is we set r = r

1for all r ∈ R. In this way, we view R as

a subring of Q. We call Q the field of fractions (or field of quotients) of the integral domainR.

Example 1.2.15. The field of fractions of Z is Q. The field of fractions of R[x] is called thefield of rational functions.

Exercises.

1.2.1. Show that the ring of gaussian integers is an integral domain.

1.2.2. Show that Z×m = {k ∈ Zm | gcd(k,m) = 1}.

1.2.3 ([Nic12, Ex. 3.2.1]). Find all the roots of x2 + 3x− 4 in the rings Z, Z6, and Z4.

1.2.4 ([Nic12, Ex. 3.2.2]). Suppose p is a prime and define

Z(p) ={ n

m∈ Q

∣∣∣ p does not divide m}.

Show that Z(p) is an integral domain and find all its units.

1.2.5 ([Nic12, Ex. 3.2.3]). Determine all the idempotents and nilpotents in a domain.

1.2.6 ([Nic12, Ex. 3.2.4]). Suppose R and S are nonzero rings. It is possible for R× S to bea domain?

1.2.7 ([Nic12, Ex. 3.2.5]). Show that Mn(R), where R is a ring, is never a domain if n ≥ 2.

1.2.8 ([Nic12, Ex. 3.2.6]). If a2 = b2 and a3 = b3 in a domain, show that a = b. Now do itfor am = bm and an = bn where gcd(m,n) = 1. Hint : 1 = xm+ yn for some x, y ∈ Z.

1.2.9 ([Nic12, Ex. 3.2.7]). Suppose that a ring R has no nonzero nilpotent elements. If ab = 0in R, show that ba = 0.

1.2.10 ([Nic12, Ex. 3.2.8]). Show that a ring R is a division ring if and only if, for eachnonzero a ∈ R, there is a unique element b ∈ R such that aba = a.

16 Chapter 1. Rings

1.2.11 ([Nic12, Ex. 3.2.11]). If F is a field with q elements, show that aq = a for all a ∈ F .Hint : Use Lagrange’s Theorem.

1.2.12 ([Nic12, Ex. 3.2.14]). Show that there is no field with 6 elements. Hint : Use Lagrange’sTheorem.

1.2.13 ([Nic12, Ex. 3.2.15]). Show that the center of a division ring is a field.

1.2.14. Show that Q as defined in (1.3) is a field.

1.2.15. Show that R′ as defined in (1.4) is a subring of Q.

1.2.16 ([Nic12, Ex. 3.2.28]). Let R be a commutative ring and call u ∈ R a nonzero-divisor ifur = 0, r ∈ R, implies r = 0. Let U ⊆ R be a set of nonzero-divisors in R such that 1 ∈ U ,and ab ∈ U whenever a, b ∈ U . Generalize the field of fractions construction by showing thata ring of quotients

Q ={ ru

∣∣∣ r ∈ R, u ∈ U}exists. Show further that R can be regarded as a subring of Q and, in this case, that eachelement of U is a unit in Q and Q = {ru−1 | r ∈ R, u ∈ U}.

1.3 Ideals and quotient rings

Recall that that if G is a group and H ⊆ G, then in order for the quotient set G/H tonaturally be a group, we need H to be a normal subgroup of G. What is the analogousnotion for rings? That is, if R is a ring and S ⊆ R, when is R/S naturally a ring?

Example 1.3.1. Let R = Z×Z, and S = {(x, x) | x ∈ Z}. It is straightforward to verify thatS is a subring of R. Note that (0, 1) + S = (−1, 0) + S. If we try to define a multiplicationon the set of cosets R/S by multiplying representatives, we have

((0, 1) + S)((0, 1) + S) = (0, 1) + S 6= (0, 0) + S = ((0, 1) + S)((−1, 0) + S),

which is a contradiction.

Definition 1.3.2 (Ideal). Let R be a ring. An additive subgroup I ⊆ R is called an ideal ofR if for every r ∈ R, we have rI ⊆ I and Ir ⊆ I. The ideal I is said to be proper if I 6= R.

We now want to define the structure of a ring on the quotient set R/I := {a+I | a ∈ R}.(Recall that a + I = a′ + I ⇐⇒ a − a′ ∈ I.) We want the addition and multiplication tobe given by

(a+ I) + (b+ I) = (a+ b) + I and (a+ I)(b+ I) = ab+ I. (1.5)

Theorem 1.3.3. Let I ⊆ R be an ideal of R. Then R/I, with addition and multiplicationdefined by (1.5), is a ring. The unity of R/I is 1 + I and the zero is 0 + I = I. If R iscommutative, then R/I is also commutative.

1.3. Ideals and quotient rings 17

Proof. We must first show that the addition and multiplication are well defined. If a+ I =a′ + I and b+ I = b′ + I, then a− a′, b− b′ ∈ I. Thus

ab− a′b′ = (a− a′)b+ a′(b− b′) ∈ Ib+ a′I ⊆ I

and so ab + I = a′b′ + I. Thus the multiplication is well defined. The proof that theaddition is also well defined and that A/I satisfies the axioms of Definition 1.1.1 is left asExercise 1.3.1.

Proposition 1.3.4. Let I be an ideal of a ring R. Then the following statements are equi-valent:

(a) 1 ∈ I.

(b) I contains a unit.

(c) I = R.

Proof. We leave the proof of this result as Exercise 1.3.2.

Definition 1.3.5 (Principal ideal). If a ∈ Z(R), then Ra = aR and this is an ideal of R,often denoted 〈a〉. It is called the principal ideal of R generated by a.

Proposition 1.3.6 (Ideals of Z). Every ideal of Z is principal.

Proof. The zero ideal {0} is a principal ideal since 0Z = {0}. If I is any nonzero ideal inZ, then I must contain some positive integer. Then there exists a least positive integer nin I by the Principle of Well-Ordering. Now let a be any element in I. Using the divisionalgorithm, we know that there exist integers q and r such that

a = nq + r

where 0 ≤ r < n. This equation tells us that r = a− nq ∈ I, but r must be 0 since n is theleast positive element in I. Therefore, a = nq and I = nZ.

Example 1.3.7. If R is any ring, then {0} and R are ideals of R. The corresponding quotientrings are R/{0} ∼= R and R/R ∼= {0}. The ideal {0} is called the zero ideal of R.

Definition 1.3.8 (Simple ring). A nonzero ring R is called simple if its only ideals are {0}and R.

Example 1.3.9. It follows from Proposition 1.3.4 that the only ideals of a division ring R are{0} and R. Hence division rings are simple.

Example 1.3.10. Let R = Z(i) be the ring of gaussian integers. Let I = 〈2 + i〉. We wish todescribe the ring R/I. Since i − (−2) = 2 + i ∈ I, we have i + I = (−2) + I. Thus, for allm,n ∈ Z,

(m+ ni) + I = (m+ I) + (−2n+ I) = (m− 2n) + I.

18 Chapter 1. Rings

Moreover, 5 = (2 + i)(2 − i) ∈ (2 + i)R = I, which implies that 5 + I = I. Thus the onlyelements of R/I, are I, 1 + I, 2 + I, 3 + I, 4 + I. Are all of these elements distinct? Supposen,m ∈ Z and n+ I = m+ I. Thus (m− n) + I = I. Then, for some a, b ∈ Z, we have

m− n = (2 + i)(a+ bi) = (2a− b) + (a+ 2b)i =⇒ a = −2b =⇒ m− n = −5b ∈ 5Z.

Therefore, R/I ∼= Z5 (a field).

Definition 1.3.11 (Prime ideal). An ideal P of a commutative ring R is called a primeideal if P 6= R and

r, s ∈ R, rs ∈ P =⇒ r ∈ P or s ∈ P.

(See Exercises 1.3.5 and 1.3.20(d) for an indication of why these ideals are called prime.)

Examples 1.3.12. (a) 2Z is a prime ideal of Z, but 4Z is not.

(b) Z× {0} is a prime ideal of Z× Z, but 2Z× {0} and {0} × {0} are not.

Theorem 1.3.13. If R is a commutative ring, an ideal P 6= R is prime if and only if R/Pis an integral domain.

Proof. Assume P is prime and (a+ P )(b+ P ) = ab+ P = 0 + P in R/P . Then ab ∈ P andso a ∈ P or b ∈ P . Thus a+P = 0 +P or b+P = 0 +P . Thus R/P is an integral domain.

Now assume that R/P is an integral domain and ab ∈ P . Then (a+P )(b+P ) = ab+P =0 + P and so either a+ P = 0 + P or b+ P = 0 + P . Hence a ∈ P or b ∈ P .

Examples 1.3.14. (a) (Z × Z)/(Z × {0}) ∼= Z. The ideal Z × {0} is prime and Z is anintegral domain.

(b) Zn is an integral domain if and only if n is prime (see Exercise 1.3.5).

(c) Since, for any ring R, we have R/{0} ∼= R, we have that R is an integral domain ifand only if its zero ideal is prime.

Theorem 1.3.15. Let I be an ideal of a ring R. Then there exists a bijective, inclusion pre-serving correspondence between the set of ideals of R/I and the set of ideals of R containingI.

Proof. We split the proof into three steps.

(a) We first show that if A is an ideal of R/I, then

A := {b ∈ R | b+ I ∈ A}

is an ideal of R containing I with A = A/I. Since 0 + I ∈ A, we have 0 ∈ A. If a, b ∈ A,then a + I ∈ A and b + I ∈ A. Since A is an additive subgroup of R/A, we then have(a− b) + I = (a+ I)− (b+ I) ∈ A. Hence a− b ∈ A. Thus A is an additive subgroup of R.

Now, if a ∈ A and r ∈ R, then

a+ I ∈ A and r + I ∈ R/I =⇒ ra+ I = (r + I)(a+ I) ∈ A (since A in an ideal of R/I)

=⇒ ra ∈ A (by the definition of A).


Similarly, one can show that ar ∈ A. Thus A is closed under multiplication by elements ofR and hence is an ideal of R.

To see that I ⊆ A, note that

a ∈ I =⇒ a+ I = 0 + I ∈ A =⇒ a ∈ A (by the definition of A).

It remains to show that A = A/I. Since

r + I ∈ A =⇒ r ∈ A =⇒ r + I ∈ A/I,

we have A ⊆ A/I. To prove the reverse inclusion, suppose x ∈ A/I. Then there is somea ∈ A ⊆ R such that x = a+ I. This implies that x = a+ I ∈ A by the definition of A.

(b) Next we show that if A is an ideal of R containing I then A := {a+ I | a ∈ A} is anideal of R/I and A = A/I := {a + I | a ∈ A} (which is clearly true). Since 0 ∈ I, we have0 + I ∈ A.

Suppose x1, x2 ∈ A. Then there exist a1, a2 ∈ A such that x1 = a1 + I and x2 = a2 + I.Since A is an ideal or R, we have a1 − a2 ∈ A. Then x1 − x2 = (a1 + I) − (a2 + I) =(a1 − a2) + I ∈ A. Thus A is an additive subgroup of R/I.

Now suppose x ∈ A and r + I ∈ R/I. Then x = a + I for some a ∈ A. Since A is anideal of R, we have ra ∈ I. Hence (r + I)x = (r + I)(a + I) = ra + I ∈ A. The proof that(a+ I)(r + I) ∈ A is similar. Hence A is an ideal of R/I.

(c) Finally, we show that if A1 and A2 are ideals of R containing I, then A1 ⊆ A2 if andonly if A1/I ⊆ A2/I. That A1 ⊆ A2 =⇒ A1/I ⊆ A2/I is obvious. We show the reverseinclusion. Suppose A1/I ⊆ A2/I. Then

a1 ∈ A1 =⇒ a1 + I ∈ A1/I

=⇒ ∃ a2 ∈ A2 such that a1 + I = a2 + I

=⇒ a1 − a2 ∈ I=⇒ a1 = a2 + a for some a ∈ I=⇒ a1 ∈ A2 (since a2 + a ∈ A2 + I ⊆ A2 because I ⊆ A2)

Theorem 1.3.16. A commutative ring is simple if and only if it is a field.

Proof. Since a field is a commutative division ring by definition, the reverse implicationfollows from Example 1.3.9. Therefore it suffices to prove the forward implication. SupposeR is a simple commutative ring and let a ∈ R \ {0}. Then aR 6= {0} is an ideal and soaR = R. Thus ab = 1 for some b ∈ R.

Definition 1.3.17 (Maximal ideal). An ideal I of a ring R is said to be maximal if I 6= Rand there is no ideal J such that I ( J ( R.

Theorem 1.3.18. Let R be any ring and let I be an ideal of R. Then R/I is simple if andonly if I is a maximal ideal.

Proof. Suppose R/I is simple and A is an ideal of R with I ⊆ A ⊆ R. By Theorem 1.3.15,A/I is an ideal of R/I with I/I ⊆ A/I ⊆ R/I. Since R/I is simple, we have A/I = R/I orA/I = I/I, which implies that A = R or A = I.

20 Chapter 1. Rings

Now suppose that I is maximal and, towards a contradiction, that R/I is not simple.Then R/I has a nonzero proper ideal of the form A/I. So I/I ( A/I ( R/I. Then, byTheorem 1.3.15, A is an ideal of R with I ( A ( R, which is a contradiction.

Corollary 1.3.19. An ideal I of a commutative ring R is maximal if and only if R/I is afield.

Proof. This follows from Theorems 1.3.16 and 1.3.18.

Corollary 1.3.20. Every maximal ideal of a commutative ring is a prime ideal.

Proof. Suppose I is a maximal ideal of a commutative ring R. Then, by Corollary 1.3.19, thequotient R/I is a field. Hence R/I is an integral domain and so I is prime by Theorem 1.3.13.

Example 1.3.21. We have that {0} is a prime ideal of Z that is not maximal. Similarly,{0} × Z is an ideal of Z× Z that is prime but not maximal.

Example 1.3.22. By Theorem 1.3.15, the ideals of Z/nZ are of the form A/nZ for some idealA of Z such that nZ ⊆ A. By Proposition 1.3.6, A = mZ for some m ∈ Z. Since nZ ⊆ mZif and only if m | n, the ideals of Z/nZ are mZ/nZ for m | n.

As an explicit example, consider n = 6. The bijective correspondence of Theorem 1.3.15is given explicitly by:

Ideal of Z/6Z Ideal of Z

Z/6Z = 〈1〉 Z{0, 2, 4} = 〈2〉 2Z{0, 3} = 〈3〉 3Z{0} = 〈0〉 = 〈6〉 6Z

Example 1.3.23. We will discuss polynomial rings in detail in Chapter 2. However, we givehere an example involving them. Consider

R = Z[x] = {a0 + a1x+ · · ·+ akxk | k ∈ N, a0, . . . , ak ∈ Z},

I = {4a0 + a1x+ a2x2 + · · ·+ akx

k | k ∈ N, a0, . . . , ak ∈ Z}.

ThenR/I = {0 + I, 1 + I, 2 + I, 3 + I} ∼= Z4.

And, for example, the ideal 〈2〉 ⊆ Z4 corresponds to the ideal

J = {2a0 + a1x+ · · ·+ akxk | k ∈ N, a0, . . . , ak ∈ Z}.

Remark 1.3.24. We have seen (Theorem 1.3.16) that if a commutative ring is simple, then itis a field. For noncommutative rings, this situation is more complicated. There are simplerings that are not division rings. For example, if F is a field, then Mn(F) is a simple ringthat is not a division ring. This follows from the following result.


Theorem 1.3.25. If R is a ring and n ∈ N+, then every ideal of Mn(R) is of the formMn(I) where I is an ideal of R.

We will not prove the above theorem (nor will we use it). The proof can be found, forinstance, in [Nic12, Lem. 3.3.3].

Exercises.

1.3.1. Complete the proof of Theorem 1.3.3.


1.3.3 ([Nic12, Ex. 3.3.1]). For each of the following, decide whether A is an ideal of the ringR. Justify your answer.

(a) R = C, A = Z.

(b) R = Z× Z, A = {(k, k) | k ∈ Z}.

(c) R =

[R R0 R

], A =

[0 R0 R

].

(d) R =

[Z Z0 Z

], A =

[Z 2Z0 Z

].

(e) R =

[R R0 R

], A =

[Z R0 Z

].

(f) R = Z(i), A = {n+ ni | n ∈ Z}.

1.3.4 ([Nic12, Ex. 3.3.2]). Suppose S is a ring. If R =

[S S0 S

]and A =

[0 S0 0

], show that

A is an ideal of R and describe the cosets in R/A.

1.3.5. Show that nZ is a prime ideal of Z if and only if n is either zero or a prime number.

1.3.6. Prove that the maximal ideals of Z are precisely the ideals pZ where p is a primenumber.

1.3.7. Suppose R is a ring and m ∈ Z. Show that mR = {mr | R ∈ R} and Am = {r ∈ R |mr = 0} are ideals of R.

1.3.8 ([Nic12, Ex. 3.3.5]). Suppose R and S are rings.

(a) If A is an ideal of R and B is an ideal of S, show that A×B is an ideal of R× S.

(b) Show that every ideal A of R×S is of the form A = A×B for some ideal A of R andsome ideal B of S. Hint : A = {a ∈ R | (a, 0) ∈ A}.

22 Chapter 1. Rings

(c) Show that the maximal ideals of R × S are either of the form A × S, where A is amaximal ideal of R, or of the form R×B, where B is a maximal ideal of S.

1.3.9 ([Nic12, Ex. 3.3.6]). If A is an ideal of R, show that M2(A) is an ideal of M2(R).

1.3.10 ([Nic12, Ex. 3.3.8]). If A and B are ideals of a ring R such that A∩B = 0, show thatab = 0 = ba for all a ∈ A and b ∈ B.

1.3.11 ([Nic12, Ex. 3.3.9]). Let R = Z(i) be the ring of gaussian integers. For each of thefollowing, find the number of elements in the factor ring R/A and describe the cosets:

(a) A = Ri

(b) A = R(1− i)(c) A = R(1 + 2i)

(d) A = R(1 + 3i)

Hint : (1 + 2i)(1− i) = 3 + i and (1 + 3i)(1− 3i) = 10.

1.3.12 ([Nic12, Ex. 3.3.10]). If R is a simple ring, show that Z(R) is a field. Show that the

converse is not true by considering R =

[F F0 F

]where F is a field.

1.3.13 ([Nic12, Ex. 3.3.12]). If X ⊆ R is a nonempty subset of a commutative ring R, definethe annihilator of X to be ann(X) = {a ∈ R | ax = 0 for all x ∈ X}.

(a) Show that ann(X) is an ideal of R.

(b) If X ⊆ Y , show that ann(Y ) ⊆ ann(X).

(c) Show that ann(X ∪ Y ) = ann(X) ∩ ann(Y ).

(d) Show that X ⊆ ann(

ann(X)).

(e) Show that ann(X) = ann(

ann(

ann(X)))

.

1.3.14 ([Nic12, Ex. 3.3.13]). Give an example of a ring R and an ideal A such that R/A iscommutative, but R is not.

1.3.15 ([Nic12, Ex. 3.3.14]). If X and Y are additive subgroups of R, define X+Y = {x+y |x ∈ X, y ∈ Y }.

(a) Show that X + Y is an additive subgroup of R that contains both X and Y .

(b) If A and B are ideals of R, show that A+B is an ideal of R.

(c) If A is an ideal of R and S is a subring of R, show that A+ S is a subring of R.

1.3.16 ([Nic12, Ex. 3.3.15]). If A is an ideal of R, show that A ∩ S is an ideal of S for allsubrings S of R.

1.3.17 ([Nic12, Ex. 3.3.16]). If A is an ideal of R, show that R/A is commutative if and onlyif rs− sr ∈ A for all r, s ∈ R.


1.3.18 ([Nic12, Ex. 3.3.17]). Let Z = Z(R) denote the center of a ring R.

(a) When is Z an ideal of R? Justify your answer.

(b) If R is simple, show that Z is a field.

(c) If R/Z is cyclic as an additive group, show that R is commutative.

1.3.19 ([Nic12, Ex. 3.3.24]). An additive subgroup L of R is called a left ideal if Ra ⊆ L forall a ∈ L. Show that R is a division ring if and only if 0 and R are the only left ideals of R(extending Theorem 1.3.16). Hint : Ra is a left ideal for each a ∈ R.

1.3.20 ([Nic12, Ex. 3.3.25]). Let R be a commutative ring. Write a | b if b = ra for somer ∈ R.

(a) Show that Rab ⊆ Ra ∩Rb for all a, b ∈ R.

(b) If Ra+Rb = R (see Exercise 1.3.15), show that Rab = Ra ∩Rb.

(c) Show that u ∈ R is a unit if and only if Ru = R.

(d) Show that Rp is a prime ideal if and only p | ab implies p | a or p | b.

(e) If R is an integral domain, show that Ra = Rb if and only if a = ub for some unitu ∈ R.

1.3.21 ([Nic12, Ex. 3.3.26]). Let A, B, and C be ideals of R and define

AB = {a1b1 + a2b2 + · · ·+ anbn | ai ∈ A, bi ∈ B, n ≥ 1}.

(a) Show that AB is an ideal of R and AB ⊆ A ∩B.

(b) Show that A(B + C) = AB + AC and (B + C)A = BA+ CA. (See Exercise 1.3.15.)

(c) Show that AR = A = RA.

(d) Show that A(BC) = (AB)C.

1.3.22 ([Nic12, Ex. 3.3.27]). If a ∈ R, let

RaR = {r1as1 + r2as2 + · · ·+ rnasn | ri, si ∈ R, n ≥ 1}.

Show that RaR is an ideal of R containing a and that it is contained in any such ideal.

1.3.23. Let R be a commutative ring. The ring R is said to be noetherian if it satisfies theascending chain condition on ideals: for every ascending chain I1 ⊆ I2 ⊆ · · · of ideals in R,there exists a positive integer N such that In = IN for all n ≥ N . An ideal I of R is said tobe finitely generated if there exist elements a1, . . . , am ∈ I (called generators of I) such thatevery element of I can be written as a1r1 + · · ·+ amrm for some r1, . . . , rm ∈ R. Prove thatR is noetherian if and only if every ideal of R is finitely generated.

1.3.24 (Bonus problem). Prove that every ring has a maximal ideal.

24 Chapter 1. Rings

1.4 Homomorphisms

We now discuss the natural maps between rings, together with some of their properties.

Definition 1.4.1 (Ring homomorphism). Let R and S be rings. A mapping f : R → S iscalled a ring homomorphism if it satisfies the following properties.

(RH1) f(1R) = 1S.

(RH2) f(a+ b) = f(a) + f(b) for all a, b ∈ R.

(RH3) f(ab) = f(a)f(b) for all a, b ∈ R.

Examples 1.4.2. (a) f : Z→ Q, f(x) = x is a ring homomorphism.

(b) f : Z→ Zn, f(x) = x is a ring homomorphism.

(c) f : Z× Z→ Z, f(x, y) = x is a ring homomorphism.

(d) f : Z× Z→ Z, f(x, y) = x+ y is not a ring homomorphism since, for example,

f((1, 1)(1, 2)) = f(1, 2) = 1 + 2 = 3 but f(1, 1)f(1, 2) = (1 + 1)(1 + 2) = 6.

Remark 1.4.3. If f : R→ S is a ring homomorphism, then f is a homomorphism of additivegroups:

f(0R) = f(0R + 0R) = f(0R) + f(0R) =⇒ f(0R) = 0S,

f(a) + f(−a) = f(a+ (−a)) = f(0R) = 0S =⇒ f(−a) = −f(a).

Here we only used axiom (RH2).

Axioms (RH2) and (RH3) are not enough to conclude that f(1R) = 1S as the followingexample illustrates.

Example 1.4.4. The mapping f : Z → Z × Z given by f(x) = (x, 0) satisfies (RH2) and(RH3), but not (RH1). Some references call mappings satisfying (RH2) and (RH3) but notnecessarily (RH1) general ring homomorphisms .

Proposition 1.4.5. Let f : R→ S be a ring homomorphism. Then the following hold.

(a) For every m ∈ Z and r ∈ R, we have f(mr) = mf(r).

(b) For every m ∈ N and r ∈ R, we have f(rm) = f(r)m.

(c) If u ∈ R×, then f(u) ∈ S× and, for every m ∈ Z, we have f(um) = f(u)m. Inparticular, f(u−1) = f(u)−1.

Proof. The proof of this result is left as Exercise 1.4.1.

Example 1.4.6. We will show that the equation m3 − 6n3 = 3 has no solutions for m,n ∈ Z.Consider the mapping f : Z → Z7 given by f(x) = x. If m3 − 6n3 = 3, then (f(m))3 +(f(n))3 = 3 in Z7. But, by explicitly considering all possible values, one can see that, fora ∈ Z7, we have a3 ∈ {0, 1, 6}. We compute

0 + 0 = 0, 0 + 1 = 1, 0 + 6 = 6, 1 + 1 = 2, 1 + 6 = 0, 6 + 6 = 5.

1.4. Homomorphisms 25

Thus, the equation m3−6n3 = 3 does not have any solutions in Z7. It follows that m3−6n3 =3 cannot have any solutions in Z.

Example 1.4.7 (Frobenius homomorphism). Let R be a commutative ring with p = charR aprime number. Then the map

f : R→ R, f(r) = rp

is a ring homomorphism, called the Frobenius homomorphism. See Exercise 1.4.3.

Definition 1.4.8 (Kernel, image). Let f : R → S be a ring homomorphism. The kernel off is ker f := {x ∈ R | f(x) = 0} and the image of f is im f = f(R) := {f(r) | r ∈ R}.

Remark 1.4.9. A ring homomorphism f is one-to-one if and only if ker f = {0}. This followsfrom the corresponding fact about group homomorphisms since f is a homomorphism ofadditive groups. By definition, f is surjective (or onto) if and only if f(R) = S

Definition 1.4.10 (Monomorphism, epimorphism, isomorphism, automorphism). An in-jective ring homomorphism is called a monomorphism. A surjective ring homomorphism iscalled an epimorphism. A ring homomorphism that is both a monomorphism and an epi-morphism is called an isomorphism. If there exists an isomorphism from a ring R to a ringS then we say that R and S are isomorphic and write R ∼= S. (Note that this definitionagrees with Definition 1.1.29.) A homomorphism (resp. isomorphism) from a ring to itself iscalled an endomorphism (resp. automorphism).

Example 1.4.11. The map σ : C→ C given by σ(z) = z is an automorphism of C.

Example 1.4.12 (Inner automorphism). If R is a ring and u ∈ R×, then

σu : R→ R, σu(r) = uru−1, r ∈ R,

is an automorphism of R called an inner automorphism. We leave the proof of this asExercise 1.4.4. Also see Example 1.1.31.

Theorem 1.4.13. Let f : R→ S be a ring homomorphism. Then:

(a) f(R) is a subring of S.

(b) ker f is an ideal of R.

Proof. (a) Since f(1R) = 1S, we have 1S ∈ f(R). Since f(0R) = 0S, we have 0S ∈ f(R).Now suppose r, s ∈ f(R). Then there exists s′, t′ ∈ R such that f(s′) = s and f(t′) = t.Thus

−s = −f(s′) = f(−s′) ∈ f(R),

s+ t = f(s′) + f(t′) = f(s′ + t′) ∈ f(R),

st = f(s′)f(t′) = f(s′t′) ∈ f(R).

26 Chapter 1. Rings

(b) We know (from MAT 2143) that ker f is an additive subgroup of R (since f is ahomomorphism of additive groups). Now,

r ∈ R, x ∈ ker f =⇒ f(rx) = f(r)f(x) = f(r)0S = 0S =⇒ rx ∈ ker f.

Similarly, one can show that r ∈ R, x ∈ ker f implies xr ∈ ker f . Thus ker f is an ideal of R.

Example 1.4.14. If I is an ideal of R, then f : R→ R/I, f(x) = x+ I, is a ring epimorphismand ker f = I.

Theorem 1.4.15 (First Isomorphism Theorem). Let f : R → S be a ring homomorphism.Then the map

f : R/(ker f)→ im f, f(r + ker f) = f(r)

is a ring isomorphism. In particular, R/(ker f) ∼= im f .

Proof. Let I = ker f . Then f is well-defined since

r + I = s+ I ⇐⇒ r − s ∈ I ⇐⇒ f(r − s) = 0 ⇐⇒ f(r)− f(s) = 0

⇐⇒ f(r) = f(s) ⇐⇒ f(r + I) = f(s+ I).

The reverse implications above also show that f is injective. It is clear that f is surjectiveand it is straightforward to verify that f is a ring homomorphism.

Remark 1.4.16. Theorem 1.4.15 is called the First Isomorphism Theorem because there arealso Second and Third Isomorphism Theorems (see Exercises 1.4.25 and 1.4.25).

Examples 1.4.17. (a) Consider f : Z→ Zn, f(x) = x. Then ker f = nZ and Z/nZ ∼= Zn.

(b) Consider f : Z → Z2 × Z2, f(x) = (x, x). Then f(Z) = {(0, 0), (1, 1)} ∼= Z2 andker f = 2Z. Thus Z/2Z ∼= f(Z) ∼= Z2.

(c) Consider f : Z4 → Z10, f(x) = 5x. Then ker f = {0, 2} ⊆ Z4 and im f = {0, 5} ⊆ Z10.We have im f ∼= Z2 and so Z4/(ker f) ∼= Z2. (Note here that f is not actually a ringhomomorphism, since it does not send the unity to the unity, but it is a general ringhomomorphism.)

Example 1.4.18. Let m and n be positive integers and let I = nZmn = {na | a ∈ Zmn}. Thenthe map f : Zmn → Zn, f(x) = x, is a ring homomorphism with im f = Zn and ker f = I.Thus Zmn/I ∼= Zn.

Remark 1.4.19. If I and J are ideals of a ring R, then

I ∩ J,IJ := {r ∈ R | r = a1b1 + · · ·+ akbk, a1, . . . , ak ∈ I, b1, . . . , bk ∈ J}, and

I + J := {r ∈ R | r = a+ b, a ∈ I, b ∈ J}

are also ideals of R. See Exercises 1.3.15(b) and 1.3.21(a).


Theorem 1.4.20. If R is any ring, then Z1R = {k1R | k ∈ Z} is a subring of R containedin the center of R. Furthermore, we have the following.

(a) If n = charR > 0, then Z1R ∼= Zn.

(b) If charR = 0, then Z1R ∼= Z.

Proof. Define a mapθ : Z→ R, θ(k) = k1R for all k ∈ Z.

By Proposition 1.4.5, θ is a ring homomorphism. Thus Z1R = θ(Z) is a subring of R byTheorem 1.4.13. Furthermore, one can verify (exercise) that Z1R is contained in the centerof R.

We have ker θ = {k ∈ Z | k1R = 0}. If n = charR > 0, then ker θ = nZ by Lemma 1.1.19.Thus Z1R = θ(Z) ∼= Z/nZ ∼= Zn by the First Isomorphism Theorem (Theorem 1.4.15).If charR = 0, then ker θ = {0} and the result again follows by the First IsomorphismTheorem.

Lemma 1.4.21. If R and S are rings and I ⊆ R and J ⊆ S are ideals, then I × J is anideal of R× S and we have (R× S)/(I × J) ∼= (R/I)× (S/J).

Proof. The map f : R × S → (R/I) × (S/J) given by f(r, s) = (r + I, s + J) is a ringhomomorphism with ker f = I × J and im f = (R/I) × (S/J) (exercise). The result thenfollows from the First Isomorphism Theorem (Theorem 1.4.15).

Lemma 1.4.22. Let R and S be rings. Then every ideal of R × S is of the form A × Bwhere A ⊆ R and B ⊆ S are ideals.

Proof. Let I ⊆ R× S be an ideal. Let

A = {r ∈ R | (r, 0) ∈ I}, B = {s ∈ S | (0, s) ∈ I}.

It is straightforward (exercise) to show that A is an ideal of R and B is an ideal of S. If(r, s) ∈ I, then (r, 0) = (r, s)(1, 0) ∈ I and (0, s) = (r, s)(0, 1) ∈ I. Thus r ∈ A and s ∈ B.Hence I ⊆ A×B. Now suppose (a, b) ∈ A×B. Then, by the definition of A and B, we have(a, 0) ∈ I and (0, b) ∈ I. Hence (a, b) = (a, 0) + (0, b) ∈ I (since I is closed under addition).Therefore A×B ⊆ I and so I = A×B.

Theorem 1.4.23 (Chinese Remainder Theorem). Let R be a ring and let A,B be two idealsof R such that A+B = R. Then R/(A ∩B) ∼= (R/A)× (R/B).

Proof. Consider the map

f : R→ (R/A)× (R/B), f(r) = (r + A, r +B).

For r, s ∈ R, we have

f(1) = (1 + A, 1 +B) = 1(R/A)×(R/Jy),

f(r + s) = (r + s+ A, r + s+B) = (r + A, r +B) + (s+ A, s+B) = f(r) + f(s)

28 Chapter 1. Rings

and

f(rs) = (rs+ A, rs+B) = ((r + A)(s+ A), (r +B)(s+B))

= (r + A, r +B)(s+ A, s+B) = f(r)f(s).

Thus f is a ring homomorphism.We next show that f is surjective. Since R = A + B, we have 1 = a + b for some a ∈ A

and b ∈ B. Now choose r1, r2 ∈ R and set r = r1b+ r2a. Then

r1 − r = r1 − (r1b+ r2a) = r1(1− b)− r2a = r1a− r2a ∈ A =⇒ r1 + A = r + A,

r2 − r = r2 − (r1b+ r2a) = r2(1− a)− r1b = r2b− r1b ∈ B =⇒ r2 +B = r +B.

Thus f(r) = (r + A, r + B) = (r1 + A, r2 + B). Since r1, r2 ∈ R were arbitrary, this impliesthat f is surjective.

Finally,

f(r) = (0 + A, 0 +B) ⇐⇒ (r + A = 0 + A and r +B = 0 +B)

⇐⇒ (r ∈ A and r ∈ B) ⇐⇒ r ∈ A ∩B.

So ker f = A ∩ B. The result then follows from the First Isomorphism Theorem (Theo-rem 1.4.15).

Example 1.4.24. Since gcd(5, 6) = 1, we have 5Z + 6Z = Z. We also have 5Z ∩ 6Z = 30Z.Thus, by the Chinese Remainder Theorem (Theorem 1.4.23), we have Z30

∼= Z5 × Z6.

More generally, we have the following result.

Lemma 1.4.25. If gcd(m,n) = 1, then Zmn ∼= Zm × Zn.

Proof. Since gcd(m,n) = 1, we have ma + nb = 1 for some a, b ∈ Z. Hence mZ + nZ = Z.Also, we have mZ∩nZ = {0} since if m|x and n|x, then mn|x (since m and n are relativelyprime). Thus the result follows from the Chinese Remainder Theorem (Theorem 1.4.23).

Example 1.4.26. How many units does Z345 have? Since 345 = 3 · 5 · 23, we have Z345∼=

Z3 × Z5 × Z23 by Lemma 1.4.25. Thus Z345 has 2 · 4 · 22 = 176 units (see Exercise 1.1.12).

Example 1.4.27. Little Mary wants to take the farm’s eggs to the market. She tries to putthem evenly in two baskets, but one is left out. She tries three baskets, but one is still leftout. Four baskets, one is left out. Five baskets. Aha! It works! How many eggs does LittleMary have?

Let x ∈ N be the number of eggs. Then we have

x = 1 in Z2,

x = 1 in Z3,

x = 1 in Z4,

x = 0 in Z5.


First note that the fourth equation implies the first. So we ignore the first equation. Now,we know that Z12

∼= Z3×Z4 by Lemma 1.4.25. Under this isomorphism x (in Z12) is mappedto (1, 1) ∈ Z3 × Z4. Thus x = 1 in Z12.

Now, again by Lemma 1.4.25, we have Z60∼= Z12×Z5

∼= Z60/(12Z60)×Z60/(5Z60). Underthis isomorphism x (in Z60) is mapped to (1, 0) ∈ Z12×Z5. Checking all the elements of Z60

which map to 1 ∈ Z12, i.e. 1, 13, 25, 37, 49, we see that 25 reduces to 0 ∈ Z5. Thus x = 25 inZ60. So Mary has 25 + 60k, k ∈ Z, eggs.

Exercises.Throughout these exercises, R is a ring.


1.4.2 ([Nic12, Ex. 3.4.1]). For each of the following determine whether the map is a ringhomomorphism. Justify your answer.

(a) θ : Z3 → Z12, where θ(r) = 4r.

(b) θ : Z4 → Z12, where θ(r) = 3r.

(c) θ : R×R→ R, where θ(r, s) = rs.

(d) θ : F(R,R)→ R, where θ(f) = f(1).

1.4.3. Show that the Frobenius homomorphism (see Example 1.4.7) is indeed a homomor-

phism. Hint: Use the binomial formula: (a+ b)p =∑p

i=0

(pi

)aibp−i.

1.4.4. Prove that the map defined in Example 1.4.12 is a ring automorphism.

1.4.5 ([Nic12, Ex. 3.4.2]). Let θ : R → S be a general ring homomorphism, where R and Sare rings. Furthermore, suppose θ is surjective, S is a domain, and θ(1) 6= 0. Prove that θis a ring homomorphism.

1.4.6 ([Nic12, Ex. 3.4.3]). Show that a general ring homomorphism θ : Z→ Z is either a ringhomomorphism or θ(k) = 0 for all k ∈ Z.

1.4.7 ([Nic12, Ex. 3.4.4]). Determine all ring homomorphisms and general ring homomor-phisms Z12 → Z6.

1.4.8 ([Nic12, Ex. 3.4.5]). If θ : R → S is a surjective ring homomorphism, show thatθ(Z(R)) ⊆ Z(S). Give an example of a (not necessarily surjective) ring homomorphismθ : R→ S where θ(Z(R)) 6= Z(S).

1.4.9 ([Nic12, Ex. 3.4.6]). If θ : R → S is a ring homomorphism and charR > 0, show thatcharS divides charR.

30 Chapter 1. Rings

1.4.10. Show that the composition of two ring homomorphisms is a ring homomorphism.

1.4.11. Complete the proof of Theorem 1.4.20 by showing that Z1R is contained in the centerof R.

1.4.12. If f is the map defined in the proof of Lemma 1.4.21, show that ker f = I × J andim f = (R/I)× (S/J).

1.4.13. Show that A and B as defined in the proof of Lemma 1.4.22 are ideals of R and Srespectively.

1.4.14 ([Nic12, Ex. 3.4.9]). Describe the homomorphic images of a division ring. More pre-cisely, if R is a division ring, what possible rings can be the image of a ring homomorphismwith domain R?

1.4.15 ([Nic12, Ex. 3.4.11–3.4.14]). (a) Show that x3 − 8x2 + 5x + 3 = 0 has no solutionx ∈ Z.

(b) Show that m3 + 14n3 = 12 has no solution in Z.

(c) Show that 7m2 + 11n2 = 9 has no solution in Z.

(d) Show that n3 + (n+ 1)3 + (n+ 2)3 = k2 + 1 has no solution in Z.

1.4.16. Prove that the inverse of a ring isomorphism is also a ring isomorphism.

1.4.17. Show that the set AutR of all ring automorphisms of R is a group under the operationof composition.

1.4.18. Show that isomorphism is an equivalence relation on the class of all rings.

1.4.19 ([Nic12, Ex. 3.4.19]). Let θ : R→ S be an onto ring homomorphism.

(a) If A is an ideal of R, show that θ(A) = {θ(a) | a ∈ A} is an ideal of S.

(b) If also ker θ ⊆ A, show that R/A ∼= S/θ(A). Hint: Use the First Isomorphism Theoremwhere α : R→ S/θ(A) is defined by α(r) = θ(r) + θ(A) for all r ∈ R.

1.4.20 ([Nic12, Ex. 3.4.20]). If n > 0 in Z, describe all the ideals of Z that contain nZ.

1.4.21 ([Nic12, Ex. 3.4.21]). Show that there is no ring homomorphism C→ R.

1.4.22 ([Nic12, Ex. 3.4.22]). Let θ : R→ S be a ring homomorphism. If θ(R) and ker θ bothcontain no nonzero nilpotent elements show that the same is true of R.

1.4.23 ([Nic12, Ex. 3.4.27]). Let R =

[S S0 S

]be the upper triangular matrix ring over a ring

S. Show that A =

[0 S0 0

]is an ideal of R and R/A ∼= S × S.

1.4.24 ([Nic12, Ex. 3.4.31]). Let R = S×T , where S and T are rings, and write S = {(s, 0) |s ∈ S}. Show that S is an ideal of R, and that R/S ∼= T and S ∼= S as rings. What is theunity of S?


1.4.25. Prove the Second Isomorphism Theorem: If A is an ideal of R and S is a subring ofR, then

(a) S + A is a subring of R,

(b) A and S ∩ A are ideals of S + A and S, respectively, and

(c) (S + A)/A ∼= S/(S ∩ A) as rings.

1.4.26. Prove the Third Isomorphism Theorem: If A ⊆ B ⊆ R, where A and B are ideals ofR, then B/A = {b+ A | b ∈ B} is an ideal of R/A and (R/A)/(B/A) ∼= R/B as rings.

1.4.27 ([Nic12, Ex. 3.4.37]). Show that Zm × Zn has a subring isomorphic to Zt, where t isthe least common multiple of m and n.

1.4.28 (Bonus problem). Let f : R → R be a ring homomorphism. Prove that f(x) = x forall x ∈ R. In other words, the identity map is the only ring homomorphism from R to R.

1.4.29 (Bonus problem). Prove that if f : R× · · · × R︸︷︷︸m times

→ R× · · · × R︸︷︷︸n times

is a ring isomorphism,

then m = n.

1.4.30 (Bonus problem). Let R be a finite commutative ring with n = |R| < ∞. Assumethat R is not a field. Prove that R has at least

√n− 1 non-units.

1.4.31 (Bonus problem). Let R = C([0, 1]) := {f ∈ F([0, 1],R) | f is continuous}. One easilychecks that this is a ring.

(a) Prove that, for every a ∈ [0, 1], the set Ia := {f ∈ R | f(a) = 0} is a maximal ideal ofR.

(b) Conversely, prove that every maximal ideal of R is of the form Ia for some a ∈ [0, 1].

(c) Prove that part (b) is false for the ring C(R) := {f ∈ F(R) | f is continuous}.

Chapter 2

Polynomials

In this chapter we turn our attention to a particularly important class of rings: polynomialrings. We first introduce polynomial rings in full generality and deduce some of their funda-mental properties. We then focus on polynomials whose coefficients lie in a field. We discussthe factorization of such polynomials and the structure of the quotient rings of the thesepolynomial rings. A reference for the material in this chapter is [Jud12, Ch. 17].

2.1 Polynomial rings

Definition 2.1.1 (Indeterminate). If R ⊆ S are rings, then an element x ∈ S is called anindeterminate over R if

a0 + a1x+ · · ·+ anxn = 0, ai ∈ R =⇒ ai = 0 ∀ i.

Lemma 2.1.2. Given a ring R, there exists a ring S satisfying the following:

(a) R ⊆ S.

(b) There exists x ∈ S such that x is an indeterminate over R.

(c) We have xa = ax for all a ∈ R.

Sketch of proof. The proof of this lemma is somewhat technical. We will only give an out-line here. The details can be found in [Nic12, §4.6] or, under the assumption that R iscommutative, in [Roy, Prop. 4.2].

Let S be the set of all sequences in R. That is, S is the set of all functions N→ R. Weoften denote an element α ∈ S by (α(0), α(1), α(2), . . . ). We define a ring structure on S bysetting

(α + β)(k) = α(k) + β(k),

(αβ)(k) =k∑`=0

α(`)β(k − `).

Then R is a subring of S if we identify a = (a, 0, 0, . . . ) for a ∈ R. This proves (a).

32

2.1. Polynomial rings 33

Now define x = (0, 1, 0, 0, . . . ). Then we can show that

a0 + a1x+ a2x2 + · · ·+ anx

n = (a0, a1, . . . , an, 0, 0, . . . )

for all a0, . . . , an ∈ R. This proves (b). Since ax = (0, a, 0, . . . ) = xa for all a ∈ R, we alsohave (c).

Definition 2.1.3 (Ring of polynomials). Let R be a ring and let S be as in Lemma 2.1.2.Then

R[x] = {a0 + a1x+ a2x2 + · · ·+ anx

n | n ≥ 0, ai ∈ R ∀ i}

is a subring of S called the ring of polynomials over R. A polynomial over R is an element

f(x) = a0 + a1x+ · · ·+ anxn, n ∈ N, a0, . . . , an ∈ R

of R[x]. The ai’s are called the coefficients of the polynomial f(x). We adopt the conventionthat am = 0 for m > n. We sometimes write f instead of f(x).

Remark 2.1.4. It follows from Definition 2.1.1 that two polynomials f(x) = a0+· · ·+akxk andg(x) = b0 + · · ·+ b`x

` are equal if and only if ai = bi for i ∈ N. Note that this is not the sameas equality of polynomial functions . For example, consider f(x) = 0, g(x) = x2 + x ∈ Z2[x].These polynomials are not equal, but the functions Z2 → Z2 given by a 7→ f(a) and a 7→ g(a)are equal (they are both the zero function). Thus, we consider polynomials to be abstractexpressions and not functions.

It also follows that R[x] is the subring of S (from Lemma 2.1.2) consisting of sequencesthat are eventually zero. That is, the elements of R[x] are sequences α ∈ S for which thereexists N ∈ N with α(k) = 0 for all k > N .

It follows from the above that if R is a ring, the addition and multiplication on R[x] iscomputed as follows. If

f(x) = a0 + · · ·+ akxk and g(x) = b0 + · · ·+ bkx

k

(note that we can write f(x) and g(x) in this form, with the same highest power xk of x byletting some of the coefficients be zero if necessary), then

f(x) + g(x) = (a0 + b0) + (a1 + b1)x+ · · ·+ (ak + bk)xk, and

f(x)g(x) =2k∑i=0

(i∑

j=0

ajbi−jxi

)= a0b0 + (a0b1 + a1b0)x+ (a0b2 + a1b1 + a2b0)x2 + · · · .

Example 2.1.5. We have

1− 3x+ x2 ∈ Z[x], 2x− 3 ∈ Z[x], but 1− 1

x6∈ Z[x].

Also

(1−3x+x2)(2x−3) = (1−3x+x2)(−3+2x) = −3+(2+9)x+(−6−3)x2+2x3 = −3+11x−9x2+2x3.

34 Chapter 2. Polynomials

Example 2.1.6. We have1− 2x, 1 + 2x2 ∈ Z4[x]

and

(1− 2x) + (1 + 2x2) = (1 + 1)− 2x− 2x2 = 2− 2x+ 2x2 = 2 + 2x+ 2x2,

2(1 + 2x2) = 2 + 0x2 = 2,

(1− 2x)(1 + 2x2) = 1 + (−2)x+ 2x2 − 0x3 = 1 + 2x+ 2x2.

Definition 2.1.7 (Degree, leading coefficient, constant coefficient, monic polynomial). Letf(x) = a0 + a1x+ · · ·+ akx

k ∈ R[x].

• If ak 6= 0, then the degree of f(x) is deg f(x) = k and ak is the leading coefficient off(x).

• a0 is called the constant coefficient of f(x).

• If ak = 1, then f(x) is called a monic polynomial .

Theorem 2.1.8. Let R be a ring. Then

(a) the center of R[x] is Z(R)[x],

(b) x ∈ Z(R[x]), and

(c) if R is commutative, then R[x] is also commutative.

Proof. We leave the proof of this theorem as Exercise 2.1.4

The zero element of R[x] is the zero polynomial 0 = 0 + 0x = 0 + 0x + 0x2 = · · · .A polynomial is called a constant polynomial if it is of the form f(x) = a0 = a0 + 0x =a0 + 0x + 0x2 = · · · for some a0 ∈ R. Note that R is the subring of R[x] consisting ofconstant polynomials.

Theorem 2.1.9. Let R be a ring. If f(x), g(x) ∈ R[x] \ {0}, then we have the following.

(a) deg(f(x) + g(x)) ≤ max{deg f(x), deg g(x)}.(b) deg(f(x)g(x)) ≤ deg f(x) + deg g(x).

(c) If R is a domain, then deg(f(x)g(x)) = deg f(x) + deg g(x).

(d) If R is a domain, then R[x] is also a domain.

(e) If R is a domain, then the only units of R[x] are the units of R.

(f) If the leading coefficient of either f(x) or g(x) is a unit in R, then f(x)g(x) 6= 0 inR[x] and deg(f(x)g(x)) = deg f(x) + g(x).

Proof. The proof of this theorem is left as Exercise 2.1.5.

Example 2.1.10. To see that the inequalities of Theorem 2.1.9 can be strict when R is not adomain, consider 1 + 2x, 2x ∈ Z4[x]. Then (1 + 2x)(2x) = 2x and so

deg(1 + 2x)(2x) = 1 < 2 = deg(1 + 2x) + deg(2x).

Also,deg((1 + 2x) + (2x)) = deg(1) = 0 < 1 = max{deg(1 + 2x), deg(2x)}.


Example 2.1.11. Let I ⊆ Z[x] be the ideal generated by x, i.e., I = 〈x〉 = xZ[x]. So

I = {xf(x) | f(x) ∈ Z[x]} = {a1x+ · · ·+ akxk | k ≥ 1, a1, . . . , ak ∈ Z}.

Consider the ring homomorphism ϕ : Z[x] → Z given by ϕ(a0 + . . . + akxk) = a0. It is a

straightforward exercise (Exercise 2.1.6) to see that ϕ is a surjective ring homomorphism withkerϕ = I. Thus, by the First Isomorphism Theorem (Theorem 1.4.15), we have Z[x]/I ∼= Z.

Theorem 2.1.12 (The Division Algorithm). Let R be a ring, f(x), g(x) ∈ R[x], f(x) 6= 0,and suppose the leading coefficient of f(x) is a unit in R. Then there exist unique polynomialsq(x), r(x) ∈ R[x] such that

(a) g(x) = q(x)f(x) + r(x) and

(b) either r(x) = 0 or deg r(x) < deg f(x).

Proof. We first prove the existence result by induction on deg g(x). If g(x) = 0 or deg g(x) <deg f(x), then we can take q(x) = 0 and r(x) = g(x). So assume m ≥ n, where m = deg g(x)and n = deg f(x). Write

f(x) = uxn + axn−1 + · · · , g(x) = bxm + cxm−1 + · · · ,

where u is a unit by hypothesis. Let

g1(x) = g(x)− bu−1xm−nf(x)

= (bxm + cxm−1 + · · · )− bu−1xm−n(uxn + axn−1 + · · · )= 0xm + (c− bu−1)xm−1 + · · · .

Then either g1(x) = 0 or deg g1(x) < m. Thus, by the inductive hypothesis, there existpolynomials q1(x) and r(x) such that g1(x) = q1(x)f(x) + r(x) with either r(x) = 0 ordeg r(x) < deg f(x). Then

g(x) = g1(x) + bu−1xm−nf(x) = (q1(x) + bu−1xm−n)f(x) + r(x),

which completes the proof of the inductive step.It remains to prove uniqueness. Suppose that we have

q1(x)f(x) + r1(x) = g(x) = q2(x)f(x) + r2(x)

with ri(x) = 0 or deg ri(x) < deg f(x) for i = 1, 2. Then r1(x)− r2(x) = (q2(x)− q1(x))f(x).If q2(x)−q1(x) 6= 0, then, since the leading coefficient of f(x) is a unit (see Theorem 2.1.9(f)),we have (q2(x)− q1(x))f(x) 6= 0 and

deg(r1(x)− r2(x)) = deg[(q2(x)− q1(x))f(x)] = deg(q2(x)− q1(x)) + deg f(x).

But this implies that deg(r1(x)− r2(x)) ≥ deg f(x), which is a contradiction. Thus q2(x)−q1(x) = 0 and so r1(x) − r2(x) = (q2(x) − q1(x))f(x) = 0. This completes the proof ofuniqueness.


Remark 2.1.13. Note that if R is in fact a field, then the leading coefficient of any nonzeropolynomial is a unit in R. Then Theorem 2.1.12 says that R[x] is a euclidean domain witheuclidean function f(x) 7→ deg f(x).

Example 2.1.14. Let’s divide x3 − x2 + 2x− 3 by x− 2.

x2 + x + 4x− 2 x3 − x2 + 2x − 3

x3 − 2x2

x2 + 2x − 3x2 − 2x

4x − 34x − 8

5

Hence,x3 − x2 + 2x− 3︸︷︷︸

g(x)

= (x2 + x+ 4)︸︷︷︸q(x)

(x− 2)︸︷︷︸f(x)

+ 5︸︷︷︸r(x)

.

Example 2.1.15. Divide x3 + 2 by 2x+ 2 in Z3[x].

2x2 + x + 22x+ 2 x3 + 0x2 + 0x + 2

x3 + x2

−x2 + 22x2 + 2x

−2x + 2x + 1

1

Thusx3 + 2︸︷︷︸g(x)

= 2x2 + x+ 2︸︷︷︸q(x)

(2x+ 2)︸︷︷︸f(x)

+ 1︸︷︷︸r(x)

.

Can we always evaluate polynomials by specializing the indeterminate? We are used todoing this in calculus, but in fact we must be careful!

Example 2.1.16. Consider the polynomial f(x) = x3 − 2x − 4 ∈ Z[x]. We have f(2) = 0,which implies that x− 2 is a factor of f(x). In fact, we have f(x) = (x2 + 2x+ 2)(x− 2).

Example 2.1.17. Suppose R is a ring and a, b ∈ R with ab 6= ba. Let f(x) = (x− a)(x− b) ∈R[x]. Then f(a) = (a− a)(b− a) = 0(b− a) = 0. But we also have f(x) = x2− ax− bx+ aband so f(a) = a2 − a2 − ba + ab = −ba + ab 6= 0. So “evaluation” of f(x) at a does notseem to be well-defined. The problem is that x lies in the center of R[x] and we are tryingto construct a (surjective) ring homomorphism that maps x to something that is not in thecenter of R.

Theorem 2.1.18 (Evaluation Theorem). Let R be a ring and a ∈ Z(R). Then the mapϕa : R[x]→ R, ϕa(f(x)) = f(a) is a surjective ring homomorphism.


Proof. Let f(x) =∑k

i=0 aixi and g(x) =

∑ki=0 bix

i be arbitrary elements of R[x]. Then wehave

ϕa(f(x) + g(x)) = ϕa

(∑(ai + bi)x

i)

=∑

(ai + bi)ai

=∑

aiai +∑

biai = ϕa(f(x)) + ϕa(g(x),

ϕa(f(x)g(x)) = ϕa

(2k∑i=0

(i∑

j=0

ajbi−jxi

))=

2k∑i=1

(i∑

j=0

a0bi−jai

)

=

(k∑i=0

aiai

)(k∑i=0

biai

)= ϕa(f(x))ϕa(g(x)),

ϕa(1R[x]) = ϕ(1R) = 1R.

The map ϕa of Theorem 2.1.18 is called evaluation of at a. The problem in Example 2.1.17was that we tried to evaluate a polynomial at an element that was not in the center of thecoefficient ring R.

Example 2.1.19. Let R be a ring. Then

ϕ : R[x]→ R, ϕ(a0 + a1x+ a2x2 + · · · ) = a0 + a1 + a2 + · · ·

is a surjective ring homomorphism since ϕ = ϕ1.

Theorem 2.1.20 (Remainder Theorem). Let R be a commutative ring and f(x) ∈ R[x].Then, for every a ∈ R, the remainder of the division of f(x) by x− a is f(a).

Proof. By Theorem 2.1.12, we can write f(x) = q(x)(x − a) + r(x) where r(x) = 0 ordeg rx < 1. Thus we have that r(x) is a constant polynomial. That is, r(x) = r for somer ∈ R. Then

f(a) = ϕa(f(x)) = ϕa(q(x))ϕa(x− a) + ϕa(r(x)) = 0 + r(a) = r.

Theorem 2.1.21 (Factor Theorem). Let R be a commutative ring and f(x) ∈ R[x]. Thena ∈ R satisfies f(a) = 0 if and only if f(x) = (x− a)q(x) for some q(x) ∈ R[x].

Proof. Clearly, f(a) = 0 if f(x) = (x − a)g(x) for some g(x) ∈ R[x]. The converse followsimmediately from the Remainder Theorem (Theorem 2.1.20).

Corollary 2.1.22. Let R be a commutative ring and ϕa : R[x]→ R, ϕa(f(x)) = f(a). Thenkerϕa = 〈x− a〉 = (x− a)R[x] and R[x]/〈x− a〉 ∼= R.

Proof. We have

f(x) ∈ kerϕa ⇐⇒ f(a) = 0 ⇐⇒ f(x) = q(x)(x− a) + 0 ⇐⇒ f(x) ∈ 〈x− a〉.

The second assertion then follows from the First Isomorphism Theorem (Theorem 1.4.15).


Definition 2.1.23 (Root of a polynomial). Let R be a commutative ring and f(x) ∈ R[x] \{0}. An element a ∈ R is called a root of f(x) if f(a) = 0.

Example 2.1.24. The roots of x2 − 1 ∈ Z8 are 1, 3, 5, 7.

Definition 2.1.25 (Multiplicity of a root). Let R be a commutative ring and f(x) ∈ R[x] \{0}. If a ∈ R is a root of f(x), we say a has multiplicity m ≥ 0 if f(x) = (x− a)mg(x) withg(a) 6= 0.

Example 2.1.26. Let’s find the multiplicity of the root 3 for f(x) = x3 + 3x+ 4 ∈ Z5[x]. Wehave f(3) = 0. Dividing f(x) by x− 3 gives f(x) = (x− 3)g(x) with g(x) = x2 +3x+2. Sinceg(3) = 0, we divide again to obtain g(x) = (x−3)(x+1). Thus we have f(x) = (x−3)2(x+1).Since 3 is not a root of x+ 1, we see that the multiplicity of the root 3 is 2.

Theorem 2.1.27. Let R be an integral domain and f(x) ∈ R[x]\{0}. If n = deg f(x), thenf(x) has at most n roots.

Proof. We prove the result by induction on n, the cases n = 0, 1 being straightforward.Suppose deg f(x) = n+ 1. Then, if f(a) = 0, we have f(x) = (x−a)g(x) with deg g(x) = n.If f has more than n + 1 roots, then we have f(a1) = f(a2) = · · · = f(an+1) = 0 for somedistinct a1, . . . , an+1 6= a. Then, for i = 1, . . . , n + 1, we have 0 = f(ai) = (ai − a)g(ai) andai − a 6= 0, which implies that g(ai) = 0. Thus g(x) has n + 1 roots, which contradicts theinductive hypothesis since deg g(x) = n.

Theorem 2.1.28 (Rational Roots Theorem). Let f(x) = a0 + a1x+ · · ·+ anxn ∈ Z[x] with

a0, an 6= 0. If r ∈ Q and f(r) = 0, then r = cd

with c | a0 and d | an.

Proof. Write r = cd

with gcd(c, d) = 1. Then

0 = f(r) = a0 + a1

( cd

)+ · · ·+ an

( cd

)n=⇒ a0d

n + a1an−1c+ · · ·+ anc

n = 0.

Thus

a0dn = −c(a1a

n−1 + a2dn−2c+ · · · ancn−1) and

ancn = −d(a0d

n−1 + a1dn−2c+ · · ·+ anc

n−1).

Since c and d are relatively prime, this implies that c | a0 and d | an.

Example 2.1.29. Although we already know that√

2 is irrational, we can give another proofbased on the Rational Roots Theorem (Theorem 2.1.28). Consider the polynomial x2 − 2.If it had a rational root, it must be one of ±2,±1. But one easily checks that none of theseis a root of x2 − 2. Thus, x2 − 2 has no rational roots. Since

√2 is a root of x2 − 2, it is not

rational.

Example 2.1.30. If f(x) ∈ Z[x] \ {0} is monic, then any rational root of f(x) is an integer.


Exercises.

2.1.1. Compute f + g and fg, where

f = 2 + 3x+ x2, g = 1 + x2 + x3 + 4x3 ∈ Z5[x].

2.1.2 ([Nic12, Ex. 4.1.2]). (a) Compute (1 + x)5 in Z5[x].

(b) Compute (1 + x)7 in Z7[x].

(c) Show that (1 + x)p = 1 + x in Zp[x] if p is a prime number.

2.1.3 ([Nic12, Ex. 4.1.3]). (a) How many polynomials of degree 3 are there in Z5[x]?

(b) How many monic polynomials of degree 3 are there in Z5[x]?

2.1.4. Prove Theorem 2.1.8.


2.1.6. Show that the map ϕ of Example 2.1.11 is a surjective ring homomorphism withkerϕ = I.

2.1.7. Give an example of a commutative ring R and a polynomial f(x) ∈ R[x]\{0} of degreen with more than n distinct roots.

2.1.8. Use the Rational Roots Theorem (Theorem 2.1.28) to show that 3√

5 is irrational.

2.1.9 ([Nic12, Ex. 4.1.4]). (a) Find all roots of (x− 4)(x− 5) in Z6 and in Z7.

(b) Find all roots of x3 − x in Z6 and in Z4.

2.1.10 ([Nic12, Ex. 4.1.5(a)]). Find the number of roots of x2−x in Z4, Z2×Z2, any integraldomain, and Z6.

2.1.11 ([Nic12, Ex. 4.1.8]). Let R be a subring of a ring S, let f(x) 6= 0 and g(x) be polyno-mials in R[x], and assume that the leading coefficient of f(x) is a unit in R. If f(x) dividesg(x) in S[x], show that f(x) divides g(x) in R[x]. Hint: The Division Algorithm.

2.1.12 ([Nic12, Ex. 4.1.9]). Show that R[x] and R have the same characteristic for any ringR.

2.1.13 ([Nic12, Ex. 4.1.11]). If a is a nonzero root of f = a0 + a1x + · · · + an−1xn−1 + anx

n,show that a−1 (if it exists) is a root of g = an + an−1x + · · · + a1x

n−1 + a0xn. Assume that

the coefficient ring R is commutative.

2.1.14 ([Nic12, Ex. 4.1.12]). If a, b, and c are real and f = x2 − (a + c)x + (ac − b2), showthat every complex root of f is real.


2.1.15 ([Nic12, Ex. 4.1.13]). Divide x3 − 4x + 5 by 2x + 1 in Q[x]. Why is it impossible inZ[x]?

2.1.16 ([Nic12, Ex. 4.1.14]). In each case, write g = qf + r in R[x], where r = 0 or deg r <deg f .

(a) g = x5 + 4x4 + x3 + 5x2 + x+ 2, f = x2 + x+ 1, R = Z6.

(b) g = x3 + x2 + 3x+ 2, f = 3x+ 1, R = Z8.

(c) g = 3x3 + 2x2 − 8x+ 1, f = x2 + 2, R = Q.

2.1.17 ([Nic12, Ex. 4.1.15]). Which of x−1, x+1, and x−2 is a factor of x4−2x3−x2 +3x−2in Z[x]?

2.1.18 ([Nic12, Ex. 4.1.16(a)]). For which primes p is x−1 a factor of f = 3x4+5x3+2x2+x+4in Zp[x]?

2.1.19 ([Nic12, Ex. 4.1.17]). In each case, factor f into linear factors in F [x].

(a) f = x4 + 12, F = Z13.

(b) f = x3 − x2 + x− 1, F = Z5.

2.1.20 ([Nic12, Ex. 4.1.18]). Let a 6= 0 in a field F . Determine the integers n ≥ 1 such thatx+ a is a factor of xn + an in F [x]. In these cases, write down the factorization.

2.1.21 ([Nic12, Ex. 4.1.19]). If F is a field, let u, v, and w be distinct roots of f = x3 +ax2 +bx+ c in F [x]. Show that a = −(u+ v + w), b = uv + uw + vw, and c = −uvw.

2.1.22 ([Nic12, Ex. 4.1.21]). (a) Show that Z4[x] has infinitely many units and infinitelymany idempotents.

(b) Find a polynomial in Z4[x] that is neither a unit nor a nilpotent.

2.1.23 ([Nic12, Ex. 4.1.23]). In each case determine the multiplicity of a as a root in f ∈ R[x].

(a) f = x3 − 2x2 − 4x+ 3, a = 3, R = Z6.

(b) f = x5 + 2x4 + x3 − x2 + 2x− 1, a = 1, R = Z4.

2.1.24 ([Nic12, Ex. 4.1.25]). In each case find all rational roots of f and factor f as much aspossible in Q[x].

(a) f = 4x4 + x3 − 3x2 + 4x− 3

(b) f = x4 − x3 − x2 − x− 2

(c) f = x4 + x3 + 3x2 + 2x+ 2

2.1.25 ([Nic12, Ex. 4.1.26]). Show that n√m is not rational unless m = kn for some integer

k.

2.1.26 ([Nic12, Ex. 4.1.27]). If f is a monic polynomial in Z[x], show that the only rationalroots (if any) are integers.

2.2. Factorization of polynomials over a field 41

2.1.27. (a) Suppose R is an infinite integral domain. Show that if f(x), g(x) ∈ R[x] satisfyf(a) = g(a) for all a ∈ R, then f(x) = g(x). In other words, two polynomials are equalif and only if they correspond to the same polynomial function.

(b) Suppose R is a finite ring. Show that there exists a polynomial f(x) ∈ R[x] \ {0} suchthat f(a) = 0 for all a ∈ R.

2.1.28. Let G be the multiplicative group of positive rational numbers. Show that G isisomorphic as a group to (Z[x],+).

2.1.29 (Bonus). Let R be a commutative ring. Prove that f(x) ∈ R[x], f(x) = a0 + · · ·+anxnis a nilpotent element of R[x] if and only if a0, . . . , an are nilpotent elements of R.

2.2 Factorization of polynomials over a field

We start this section with a few examples.

Example 2.2.1. Let’s factor f(x) = 3x3 − x2 − x − 4 ∈ Q[x] as much as possible. By theRational Roots Theorem (Theorem 2.1.28), the only possible rational roots are of the formcd

where c | 4 and d | 3. We see that x = 43

is a root. Polynomial division then gives thatf(x) =

(x− 4

3

)(3x2 + 3x + 3) = (3x − 4)(x2 + x + 1). Again, using the Rational Roots

Theorem, we can see that x2 + x+ 1 has no rational roots and so we can factor no further.

Example 2.2.2. Let’s factor f(x) = x4+25x+24 ∈ Q[x] as much as possible. Since f(−1) = 0,we know that x + 1 is a factor of f(x). Polynomial division then gives f(x) = (x + 1)(x3 −x2 + x + 24). By the Rational Roots Theorem (Theorem 2.1.28), any rational roots ofx3 − x2 + x + 24 must be in the set {±1,±2,±3,±4,±6,±8,±12,±24}. But none of theseis a root and therefore f(x) cannot be factored any further (since if it factored, at least onefactor would need to be of degree one).

Example 2.2.3. Let’s factor f(x) = 2x5 +x4 + 4x3 + 2x2 + 2x+ 1 ∈ Q[x] as much as possible.Since f(−1/2) = 0, we get f(x) = (2x+ 1)(x4 + 2x2 + 1) = (2x+ 1)(x2 + 1)2. Since x2 + 1has no rational (or even real) roots, we cannot factor f(x) any further (in Q[x]).

Definition 2.2.4 (Irreducible polynomial). Let F be a field. A polynomial f(x) ∈ F[x] iscalled irreducible over F if

(a) f(x) 6= 0 and deg f(x) ≥ 1 (i.e. f(x) is a nonconstant polynomial), and

(b) if f(x) = p(x)q(x) in F[x], then either deg p(x) = 0 or deg q(x) = 0.

A nonzero polynomial of positive degree is called reducible (or we say that it factors) if it isnot irreducible.

Example 2.2.5. If deg f(x) = 1 then f(x) is irreducible.

Example 2.2.6. If f(x) is irreducible, then, for every a 6= 0, the polynomial af(x) is alsoirreducible.


Theorem 2.2.7. Let F be a field and consider f(x) ∈ F[x] \ {0} such that deg f(x) ≥ 2.

(a) If f(x) is irreducible, then it does not have any roots in F.

(b) Assume deg f(x) ≤ 3. Then f(x) is irreducible if and only if f(x) does not have anyroots in F.

Proof. (a) If f(x) has a root a ∈ F, then p(x) = (x − a)q(x) for some q(x) ∈ F[x] bythe Factor Theorem (Theorem 2.1.21). Since deg p(x) ≥ 2, this means that p(x) is notirreducible, contradicting the hypothesis. Thus p(x) has no root in F.

(b) Assume p(x) has no root in F. If p(x) is reducible, then p(x) = f(x)g(x) and so f(x)and g(x) have no roots in F. Thus deg f(x) 6= 1 and deg g(x) 6= 1. But this contradicts thefact that deg f(x) + deg g(x) = deg p(x) is equal to 2 or 3. Thus p(x) is irreducible. Theconverse follows from part (a).

Example 2.2.8. The polynomial x2− 2 is irreducible over Q but reducible over R since it hasroots in R but not in Q.

Example 2.2.9. The polynomial x2 + 1 is irreducible in R[x] but reducible in C[x].

Example 2.2.10. The polynomial x3 + 2x2 + x + 2 is irreducible in Z5[x] because it has noroots.

Example 2.2.11. The polynomial x2 − 2 is reducible in R[x] but irreducible in Z3[x]. Thepolynomial x2 + 2 is irreducible in R[x] but reducible in Z3[x] (in particular, x2 + 2 =(x− 1)(x+ 1) in Z3[x]).

Theorem 2.2.12 (Fundamental Theorem of Algebra). If f(x) ∈ C[x] is a nonconstantpolynomial, then f(x) has a root in C.

We will accept the Fundamental Theorem of Algebra without proof. For those interestedin reading the proof, it can be found in [Jud12, Th. 23.33].

Corollary 2.2.13. Suppose f(x) ∈ C[x] with deg f(x) ≥ 1. If we write f(x) = a0 + a1x +· · ·+anxn, an 6= 0, then f(x) factors as f(x) = an(x−z1)(x−z2) · · · (x−zn), where z1, . . . , znare the roots of f(x), counted according to multiplicity. In particular, the only irreduciblepolynomials in C[x] are linear.

Proof. The proof is by induction on the degree of f(x), using the Fundamental Theorem ofAlgebra (Theorem 2.2.12). The details are left as Exercise 2.2.1.

Theorem 2.2.14. If f(x) ∈ R[x] is a nonconstant polynomial, then it factors (in R[x]) asa product of polynomials of degree at most two. In particular, the irreducible polynomials inR[x] are either linear or quadratic.

Proof. The proof is by induction on deg f(x). The case deg f(x) ≤ 2 is immediate. Assumethe result holds for polynomials of degree ≤ n and consider f(x) with deg f(x) = n + 1.There are two cases.


(a) If f(z) = 0 for some z ∈ R, then f(x) = (x− z)g(x) and deg g(x) = n. The result thenfollows by the inductive hypothesis applied to g(x).

(b) Otherwise, f(x) has no real roots. By the Fundamental Theorem of Algebra (Theo-rem 2.2.12), it has a complex root z. Then f(z) = 0 and so f(z) = f(z) = 0 = 0.(Here z denotes the complex conjugate of z.) Therefore,

f(x) = (x− z)(x− z)g(x) = (x2 − (z + z)x+ zz)︸︷︷︸∈R[x]

g(x),

and the result again follows by applying the inductive hypothesis to g(x).

Theorem 2.2.15 (Gauss’ Lemma). Let f(x) = g(x)h(x) in Z[x]. If a prime p ∈ Z dividesevery coefficient of f(x), then either p divides every coefficient of g(x) or p divides everycoefficient of h(x).

Proof. Consider the ring homomorphism ϕ : Z[x] → Zp[x], ϕ(∑aix

i) =∑aix

i (called re-duction modulo p). We let f(x) = ϕ(f(x)). If a prime number p divides every coefficient off(x), then f(x) = 0 in Zp[x]. Thus 0 = g(x)h(x). Since Zp is a field, Zp[x] is an integraldomain by Theorem 2.1.9(e). Thus f(x) = 0 or h(x) = 0. But this implies that either everycoefficient of g(x) is zero in Zp or every coefficient of h(x) is zero in Zp.

Definition 2.2.16 (Proper factorization). Let f(x) ∈ Z[x] be a nonconstant polynomial. Aproper factorization of f(x) is a way of writing f(x) = g(x)h(x), where g(x), h(x) ∈ Z[x]with deg g(x), deg h(x) > 0.

Theorem 2.2.17. Suppose f(x) is a nonconstant polynomial in Z[x].

(a) If f(x) = g(x)h(x) with g(x), h(x) ∈ Q[x], then f(x) = g0(x)h0(x) for some g0(x), h0(x) ∈Z[x] with deg g0(x) = deg g(x) and deg h0(x) = deg h(x).

(b) The polynomial f(x) is irreducible in Q[x] if and only if it has no proper factorizationin Z[x].

Proof. (a) Let a and b be the least common multiples of the denominators of the coeffi-cients of g(x) and h(x), respectively. Then g1(x) := ag(x) ∈ Z[x] and h1(x) := bh(x) ∈ Z[x].Thus, we have

abf(x) = g1(x)h1(x)

in Z[x]. Suppose p is a prime number dividing ab. Then, by Gauss’ Lemma (Theorem 2.2.15),either p divides all the coefficients of g1(x) or it divides all the coefficients of h1(x). Thus wecan cancel p to give

ab

pf(x) = g2(x)h2(x)

in Z[x]. Continuing in this manner, we can cancel every prime factor of ab to obtain afactorization

f(x) = gk(x)hk(x)

in Z[x]. The result then follows since deg g(x) = deg g1(x) = · · · = deg gk(x) and similarlydeg h(x) = deg hk(x).


(b) if f(x) is irreducible in Q[x], then it has no proper factorization in Q[x] and henceno proper factorization in Z[x]. The converse follows from part (a).

If ϕ : R→ S is a ring homomorphism, then we have an induced ring homomorphism

R[x]→ S[x],∑

aixi 7→

∑α(ai)x

i, ai ∈ R.

In particular, if p is a prime number, taking ϕ to be usual ring homomorphism Z → Zp,ϕ(x) = x, gives a ring homomorphism Z[x]→ Zp[x] called reduction modulo p.

Theorem 2.2.18 (Modular Irreducibility Test). Let f(x) = a0 + a1x + · · · + anxn ∈ Z[x],

n ≥ 1, an 6= 0. Suppose that there exists a prime number p such that

(a) p - an and

(b) the reduction of f(x) modulo p is irreducible in Zp[x].

Then f(x) is irreducible in Q[x].

Proof. The first condition tells us that deg f(x) = deg f(x). Suppose f(x) is reducible inQ[x]. Then there is a proper factorization f(x) = g(x)h(x) in Z[x] by Theorem 2.2.17. Then

deg g(x) ≤ deg g(x) < deg f(x) = deg f(x).

Similarly, deg h(x) < deg f(x). Since f(x) = g(x)h(x), this contradicts the irreducibility off(x) in Zp[x].

Example 2.2.19. We show that f(x) = 4x4 − 3x2 − 2x − 1 is irreducible in Q[x]. We applyTheorem 2.2.18 with p = 3. Suppose f(x) = x4 + x + 2 = g(x)h(x) in Z3[x]. Thendeg g(x) + deg h(x) = 4. Assume, without loss of generality, that deg g(x) ≤ deg h(x). Thendeg g(x) ≤ 2.

It is not possible to have deg g(x) = 1, since x4 + x+ 2 does not have a root in Z3. So itremains to consider the case that x4 + x+ 2 factors as a product of two quadratics in Z3[x].So assume that in Z3[x] we have

x4 + x+ 2 = (x2 + ax+ b)(x2 + cx+ d).

This implies that

bd = 2, ad+ bc = 1, b+ d+ ac = 0, a+ c = 0. (2.1)

The first equation implies that b = 2d−1 and the fourth implies that a = −c. Substitutinginto the third equation then gives

2d−1 + d− c2 = 0 =⇒ 2 + d2 = c2d. (2.2)

On the other hand, substituting into the second equation of (2.1) gives

ad+ 2cd−1 = 1 =⇒ ad2 + 2c = d =⇒ (2− d2)c = d.

Since bd = 2, we have d 6= 0, and so the above implies that c 6= 0. Thus c is equal to 1 or 2.In either case we have c2 = 1 and so (2.2) becomes 2 + d2 = d. Hence d2 − d + 2 = 2. Butthis equation has no roots in Z3. This contradiction completes the proof.


Theorem 2.2.20 (Eisenstein Criterion). Let f(x) = a0 + a1x + · · · + anxn ∈ Z[x], n ≥ 1,

an 6= 0. Suppose that there exists a prime p ∈ Z such that

(a) p divides each of a0, . . . , an−1,

(b) p does not divide an, and

(c) p2 does not divide a0.

Then f(x) is irreducible in Q[x].

Proof. Suppose, towards a contradiction, that f(x) is not irreducible in Q[x]. Then, byTheorem 2.2.17, we have a proper factorization f(x) = g(x)h(x) in Z[x]. Let

g(x) = b0 + b1x+ · · ·+ bsxs and h(x) = c0 + c1x+ · · ·+ ctx

t,

with bs, ct 6= 0.

• Since p | b0c0, but p2 - a0 = b0c0, we have that p divides exactly one of b0 or c0. Withoutloss of generality, assume that p | b0 but p - c0.

• Since p - an = bsct, we have p - bs.• Let k be the smallest integer such that p - bk. So 0 < k < s ≤ n.

• Equating the coefficients of xk in f(x) = g(x)h(x) gives

ak = bkc0 + bk−1c1 + · · ·+ b0ck.

We know that p divides ak by assumption. Also, by our choice of k, p divides everyterm in the sum after the first. Thus p also divides bkc0. Since p is prime, it musttherefore divide either bk or c0, a contradiction.

Example 2.2.21. The polynomial 2x7 − 5x4 + 10x2 − 15 is irreducible in Q[x]. This can beseen by applying the Eisenstein Criterion (Theorem 2.2.20) with p = 5.

Example 2.2.22. The polynomial f(x) = x4 + x3 + x2 + x + 1 is irreducible in Q[x]. To seethis, first note that f(x)(x− 1) = x5 − 1. Thus,

f(x+ 1) =(x+ 1)5 − 1

(x+ 1)− 1=

(x+ 1)5 − 1

x= x4 + 5x3 + 10x2 + 10x+ 5.

Then we see that f(x+ 1) does not factor by the Eisenstein Criterion (Theorem 2.2.20) withp = 5. Hence f(x) does not factor either.

Example 2.2.23. If p is a prime number, then xp−1 + xp−2 + · · ·+ x+ 1 is called a cyclotomicpolynomial . Using the argument of Example 2.2.22, one can show that this polynomial isirreducible.

Theorem 2.2.24. Let F be a field and f(x), g(x) ∈ F[x] \ {0}. Then there exists a uniquepolynomial d(x) ∈ F[x] \ {0} such that

(a) d(x) is monic,


(b) d(x) | f(x) and d(x) | g(x),

(c) if h(x) ∈ F[x] satisfies h(x) | f(x) and h(x) | g(x), then h(x) | d(x),

(d) there exists u(x), v(x) ∈ F[x] such that d(x) = u(x)f(x) + v(x)g(x).

Proof. Let d(x) be a monic element of {u(x)f(x) + v(x)g(x) | u(x), v(x) ∈ F[x]} of smallestdegree. Then (a), (c) and (d) are obvious. To see (b), divide f(x) by d(x) to obtainf(x) = q(x)d(x) + r(x) with r(x) = 0 or deg r(x) < deg d(x). Then

r(x) = f(x)− q(x)d(x) = f(x)(1− q(x)u(x)) + g(x)(−q(x)v(x)).

Thus, by our choice of d(x), we cannot have deg r(x) < deg d(x). Thus r(x) = 0, whichproves that d(x) | f(x). The proof that d(x) | g(x) is analogous. Thus (b) is proved.

It remains to show the uniqueness assertion. Suppose d1(x) and d2(x) both satisfy thegiven conditions, Then, d1(x) | d2(x) and d2(x) | d1(x). Hence d1(x) = d2(x)k1(x) =d1(x)k2(x)k1(x) for some k1(x), k2(x) ∈ F[x]. Thus k1(x)k2(x) = 1, which implies that k1(x)and k2(x) are constant polynomials. But since d1(x) and d2(x) are both monic, we muchhave k1(x) = k2(x) = 1. Thus d1(x) = d2(x).

Definition 2.2.25 (Greatest common divisor of polynomials). The polynomial d(x) whoseexistence is asserted in Theorem 2.2.24 is called the greatest common divisor of f(x) andg(x) and is denoted gcd(f(x), g(x)).

Algorithm 2.2.26 (Euclidean Algorithm). Let f(x), g(x) ∈ F[x] with g(x) 6= 0. Set a1(x) =f(x) and a2(x) = g(x). By repeated polynomial division, we have

a1(x) = q1(x)a2(x) + a3(x) with a3(x) 6= 0 and deg a39x) < deg a2(x),

a2(x) = q2(x)a3(x) + a4(x) with a4(x) 6= 0 and deg a4(x) < ϕa3(x),

. . . . . .

ak−2(x) = qk−2(x)ak−1(x) + ak(x) with ak(x) 6= 0 and deg ak(x) < deg ak−1(x),

ak−1(x) = qk−1(x)ak(x).

Then gcd(f(x), g(x)) = ak(x). A proof of this result is outlined in Exercise 2.2.3.

Example 2.2.27. In Q[x], we have gcd(x2+1, 2x−1) = 1 and gcd(x4−2x2+1, 2x2−2) = x2−1.

Proposition 2.2.28. Let F be a field, p(x) ∈ F[x] an irreducible polynomial, and f1(x), . . . , fn(x) ∈F[x]. If p(x) | f1(x) · · · fn(x), then p(x) | fi(x) for some 1 ≤ i ≤ n.

Proof. We prove the result by induction. The case n = 1 is obvious. Now consider the casen = 2. Suppose p(x) divides f1(x)f2(x). Let d(x) = gcd(p(x), f1(x)). Then d(x) | p(x) andthus, since p(x) is irreducible, we have deg d(x) = 0 (so d(x) = 1) or deg d(x) = deg p(x).If deg d(x) = deg p(x), then p(x) = ad(x) for some nonzero a ∈ F and so p(x) divides f1(x)since d(x) does. On the other hand, if d(x) = 1, then we have u(x)p(x) + v(x)f1(x) = 1 forsome u(x), v(x) ∈ F[x] by Theorem 2.2.24. Thus u(x)p(x)f2(x) + v(x)f1(x)f2(x) = f2(x).Therefore p(x) divides f2(x) since it divides f1(x)f2(x). Thus the result holds for n = 2.

If n > 2, then p(x) | f1(x)f2(x) · · · fn(x) implies p(x) | f1(x) or p(x) | f2(x) · · · fn(x) (bythe n = 2 case). The result then follows by the inductive hypothesis.


Corollary 2.2.29. If F is a field and p(x) ∈ F[x] is irreducible, then 〈p(x)〉 is a prime idealof F[x].

Proof. Suppose p(x) is irreducible and f(x)g(x) ∈ 〈p(x)〉. Then p(x) | f(x)g(x). Hence, byProposition 2.2.28, p(x) divides f(x) or g(x), which implies that f(x) ∈ 〈p(x)〉 or g(x) ∈〈p(x)〉.

Example 2.2.30. We have that 〈x2 + x + 1〉 is a prime ideal of Z2[x]. We will see later that〈p(x)〉 is in fact a maximal ideal of F[x] whenever p(x) is irreducible (see Theorem 2.3.6).

Theorem 2.2.31 (Unique Factorization Theorem). Let F be a field and f(x) ∈ F[x] be anonconstant polynomial. Then

(a) f(x) = ap1(x) · · · pm(x) where a ∈ F× and p1(x), . . . , pm(x) ∈ F[x] are irreducible andmonic, and

(b) the factorization of part (a) is unique up to a reordering of the pi(x)’s.

Proof. (a) We prove the result by induction on n = deg f(x). The case n = 1 is obvioussince degree one polynomials are irreducible. Assume the result holds for n and considera polynomial f(x) of degree n + 1. If f(x) is irreducible, we are done. Otherwise, wecan factor f(x) = g(x)h(x) with deg g(x), deg h(x) ≤ n. The result then follows from theinductive hypothesis.

(b) Suppose we have two factorizations

f(x) = ap1(x) · · · pm(x) = bq1(x) · · · q`(x), (2.3)

with a, b ∈ F× and p1(x), . . . , pm(x), q1(x), · · · , q`(x) monic irreducible. Equating the leadingterms, we see that a = b. Now, p1(x) | q1(x) · · · q`(x), and so p1(x) | qi(x) for some 1 ≤ i ≤ `by Proposition 2.2.28. Thus p1(x) = qi(x) since both are monic and irreducible. Cancelingp1(x) and qi(x) in (2.3) gives

p2(x) · · · pm(x) = q1(x) · · · qi−1(x)qi(x) · · · q`(x).

The theorem follows by repeating the above argument.

Exercises.

2.2.1. Prove Corollary 2.2.13.

2.2.2. Show that if p is a prime number, then the cyclotomic polynomial xp−1+xp−2+· · ·+x+1is irreducible. See Example 2.2.23.

2.2.3. Let f(x), g(x) ∈ F[x], with g(x) 6= 0.

(a) Show that Algorithm 2.2.26 terminates after a finite number of steps.


(b) In the notation of this algorithm, show that gcd(ai(x), ai+1(x)) = gcd(ai+1(x), ai+2(x))for all i = 1, . . . , k − 2.

(c) In the same notation, show that gcd(ak−1(x), ak(x)) = ak(x).

(d) Conclude that gcd(f(x), g(x)) = ak(x).

2.2.4 ([Nic12, Ex. 4.2.1]). (a) If F is a field and a ∈ F , a 6= 0, show that a divides f forevery f ∈ F [x].

(b) If p ∈ F [x] divides f for every f ∈ F [x], show that p ∈ F , p 6= 0.

2.2.5 ([Nic12, Ex. 4.2.2]). If f, g ∈ F [x] and F is a field, consider the following statements:

(a) f = ag for some 0 6= a ∈ F ;

(b) f and g have the same roots in F .

Prove that (a) =⇒ (b). Does (b) =⇒ (a)? Justify your answer.

2.2.6 ([Nic12, Ex. 4.2.3]). In each case, explain why f is reducible over any field.

(a) f = x3 − 2x2 + 2x− 2

(b) f = x3 + x2 + 4

2.2.7 ([Nic12, Ex. 4.2.4]). In each case determine whether the polynomial is irreducible. Givereasons.

(a) x3 + 5 in Z7[x]

(b) x2 − 2 in R[x]

(c) x2 + 11 in C[x]

(d) x3 − 4 in Z11[x]

(e) x3 + x+ 1 in Z5[x]

(f) x2 + x+ 1 in Z17[x]

2.2.8 ([Nic12, Ex. 4.2.5]). In each case, determine whether the polynomial is irreducible overeach of the fields Q, R, C, Z2, Z3, Z5, and Z7.

(a) x2 − 3

(b) x2 + x+ 1

(c) x3 + x+ 1

(d) x3 − 2

2.2.9 ([Nic12, Ex. 4.2.6]). Let R be an integral domain and let f ∈ R[x] be monic. If ffactors in R[x], show that it has a factorization f = gh, where g and h are both monic (andof positive degree).

2.2.10 ([Nic12, Ex. 4.2.8]). (a) If x2 + ax + b has roots u and v in a field F , show thatb = uv and a = −(u+ v).


(b) Show that 1 + i is a root of x2 + (1− 2i)x− (3 + i) ∈ C[x]. Find the other root.

2.2.11 ([Nic12, Ex. 4.2.9]). Show that an odd degree polynomial in R[x] has a real root.(Requires calculus.)

2.2.12 ([Nic12, Ex. 4.2.10]). Find all monic irreducible cubics in Z2[x].

2.2.13 ([Nic12, Ex. 4.2.12]). Let p be a monic quartic in Z2[x]. Show that p is irreducible inZ2[x] if and only if

(a) p has no root in Z2, and

(b) p 6= x4 + x2 + 1.

Hint : Exercise 2.2.8.

2.2.14 ([Nic12, Ex. 4.2.13]). Show that a monic quintic p in Z2[x] in irreducible if and only if

(a) p has no root in Z2, and

(b) p is neither x5 + x4 + 1 nor x5 + x+ 1.

Hint : Exercise 2.2.12.

2.2.15 ([Nic12, Ex. 4.2.14]). Find all monic irreducible quadratics in Z3[x].

2.2.16 ([Nic12, Ex. 4.2.15]). Find a list of six quartics in Z3[x] such that a monic quartic pis irreducible if and only if it has no root in Z3 and is not in the list. Hint : Exercise 2.2.15.

2.2.17 ([Nic12, Ex. 4.2.18]). In each case, factor f as a product of irreducible polynomials inF [x].

(a) f = 3x4 + 2, F = Z5

(b) f = 3x4 + 2, F = Z11

(c) f = x3 + 2x2 + 2x+ 1, F = Z7

(d) f = x3 + 2x2 + 2x+ 1, F = Z3

(e) f = x4 − x2 + x− 1, F = Z13.

(f) f = x4 − x2 + x− 1, F = Z17.

2.2.18 ([Nic12, Ex. 4.2.19]). Factor x5 + x4 + 1 as a product of irreducible polynomials inZ2[x].

2.2.19 ([Nic12, Ex. 4.2.20]). Factor x5 + x2 − x + 1 as a product of irreducible polynomialsin Z3[x].

2.2.20 ([Nic12, Ex. 4.2.21]). Show that each polynomial is irreducible in Q[x].

(a) 3x3 + 5x2 + x+ 2

(b) 5x3 + 2x+ 3


(c) x3 + 9x2 + x+ 6

(d) x3 + x2 + 10x+ 8

2.2.21 ([Nic12, Ex. 4.2.22]). Show that each polynomial is irreducible in Q[x].

(a) x5 + 6x4 + 12x+ 15

(b) 4x5 + 18x4 + 7x3 − 28x2 + 14

2.2.22 ([Nic12, Ex. 4.2.24]). Show that f(x) = x4 + 4x3 + 4x2 + 4x+ 5 is irreducible over Qby considering f(x− 1).

2.2.23 ([Nic12, Ex. 4.2.29]). Show that xn − p is irreducible in Q[x] for all n ≥ 2 and allprimes p ∈ Z. (Hence Q[x] has infinitely many irreducible polynomials of every degree ≥ 2.)

2.2.24 ([Nic12, Ex. 4.2.30]). Show that xp − a is reducible in Zp[x] for every a ∈ Zp.

2.2.25 ([Nic12, Ex. 4.2.31]). Let F ⊆ K be fields and f, g ∈ F [x].

(a) If f is irreducible in K[x], show that it is irreducible in F [x].

(b) If f and g are relatively prime in F [x] (i.e. their greatest common divisor is 1), showthat there are relatively prime in K[x]. Hint : Theorem 2.2.24(d).

2.2.26 ([Nic12, Ex. 4.2.34]). Let f = x3 − 42x2 + 35x + m. Show that there are infinitelymany integers m for which f is irreducible in Q[x]. Hint : Eisenstein Criterion.

2.2.27 ([Nic12, Ex. 4.2.35]). In each case, factor f into irreducible polynomials in Q[x].

(a) f = x4 + 3x3 + x2 + 3x+ 1

(b) f = x4 + x3 − 7x2 + 3x− 2

(c) f = x4 + 2x3 − 2x2 + 7x− 2

(d) f = x4 − x3 + 2x2 − 3x+ 2

2.2.28 ([Nic12, Ex. 4.2.36]). If m and p are integers with p prime, show that x4 +mx+ p isirreducible in Q[x] if and only if it has no roots in Q.

2.2.29 ([Nic12, Ex. 4.2.39]). In each case compute d = gcd(f, g), and express it in F [x] as alinear combination of f and g.

(a) f = x2 + 2, g = x3 + 4x2 + x+ 1, F = Z5

(b) f = x2 + 1, g = x5 + x4 + x3 + x2 + x+ 1, F = Z2

(c) f = x2 − x− 2, g = x5 − 4x3 − 2x2 + 7x− 6, F = Q(d) f = x3 + x− 2, g = x5 − x4 + 2x2 − x− 1, F ∈ Q

2.2.30 (Bonus). Prove that C[x]/〈x2 + 1〉 ∼= C× C (as rings).

2.2.31 (Bonus problem). (a) Let R be a commutative ring. Prove that f(x) ∈ R[x],f(x) = a0 + · · · + anx

n is a nilpotent element of R[x] if and only if a0, . . . , an are nilpotentelements of R.

2.3. Quotient rings of polynomials over a field 51

(b) Let θ : M2(R)→M2(R) be a map such that

(i) for all c ∈ R, θ

([c 00 c

])=

[c 00 c

],

(ii) for all A,B ∈M2(R), we have θ(AB) = θ(A)θ(B),

(iii) for all A,B ∈M2(R) and a, b ∈ R, we have θ(aA+ bB) = aθ(A) + bθ(B).

Prove that θ(X) = PXP−1 for some invertible P ∈M2(R).

2.3 Quotient rings of polynomials over a field

Theorem 2.3.1. Let F be a field and let I be a nonzero ideal of F[x]. Then there is a uniquemonic polynomial h(x) ∈ F[x] such that I = 〈h(x)〉 = h(x)F[x] = {h(x)g(x) | g(x) ∈ F[x]}.

Proof. Let I be a nonzero ideal of F[x]. Then I contains nonzero polynomials and hence itcontains monic polynomials (since it is an ideal, we can multiply any polynomials by theinverse of the leading coefficient to obtain a monic polynomial in I). Among all the monicpolynomials in I, choose h(x) of minimal degree. Then 〈h(x)〉 ⊆ I.

Now suppose f(x) ∈ I. By the Euclidean Algorithm (Algorithm 2.2.26), we have f(x) =q(x)h(x) + r(x) for q(x), r(x) ∈ F[x] with r(x) = 0 or deg r(x) < deg h(x). Suppose r(x) 6= 0and let a be the leading coefficient of r(x). Then a−1r(x) is monic and

a−1r(x) = a−1(h(x)− q(x)f(x)) ∈ I.

But deg(a−1r(x)) = deg r(x) < deg h(x), which contradicts our choice of h(x). Thus r(x) = 0and so I = 〈h(x)〉.

Example 2.3.2. Consider the ring R[x] and let I = 〈x2+1〉. We can divide any f(x) ∈ R[x] byx2 +1 to obtain f(x) = q(x)(x2 +1)+(ax+b) for unique q(x) ∈ R[x] and a, b ∈ R. Thus, anyelement of R[x]/I can be written in the form (ax+b)+I. We have (ax+b)+I = (a′x+b′)+Iif and only if a = a′ and b = b′. Furthermore we have

((ax+ b) + I) + ((a′x+ b′) + I) = ((a+ a′)x+ (b+ b′)) + I,

((ax+ b) + I)((a′x+ b′) + I) = (aa′x2 + ab′x+ a′bx+ bb′) + I = ((a′b+ ab′)x+ (bb′ − aa′)) + I.

Thus the map R[x]/I ∼= C via the map ax+ b 7→ ai+ b.

The above example is a special case of the following general theorem. Its proof can befound in [Nic12, §4.3].

Theorem 2.3.3. Let F be a field, h(x) ∈ F[x], and I = 〈h(x)〉. Suppose n = deg h(x). Thenevery element of R = F[x]/I has a unique representative of degree ≤ n− 1. Thus the factorring R = F[x]/I is given by

R ∼= {a0 + a1t+ · · ·+ an−1tn−1 | a0, . . . , an−1 ∈ F}

(isomorphism of vector spaces).


The addition in the representation of R given in Theorem 2.3.3 is given by adding coef-ficients. The multiplication is computed by using the fact that h(t) = 0. This allows us toexpress tn in terms of lower powers of t.

Example 2.3.4. Consider the quotient ring Z2[x]/〈x2〉. By Theorem 2.3.3, we have

Z2[x]/〈x2〉 ∼= {at+ b | a, b ∈ Z2} = {0, 1, t, 1 + t}.

When computing in this ring, we use the fact that t2 = h(t) = 0. Thus we have the followingaddition and multiplication tables.

+ 0 1 t 1 + t0 0 1 t 1 + t1 1 0 1 + t tt t 1 + t 0 1

1 + t 1 + t t 1 0

× 0 1 t 1 + t0 0 0 0 01 0 1 t 1 + tt 0 t 0 t

1 + t 0 1 + t t 1

Note that this ring is not a field since, for example, the nonzero element t is not invertible.

Example 2.3.5. Consider the quotient ring Z2[x]/〈x2 + x+ 1〉. By Theorem 2.3.3, we have

Z2[x]/〈x2 + x+ 1〉 ∼= {a+ bt | a, b ∈ Z2}.

When computing in this ring, we use the fact that h(t) = 0 and so t2 = −t− 1 = t+ 1. Themultiplication table is therefore as follows:

× 0 1 t 1 + t0 0 0 0 01 0 1 t 1 + tt 0 t 1 + t 1

1 + t 0 1 + t 1 t

This is a field with four elements. (Recall that Z4 is not a field, so this quotient is notisomorphic to Z4.)

Theorem 2.3.6. Let F be a field and h(x) ∈ F[x] with deg h(x) ≥ 1. Then the followingstatements are equivalent:

(a) h(x) is irreducible.

(b) 〈h(x)〉 is a prime ideal.

(c) 〈h(x)〉 is a maximal ideal.

Proof. (a) ⇒ (b): This is Corollary 2.2.29.(b) ⇒ (c): Suppose 〈h(x)〉 is prime but not maximal. Then there exists some ideal I of

F[x] such that 〈h(x)〉 ( I ( F[x]. Now, we know by Theorem 2.3.1 that I = 〈p(x)〉 for somep(x) ∈ F[x]. Thus h(x) = p(x)q(x) for some q(x) ∈ F[x]. But p(x) 6∈ 〈h(x)〉, and so q(x) ∈〈h(x)〉, i.e., q(x) = h(x)r(x) for some r(x) ∈ F[x]. This implies that h(x) = p(x)h(x)r(x),and so deg p(x) = 0, which contradicts the fact that I 6= F[x].

(c)⇒ (a): If h(x) is not irreducible, then h(x) = h1(x)h2(x) for some h1(x), h2(x) ∈ F[x]with deg h1(x), deg h2(x) < deg h(x). Thus 〈h(x)〉 ( 〈h1(x)〉 ( F[x], which implies that〈h(x)〉 is not maximal.

2.3. Quotient rings of polynomials over a field 53

One of the important consequences of Theorem 2.3.6 is that fields of order pm existfor all prime numbers p and positive integers m. More precisely, it can be shown thata monic irreducible polynomial h(x) of degree m exists in Zp[x] for every such p and m.Then, by Theorem 2.3.6, the ideal 〈h(x)〉 is maximal and hence Zp[x]/〈h(x)〉 is a field byCorollary 1.3.19. It is straightforward to verify that this field has pm elements.

Example 2.3.7. The polynomial x2 + x + 1 has no root in Z2 and so it is irreducible byTheorem 2.2.7. Thus

Z2[x]/〈x2 + x+ 1〉 = {a+ bt | a, b ∈ Z2, t2 + t+ 1 = 0}

is a field with four elements.

Exercises.

2.3.1. Prove Theorem 2.3.1. Hint: Consider a monic element f(x) ∈ I of smallest degreeand follow the approach of Proposition 1.3.6.

2.3.2. Use the First Isomorphism Theorem (Theorem 1.4.15) to give an alternative proofthat R[x]/〈x2 + 1〉 ∼= C (see Exercise 2.3.2).

2.3.3 (Bonus Problem). Prove that C[x]/〈x2 + 1〉 ∼= C× C (as rings).

Chapter 3

Integral domains

In this chapter we discuss integral domains in further detail. In particular, we introduce thenotions of unique factorization domains, principal ideal domains and euclidean domains. Areference for the material in this chapter is [Jud12, Ch. 18]. Throughout this chapter, unlessotherwise noted, we assume that R is an integral domain.

3.1 Unique factorization domains

Recall that, for a, b ∈ R, we say that a divides b, and we write a | b if b = ac for some c ∈ R.

Definition 3.1.1 (Factorization). A factorization of an element a ∈ R is way of writinga = bc with b, c ∈ R. Such a factorization is said to be trivial if a or b is a unit in R.

Trivial factorizations are not so interesting. We will consider two factorizations r = aband r = (ua)(u−1b) for some unit u ∈ R× to be essentially the same.

Example 3.1.2. In Z we have 8 = 4 · 2 = (−4) · (−2). In R[x] we have (x3− 1) = (x− 1)(x2 +x+ 1) = (3x− 3)

(13x2 + 1

3x+ 1

3

).

Theorem 3.1.3. Let a, b ∈ R. The following are equivalent:

(a) a | b and b | a,

(b) a = ub for some u ∈ R×,

(c) 〈a〉 = 〈b〉.

Proof. Exercise.

Definition 3.1.4 (Associates). Two elements a, b ∈ R are called associates if a | b and b | a.When a and b are associates, we write a ∼ b.

Lemma 3.1.5. The relation ∼ (i.e. being associates) is an equivalence relation.

Proof. Exercise.

Example 3.1.6. The equivalence classes of ∼ in Z are {0}, {±1} {±2}, . . ..

54

3.1. Unique factorization domains 55

Example 3.1.7. Consider the ring Z(√

3) := {a+b√

3 | a, b ∈ Z}. Since (2+√

3)(2−√

3) = 1,we see that 2+

√3 ∈ Z(

√3)×. Then, since

√3(2+

√3) = 3+2

√3, we see that

√3 ∼ 3+2

√3.

Example 3.1.8. Let’s find all the units in Z(√−5) := {a+ b

√−5 | a, b ∈ C}. We have

(a+ b√−5)(c+ d

√−5) = 1 =⇒ (a− b

√−5)(c− d

√−5) = 1 =⇒ (a2 + 5b2)(c2 + 5d2) = 1.

Since both a2 + 5b2 and c2 + 5d2 are nonnegative integers, we must have a2 + 5b2 = 1 and soa = ±1 and b = 0. Therefore Z(

√−5)× = {±1}. So, the only associates of z ∈ Z(

√−5) are

±z. In contrast, recall that Z(i)× = {±1,±i} (see Example 1.1.25).

The next definition generalizes the concepts of irreducible and prime elements to arbitraryintegral domains.

Definition 3.1.9 (Irreducible, prime). Let r ∈ R. We say that r is irreducible if the followingconditions are satisfied:

(a) r 6= 0,

(b) r 6∈ R×,

(c) if r = ab, then a ∈ R× or b ∈ R×.

We say that r is prime if the following conditions are satisfied:

(a) r 6= 0,

(b) r 6∈ R×,

(c) for every a, b ∈ R, if r | ab, then r | a or r | b.

Example 3.1.10. Every prime number p ∈ Z is both irreducible and prime.

Example 3.1.11. Every irreducible polynomial f(x) ∈ F[x] (where F is a field) is both irre-ducible (by definition) and prime (by Proposition 2.2.28).

Theorem 3.1.12. Any prime element of an integral domain is irreducible.

Proof. Suppose r ∈ R is prime and r = ab for some a, b ∈ R. Then r | ab and so r | a orr | b. Without loss of generality (i.e. interchanging a and b if necessary), we can assumer | a. Then a = rc for some c ∈ R. Then

r = ab = rcb =⇒ 1 = cb =⇒ b is a unit,

where in the first implication we used the cancelation law in domains (Proposition 1.2.7).

The converse of the above theorem is false, as the next example illustrates.

Example 3.1.13. We will show that 1 +√−5 is an irreducible element of Z(

√−5) that is not

prime.To see that it is irreducible, suppose that we have a factorization

1 +√−5 = (a+ b

√−5)(c+ d

√−5).

56 Chapter 3. Integral domains

Then, taking complex conjugates, we have

1−√−5 = (a− b

√−5)(c− d

√−5).

Multiplying the sides of this equation by the corresponding sides of the previous one gives

6 = (a2 + 5b2)(c2 + 5d2).

Since each of the factors on the right is positive, we must have (interchanging the two factorsif necessary)

• a2 + 5b2 = 1 and c2 + 5d2 = 6, in which case a = 1, b = 0; or

• a2 + 5b2 = 2 and c2 + 5d2 = 3, which is impossible.

Thus the factorization is trivial.To see that 1 +

√−5 is not prime, note that (1 +

√−5)(1−

√−5) = 6 = 3 · 2. However,

we claim that 1 +√−5 does not divide 2 or 3. If it divided 2, we could find x, y ∈ Z with

2 = (1 +√−5)(x+ y

√−5) =⇒ 2 = (1−

√−5)(x− y

√−5) =⇒ 4 = 6(x2 + 5y2),

which is a contradiction. Similarly, if 1 +√−5 divided 3, we would have x, y ∈ Z with

3 = (1 +√−5)(x+ y

√−5) =⇒ 3 = (1−

√−5)(x− y

√−5) =⇒ 9 = 6(x2 + 5y2),

which is again a contradiction. Thus 1 +√−5 is not prime.

Definition 3.1.14 (Unique factorization domain). An integral domain R is called a uniquefactorization domain (or UFD) if it satisfies the following conditions:

(a) If r ∈ R such that r 6= 0 and r 6∈ R×, then r can be written as a product of irreducibleelements.

(b) If r ∈ R, r 6= 0 and r 6∈ R× with r = p1 · · · ps = q1 · · · qt (where the pi and qi areirreducibles) then s = t and, after relabeling if necessary, pi ∼ qi for i = 1, . . . , s.

Example 3.1.15. The ring Z of integers is a UFD. Also, by Theorem 2.2.31, the ring F[x] isa UFD for any field F.

Lemma 3.1.16. Every irreducible element in a UFD is prime.

Proof. Suppose R is a UFD and r ∈ R is irreducible with r | ab for some a, b ∈ R. Thenwe have ab = rs for some s ∈ RR. Since R is a UFD, we can factor a, b, s as products ofirreducibles:

a = p1 · · · pk, b = q1 · · · q`, s = t1 · · · tn.

Thus p1 · · · pkq1 · · · q` = rt1 · · · tn. By the uniqueness of factorization in a UFD, we haver ∼ pi for some i = 1, . . . , k or r ∼ qj for some j = 1, . . . , `. Thus r | a or r | b.

Example 3.1.17. By Example 3.1.13 and Lemma 3.1.16, we see that Z(√−5) is not a UFD.


In a UFD, one can use the factorization of an element to find all its divisors (up toassociates). Suppose R is a UFD and a ∈ R is a nonzero nonunit. Then we can writeuniquely

a ∼ pa11 pa22 · · · parr ,

where ai ≥ 1 and pi is a prime in R for i = 1, . . . , r and the pi are nonassociates (i.e.pi 6∼ pj for i 6= j). The uniqueness here means that the primes are uniquely determined upto associates (and reordering) as are the exponents ai. The divisors of a are then determineduniquely (up to associates). Precisely, we have

d | a ⇐⇒ d ∼ pd11 pd22 · · · pdrr , for 0 ≤ di ≤ ai, i = 1, . . . , r.

The proof of this is left as an exercise (or see [Nic12, §5.1, p. 258]).

Definition 3.1.18 (Greatest common divisor, least common multiple). Suppose s1, . . . , sn ∈R. An element d ∈ R is called a greatest common divisor (gcd) of s1, . . . , sn, and is denotedgcd(s1, . . . , sn), if it satisfies the following conditions:

(a) d | si for i = 1, 2, . . . , n.

(b) If r ∈ R satisfies r | si for i = 1, 2, . . . , n, then r | d.

An element m ∈ R is called a least common multiple (lcm) of s1, . . . , sn, and is denotedlcm(s1, . . . , sn) if it satisfies:

(a) si | m for i = 1, 2, . . . , n.

(b) If r ∈ R satisfies si | r for i = 1, 2, . . . , n, then m | r.

Note that the definition of gcd given above agrees with usual definition of a gcd of integersand Definition 2.2.25 for polynomials, except that we require the gcd in Z to be positive andthe gcd in F[x] to be monic. These extra conditions ensure uniqueness of the gcd. In ageneral UFD, we do not have any analogous condition to ensure uniqueness. Thus, in anarbitrary UFD, greatest common divisors (and least common multiples) are defined only upto associates. We write gcd(s1, . . . , sn) and lcm(s1, . . . , sn) to denote any gcd and lcm. Inparticular, we have:

if si ∼ s′i for i = 1, . . . , n then gcd(s1, . . . , sn) ∼ gcd(s′1, . . . , s′n).

Note that gcd’s and lcm’s need not exist in an arbitrary integral domain, as the followingexample illustrates.

Example 3.1.19. Consider the elements x5, x6 in the subring Z[x2, x3] of Z[x] consisting of allfinite sums of the form

∑aij(x

2)i(x3)j, aij ∈ Z. We see that x2 and x3 are both commondivisors of x5 and x6, but neither of x2 or x3 divides the other (in Z[x2, x3]). Thus, the gcdof x5 and x6 in Z[x2, x3] does not exist.

Proposition 3.1.20. Gcds and lcms exists in UFDs. Suppose R is a UFD and a, b, c, . . .is a finite list of nonzero elements in R. Let p1, . . . , pr be the nonassociated primes dividingone of a, b, c, . . . and write

a ∼ pa11 pa22 · · · parr , ai ∈ N,


b ∼ pb11 pa22 · · · pbrr , bi ∈ N,

c ∼ pb11 pa22 · · · pcrr , ci ∈ N.

...

For each i = 1, 2, . . . , r, let di = min(ai, bi, ci, . . . ) and mi = max(ai, bi, ci, . . . ). Then

gcd(a, b, c, . . . ) ∼ pd11 pd22 · · · pdrr and lcm(a, b, c, . . . ) ∼ pm1

1 pm22 . . . pmrr .

Proof. The proof when R = Z carries over to the general case. The details are left as anexercise.

Definition 3.1.21 (Ascending chain condition on principal ideals). We say an integral dom-ain R satisfies the ascending chain condition on principal ideals (or ACCP) if R contains nostrictly increasing infinite chain

〈a1〉 ( 〈a2〉 ( 〈a3〉 · · ·

of principal ideals.

Example 3.1.22. Suppose we have an chain of strictly increasing chain of principal ideals inZ:

{0} 6= 〈m1〉 ( 〈m2〉 ( m3〉 ( · · · .

Then |m1| > |m2| > |m3| > · · · . Thus the chain must terminate after finitely many ideals.So Z satisfies the ACCP.

Example 3.1.23. Consider an ascending chain of principal ideals in F[x], where F is a field:

{0} 6= 〈p1(x)〉 ( 〈p2(x)〉 ( p3(x)〉 ( · · · .

Then deg p1(x) > deg p2(x) > · · · and so the chain must be finite. Therefore F[x] satisfiesthe ACCP.

Example 3.1.24. Let R = {n + xf(x) | n ∈ Z, f(x) ∈ Q[x]}, the set of polynomials in Q[x]whose constant term is an integer. Since R is a subring of Q[x], it is an integral domain.However,

〈x〉 ( 〈12x〉 ( 〈 1

22x〉 ( · · ·

is an infinite ascending chain of principal ideals in R. Thus R does not satisfy the ACCP.

Theorem 3.1.25. Suppose R is an integral domain that satisfies the ACCP. Then everynonzero nonunit in R is a product of irreducibles.

Proof. Suppose a nonzero nonunit a ∈ R cannot be written as a product of irreducibles.Then a is not irreducible and so we have

a = r1a1, a 6∼ r1, a 6∼ a1.


Now, at least one of r1, a1 cannot be written as a product of irreducibles (since if they bothcan, then so can a). Without loss of generality, we may assume that a1 cannot be writtenas a product of irreducibles. Then we again have

a1 = r2a2, a1 6∼ r2, a1 6∼ a2,

where a2 cannot be written as a product of irreducibles. Continuing in this manner, we have

〈a〉 ( 〈a1〉 ( 〈a2〉 ( · · · .

Thus R does not satisfy the ACCP.

Theorem 3.1.26. The following are equivalent:

(a) R is a UFD.

(b) R satisfies the ACCP and gcd(a, b) exists for all a, b ∈ R.

(c) R satisfies the ACCP and every irreducible element of R is prime.

Proof. (a) ⇒ (b): Assume R is a UFD. Then Proposition 3.1.20 shows that gcds exist in R.Now suppose we have an ascending chain of principal ideals in R:

{0} ( 〈a1〉 ( 〈a2〉 ( · · · .

Consider a factorization a1 = p1 · · · pm of a into irreducibles. Since 〈a1〉 ( 〈a2〉, we havea2 ∼ pi1 · · · pik , where {i1, . . . , ik} ( {1, . . . ,m}. Similarly, we have a3 ∼ pj1 · · · pj` , where{j1, . . . , j`} ( {i1, . . . , ik}. It follows that the chain is finite. Thus R satisfies the ACCP.

(b) ⇒ (c): Suppose (b) holds. Suppose p ∈ R is irreducible and p | ab in R. Letd = gcd(a, p). Then d | p and so d ∼ p or d ∼ 1 (since p is irreducible). If p ∼ d then, sinced | a, we have p | a. On the other hand, if d ∼ 1, then gcd(a, p) ∼ 1. We claim that thisimplies that gcd(ab, pb) ∼ b. Assuming this claim, we have p | b since p | ab and p | pb. Sowe have shown that p | a or p | b and so p is prime.

It remains to prove the claim. Let d′ = gcd(ab, pb). We have b | ab and b | pb. Thusb | d′. Say d′ = bu. We show that u is a unit. Write ab = d′x for x ∈ R. So ab = bux. Thusa = ux, since b 6= 0. Thus u | a. Similarly, u | p. So u | gcd(a, p) ∼ 1. Hence u is a unit.

(c) ⇒ (a): Suppose (c) holds. By Theorem 3.1.25, every element is a product of irredu-cibles. So it remains to show the uniqueness of such factorizations. Suppose

p1p2 · · · pr ∼ q1q2 · · · qs

are distinct factorizations, where the pi and qi are irreducibles and r+s is minimal. If r = 1,then we have p1 = q1 · · · qs. Since p1 is irreducible, this implies that s = 1, which contradictsthe assumption that the factorizations are distinct. An analogous argument shows that wecannot have s = 1. Thus we may assume that r, s ≥ 2. Since p1 | q1 · · · qs, we must havep1 | qj for some j since p1 is prime by assumption. Relabeling if necessary, we may assumethat p1 | q1. Since q1 is irreducible, this implies that p1 ∼ q1. Thus p2 · · · pr ∼ q2 · · · qs aredistinct factorizations, contradicting the minimality of r + s


Example 3.1.27. The ring R of Example 3.1.24 is not a UFD since it does not satisfy theACCP.

In the remainder of this section is devoted to the proof of the important result that if Ris a UFD, then so is R[x] (see Theorem 3.1.33). Before proving this, we will need to definea few terms and prove some preliminary results.

Definition 3.1.28 (Content, primitive polynomial). Suppose R is a UFD and f(x) ∈ R[x]\{0}. Then the content of f(x) is the gcd of the nonzero coefficients of f and is denotedc(f(x)). We say that f(x) is a primitive polynomial if c(f(x)) ∼ 1.

Example 3.1.29. In Z[x], c(12 + 6x + 9x4) = 3. However, in Q[x], c(12 + 6x + 9x4) = 1 (ofcourse, it is also 3, since 1 and 3 are associates in Q). Thus 12 + 6x + 9x4 is primitive inQ[x] but not in Z[x].

Lemma 3.1.30. Let R be a UFD and let f(x) ∈ R[x] \ {0}.

(a) f(x) can be written as f(x) = c(f(x))f1(x), where f1(x) ∈ R[x] is primitive.

(b) If a ∈ R \ {0}, then c(a(f(x))) = ac(f(x)).

(c) If f(x) is irreducible and deg f(x) ≥ 1, then f(x) is primitive.

Proof. Exercise.

The following theorem is a generalization of Theorem 2.2.15

Theorem 3.1.31 (Gauss’ Lemma). Suppose R is a UFD and f(x), g(x) ∈ R[x] \ {0}. Then

c(f(x)g(x)) = c(f(x))c(g(x)).

In particular, the product of primitive polynomials is primitive.

Proof. Write f(x) = c(f(x))f1(x) and g(x) = c(g(x))g1(x), where f1(x), g1(x) are primitive.Then

c(f(x)g(x)) ∼ c(c(f(x))c(g(x))f1(x)g1(x)) ∼ c(f(x))c(g(x))c(f1(x)g1(x))

by Lemma 3.1.30. Thus, it suffices to prove the result when f(x) and g(x) are primitive.Let f(x) =

∑mi=0 aix

i and g(x) =∑n

i=0 bixi. Suppose that p is a prime dividing the

coefficients of f(x)g(x). Let r be the smallest integer such that p - ar and s be the smallestinteger such that p - bs. The coefficient of xr+s in f(x)g(x) is

cr+s = a0br+s + a1br+s−1 + · · ·+ ar+s−1b1 + ar+sb0.

Since p divides a0, . . . , ar−1 and b0, . . . , bs−1, p divides every term of cr+s except for the termarbs. However, since p | cr+s, either p divides ar or p divides bs. This contradiction completesthe proof.

Proposition 3.1.32. Suppose R is a UFD and let F be its field of quotients. We regard Ras a subring of F in the usual way (see Section 1.2). Suppose p(x) is a primitive element ofR(x). Then p(x) is irreducible in R[x] if and only if it is irreducible in F[x].


Proof. First suppose that p(x) is irreducible in F[x]. Let p(x) = f(x)g(x) be a factorizationin R[x]. If deg f(x), deg g(x) ≥ 1, then this is a nontrivial factorization in F[x], whichcontradicts the fact that p(x) is irreducible in F[x]. Thus we can assume, without loss ofgenerality, that deg f(x) = 0. Hence f(x) = u ∈ R. Then, since p(x) is primitive, we have

1 ∼ c(p(x)) = c(f(x)g(x)) = c(f(x))c(g(x)) = uc(g(x)).

Hence u is a unit in R and so the factorization p(x) = f(x)g(x) is trivial in R[x]. Hence p(x)is irreducible in R[x].

Now suppose p(x) is primitive and irreducible in R[x] and p(x) = g(x)h(x) in F[x]. Let aand b be the products of the denominators of the coefficients of g(x) and h(x) respectively.Then g1(x) := ag(x) and h1(x) := bh(x) are in R[x] and

abp(x) = g1(x)h1(x)

is a factorization in R[x]. Thus, by Gauss’ Lemma (Theorem 3.1.31), we have

ab ∼ abc(p(x)) = c(abp(x)) = c(g1(x)h1(x)) ∼ c(g1(x))c(h1(x)). (3.1)

Now write g1(x) = c(g1(x))g2(x) and h1(x) = c(h1(x))h2(x), where g2(x) and h2(x) areprimitive in R[x]. Thus

abp(x) = g1(x)h1(x) = c(g1(x))c(h1(x))g2(x)h2(x).

Therefore, by (3.1), we have p(x) ∼ g2(x)h2(x) in R[x]. Since p(x) is irreducible in R[x], thisimplies that either g2(x) or h2(x) is a unit in R[x] (hence a unit in R). If g2(x) = u ∈ R×,then

ag(x) = g1(x) = c(g1(x))g2(x) = c(g1(x))u =⇒ g(x) = a−1c(g1(x))u ∈ F[x]×.

Similarly, if h2(x) ∈ R×, then h(x) ∈ F[x]×.

Theorem 3.1.33. If R is a UFD, then so is R[x].

Proof. Let p(x) be a nonzero polynomial in R[x]. If p(x) is a constant polynomial, then itmust have a unique factorization since R is a UFD. Now suppose that p(x) is a polynomialof positive degree in R[x]. Let F be the field of fractions of R, and let

p(x) = f1(x)f2(x) · · · fn(x)

be a factorization of p(x) in F[x], where each fi(x) is irreducible in F[x]. Choose ai ∈ R suchthat aifi(x) is in R[x] (e.g. take ai to be the product of the denominators of the coefficientsof fi(x)). Then, for i = 1, . . . , n, let bi = c(aifi(x)) ∈ R, so that aifi(x) = bigi(x), wheregi(x) is a primitive polynomial in R[x]. Since gi(x) = ai

bifi(x), the polynomials fi(x) and

gi(x) are associates in F[x]. Thus, since fi(x) is irreducible in F[x], so is gi(x). Then, byProposition 3.1.32, each gi(x) is irreducible in R[x]. Now, we have

a1 · · · anp(x) = b1 · · · bng1(x) · · · gn(x). (3.2)


Since g1(x) · · · gn(x) is primitive, we have

a1 · · · anc(p(x)) = c(a1 · · · anp(x)) = c(b1 · · · bng1(x) · · · gn(x)) = b1 · · · bn.

Thus a1 · · · an divides b1 · · · bn. Therefore, dividing both sides of (3.2) by a1 · · · an, we havep(x) = ag1(x) · · · gn(x), where a ∈ R. Since R is a UFD, we can factor a as c1 · · · ck, whereeach of the ci’s is irreducible in R (hence in R[x]). Thus we have proven the existence offactorizations into irreducibles.

We will now show the uniqueness of such factorizations. Let

p(x) = a1 · · · amf1(x) · · · fn(x) = b1 · · · brg1(x) · · · gs(x)

be two factorizations of p(x), where all of the factors are irreducible in R[x]. By Proposi-tion 3.1.32, each of the fi(x)’s and gi(x)’s is irreducible in F[x]. The ai’s and the bi’s areunits in F. Since F[x] is a PID, it is a UFD; therefore, n = s. Now rearrange the gi(x)’s sothat fi(x) and gi(x) are associates for i = 1, . . . , n. Then there exist c1, . . . , cn and d1, . . . , dnin R such that (ci/di)fi(x) = gi(x) or cifi(x) = digi(x). The polynomials fi(x) and gi(x) areprimitive; hence, ci and di are associates in R. Thus, a1 · · · am = ub1 · · · br in R, where u isa unit in R. Since R is a unique factorization domain, m = s. Finally, we can reorder thebi’s so that ai and bi are associates for each i. This completes the uniqueness part of theproof.

Corollary 3.1.34. If R is a UFD, then so is R[x1, . . . , xn]

Proof. This follows by an easy induction using Theorem 3.1.33 and the fact that

R[x1, . . . , xn] = (R[x1, . . . , xn−1])[xn].

Remark 3.1.35 (Noetherian and artinian rings). A commutative ring that satisfies the as-cending chain condition on ideals (the analogue of the ACCP, but where we do not specifythat the ideals are principal) is called a noetherian ring . A commutative ring that satisfiesthe descending chain condition on ideals is called an artinian ring . Artinian rings are alsonoetherian (but not vice versa). Noetherian and artinian rings are important classes of rings,but we will not study these classes of rings (per se) in this course. In noncommutative rings,one can talk of left and right noetherian (or artinian) rings.

Exercises.


3.1.2. Prove Lemma 3.1.5.


3.1.4. Prove Lemma 3.1.30.

3.2. Principal ideal domains 63

3.1.5. Suppose R is an integral domain and a, b ∈ R \ {0}. Show that gcd(a, b) exists if andonly if lcm(a, b) exists. Furthermore, show that if they exist, we have gcd(a, b) lcm(a, b) ∼ ab.

3.1.6 (Bonus problem). Let p(x) ∈ Z[x]. Assume that for every root α ∈ C of p(x) we have|α| = 1. Prove that p(x) | (xm − 1) for some m > 0.

3.2 Principal ideal domains

Definition 3.2.1 (Principal ideal domain). An integral domain R is called a principal idealdomain (PID) if every ideal of R is principal (i.e. generated by a single element).

Example 3.2.2. By Proposition 1.3.6 and Theorem 2.3.1, the rings Z and F[x] (where F is afield) are PIDs. Any field is also a PID since its only ideals are 〈0〉 and 〈1〉, which are bothprincipal.

Theorem 3.2.3. Every PID is a UFD.

Proof. Suppose R is a PID. By Theorem 3.1.26, it suffices to show that every irreducibleelement of R is prime and that R satisfies the ACCP.

Suppose p ∈ R is irreducible and p | ab in R. Let I be the ideal Ra + Rp. Then, sinceR is a PID, we have I = 〈d〉 for some d ∈ R. Thus d | a and d | p. Since p is irreducible,this implies that d ∼ p or d ∼ 1. If d ∼ p, then I = 〈p〉 and so p | a (since a ∈ I). On theother hand, if d ∼ 1, then 1 ∈ I. Thus there exist r, s ∈ R such that ra + sp = 1. Thenrab+ spb = b. Since p | ab, this implies that p | b. Hence p is prime.

It remains to show that R satisfies the ACCP. Suppose

〈a1〉 ⊆ 〈a2〉 ⊆ · · · .

Let A =⋃∞i=1〈ai〉. Then A is an ideal of R (exercise). Thus, since R is a PID, we have

A = 〈a〉 for some a ∈ R. Then a ∈ 〈ai〉 for some i. Hence 〈ai〉 = 〈a〉 = A and so 〈an〉 = Afor n ≥ i. So R satisfies the ACCP.

The converse of Theorem 3.2.3 is false, as the following example illustrates.

Example 3.2.4. The ring Z[x] is a UFD (by Theorem 3.1.33) but not a PID since, for example,the ideal

I = 〈2, x〉 := {a0 + a1x+ · · ·+ anxn | a0 ∈ 2Z, a1, . . . , an ∈ N}

is not a principal ideal (exercise).

Remark 3.2.5. The fact that Z is a PID but Z[x] is not shows that the analogue of Theo-rem 3.1.33 for PIDs does not hold.

Theorem 3.2.6. If R is a PID, then the gcd of any nonzero elements a1, . . . , an ∈ R existsand it can be written in the form d = r1a1 + · · ·+ rnan for some r1, . . . , rn ∈ R.

Proof. Let I = 〈a1, . . . , an〉 := 〈a1〉 + · · · + 〈an〉. Since R is a PID, we have I = 〈d〉 forsome d ∈ R. Then d = r1a1 + · · · + rnan for some r1, . . . , rn ∈ R by the definition ofI. Since a1, . . . , an ∈ 〈d〉, we have d | ai for all i. Furthermore, if c | ai for all i, thenc | (r1a1 + · · ·+ rnan) and so c | d. Thus d ∼ gcd(a1, . . . , an).


Theorem 3.2.7. Suppose R is a PID and p ∈ R, p 6= 0, p 6∈ R×. Then the followingstatements are equivalent:

(a) p is a prime element.

(b) 〈p〉 is a prime ideal.

(c) 〈p〉 is a maximal ideal.

Proof. (a) ⇒ (b): Suppose p is a prime element and ab ∈ 〈p〉. Then p | ab, which impliesthat p | a or p | b. Thus a ∈ 〈p〉 or b ∈ 〈p〉. Thus 〈p〉 is a prime ideal.

(b) ⇒ (c): Suppose 〈p〉 is a prime ideal. Then p is prime since

p | ab =⇒ ab ∈ 〈p〉 =⇒ a ∈ 〈p〉 or b ∈ 〈p〉 =⇒ p | a or p | b.

Thus p is irreducible by Theorem 3.1.12. Now suppose 〈p〉 ⊆ 〈q〉 ( R. Then q | p. Since〈q〉 6= R, we have q 6∼ 1. Since p is irreducible, this implies that p ∼ q. Thus 〈p〉 = 〈q〉.

(c) ⇒ (a): Suppose 〈p〉 is a maximal ideal. Then it is a prime ideal by Corollary 1.3.20.Hence, by the above, we see that p is prime.

Example 3.2.8. In Z[x], the ideal 〈x〉 is prime but 〈x〉 ( 〈2, x〉 ( Z[x] and so this ideal is notmaximal. Thus Z[x] is not a PID (see Example 3.2.4).

Example 3.2.9. Let F be a field. In F[x, y], the ideal 〈x〉 is prime, but 〈x〉 ( 〈x, y〉 ( F[x, y].Thus F[x, y] is not a PID. However, F[x, y] is a UFD by Corollary 3.1.34.

Proposition 3.2.10. Suppose R is a PID and a1, . . . , an ∈ R \ {0}.

(a) d ∼ gcd(a1, . . . , an) if and only if 〈a1〉+ 〈a1〉+ · · ·+ 〈an〉 = 〈d〉.

(b) m ∼ lcm(a1, . . . , an) if and only if 〈a1〉 ∩ 〈a1〉 ∩ · · · ∩ 〈an〉 = 〈m〉.

Proof. Part (a) follows from Theorem 3.2.6. The proof of part (b) is left as an exercise.

Exercises.

3.2.1. Show that if I1 ⊆ I2 ⊆ · · · is a chain of ideals in a ring R, then I :=⋃∞i=1 Ii is an ideal

of R.

3.2.2. Show that 〈2, x〉 ⊆ Z[x] is not a principal ideal (see Example 3.2.4).

3.2.3. Prove Proposition 3.2.10(b).

3.3. Euclidean domains 65

3.3 Euclidean domains

Definition 3.3.1 (Division algorithm, divisor function). If R is an integral domain, we saythat R has a division algorithm if there exists a map ν : R \ {0} → N (called a divisorfunction) such that, for all a, b ∈ R with b 6= 0, there exist q, r ∈ R with

a = qb+ r and either r = 0 or ν(r) < ν(b).

Definition 3.3.2 (Euclidean domain). An integral domain R is called a euclidean domainif it has a divisor function ν : R \ {0} → N that satisfies the following condition:

a, b ∈ R \ {0} =⇒ ν(ab) ≥ ν(a). (3.3)

Such a divisor function is called a euclidean valuation (or euclidean function).

Remark 3.3.3. Condition (3.3) is actually superfluous in the following sense. Any domainR that has a division algorithm possesses a euclidean valuation (i.e. a divisor function sa-tisfying (3.3)). Indeed, if ν is a divisor function for R, then

ν ′(a) = minr∈R\{0}

ν(ra), a ∈ R,

is a euclidean valuation on R.

Examples 3.3.4. (a) The ring Z is a euclidean domain with euclidean valuation ν : Z→ N,ν(n) = |n|.

(b) For any field F, the ring F[x] is a euclidean domain with divisor function ν : F[x] \{0} → N, ν(f(x)) = deg f(x) (see Theorem 2.1.12).

Theorem 3.3.5. Every euclidean domain is a PID (and hence a UFD).

Proof. Suppose R has a divisor function ν, and let I be an ideal of R. If I is the zero ideal,then it is clearly principal. So we assume I 6= {0}. Choose b ∈ I \ {0} with ν(b) minimal.We claim that I = 〈b〉. Since b ∈ I, we have 〈b〉 ⊆ I. So it remains to show that I ⊆ 〈b〉.Suppose a ∈ I. Then we have

a = qb+ r where r = 0 or ν(r) < ν(a).

Then r = a − bq ∈ I and so, by the minimality of ν(b), we must have r = 0. Thus a = qband so a ∈ 〈b〉.

Lemma 3.3.6. The ring Z(i) of gaussian integers is a euclidean domain, hence a PID anda UFD.

Proof. (See [Jud12, §18.2, Example 8].) For a, b ∈ Z, define ν(a + bi) = a2 + b2 = |a + bi|2.We claim that ν is a euclidean valuation on Z(i).

Let z, w ∈ Z(i) \ {0}. Then ν(zw) = |zw|2 = |z|2|w|2 = ν(z)ν(w). Since ν(z) ≥ 1 forevery nonzero z ∈ Z(i), we have ν(zw) ≥ ν(w).


Next, we must show that for any z = a + bi and w = c + di in Z(i) with w 6= 0, thereexist elements q and r in Z(i) such that z = qw + r with either r = 0 or ν(r) < ν(w). Wecan view z and w as elements in Q(i) = {p + qi : p, q ∈ Q}, the field of fractions of Z(i).Observe that

zw−1 = (a+ bi)c− dic2 + d2

=ac+ bd

c2 + d2+bc− adc2 + d2

i

=

(m1 +

n1

c2 + d2

)+

(m2 +

n2

c2 + d2

)i

= (m1 +m2i) +

(n1

c2 + d2+

n2

c2 + d2i

)= (m1 +m2i) + (s+ ti)

in Q(i). In the last steps we are writing the real and imaginary parts as an integer plus aproper fraction. That is, we take the closest integer mi such that the fractional part satisfies|ni/(c2 + d2)| ≤ 1/2. For example, we write

9

8= 1 +

1

815

8= 2− 1

8.

Thus, s and t are the “fractional parts” of zw−1 = (m1 +m2i) + (s+ ti). We also know thats2 + t2 ≤ 1/4 + 1/4 = 1/2. Multiplying by w, we have

z = zw−1w = w(m1 +m2i) + w(s+ ti) = qw + r,

where q = m1 + m2i and r = w(s + ti). Since z and qw are in Z(i), r must be in Z(i).Finally, we need to show that either r = 0 or ν(r) < ν(w). However,

ν(r) = ν(w)ν(s+ ti) ≤ 1

2ν(w) < ν(w).

Remark 3.3.7. It is not easy to find examples of PIDs that are not euclidean domains, butsuch rings exist. The first such example was given by T. Motzkin in 1949: Z

(12

(1 +√−19

))is a noneuclidean PID.

Summarizing some of the results proven in this chapter, we have the following chain ofclass inclusions:

commutative rings ) integral domains ) UFDs ) PIDs ) euclidean domains ) fields.

Chapter 4

Fields

In this chapter we delve into the theory of fields. One of our motivating ideas will be rootsof polynomials. We have seen many examples of polynomials f(x) ∈ F[x], where F is afield, that do not factor completely into linear factors. However, we will see that it is alwayspossible to enlarge the field F so that f(x) does completely factor. Furthermore, we will seethat, up to isomorphism, a unique minimal such field exists, called the splitting field of f(x).This construction will also allow us to completely classify all finite fields. A reference for thematerial in this chapter is [Jud12, Ch. 21].

4.1 Brief review of vector spaces

In this section we give a very brief review of the definition of a vector space over a field andsome important results on vector spaces. For more details, see [Sav]. Through this section,F will denote a field.

Definition 4.1.1 (Vector space). A vector space over F is an abelian group V (writtenadditively and whose operation is called vector addition) and a scalar multiplication F×V →V , (a, v) 7→ av such that

• a(v + w) = av + aw for all a ∈ F and v, w ∈ V .

• (a+ b)v = av + bv for all a, b ∈ F and v ∈ V .

• a(bv) = (ab)v for all a, b ∈ F and v ∈ V .

• 1v = v for all v ∈ V .

Example 4.1.2. The set Fn := {(a1, . . . , an) | ai ∈ F ∀ i} is a vector space over F with thecomponentwise vector addition and the usual scalar multiplication.

Example 4.1.3. Any ring R such that F is a subring of R is vector space over F. For example,C is a vector space over R (and over Q).

Definition 4.1.4 (Subspace, linear combination, linear dependence, span, basis). SupposeV is a subspace over F. If W ⊆ V and W is also a vector space over F (under the restrictedoperations), then W is a called a subspace of V .

67

68 Chapter 4. Fields

If v1, . . . , vn ∈ V , then a linear combination of v1, . . . , vn is an element of V of the formc1v1 + · · · cnvn, where c1, . . . , cn ∈ F.

The vectors v1, . . . , vn ∈ V are called linearly dependent if there exist c1, . . . , cn ∈ F, withat least one of the ci nonzero, such that c1v1 + · · · + cnvn = 0. If v1, . . . , vn are not linearlydependent, then they are called linearly independent .

If v1, . . . , vn ∈ V , then Span{v1, . . . , vn} := {c1v1 + · · ·+ cnvn | c1, . . . , cn ∈ F}. We writeSpanF when we wish to emphasize the field in question. We say V is spanned (or generated)by v1, . . . , vn if V = Span{v1, . . . , vn}.

We say v1, . . . , vn is a basis for V (over F) if

(a) V = SpanF{v1, . . . , vn}, and

(b) v1, . . . , vn are linearly independent.

Theorem 4.1.5. Let V be a vector space over F. Suppose that

(a) V = Span{v1, . . . , vn} for some v1, . . . , vn ∈ V , and

(b) the vectors w1, . . . , wm ∈ V are linearly independent.

Then m ≤ n.

Corollary 4.1.6. If v1, . . . , vn and w1, . . . , wn are two bases of a vector space V over F, thenm = n.

Definition 4.1.7 (Dimension). A vector space V over F is called finite dimensional if it hasa finite basis. The number of elements of such a basis is called the dimension of V (over F)and is denoted dimV , or dimF V when we wish to emphasize the field F. If V does not havea finite basis, we say it is infinite dimensional .

Example 4.1.8. The vectors 1, i form a basis of C over R. Thus dimR C = 2. The vector 1 isa basis of C over C, so dimC C = 1. The space C[x] is infinite dimensional over C with basis1, x, x2, x3, . . . .

Theorem 4.1.9. Let V 6= {0} be a finite dimensional vector space.

(a) Every linearly independent subset of V can be extended (i.e. enlarged) to a basis.

(b) Every spanning set of V has a subset which is a basis of V .

Exercises.

4.1.1 (Bonus problem). Show that R is infinite dimensional as a vector space over Q.

4.2. Field extensions 69

4.2 Field extensions

We know many examples of fields that are subfields of larger fields. In this section, we willexamine these inclusions of fields and some of their important properties.

Definition 4.2.1 (Field extension, finite extension). A field extension is a pair of fieldsF ⊆ E. We say that F is a subfield of E and that E is an extension of F. If E, as a vectorspace over F, is finite dimensional, then E is called a finite extension of F, and dimF E isdenoted by [E : F].

Example 4.2.2. We have that R ⊆ C is a finite extension, but Q ⊆ R is not a finite extension(see Exercise 4.1.1).

Theorem 4.2.3. Let F ⊆ E be a finite extension and [E : F] = n. If u ∈ E, then there existsa nonzero polynomial f(x) ∈ F[x] such that f(u) = 0 and deg f(x) ≤ n.

Proof. Consider the elements 1, u, u2, . . . , un ∈ E. Since dimF E = n, this list of n + 1elements must be linearly dependent by Theorem 4.1.5. Thus, there exist a0, a1, . . . , an, withat least one ai 6= 0, such that a0 + a1u+ · · ·+ anu

n = 0.

Definition 4.2.4 (Algebraic, transcendental). Let F ⊆ E be a field extension. An elementu ∈ E is called algebraic over F if f(u) = 0 for some nonzero polynomial f(x) ∈ F[x]. Ifu ∈ E is not algebraic over F, then it is called transcendental over F. If every element of Fis algebraic over E, then we say F ⊆ E is an algebraic extension.

Example 4.2.5. We see that R ⊆ C is an algebraic extension since a+ bi, a, b ∈ R, is the rootof x2 − 2ax+ a2 + b2.

Example 4.2.6. (a) The real number 3√

2 is algebraic over Q since it is a root of x3 − 2.

(b) The real number√

5 +√

2 is algebraic over Q since it is a root of x4 − 10x2 + 23.

(c) The real number√

2−√

3 is algebraic over Q since it is a root of x4 − 10x2 + 1.

It is natural ask if Q ⊆ R is an algebraic extension. The answer is no. This follows fromthe next theorem.

Theorem 4.2.7 (Hermite 1873, Lindemann 1882). The real numbers e and π are transcen-dental over Q.

Remark 4.2.8. Since the set of polynomials with integer (or rational) coefficients is coun-table and each polynomial has a finite number of roots, the set of real numbers that arealgebraic over Q is countable. Therefore, there are uncountably many real numbers that aretranscendental over Q.

Definition 4.2.9 (Field generated by elements). If F ⊆ E is a field extension and u1, . . . , un ∈E, then

F(u1, . . . , un) :=⋂

K fieldF⊆K⊆E

u1,...,un∈K

K.


In other words, F(u1, . . . , un) is the smallest subfield of E containing F and u1, . . . , un. It iscalled the field generated by u1, . . . , un over F.

Example 4.2.10. We have R(i) = C and Q(i,−i) = Q(i) = {a+ bi | a, b ∈ Q}.

Example 4.2.11. We have Q(√

2) = {a + b√

2 | a, b ∈ Q} (as a subfield of R or C) since theright hand side is contained in the left hand side and the right hand side is a field. Thus thisnotation matches our previous notation (see Example 1.2.9).

It follows from Theorem 4.2.3 that all finite extensions are algebraic. If F ⊆ E is analgebraic extension, is it necessarily true that [E : F] < ∞? The answer is no since, forexample, Q ⊆ Q(

√2, 3√

2, 4√

2, . . . ) is an algebraic extension, but is not a finite extension (seeExercise 4.2.4).

Lemma 4.2.12. Let F ⊆ E be a field extension and u, v ∈ E. Then F(u)(v) = F(u, v).

Proof. We have

F(u)(v) =⋂

F(u)⊆K⊆Ev∈K

K and F(u, v) =⋂

F⊆K⊆Eu,v∈K

K.

Now, if K satisfies F(u) ⊆ K ⊆ E and v ∈ K, then it satisfies F ⊆ K ⊆ E and u, v ∈ K. SoF(u)(v) ⊇ F(u, v). On the other hand, if K satisfies F ⊆ K ⊆ E and u, v ∈ K, then F(u) ⊆ Kand v ∈ K. So F(u)(v) ⊆ F(u, v).

Corollary 4.2.13. Let F ⊆ E be a field extension and u1, . . . , un ∈ E. Then for every1 ≤ k ≤ n− 1, we have F(u1, . . . , uk)(uk+1, . . . , un) = F(u1, . . . , un).

Proof. This follows from Lemma 4.2.12 by induction.

Remark 4.2.14. Corollary 4.2.13 implies that F(u1, . . . , un) can be constructed via a chain ofsimple extensions:

F(u1, . . . , un) = F(u1)(u2) · · · (un).

Definition 4.2.15 (Minimal polynomial, degree). Let F ⊆ E be a field extension and u ∈ Ebe algebraic over F. The minimal polynomial of u over F is the nonzero monic polynomialm(x) ∈ F[x] of smallest degree such that m(u) = 0. The degree of u over F, denoted degF(u),is the degree of m(x).

Examples 4.2.16. (a) Consider the field extension R ⊆ C and the element i =√−1 ∈ C.

Then m(x) = x2 + 1 and degR(i) = 2.

(b) Consider the field extension Q ⊆ R and the element 3√

2 ∈ R. Then m(x) = x3 − 2and degQ(m(x)) = 3. How do we know that m(x) is the monic polynomial of minimal degree

such that m( 3√

2) = 0? Suppose f(x) ∈ Q[x] with f( 3√

2) = 0. Let d(x) = gcd(f(x), x3 − 2)(in Q[x]). Then

d(x) = a(x)f(x) + b(x)(x3 − 2) for some a(x), b(x) ∈ Q[x].

Thus d( 3√

2) = 0 and so d(x) 6= 1. Since d(x) | (x3 − 2) and x3 − 2 is irreducible by theEisenstein Criterion, we therefore have d(x) = x3 − 2. Thus (x3 − 2) | f(x).


(c) Suppose F ⊆ E is a field extension and u ∈ F. Then the minimal polynomial of u isx− u.

Theorem 4.2.17. Let F ⊆ E be a field extension and u ∈ E be algebraic over F with minimalpolynomial m(x) ∈ F[x]. Then the following statements hold:

(a) m(x) is irreducible in F[x].

(b) For every f(x) ∈ F[x], we have f(u) = 0 ⇐⇒ m(x) | f(x).

(c) m(x) is uniquely determined by u.

Proof. (a) Suppose m(x) = a(x)b(x) in F[x]. Then 0 = m(u) = a(u)b(u) and so a(u) = 0or b(u) = 0. Suppose, without loss of generality, that a(u) = 0. Then, by the minimality ofdegm(x), we must have deg a(x) = degm(x). Hence b(x) ∈ F× (since it is nonzero and hasdegree zero). So m(x) is irreducible.

(b) Clearly, if m(x) | f(x), then f(u) = 0. Now suppose that f(u) = 0. Let d(x) =gcd(f(x),m(x)). Then d(x) is an F[x]-linear combination of f(x) and m(x). So d(u) = 0,hence d(x) 6= 1. Thus d(x) = m(x) (since d(x) is a monic divisor of m(x), which is irreducibleby the above). Therefore m(x) = d(x) | f(x).

(c) Suppose m′(x) is another minimal polynomial for u. Then, by the above, we havem(x) | m′(x) and m′(x) | m(x). Since m(x) and m′(x) are both monic, this implies thatm(x) = m′(x).

Example 4.2.18. Let’s find the minimal polynomial of u =√

1 +√

3 over Q. We have

(u2 − 1)2 = 3 =⇒ u4 − 2u2 − 2 = 0.

Thus the minimial polynomial of u divides u4−2u2−2 by Theorem 4.2.17(b). Since x4−2x2−2is irreducible (apply the Eisenstein Criterion with p = 2), it must be the minimal polynomialof u.

Theorem 4.2.19. Let F ⊆ E be a field extension, u ∈ E be algebraic over F and d = degF(u).

(a) F(u) = {a0 + a1u+ · · ·+ ad−1ud−1 | a0, . . . , ad−1 ∈ F}.

(b) 1, u, . . . , ud−1 is a basis of F(u) over F. In particular, [F(u) : F] = degF(u).

(c) F(u) ∼= F[x]/〈m(x)〉, where m(x) is the minimal polynomial of u over F.

Proof. Set θ : F[x] → E, θ(f(x)) = f(u). Then θ is a ring homomorphism since it is thecomposition of the ring homomorphisms

F[x] ↪→ E[x]ϕu−→ E,

where ϕu is the evaluation map at u. Since

θ(f(x)) = 0 ⇐⇒ f(u) = 0 ⇐⇒ m(x) | f(x) ⇐⇒ f(x) ∈ 〈m(x)〉,


we have ker θ = 〈m(x)〉. We also see that

im θ = {a0 + a1u+ · · ·+ ad−1ud−1 | ai ∈ F} (4.1)

since

f(x) ∈ F[x] =⇒ f(x) = m(x)q(x) + r(x)

for some r(x) = a0 + a1x+ · · ·+ ad−1xd−1, ai ∈ F. Thus f(u) = r(u) = a0 + · · ·+ ad−1u

d−1.By the First Isomorphism Theorem (Theorem 1.4.15), we have im θ ∼= F[x]/〈m(x)〉. Since

m(x) is irreducible and F[x] is a PID, the ideal 〈m(x)〉 is maximal by Lemma 3.1.16 andTheorem 3.2.7. Thus im θ is a field by Corollary 1.3.19.

We claim that im θ = F(u). We have u ∈ im θ and F ⊆ im θ, so F(u) ⊆ im θ. By (4.1), wealso have im θ ⊆ K for every field K such that F ⊆ K ⊆ E with u ∈ K. Hence im θ ⊆ F(u).

Finally, if 1, u, . . . , ud−1 were linearly dependent, then we would have degm(x) < d, whichis a contradiction. Hence 1, u, . . . , ud−1 is a basis of F(u) over F.

Example 4.2.20. Let u = 2− i. Let’s describe the multiplication in Q(u). We have

(u− 2)2 = (−i)2 = −1 =⇒ u2 − 4u+ 5 = 0.

Let m(x) = x2−4x+5. Since m(x) has degree two and has no rational roots, it is irreducibleover Q. Thus m(x) is the minimal polynomial of u over Q. Therefore, by Theorem 4.2.19,we have

Q(u) = {a+ bu | a, b ∈ Q}.

Since u2 = 4u− 5, the multiplication in Q(u) is given by

(a1 + b1u)(a2 + b2u) = a1a2 + (a1b2 +a2b1)u+ b1b2u2 = (a1a2−5b1b2) + (a1b2 +a2b1 + 4b1b2)u.

Corollary 4.2.21. Let F ⊆ E be a field extension. If u, v ∈ E have the same minimalpolynomial, then F(u) ∼= F(v).

Example 4.2.22. We have Q( 3√

2) = {a0 + a13√

2 + a23√

4 | a0, a1, a2 ∈ Q} and Q( 3√

2) ∼=Q[x]/〈x3−2〉. Furthermore, [Q( 3

√2) : Q] = 3. We also have Q(ζ 3

√2) ∼= Q( 3

√2) for any ζ ∈ C

satisfying ζ3 = 1, since 3√

2 and ζ 3√

2 have the same minimal polynomial.

Example 4.2.23. Note that Q ⊆ Q( n√

2) ⊆ R and the minimal polynomial of n√

2 over Qis xn − 2 (using the Eisenstein Criterion with p = 2). Thus [R : Q] is not finite (seeExercise 4.1.1).

Theorem 4.2.24 (Multiplication Theorem). Let F ⊆ E ⊆ K be field extensions.

(a) If e1, . . . , em is a basis of E over F and k1, . . . , kn is a basis of K over E, then{eikj}1≤i≤m, 1≤j≤n is a basis of K over F.

(b) [K : F] <∞ if and only if [K : E] <∞ and [E : F] <∞.

(c) If [K : F] <∞, then [K : F] = [K : E][E : F].


Proof. (a) Let v ∈ K. Then

v =n∑i=1

ciki, ci ∈ E =⇒ v =n∑i=1

(m∑j=1

dijej

)ki =

∑1≤i≤n1≤j≤m

dijejki, dij ∈ F.

Thus {eikj}1≤i≤m, 1≤j≤n spans K over F. Since

∑i,j

fijeikj = 0 =⇒∑j

(∑i

fijei

)kj = 0 =⇒

∑i

fijei = 0 ∀ j =⇒ fij = 0 ∀ i, j,

we see that the set {eikj}1≤i≤m, 1≤j≤n is linearly independent over F.

(b) If [K : F] < ∞, then [K : E] < ∞ since any basis for K over F is a spanning set forK over E. Also [E : F] < ∞ since E is a subspace of K (over F). The reverse implicationfollows from part (a).

(c) This follows from part (a).

Corollary 4.2.25. Let F ⊆ E be a field extension, let u ∈ E be algebraic over F, and letv ∈ F(u). Then v is also algebraic over F and degF(v) | degF(u).

Proof. We have F ⊆ F(v) ⊆ F(u). Since [F(u) : F] is finite, so is [F(v) : F]. Hence v isalgebraic by Theorem 4.2.3. Thus

degF(u) = [F(u) : F] = [F(u) : F(v)][F(v) : F] = [F(u) : F(v)] degF(v).

Example 4.2.26. Let u = 3√

2. Then Q ⊆ Q(u2) ⊆ Q(u) and so, by Corollary 4.2.25, we have[Q(u2) : Q] | 3. Since Q(u2) 6= Q, we must have [Q(u2) : Q] = 3 and so [Q(u) : Q(u2)] = 1.Thus Q(u) = Q(u2).

Note, however, that if u = 4√

2, then Q(u2) ( Q(u) since [Q(u) : Q] = 4, but [Q(u2) :Q] = 2.

Lemma 4.2.27. Let F ⊆ E be a field extension. If u, v ∈ E are algebraic over F, then uvand u+ v are algebraic over F.

Proof. We have F ⊆ F(u) ⊆ F(u, v) = F(u)(v). Since v is algebraic over F, it is alsoalgebraic over F(u). Thus [F(u, v) : F(u)] < ∞. We also have [F(u) : F] < ∞. Therefore,by the Multiplication Theorem (Theorem 4.2.24), we have that [F(u, v) : F] <∞ (since u isalgebraic over F). Therefore, by Theorem 4.2.3, any element of F(u, v) is algebraic over F.In particular, u+ v and uv are algebraic over F.

Example 4.2.28. Let’s find the minimal polynomial of u =√

2 +√

5 over Q. First note that

u2 = 7 + 2√

10 =⇒ u2 − 7 = 2√

10 =⇒ (u2 − 7)2 = 40 =⇒ u4 − 14u2 + 9 = 0.

Thus, if we set f(x) = x4 − 14u2 + 9, we have f(u) = 0. Is this the monic polynomial ofminimal degree with this property?


Note that(√

2 +√

5)3 = 17√

2 + 11√

5.

Thus,√

2 =1

6

((√

2 +√

5)3 − 11(√

2 +√

5))∈ Q(

√2 +√

5).

Therefore we also have√

5 = (√

2 +√

5)−√

2 ∈ Q(√

2 +√

5). It follows that Q(√

2,√

5) ⊆Q(√

2 +√

5). Since the reverse inclusion is obvious, we have Q(√

2 +√

5) = Q(√

2,√

5).Note that degQ(

√2 +√

5) = [Q(√

2 +√

5) : Q] = [Q(√

2,√

5) : Q]. Now, [Q(√

2) : Q] = 2

(since x2 − 2 is irreducible). We claim that [Q(√

2,√

5) : Q(√

2)] = 2. Note that√

5 is aroot of x2 − 5 ∈ Q(

√2)[x], so the minimal polynomial of

√5 over Q(

√2) divides x2 − 5. It

suffices to show that this polynomial is irreducible. Since it is of degree two, it is enough toshow that it has no roots. But, for a, b ∈ Q, we have

(a+ b√

2)2 = 5 =⇒ a2 + 2b2 + 2ab√

2 = 5 =⇒ a2 + 2b2 − 5

2ab=√

2

(or a = 0 or b = 0, which would contradict the fact that√

5 is irrational) which contradictsthe fact that

√2 is irrational. Therefore we have

degQ(√

2 +√

5) = [Q(√

2 +√

5) : Q] = [Q(√

2 +√

5) : Q(√

2)][Q(√

2) : Q] = 2 · 2 = 4.

So f(x) is indeed minimal.

Remark 4.2.29. In Example 4.2.28, we showed that Q(√

2 +√

5) = Q(√

2,√

5). In fact,one can show that if F is a field of characteristic zero and u, v are algebraic over F, thenF(u, v) = F(w) for some w ∈ F(u, v). This is called the Primitive Element Theorem (seeExercise 4.2.5).

Theorem 4.2.30. A field extension F ⊆ E is finite if and only if E = F(u1, . . . , un), whereeach ui ∈ E is algebraic over F.

Proof. First suppose that E = F(u1, . . . , un), where each ui ∈ E is algebraic over F. Considerthe chain of extensions

F ⊆ F(u1) ⊆ F(u1, u2) ⊆ · · · ⊆ F(u1, . . . , un).

For i = 1, . . . , n, the element ui is algebraic over F(u1, . . . , ui−1) (we interpret this to be Fif i = 1) since it is algebraic over F. Thus [F(u1, . . . , un) : F(u1, . . . , un−1)] is finite. It thenfollows from the Multiplication Theorem (Theorem 4.2.24) that [E : F] is finite.

Now suppose that F ⊆ E is a finite extension. We prove the result by induction ond = [F : E]. If d = 1, then F = E and we are done. So assume d > 1. Choose u ∈ E \ F.Then [F(u) : F] > 1. Thus, by the Multiplication Theorem (Theorem 4.2.24), we have

[E : F(u)] =[E : F]

[F(u) : F]< [E : F].

Thus, by the inductive hypothesis, we have E = F(u)(u1, . . . , un) = F(u, u1, . . . , un). Theelements u, u1, . . . , un are algebraic over F by Theorem 4.2.3.


Theorem 4.2.31. If F ⊆ E ⊆ K are fields, then F ⊆ K is an algebraic extension if and onlyif both F ⊆ E and E ⊆ K are algebraic extensions.

Proof. If F ⊆ K is an algebraic extension, then all elements of K are algebraic over F, henceover E. Also, all elements of E, being also elements of K, are algebraic over F. So F ⊆ Eand E ⊆ K are both algebraic extensions.

Now assume that F ⊆ E and E ⊆ K are algebraic extensions. Let u ∈ K. SinceE ⊆ K is algebraic, there exists a nonconstant f(x) ∈ E[x] such that f(u) = 0. Letf(x) = a0 + a1x+ · · ·+ amx

m (so a0, . . . , am ∈ E) and consider the chain of field extensions

F ⊆ F(a0) ⊆ F(a0, a1) ⊆ · · · ⊆ F(a0, . . . , am) ⊆ F(a0, . . . , am, u).

Then we have:

• [F(a0, . . . , am, u) : F(a0, . . . , am)] ≤ deg f(x) <∞.

• Since a0 ∈ E, we know that a0 is algebraic over F, and so [F(a0) : F] <∞.

• Similarly, a1 is algebraic over F and so [F(a0, a1) : F(a0)] ≤ [F(a1) : F] <∞.

• ...

• am is algebraic over F, and so [F(a0, . . . , am) : F(a0, . . . , am−1)] ≤ [F(am) : F] <∞.

Thus [F(a0, . . . , am, u) : F] <∞. Therefore, by Theorem 4.2.3, u is algebraic over F.

Remark 4.2.32. The extension Q ⊆ Q(π) is not algebraic.

Example 4.2.33. Let’s find the minimal polynomial of 4√

2 over Q(√

2). Let f(x) = x2−√

2 ∈Q(√

2)[x]. Then f( 4√

2) = 0. Now consider

Q ⊆ Q(√

2) ⊆ Q(4√

2).

We have [Q( 4√

2) : Q] = 4 since x4 − 2 is irreducible over Q. Also, [Q(√

2) : Q] = 2since x2 − 2 is irreducible over Q. Thus [Q( 4

√2) : Q(

√2)] = 2. Now, we clearly have

Q( 4√

2) = Q(√

2)( 4√

2). By Theorem 4.2.19(b), we thus have degQ(√

2)(4√

2) = 2. Thus f(x) is

the minimal polynomial of 4√

2 over Q(√

2).

Remark 4.2.34. It follows from the above example that x2 −√

2 is irreducible in Q(√

2)[x].

Exercises.

4.2.1. Show that√

3 +√

5 is algebraic over Q.

4.2.2. Show that the intersection in Definition 4.2.9 is in fact a field. More generally, showthat if F is a field, then the intersection of an arbitrary collection of subfields of F is a field.


4.2.3. Find the minimal polynomial of u =√

3 +√

6 over Q.

4.2.4 (Bonus problem). Show that Q ⊆ Q(√

2, 3√

2, 4√

2, . . . ) is an algebraic extension, but isnot a finite extension.

4.2.5 (Bonus problem). Assume that charF = 0. Let F ⊆ E be an algebraic extension andα1, . . . , αn ∈ E. Prove that there exists a β ∈ E such that F(α1, . . . , αn) = F(β).

4.3 Splitting fields

We know that, if F is a field, then a polynomial f(x) ∈ F[x] may not have any roots in F,but may have roots in some extension E of F. In this section, we will discuss this idea infurther detail.

Theorem 4.3.1 (Kronecker’s Theorem). If F is any field and f(x) ∈ F[x] is a nonconstantpolynomial, then there exists an extension F ⊆ E such that f(x) has a root in E.

Proof. Let p(x) ∈ F[x] be irreducible such that p(x) | f(x) (note that such a p(x) exists sinceF[x] is a UFD). Then set E = F[t]/〈p(t)〉. Since p(t) is irreducible and F[t] is a PID, the ideal〈p(t)〉 is maximal by Lemma 3.1.16 and Theorem 3.2.7. Thus E is a field by Corollary 1.3.19.We also have F ⊆ E (viewing the elements of F as the classes of the constant polynomials).Since f(t) ∈ 〈p(t)〉, we have f(t) = 0 in E.

Example 4.3.2. The polynomial f(x) = x3 + x+ 1 ∈ Z2[x] does not have a root in Z2. Sincef(x) has degree three, this implies that f(x) is irreducible. Then f(x) has a root in the field

E = Z2[t]/〈f(t)〉 = {a0 + a1s+ a2s2 | a0, a1, a2 ∈ Z2},

where s denotes the image of t in E. In fact, f(x) factors completely in E[x]:

f(x) = (x+ s)(x2 + sx+ (1 + s2)) = (x+ s)(x+ s2)(x+ s+ s2).

Definition 4.3.3 (Splitting field). Let F be a field and f(x) ∈ F[x] be a nonconstant poly-nomial. An extensions F ⊆ E is called a splitting field of f(x) over F if

(a) f(x) = a(x− u1) · · · (x− un), a, u1, . . . , un ∈ E, and

(b) E = F(u1, . . . , un).

When (a) holds, we say that f(x) splits over E (or in E[x]).

Example 4.3.4. In Example 4.3.2, E is a splitting field of x3 + x+ 1 ∈ Z2[x] over Z2

Example 4.3.5. We have that Q(√

2) is a splitting field of x2 − 2 ∈ Q[x] over Q, whereasQ( 4√

2) is not.

Example 4.3.6. If f(x) = x− a, a ∈ F, then F is a splitting field of f(x) over F.

4.3. Splitting fields 77

Example 4.3.7. We have that Q(√−1) is a splitting field of x2 + 1 over Q, but not of x2 + 2.

However, R(√−1) is a splitting field of x2 + 1 over R and is also a splitting field of x2 + 2

over R.

Theorem 4.3.8. Let F be a field and f(x) ∈ F[x] be a nonconstant polynomial. Then thereexists a splitting field E of f(x) such that [E : F] ≤ n!, where n = deg f(x).

Proof. We prove the result by induction on n. If n = 1, then E = F and we are done. Nowassume n > 1. Set E1 = F[t]/〈p(t)〉 where p(x) is an irreducible polynomial dividing f(x).So f(x) has a root in E and so we have

f(x) = (x− a)g(x) for some g(x) ∈ E1[x], a ∈ E1.

Since deg g(x) ≤ n − 1, by the induction hypothesis, g(x) has a splitting field E such that[E : E1] ≤ (n− 1)!. Since [E1 : F] = deg p(x) ≤ deg f(x) = n, we have

[E : F] = [E : E1][E1 : F] ≤ n · (n− 1)! = n!.

Example 4.3.9. Let’s find a splitting field E of f(x) = x4 − 2x2 − 3 ∈ Q[x] and find [E : Q].Since f(x) = (x2 − 3)(x2 + 1), we have that E = Q(

√3,√−1) is a splitting field. Consider

the field extensionsQ ⊆ Q(

√−1) ⊆ Q(

√−1,√

3) = E.

We have [Q(√−1) : Q] = 2 since x2 + 1 is irreducible in Q[x]. Also, [Q(

√−1,√

3) :Q(√−1)] = 2 since x2−3 is irreducible in Q(

√−1)[x] (since

√3 6∈ Q(

√−1)). So [E : Q] = 4.

Example 4.3.10. Let’s find a splitting field E of f(x) = x3 − 5 ∈ Q[x] and find [E : Q]. Wehave

x3 − 5 = (x− 3√

5)(x− ω 3√

5)(x− ω2 3√

5), ω = e2πi/3.

So E = Q( 3√

5, ω 3√

5, ω2 3√

5) = Q( 3√

5, ω 3√

5). Consider the extensions Q ⊆ Q( 3√

5) ⊆Q( 3√

5, ω 3√

5). We have[Q(

3√

5) : Q] = deg(x3 − 5) = 3.

Since Q( 3√

5) ( Q( 3√

5, ω 3√

5), we have

2 ≤ [Q(3√

5, ω3√

5) : Q(3√

5)] = degm(x), (4.2)

where m(x) is the minimal polynomial of ω 3√

5 in Q( 3√

5)[x]. Now, ω 3√

5 is a root of x3 − 5.Recalling the difference of cubes factorization x3 − a3 = (x− a)(x2 + ax+ a2), we have

x3 − 5 = (x− 3√

5)(x2 +3√

5x+ (3√

5)2).

Since ω 3√

5 is not a root of the first factor, it is a factor of the second. Thus, the degree ofthe minimal polynomial is at most two. Since, by (4.2), it is at least two, it must be equalto two (and m(x) = x2 + 3

√5x+ ( 3

√5)2). Thus

[E : Q] = [Q(3√

5, ω3√

5) : Q(3√

5)][Q(3√

5) : Q] = 2 · 3 = 6.

The above example illustrates the fact that the degree of a splitting field of a polynomial(over the base field) can be strictly greater than the degree of the polynomial.


Example 4.3.11. Let’s find a splitting field E ⊇ Q for f(x) = x3 − 1 and find [E : Q]. Wehave

x3 − 1 = (x− 1)(x− ω)(x− ω2), ω = e2πi/3.

Thus E = Q(1, ω, ω2) = Q(ω). Now x3 − 1 = (x − 1)(x2 + x + 1). Since ω is not a root ofthe first factor, it is a root of the second. Because the roots of x2 + x + 1 are not rational,this polynomial is irreducible over Q. Thus x2 + x + 1 is the minimal polynomial of ω overQ. Hence [Q(ω) : Q] = 2.

A natural question to ask is whether or not splitting fields are unique. Our next goal inthis section is to show that they are unique up to isomorphism.

Suppose F ⊆ R and F ⊆ R, where F and F are fields that are subrings of the ringsR and R, respectively. If σ : F → F is a ring homomorphism, then a ring homomorphismσ : R → R is said to extend σ if σ(a) = σ(a) for all a ∈ F. In other words, the followingdiagram commutes (where the vertical maps are inclusions):

Rσ−−−→ Rx x

F σ−−−→ F

Example 4.3.12. Suppose σ : F→ F is an isomorphism of fields. For f(x) = a0 + a1x+ · · ·+anx

n ∈ F[x], define fσ(x) := σ(a0) + σ(a1)x + · · · + σ(an)xn ∈ F[x]. Then the mappingF[x]→ F[x], f(x) 7→ fσ(x) is a ring isomorphism that extends σ (exercise).

Suppose p(x) is a monic irreducible polynomial in F[x]. Then pσ(x) is monic in F[x] (sinceσ(1) = 1) and is irreducible (exercise). Consider the composition of ring homomorphisms

F[x]f(x)7→fσ(x)−−−−−−−→ F[x] � F[x]/〈pσ(x)〉,

where the second map is the quotient homomorphism. It is easy to show that the kernel ofthis composition is 〈p(x)〉. Thus, by the First Isomorphism Theorem (Theorem 1.4.15), wehave an induced ring isomorphism

ϕ : F[x]/〈p(x)〉 → F[x]/〈pσ(x)〉, ϕ(f(x) + 〈p(x)〉) = fσ(x) + 〈pσ(x)〉 for f(x) ∈ F[x]. (4.3)

Theorem 4.3.13. Suppose σ : F→ F is an isomorphism of fields and p(x) ∈ F[x] is monicand irreducible. Let u be a root of p(x) in an extension field E ⊇ F and let v be a root ofpσ(x) in an extension field E ⊇ F. Then there is a unique isomorphism

F(u)→ F(v), f(u) 7→ fσ(v), f ∈ F[x],

that extends σ and maps u to v.

Proof. Since p(x) is monic and irreducible, it is the minimal polynomial of u over F. Asin the proof of Theorem 4.2.19, we have a ring isomorphism F(u) ∼= F[x]/〈p(x)〉 given by

4.3. Splitting fields 79

f(u) 7→ f(x) + 〈p(x)〉. Similarly, F[x]/〈pσ(x)〉 ∼= F(v). Composing these isomorphisms withthe isomorphism (4.3), we have

F(u) → F[x]/〈p(x)〉 → F[x]/〈pσ(x)〉 → F(v),f(u) 7→ f(x) + 〈p(x)〉 7→ fσ(x) + 〈pσ(x)〉 7→ fσ(v).

Thus the composite map F(u) → F(v), f(u) 7→ fσ(v) is an isomorphism. Taking f(x) = x,we see that this map sends u to v. Considering elements of F as a constant polynomials, wealso see that the map extends σ.

Example 4.3.14. Let’s consider Theorem 4.3.13 in the special case that F = F. Let p(x) be amonic irreducible polynomials of degree n and let u and v be two roots of p(x) in extensionfields E ⊇ F and E ⊇ F respectively. Then the map

σ : F(u)→ F(v), σ(a0 + a1u+ · · ·+ an−1un−1) = a0 + a1v + · · ·+ an−1v

n−1,

is an isomorphism that fixes F (i.e. σ(a) = a for all a ∈ F) and maps u to v. For exampleQ( 3√

5) ∼= Q(e2πi/3 3√

5) since 3√

5 and e2πi/3 3√

5 are both roots of the irreducible polynomialx3 − 5 ∈ Q[x].

The following theorem shows that splitting fields are unique.

Theorem 4.3.15. Suppose σ : F → F is an isomorphism of fields and let f(x) ∈ F[x] be anonconstant polynomial. If E ⊇ F is a splitting field for f(x) and E ⊇ F is a splitting fieldfor fσ(x), then there is an isomorphism E→ E that extends σ.

Proof. We prove the result by induction on n = deg f(x) = deg fσ(x). If n = 1, then E = Fand E = F, so σ itself has the desired properties.

Now assume n > 1. Let p(x) be a monic irreducible divisor of f(x). Let u ∈ E be a rootof f(x) and let v ∈ E be a root of pσ(x). By Theorem 4.3.13, σ extends to an isomorphismτ : F(u)→ F(v) such that τ(u) = v.

E ψ−−−→ Ex xF(u)

τ−−−→ F (v)x xF σ−−−→ F

Writef(x) = a(x− u)g(x), g(x) ∈ F(u)[x], deg g(x) = n− 1.

Then we have

fσ(x) = f τ (x) = σ(a)(x− v)gτ (x), gτ (x) ∈ F(v)[x], deg gτ (x) = n− 1.

It follows that E is a splitting field for gτ (x) (since it is a splitting field for fσ(x)). We alsohave that E is a splitting field for g(x) (since it is a splitting field for f(x)). Thus, by theinduction hypothesis, there is an isomorphism ψ : E→ E that extends τ and hence extendsσ.


Recall that the Fundamental Theorem of Algebra states that every polynomial in C[x]splits. In the remainder of this section, we discuss this property of fields.

Theorem 4.3.16. If F is a field, then the following conditions are equivalent:

(a) Every nonconstant polynomial in F[x] has a root in F.

(b) Every irreducible polynomial in F[x] has degree one.

(c) Every nonconstant polynomial in F[x] splits in F[x].

(d) If F ⊆ E is an algebraic extension, then E = F.

Proof. (a) ⇒ (b): This follows immediately from the Factor Theorem (Theorem 2.1.21).(b) ⇒ (c): Since F[x] is a UFD, every polynomial is a product of irreducibles, hence a

product of degree one polynomials by (b).(c) ⇒ (d): If u ∈ E, let f(x) ∈ F[x] be a nonzero polynomial such that f(u) = 0 (such a

polynomial exists since F ⊆ E is an algebraic extension). Then f(x) in nonconstant and so,by (c),

f(x) = a(x− b1)(x− b2) · · · (x− bn) for some a, b1, . . . , bn ∈ F.

Since f(u) = 0, we have u = bi for some i. Thus u ∈ F.(d) ⇒ (a): Suppose f(x) ∈ F[x] is nonconstant. By Theorem 4.3.1, we can find a root u

of f(x) in some extension field E. Thus F ⊆ F(u) is an algebraic extension (since it is finiteby Theorem 4.2.19). Thus F(u) = F by (d) and so u ∈ F.

Definition 4.3.17 (Algebraically closed). A field is algebraically closed if it satisfies theconditions of Theorem 4.3.16.

Example 4.3.18. The field C of complex numbers is algebraically closed.

Proposition 4.3.19. Suppose F ⊆ E is a field extension. Let

A = {u ∈ E | u is algebraic over F}.

Then A is a subfield of E and so is an algebraic extension of F.

Proof. For any u, v ∈ A, the field F(u, v) is a finite field extension of F by Theorem 4.2.30.Thus, all elements of F(u, v) are algebraic over F by Theorem 4.2.3. So F(u, v) ⊆ A. Inparticular, u+ v, uv,−u ∈ A. Furthermore, if u 6= 0, then u−1 ∈ A. So A is a field.

Definition 4.3.20 (Algebraic closure). If F ⊆ E is a field extension, then the field A inProposition 4.3.19 is called the algebraic closure of F in E.

If F is a field, a field extension F ⊆ E is called an algebraic closure of F if E is an algebraicextension of F and E is algebraically closed. (Note the subtle difference in terminology –here the closure is not in any larger field.)

Proposition 4.3.21. Suppose F ⊆ E is a field extension and E is algebraically closed. Thenthe algebraic closure of F in E is itself algebraically closed.

4.4. Finite fields 81

Proof. Let A = {u ∈ E | u is algebraic over F} be the algebraic closure of F in E. Suppose

f(x) = a0 + a1x+ · · ·+ anxn

is a nonconstant polynomial in A[x]. Then f(x) ∈ E[x] and so f(x) has a root u in E.Let K = F(a0, a1, . . . , an). Then [K : F] is a finite extension by Theorem 4.2.30, since eachai ∈ A, and so is algebraic over F. Furthermore [K(u) : K] is finite because u is algebraicover K. Thus [K(u) : F] is finite and hence algebraic. Since u ∈ K(u), it follows that uis algebraic over F. Thus every noncontant polynomial in A[x] has a root in A. So A isalgebraically closed.

Definition 4.3.22 (Field of algebraic numbers). The field of algebraic numbers is

A = {u ∈ C | u is algebraic over Q}.

By Proposition 4.3.21, A is an algebraic closure of Q.

Example 4.3.23. We see from Example 4.2.23 that A is an algebraic extension of Q that isnot finite.

Theorem 4.3.24. Every field F has an algebraic closure E ⊇ F. Furthermore, if E ⊇ F isanother algebraic closure, then there is an isomorphism σ : E→ E that fixes F.

The proof of Theorem 4.3.24 requires Zorn’s Lemma and will be omitted.

Exercises.

4.3.1. Prove that the map f(x) 7→ fσ(x) defined in Example 4.3.12 is a ring isomorphism.

4.4 Finite fields

In this final section, we look at finite fields. In particular, we will completely classify themup to isomorphism.

Suppose F is a finite field. Then its additive group is finite and thus, by Lemma 1.1.19, itscharacteristic is positive. Thus, by Proposition 1.2.12, its characteristic is a prime numberp. Therefore, by Theorem 1.4.20, Zp ⊆ F (the map ι : Zp ↪→ F, ι(k) = k · 1 is an injectivering homomorphism).

Theorem 4.4.1. If F is a finite field, then |F| = pn for some n, where p = charF.

Proof. Since F is a finite field, [F : Zp] is finite. If n = [F : Zp], then F ∼= (Zp)n as a vectorspace and the result follows.

Theorem 4.4.2. Let p be a prime number and let F ⊇ Zp be a splitting field of f(x) = xpn−x.

Then |F| = pn.


Proof. Let S = {a ∈ F | apn − a = 0}. We will prove that Zp ⊆ S and S is a field (henceS = F). Fermat’s Little Theorem implies that for any integer k, kp ∼= k mod p. Thus, forall k ∈ Zp, we have

kpn

=(kp

n−1)p

= kpn−1

= kpn−2

= · · · = k.

Therefore Zp ⊆ S.Now,

a, b ∈ S =⇒ (a+ b)p =

p∑i=0

(pi

)aibp−i = ap + bp.

Therefore

(a+ b)pn

= ((a+ b)p)pn−1

= (ap + bp)pn−1

=(ap

2

+ bp2)pn−2

= · · · = apn

+ bpn

= a+ b.

Thus a+ b ∈ S. So S is closed under addition. Also

a ∈ S =⇒ (−a)pn

= (−1)pn

apn

= (−1)pn

a = −a,

since if p > 2, then p is odd, and if p = 2, then −1 = 1.We also have

a, b ∈ S =⇒ (ab)pn

= apn

bpn

= ab =⇒ ab ∈ S,and

a ∈ S =⇒(

1

a

)pn=

1

apn=

1

a=⇒ 1

a∈ S.

Thus S is a field and so S = F.It remains to find |F| = |S|. In F[x] we have

f(x) = (x− a1)m1(x− a2)m2 · · · (x− ar)mr , r = |S|, a1, . . . , ar distinct.

If we can show that m1 = m2 = · · · = mr = 1, then it will follow that r = pn. Assume,towards a contradiction, that mi > 1 for some i. Then f(x) = (x − ai)

2g(x) for someg(x) ∈ F[x]. Thus f(ai) = 0 and

f ′(x) = 2(x− ai)g(x) + (x− ai)2g′(x)

(using the usual rules of differentiation: the derivative is F-linear and the derivative of xn

is nxn−1). Thus f ′(ai) = 0. However, since f(x) = xpn − x, we have f ′(x) = −1, which is

never zero. This contradiction completes the proof.

Theorem 4.4.3. If F is a field with pn elements, where p = charF, then F is a splitting fieldof f(x) = xp

n − x.

Proof. We have |F×| = pn − 1. Thus, by Lagrange’s Theorem, we have apn−1 = 1 for all

a ∈ F×. Thus apn

= a for all a ∈ F and so every element of F is a root of f(x).Since deg f(x) = pn, f(x) has at most pn roots in F. Thus

pn = |F| ≤ # of roots of f(x) ≤ pn.

Therefore, F is the set of all roots of f(x). In particular, it is a splitting field of f(x).

4.4. Finite fields 83

Corollary 4.4.4. If p is a prime number and n ≥ 1, then, up to isomorphism, there existsa unique field with pn elements.

Proof. This follows from Theorems 4.4.3 and 4.3.15.

Definition 4.4.5 (Galois field). The unique field with pn elements is called the Galois fieldof order pn and is denoted GF(pn).

Theorem 4.4.6. (a) If K ⊆ GF(pn) is a subfield, then K ∼= GF(pm), where m | n.

(b) If m | n, then GF(pn) has a unique subfield of pm elements (which is therefore isomor-phic to GF(pm)).

Proof. (a) We have

[GF(pn) : K][K : Zp] = [GF(pn) : Zp] = n =⇒ [K : Zp] | n.

(b) If K ⊆ GF(pn) is a subfield with |K| = pm, then K is the splitting field of xpm − x by

Theorem 4.4.3 and hence is unique. It remains to prove existence. Note that

apm

= a =⇒ ap2m

= (apm

)pm

= apm

= a.

Continuing in this manner, we see that apn

= a. Thus xpm − x splits in GF(pn). Set

E = {u ∈ GF(pn) | u is a root of xpm − x}.

Then |E| = pm because the roots of xpm − x are distinct. Furthermore, E is a field as in the

proof of Theorem 4.4.2.

Example 4.4.7. The lattice of subfields of GF(p24) is as follows:

GF(p24)

GF(p12)

GF(p6)

GF(p3)

GF(p8)

GF(p4)

GF(p2)

GF(p)

Theorem 4.4.8. Suppose F is a field. If G is a finite subgroup of F×, then G is cyclic.

Proof. Let G be a finite subgroup of F× with n elements. By the Fundamental Theorem ofFinite Abelian Groups (from MAT 2143),

G ∼= Zpe11 × · · · × Zpekk .


for some (not necessarily distinct) primes pi such that n = pe11 · · · pekk . Let m be the least

common multiple of pe11 , . . . , pekk . Then G contains an element of order m (the product of

generators of the cyclic factors above). Since every α in G satisfies αm = 1, we see thatevery element of G is a root of xm − 1. Since xm − 1 has at most m roots in F, n ≤ m. Onthe other hand, we know that m ≤ |G|; therefore, m = n. Thus, G contains an element oforder n and must be cyclic.

Corollary 4.4.9. The multiplicative group of all nonzero elements of a finite field is cyclic.

Definition 4.4.10 (Primitive element, primitive root of unity). If F is a finite field, thena generator for F× is called a primitive element of F. If a (possibly infinite) field F has amultiplicative subgroup G of order n, then a generator of G (which exists by Theorem 4.4.8)is called a primitive nth root of unity in F. For example, e2πi/n is a primitive nth root ofunity in C for each n ≥ 2.

Index

ACCP, 58algebraic closure, 80algebraic element, 69algebraic extension, 69algebraic numbers, 81algebraically closed, 80ann(X), 22annihilator, 22artinian ring, 62ascending chain condition on ideals, 62ascending chain condition on principal ideals,

58associates, 54automorphism, 25

inner, 25

basis, 68

c(f(x)), 60center, 8centralizer, 10characteristic, 7Chinese Remainder Theorem, 27coefficient, 33commutative ring, 4constant coefficient, 34constant polynomial, 34content, 60cyclotomic polynomial, 45, 47

degreeof a polynomial, 34of an element in a field extension, 70

descending chain condition on ideals, 62difference of elements in a ring, 6dimension, 68direct product of rings, 5division, 54

of polynomials, 35Division Algorithm, 35division algorithm, 65division ring, 11divisor function, 65domain, 11

Eisenstein Criterion, 45endomorphism, 25epimorphism, 25equality of polynomials, 33euclidean domain, 36, 65euclidean function, 36, 65euclidean valuation, 65evaluation, 37Evaluation Theorem, 36extending a ring homomorphism, 78extension, 69

F(X,R), 5Factor Theorem, 37factorization, 54

proper, 43trivial, 54

field, 11field extension, 69field generated by elements, 70field of fractions, 15field of quotients, 15finite dimensional, 68finite extension, 69First Isomorphism Theorem, 26Frobenius homomorphism, 25Fundamental Theorem of Algebra, 80

Galois field, 83Gauss’ Lemma, 43, 60gaussian integers, 8

85

86 Index

are a euclidean domain, 65gcd, 57general ring, 4general ring homomorphism, 24GF(pn), 83greatest common divisor, 57

of polynomials, 46group of units, 6

homomorphismof general rings, 24of rings, 24

ideal, 16zero, 17

idempotent, 7identity element, 4image, 25indeterminate, 32infinite dimensional, 68inner automorphism, 25integral domain, 11irreducible, 55

polynomial, 41isomorphic rings, 8, 25isomorphism, 8, 25

kernel, 25Kronecker’s Theorem, 76

lcm, 57leading coefficient, 34least common multiple, 57left ideal, 23Lie algebra, 5linear combination, 68linearly dependent, 68linearly independent, 68

maximal ideal, 19minimal polynomial, 70Modular Irreducibility Test, 44monic polynomial, 34monomorpihsm, 25Multiplication Theorem, 72multiplicative inverse, 5

multiplicity of a root, 38

N, 3nilpotent, 7noetherian ring, 62nonassociative ring, 5

opposite ring, 9

PID, 63polynomial, 33

function, 33polynomial ring, 5prime element, 55prime ideal, 18primitive element, 84Primitive Element Theorem, 74primitive polynomial, 60primitive root of unity, 84principal ideal, 17principal ideal domain, 63proper factorization, 43proper ideal, 16

quaternions, 13

R×, 5rational function, 15Rational Roots Theorem, 38reducible polynomial, 41reduction modulo p, 44Remainder Theorem, 37ring, 4

automorphism, 25endomorphism, 25epimorphism, 25homomorphism, 24isomorphism, 25monomorphism, 25of polynomials, 33

ring isomorphism, 8root of a polynomial, 38

scalar multiplication, 67Second Isomorphism Theorem, 31simple ring, 17skew field, 11

Index 87

span, 68split, 76splitting field, 76subfield, 69subring, 7Subring Test, 7subspace, 67subtraction in a ring, 6

Third Isomorphism Theorem, 31transcendental element, 69

UFD, 56unique factorization domain, 56unit, 5unity, 4

vector addition, 67vector space, 67

Zn, 5zero divisor, 11zero ideal, 17zero polynomial, 34zero ring, 5

Bibliography

[Jud12] Thomas W. Judson. Abstract algebra. 2012 annual edition, 2012. Available athttp://abstract.ups.edu/index.html.

[Nic12] W. Keith Nicholson. Introduction to abstract algebra. Wiley-Interscience [JohnWiley & Sons], Hoboken, NJ, fourth edition, 2012.

[Roy] Damien Roy. MAT 3143 – Linear Algebra II, Lecture notes (transla-ted by A. Savage). Available at http://alistairsavage.ca/mat3141/notes/

MAT3141-LinearAlgebraII.pdf.

[Sav] Alistair Savage. MAT 2141 – Linear Algebra I, Lecture notes. Available at http:

//alistairsavage.ca/mat2141/notes/MAT2141-LinearAlgebraI.pdf.

88


http://alistairsavage.ca/mat3141/notes/MAT 3141 - Linear Algebra II.pdf

http://alistairsavage.ca/mat3141/notes/MAT 3141 - Linear Algebra II.pdf

http://alistairsavage.ca/mat2141/notes/MAT 2141 - Linear Algebra I.pdf

http://alistairsavage.ca/mat2141/notes/MAT 2141 - Linear Algebra I.pdf

Documents

Ring Theory Course notes for MAT 3143 (Winter 2013)