Lecture Notes

in

Galois Theory

Gunnar Traustason(2nd semester 2017)

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1 The polynomial ring K[x]

I. Some general recap on rings

Definition. A ring is a set equipped with two binary operations + (addition) and ·(multiplication) such that

a) R is an abelian group under addition:

1) (a+ b) + c = a+ (b+ c) for all a, b, c ∈ R2) a+ b = b+ a for all a, b ∈ R3) There exists 0 ∈ R such that a+ 0 = a for all a ∈ R4) For all a ∈ R there exists an element −a ∈ R such that a+ (−a) = 0.

b) R is associative with 1 under multiplication:

5) (a · b) · c = a · (b · c) for all a, b, c ∈ R6) There exists 1 ∈ R such that a · 1 = 1 · a = a for all a ∈ R.

c) R satisfies the distributive laws:

7) a · (b+ c) = a · b+ a · c for all a, b, c ∈ R8) (b+ c) · a = b · a+ c · a for all a, b, c ∈ R.

Remarks. 1) Recall that 0, 1 are unique, called the additive and multiplicative iden-tities.2) Recall that a · 0 = 0 · a = 0 and a(−b) = (−a)b = −ab.

Definition. A ring R is said to be commutative if ab = ba for all a, b ∈ R.

Definition A subset S of a ring R is said to be a subring, if 1R ∈ S and a+ b, ab,−a ∈ Swhenever a, b ∈ S. We use the notation S ≤ R for ‘S is a subring of R’.

Remark. Notice that if S ≤ R, then 0R = 1R + (−1R) ∈ S.

Definition. 1) A nonempty subset of a commutative ring R is an ideal if −a, a+b, ra ∈ Iwhenever a, b ∈ I and r ∈ R. We then sometimes write I �R for ‘I is an ideal of R’.

2) An ideal I of a commutative ring is a principal ideal if I = Ra for some a ∈ R.

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Definition. Let R be a commutative ring.

1) R is an integral domain (ID) if R 6= {0} and satisfies

ab = 0⇒ a = 0 or b = 0.

2) R is a principal ideal domain (PID) if R is an integral domain and every ideal of R isa principal ideal.

Quotient rings. Let R be a commutative ring with an ideal I. The quotient ringR/I is

R/I = {a = a+ I : a ∈ R}where the addition and multiplication are given by

(a+ I) + (b+ I) = a+ b+ I, (a+ I)(b+ I) = ab+ I.

R/I then becomes a ring with 0 = 0 + I and 1 = 1 + I as the additive and multiplicativeidentities. The additive inverse of a+ I is −a+ I.

The characteristic of a ring. The characteristic, charR, of R is a non-negative integerdefined as follows:

If there exists a postive integer n such that

n1R = 1R + · · ·+ 1R︸ ︷︷ ︸n

= 0R,

then charR is the smallest such positive integer. If no such positive integer exists thencharR = 0.

Remarks. (1) If charR = 1 then 1R = 0R and a = 1R · a = 0R · a = 0R for alla ∈ R. Thus R = {0} is the only ring with characteristic 1.(2) If charR = n > 0 then na = n1R · a = 0R · a = 0R for all a ∈ R. That is nR = 0.

Example. char Q = 0 and the characteristic of Zn = Z/nZ is n.

Definition. Let R, S be rings. A map φ : R→ S is said to be a ring homomorphism if:

φ(a+ b) = φ(a) + φ(b)

φ(ab) = φ(a) · φ(b)

φ(1R) = 1S

If φ is bijective, then we say that φ is a ring isomorphism. We often write R ∼= S for ‘Risomorphic to S’.

Remarks. Recall that φ(0R) = 0S and φ(−a) = −φ(a).

First ring isomorphism theorem. If φ : R→ S is a ring homomorphism the imφ ≤ S,kerφ�R and

imφ ∼= R/kerφ.

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Here imφ = {φ(a) : a ∈ R} and kerφ = {a ∈ R : φ(a) = 0}.

Remark. Recall that φ is injective if and only if kerφ = {0}.

The smallest subring of R. A subring S of a given ring R contains 1R and thusZ1R = {n1R : n ∈ Z}. Notice that we we have a ring homomorphism

φ : Z→ R, n 7→ n1R.

In particular Z1R = imφ is a subring of R and clearly the smallest subring. There aretwo possible scenarios.

If charR = n > 0 then kerφ = nZ and by the first isomorphism theorem we haveZ1R = imφ ∼= Z/nZ = Zn.

If charR = 0 then kerφ = {0} and φ is injective. Hence in this case Z1R = imφ ∼= Z.

II. Some basic facts about fields

Definition. A commutative ring R 6= {0} is a field if for every 0 6= a ∈ R there isb ∈ R such that ab = 1.

Remarks. (1) The element b is unique and is called the multiplicative inverse of a.It is normally denoted a−1 or 1/a.(2) One could also say that a commutative ring R 6= {0} is a field if R \ {0} is a group.(3) Every field is an ID. If ab = 0 and a 6= 0 then b = a−1(ab) = a−1 · 0 = 0.

Definition. A subset K of a field F is a subfield if it is a subring that is furthermoreclosed under taking inverses of non-zero elements. Thus K is a subfield if 1F ∈ K anda+ b, ab,−a ∈ K whenever a, b ∈ K and if furthermore a−1 ∈ K whenever 0 6= a ∈ K.

Remark. In fact K is a subfield of F if and only if K is a field in its own right withrespect to the induced addition and multiplication from F.

Remark. Notice that in the last remark, we do not have do include the requirementthat 1F ∈ K. The reason for this is that we are already assuming that K is a field andfrom

1K · 1F = 1K = 1K · 1K,

cancellation by 1K 6= 0 gives 1F = 1K ∈ K. However without the assumption that K isalready a field this argument doesn’t work. For example {0F} is not a subfield of F. Hencethe requirement 1F ∈ K is needed in our definition of a subfield above.

Proposition 1.1 If R is an integral domain then charR is either 0 or a prime number.

Proof As R 6= {0} we have that charR 6= 1. We argue by contradiction and supposethat charR = rs where 1 < r, s < charR. But then

r1R · s1R = rs1R = 0R

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whereas r1R, s1R are non-zero. This contradicts the fact that R is an ID. 2

Remark. As every field is an ID the last proposition holds in particular for fields.

Definition. Let K be a field. The prime subfield of K is the intersection of all thesubfields of K

Remark. It is not difficult to see that the intersection of all the subfields is a sub-field as well. Thus the prime subfield of K is the unique smallest subfield of K.

From last proposition and the discussion at the end of Section I, we thus know thatif K is a field, whose characteristic is a prime number p, then its smallest subring Z1K isisomorphic to Zp that is a field and thus the prime subfield of K.

We are thus only left with the situation when char K = 0. In this case we have seenthat the smallest subring Z1K of K is isomorphic to Z that is not a field. Here the primesubfield is

{ n1K

m1K: n,m ∈ Z where m 6= 0}

that is isomorphic to the field Q. Thus every field of characteristic 0 contains a copy ofQ as a subfield. (See exercise sheet 1 for the details).

Definition. We say that a field K is a prime field if it has no proper subfields.

Remark. From the discussion above we see that up to isomorphisms the prime fields areQ and Zp, p a prime number.

III. Factorization in the polynomial ring K[x]

Let K be any field and consider the ring K[x] of all polynomials

f = a0 + a1x+ · · ·+ anxn

over K where n ≥ 0 and a0, . . . , an ∈ K.

Definition. If f is a non-zero polynomial then the degree of f , deg f, is the largestnon-negative integer m such that am 6= 0. If f = 0 then deg f = −∞.

Remark. Recall that deg fg = deg f + deg g.

More generally we can consider polynomial rings over any commutative ring. Let Rand S be commutative rings and suppose we have a homomorphism φ : R → S. Wecan use this homomorphism to extend it to an induced homomorphism φ∗ : R[x]→ S[x].Where

φ∗(a0 + a1x+ · · ·+ anxn) = φ(a0) + φ(a1)x+ · · ·+ φ(an)xn.

(See exercise sheet 1 for the details).

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Definition. Let R be an integral domain and let Let f, g ∈ R. We say that f di-vides g in R, and write f |g, if there exists h ∈ R such that g = fh. We then also say thatf is a factor of g in R.

Remark. Notice that f |g in R if and only if Rg ⊆ Rf .

Definition. Let R be an integral domain. An element u ∈ R is a unit if it has amultiplicative inverse in R, i.e. if there exists v ∈ R such that uv = 1.

Remark. Recall that the set of units of K[x] is the group K∗ = K \ {0} of nonzeroelements of the field K.

Definition. Let R be an integral domain. Let r be a nontrivial element of R that isnot a unit.

(1) We say that r is irreducible in R if for any r, s ∈ R

r = st⇒ either s or t is a unit in R.

(2) We say that r is a prime in R, if for any s, t ∈ R

r|st (in R)⇒ r|s or r|t (in R).

Remark. When R = K[x] we know that K \ {0} is the set of units. We can thus restatethe definition of irreducibility as follows. Let f be a polynomial in K[x] where deg f ≥ 1.We have that f is irreducible in K[x] if for g, h ∈ K[x]

f = gh⇒ deg g = 0 or degh = 0.

Remark. From Algebra 2B you know that K[x] is a principal ideal domain and that as aresult the properties of being irreducible and a prime are equivalent. You also know thatK[x] is a unique factorization domain.

Recall that a polynomial 0 6= f ∈ K[x] is said to be monic if the highest degree nonzerocoefficient is 1.

Remark. Notice that if K is a field then all polynomials ax + b ∈ K[x] of degree 1(and thus with a 6= 0) are irreducible.

Unique factorization in K[x]. Let 0 6= f ∈ K[x]. Then

f = uf1f2 · · · fn

where n ≥ 0, u ∈ K \ {0} and f1, . . . , fn irreducible monic polynomials in K[x]. Further-more u is unique and f1, . . . , fn are also unique up to order.

Examples. (1) 4x4 − 16 = 4(x2 − 2)(x2 + 2) is the unique factorization of 4x4 − 4in Q[x].(2) The unique factorization in R[x] is

4x4 − 16 = 4(x−√

2)(x+√

2)(x2 + 2).

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(3) The unique factorization in C[x] is

4x4 − 16 = 4(x−√

2)(x+√

2)(x− i√

2)(x+ i√

2).

Division with remainder. If 0 6= f, g ∈ K[x]. Then there exist r, s ∈ K[x] such that

f = gs+ r

and deg r < deg g.

Definition. We say that an element t ∈ K is a root of the polynomial f = a0 + a1x +· · ·+ anx

n in K[x] iff(t) = a0 + a1t+ · · ·+ ant

n = 0.

Lemma 1.2 Let K be a field and let f ∈ K[x]. An element t ∈ K is a root of f if andonly if x− t divides f in K[x].

Proof Suppose f(t) = 0. Using division with remainder we have

f = (x− t)g + r

where r is of degree less than 1 and thus in K. Now r = (t − t)g(t) + r = f(t) = 0 andthus f = (x− t)g.

Conversely, if x − t divides f , say f = (x − t)g, then f(t) = (t − t)g(t) = 0 and t isa root of f . 2

Greatest common divisor and lowest common multiple. Let 0 6= f, g ∈ K[x].

(1) An element d ∈ K[x] is called a greatest common divisor (gcd) of f and g if:

(i) d|f and d|g(ii) if r|f and r|g then r|d.

2) An element c ∈ K[x] is called a lowest common multiple (lcm) of f and g if:

(i) f |c and g|c(ii) if f |r and g|r then c|r.

Remark. (1) If d1, d2 are both greatest common divisors of f and g, then d2 = ud1

for some 0 6= u ∈ K. In particular there is a unique monic polynomial that is the greatestcommon divisor of f and g.

(2) If c1, c2 are both lowest common multiples of f and g, then c2 = uc1 for some0 6= u ∈ K. In particular there is a unique monic polynomial that is the lowest com-mon multiple of f and g.

(3) From Algebra 2B you know that

K[x]f + K[x]g = K[x]d,

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andK[x]f ∩K[x]g = K[x]c.

Remark. In particular there exists polynomials r, s ∈ K[x] such that

rf + sg = d.

IV. Criteria for irreducibility in Q[x]

Remark Everything we do in this in this section works also if we replace Z by anyprincipal ideal domain R and Q by the field of fractions F for R.

Lemma 1.3 (Gauss). Let p be a prime in Z and f, g ∈ Z[x]. Then (in Z[x])

p|fg ⇒ p|f or p|g.

Proof Suppose that

fg = a0 + a1x+ · · ·+ anxn

f = b0 + b1x+ · · · brxr

g = c0 + c1x+ · · · csxs.

We argue by contradiction and suppose that p divides neither f nor g. Let bi, cj be theearliest coefficients of f and g respectively that are not divisible by p. Notice that

ai+j = b0ci+j + b1ci+j−1 + · · ·+ bi−1cj+1 + bicj + bi+1cj−1 + · · ·+ bi+jc0.

Since p divides ai+j, b0, . . . , bi−1, c0, . . . , cj−1 it follows that p divides bicj. But as p is aprime we then get the contradiction that p divides either bi or cj. 2

Theorem 1.4 Let f ∈ Z[x] be a polynomial where deg f ≥ 1. If f is irreducible in Z[x],then f is also irreducible in Q[x].

Proof We argue by contradiction and suppose that f is irreducible over Z but not overQ. It follows that

f = gh

for some polynomials g, h ∈ Q[x] where g, h are of both of degree at least 1. By multiplyingby a suitable positive integer

n = p1 · · · pr,

where p1, . . . , pr are prime numbers, we get an equation

p1 · · · prf = g0h0

where g0, h0 ∈ Z[x]. We now use last lemma. As pr divides g0h0 it must divide either g0

or h0. Cancelling out this prime on both sides we get

p1 · · · pr−1f = g1h1

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for some polynomials g1, h1 over Z. (Notice that the degrees of g1, h1 are the same as thedegrees of g0, h0 respectively). By repeating this we can eliminate the primes one by oneand we arrive at

f = grhr

where gr, hr are polynomials over Z (and where gr, hr have the same degrees as g, h). Butthis contradicts the assumption that f was irreducible over Z.

Remark. The converse is not true. For example the polynomial f = 2x is irreducible inQ[x] but is not irreducible in Z[x] as 2 is not a unit in Z. From the proof of last resultwe can also deduce the following.

Corollary 1.5 Let f ∈ Z[x] and suppose that f = g1g2 with g1, g2 ∈ Q[x]. Then thereexist polynomials h1, h2 ∈ Z[x] of same degrees as g1, g2 such that

f = h1h2.

Theorem 1.6 (Eisenstein irreducibility criterion) Let

f = a0 + a1x+ · · ·+ anxn

be a polynomial over Z of degree n ≥ 2. Suppose there is a prime p ∈ Z such that

(1) p does not divide an(2) p divides a0, a1, . . . , an−1

(3) p2 does not divide a0.

Then f is irreducible over Q.

Proof We argue by contradiction and suppose that f is not irreducible over Q. We thenhave

a0 + a1x+ · · ·+ anxn = (b0 + · · ·+ brx

r)(c0 + · · ·+ csxs)

where r + s = n, 1 ≤ r, s < n. By Corollary 1.5 we can assume that all the coefficientsare in Z.

As a0 = b0c0 is divisible by p but not by p2, it follows that exactly one of b0, c0 is di-visible by p. W.l.o.g. we can suppose that

p divides b0 but p does not divide c0. (∗)

As p does not divide an = brcs it follows that p does not divide br. Let i be the smallestpositive integer such that p does not divide bi. Notice that 0 < i ≤ r < n. Then

ai = b0ci + b1ci−1 + · · ·+ bi−1c1 + bic0

and since p divides ai, b0, . . . , bi−1 but p does not divide bi, it follows that p divides c0.This however contradicts (*).

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Example. The polynomial f = 29x5 + 5

3x4 + x3 + 1

3is irreducible over Q iff 9f =

2x5 + 15x4 + 9x3 + 3 is irreducible over Q. By Eisenstein, using p = 3, the latter is irre-ducible.

We now turn to yet another irreducibility criteria. For a prime number p in Z we considerthe homomorphism

φ : Z→ Zp, a 7→ a = a+ Zp.

Theorem 1.7 Let p be a prime number and let

f = a0 + a1x+ · · ·+ anxn ∈ Z[x]

where deg(f) = n ≥ 1 and where p does not divide an. Consider the correspondingpolynomial

φ∗p(f) = a0 + a1x+ · · ·+ anxn

over Zp. Then if φ∗pf is irreducible in Zp[x], f is irreducible in Q[x].

Proof Argue by contractions and suppose that there exist g, h ∈ Q[x] of degrees at least1 and where f = gh. By Corollary 1.5 we can assume that g, h ∈ Z[x]. Then

φ∗p(f) = φ∗p(g) · φ∗p(h)

is a factorization of φ∗p(f) in Zp[x] (notice that φ∗p(f), φ∗p(g), φ∗p(h) have same degrees asf, g, h) that contradicts our assumption that φ∗p(f) was irreducible in Zp[x]. 2.

Example. Show that f = x3 + 11x+ 100 is irreducible over Q.

Solution. We have that g = φ∗3(f) = x3 + 2x + 1 ∈ Z3[x] is irreducible as it has noroot (and thus no linear factor). In fact g(0) = g(1) = g(2) = 1. By last proposition itthus follows that f is irreducible over Q.

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2 Field Extensions

I. Field extensions and their associated vector space and group

Definiton. A field extension is a pair of two fields K,F where K is contained in F.We will use

K ⊆ F

to denote the field extension.

The associated vector space. For a field extension K ⊆ F we can view the largerfield F as a vector space over the subfield K. Here the vector space addition is the sameas the field addition in F and the scalar multipliation from K is inherited from the fieldmultiplication in F. One readily checks that all the vector space axioms hold (see ExerciseSheet 3).

Definition. Let K ⊆ F be a field extension. The degree of the field extension, de-noted [F : K], is the dimension of F viewed as a vector space over K.

Examples. (1) C = R⊕ Ri and thus [C : R] = 2.(2) R has uncountable basis as a vector space over Q.(3) Notice that every field can be viewed as a vector space over its prime subfield. Thuseither a vector space over Q or a vector space over Zp.

Lemma 2.1 (The tower law). Let K ⊆ M ⊆ F be an ascending chain of fields. Supposethat (xi)i∈I is a basis for M over K and (yj)j∈J is a basis for F over M. Then (xiyj)i∈I,j∈Jis a basis for F over K. In particular we have that [F : K] = [F : M] · [M : K].

Proof To start with, we have F =∑

j∈J Myj and M =∑

i∈I Kxi. Therefore

F =∑j∈J

(∑i∈I

Kxi)yj =∑

i∈I,j∈J

Kxiyj

and F is generated by (xiyj)i∈I,j∈J as a vector space over K. We need to prove linearindependence. But if

0 =∑

i∈I,j∈J

aijxiyj =∑j∈J

(∑i∈I

aijxi)yj,

then∑

i∈I aijxi = 0 for all j ∈ J as (yj)j∈J is a basis for F over M. But as (xi)i∈I is abasis for M over K it then follows that aij = 0 for all i ∈ I and j ∈ J . This finishes theproof. 2

The Galois group of a field extension. We will see that there is a natural groupassociated to any field extension. Let us first recall the definition of a group.

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Definition. A group is a set G equipped with a binary operation · such that

(1) (a · b) · c = a · (b · c) for all a, b, c ∈ G.(2) There exists 1 ∈ G such that a · 1 = 1 · a = a for all a ∈ G.(3) For each a ∈ G there exists a−1 ∈ G such that a · a−1 = a−1 · a = 1.

Definition. A subset H of G is a subgroup if 1 ∈ H and ab, a−1 ∈ H whenever a, b ∈ H.We often use H ≤ G for ‘H is a subgroup of G’.

Let F be a field and let Aut F be the set of all ring automorphisms (that is isomorphisms)φ : F→ F. Recall from Algebra 2B, that Aut F is a group with respect to composition (asubgroup of the group of all bijections F→ F).

Let K ⊆ F be a field extension. Let

Aut KF = {φ ∈ Aut F : φ(a) = a for all a ∈ K}.

Lemma 2.2 Aut KF is a subgroup of Aut F.

Proof Clearly id : F → F is in Aut KF. It remains to see that Aut KF is closed undercomposition and taking inverses. Let φ, ψ ∈ Aut KF then for every a ∈ K we have

ψφ(a) = ψ(φ(a)) = ψ(a) = a.

anda = φ−1φ(a) = φ−1(φ(a)) = φ−1(a).

This shows that ψφ and φ−1 ∈ Aut KF that finishes the proof. 2

Defintion. Let K ⊆ F be a field extension. The group AutKF is called the Galoisgroup of the field extension.

Example. Consider the field extension R ⊆ C. Here C = R⊕Ri. Let σ ∈ Aut RC. Now σfixes all the elements in R pointwise. As σ(i)2 = σ(i2) = σ(−1) = −1 we have σ(i) = ±i.Thus we have that Aut RC = 〈τ〉, a cyclic group of order 2 where τ : C→ C, a+bi 7→ a−ibis the conjugation map. This is a homormorphism as zw = z ·w as well as z + w = z+w.

Lemma 2.3 Let K ⊆M ⊆ F be an ascending chain of fields. Then Aut MF is a subgroupof Aut KF.

Proof As both Aut KF and Aut MF are groups it suffices to show that Aut MF ⊆ Aut KF.If σ ∈ Aut MF then σ fixes M pointwise and hence also the subfield K. Thus σ ∈ Aut KF. 2

Let H be a subgroup of G = Aut KF. For every subgroup H of G we let

FixH = {a ∈ F |σ(a) = a for all σ ∈ H}.

Lemma 2.4 FixH is a subfield of F containing K.

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Proof To show that the subset FixH of F is a subfield we need only to check that all therelevant closure properties hold. Clearly 1 ∈ FixH (as 1 ∈ K). If a, b ∈ FixH and σ ∈ H,then σ(a + b) = σ(a) + σ(b) = a + b, σ(ab) = σ(a)σ(b) = ab and σ(−a) = −σ(a) = −a.Hence FixH is closed under addition and multiplication as well as with respect to takingadditive inverses. Futhermore if 0 6= a ∈ FixH, then σ(a−1) = σ(a)−1 = a−1 and thusFixH is also closed under taking multiplicative inverses of non-zero elements. This showsthat FixH is a subfield of F. Finally as all elements in G fix K pointwise (and thus allelements in H in particular), it is clear that FixH contains K. 2

Let K ⊆ F be a field extension. Let

F = {M : K ≤M ≤ F}, and G = {H : H ≤ G = Aut KF}.

We introduce two mapsΦ : F → G,M 7→ Aut MF

andΨ : G → F , H 7→ Fix(H).

Notice that by Lemma 2.3 and Lemma 2.4 we know that the two maps are well defined.We will show later in the course that under certain restrictions these two maps will provideus with a close remarkable correspondence beween the intermediate subfield structure ofthe field extension K ⊆ F and the subgroup structure of its Galois group Aut KF. Nextlemma gives the first indication that there is link between the two.

Lemma 2.5 The maps Ψ and Φ satisfy the following properties.

(1) Φ(M1) ≥ Φ(M2) when M1 ≤M2. Also Ψ(H1) ≥ Ψ(H2) when H1 ≤ H2.(2) H ≤ Φ(Ψ(H)) for all H ∈ G and M ≤ Ψ(Φ(M)) for all M ∈ F .(3) Φ(M) = Φ(Ψ(Φ(M))) for all M ∈ F and Ψ(H) = Ψ(Φ(Ψ(H))) for all H ∈ G.

Proof. See Exercise Sheet 3. 2.

II. Algebraic and transcendental elements

Definition. Let K ⊆ F be a field extension. An element a ∈ F is said to be alge-braic over K if there is a non-zero polynomial f ∈ K[X] such that f(a) = 0. Otherwisewe say that a is transcendental over K.

Examples. As√

2 is the root in R of f = x2 − 2 ∈ Q[x],√

2 is algebraic over Q.On the other hand π ∈ R isn’t the root of any polynomial in Q[x] and thus π is transcen-datal over Q.

The smallest subring of F containing K and a ∈ F.

Let K ⊆ F be a field extension and let a ∈ F. Any subring of F that contains K and amust contain all polynomial expressions

K[a] = {f(a) : f ∈ K[x]}.

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Notice that K[a] is the image of the ring homomorphism φ : K[x] → F, f 7→ f(a) andthus K[a] is a subring of F. This is clearly the smallest subring of F that contains K anda. We are now going to analyse the structure of the smallest subfield of F containing Kand a. Denote this subfield by K(a).

Case 1(a transcendental over K). This is quite straightforward. As a is not a rootof any non-zero polynomial we have that the ring homomorphism φ above is injective.Thus K[a] ∼= K[x], the ring of polynomials in one variable over K. Let K(x) = {f/g :f, g ∈ K[x] and g 6= 0}. be the field of all rational expressions in one variable. We canuse φ to obtain an injective ring homormorphism

ψ : K(x)→ F, f/g 7→ f(a)/g(a).

The image is clearly the smallest subfield of F containing K and a. Thus K(a) ∼= K(x).(See Exercise sheet 3 for more details).

Case 2. (a algebraic over K).

Proposition 2.6 Let K ⊆ F be a field extension and let a ∈ F be algebraic over K.

(1) K(a) = K[a] ∼= K[x]/K[x]f for a unique irreducible monic polynomial f ∈ K[x].(2) Let n = deg f . Then (1, a, a2, . . . , an−1) is a basis for K(a) as a vector space over K.Thus

K(a) = K⊕Ka⊕ · · ·Kan−1

and [K(a) : K] = deg f .

Proof (1) Consider the ring homomorphism φ : K[x]→ F, f 7→ f(a). As K[x] is a PID,kerφ = K[x]f for some polynomial f ∈ K[x] that is of degree at least 1 as a is algebraicover K. Furthermore f is unique provided we further require that it is monic (if there wasanother such polynomial monic polynomial then g− f would be a non-trivial polynomialin K[x]f of degree less than deg f but this is absurd). To see why f must be irreducible inK[x], we argue by contradiction and suppose that f = gh where 1 ≤ deg g, degh < deg f .But then

0 = f(a) = g(a)h(a)

and as F is a field we must have either g(a) = 0 or h(a) = 0. But then one of h, g is inkerφ = K[x]f and thus divisible by f . Again this is absurd. By the first isomorphismtheorem K[a] ∼= K[x]/K[x]f .

From this one can deduce that K[a] is a field. Let 0 6= g(a) ∈ K[a]. As f is irre-ducible and as f does not divide g in K[x], we have that f and g are coprime and we thushave

1 = fr + gs

for some r, s ∈ K[x]. But then 1 = f(a)r(a) + g(a)s(a) = 0 · r(a) + g(a)s(s) = g(a)s(a).This shows that g(a has a multiplicative inverse s(a) ∈ K[a]. We have thus shown thatK[a] is a field and thus the smallest subfield containing a and K. Hence K(a) = K[a].

(2) Let g(a) ∈ K[a] for some g ∈ K[x]. Using division with remainder there exist

13

q, r ∈ K[x] such that g = fq+r and where deg r < deg f , say r = b0 +b1x+ · · ·+bn−1xn−1.

Theng(a) = f(a)q(a) + r(a) = 0 · q(a) + r(a) = b0 + b1a+ · · ·+ bn−1a

n−1

and we thus have g(a) ∈ K+Ka+ · · ·+Kan−1. It remains only to show that 1, a, . . . , an−1

are linearly independent. Suppose

b0 + b1a+ · · ·+ bn−1an−1 = 0.

Then g = b0 + b1x + · · · + bn−1xn−1 ∈ kerφ = K[x]f . As deg f > deg g, this can only

happen if g = 0 that is if b0, . . . , bn−1 = 0. This shows that 1, a, . . . , an−1 are linearlyindepdendent. 2

Remark. Notice that the structure of K[a] only depends on K[x] and f . The addi-tion is just like addtion of polynomials

(b0+b1a+· · ·+bn−1an−1)+(c0+c1a+· · ·+cn−1a

n−1) = (b0+c0)+(b1+c1)a+· · ·+(bn−1+cn−1)an−1

and the multiplication is just like multiplication of polynomials but where we then usef(a) = 0, that is

an = −(a0 + a1a+ · · ·+ an−1an−1)

to reduce the polynomial expression to something of degree at most n− 1.

Examples. Consider the field extension Q ⊆ C. We have that the relevant polyno-mials for i,

√2 and 3

√2 are x2 + 1, x2 − 2 and x3 − 2. Here thus

Q(i) = Q⊕QiQ(√

2) = Q⊕Q√

2

Q(3√

2) = Q⊕Q 3√

2⊕Q(3√

2)2.

Here for example i2 = −1 and thus (2 − i)(1 + 2i) = 2 + 3i − 2i2 = 2 + 3i + 2 = 4 + 3i.For a = 3

√2, we have a3 = 2 and thus

(1+3√

2)(1− (3√

2)2) = (1 +a)(1−a2) = 1 +a−a2−a3 = −2+a−a2 = −2+3√

2− (3√

2)2.

Definition. Let K ⊆ F be a field extension and let a be an element in F that is algebraicover K. The minimal polynomial of a over K is the monic polynomial ma ∈ K[x] ofsmallest degree such that ma(a) = 0.

Remark. We have shown above that ma is irreducible over K and unique and thatma divides any polynomial g ∈ K[x] where g(a) = 0.

Example. The minimal polynomial of 3√

3−1 in R over Q is (x+1)3−3 = x3+3x2+3x−2.

Definition. Let K ⊆ F be a field extension. If there exists a ∈ F such that F = K(a),then we say that K ⊆ F is a simple extension. The degree of the extension [K(a) : K] issometimes also called the degree of a over K.

14

Remark. We have just observed that the structure of K(a) = K[a] just depends onK[x] and the minimal polynomial of a over K. In fact this is also clear from the iso-morphism K(a) ∼= K[x]/K[x]ma. This provides us with a clue as to how we can, for anyirreducible polynomial f , construct a field F such that F = K(a) and f(a) = 0. Let usfirst recall why K[x]/K[x]f is a field in this case.

Lemma 2.7 Let f be an irreducible polynomial in K[x]. We have that K[x]/K[x]f a field.

Proof Let 0 6= a ∈ K[x]/K[x]f . Then f does not divide a and thus a and f have nocommon irreducible factor. It follows that a and f are coprime and 1 = ra+ sf for somepolynomials r, s ∈ K[x]. We thus have

1 = ra+ sf = ra+ s0 = ra.

So a has an inverse r. This shows that any non-zero element in K[x]/K[x]f has a multi-plicative inverse. Hence K[x]/K[x]f is a field. 2

Remark. Recall from Algebra 2B that we can think of K[x]/K[x]f naturally as a vectorspace over K where the scalar multiplication is

ag = ag

for a ∈ K and g ∈ K[x].

Proposition 2.8 Let f be an irreducible polynomial in K[x] of degree n and let F be thefield K[x]/K[x]f . Let 1F = 1 be the multiplicative identity of F and let t = x = x+ K[x]f .Then

F = K1F ⊕Kt⊕ · · · ⊕Ktn−1.

furthermore f(t) = 0.

Proof Let g = g + K[x] be an arbitrary element in F where g ∈ K[x]. Using divisionwith remainder we get

g = fs+ r

for some polynomial b0 + b1x + · · · + bn−1xn−1 ∈ K[x] of degree less than n. Hence (as

f = 0)

g = f s+ b01 + b1x+ · · ·+ bn−1(x)n−1 = b01F + b1t+ · · ·+ bn−1tn−1.

This shows that 1F, t, t2, . . . , tn−1 generate F as a vector space over K. Let us see why

they are linearly independent. Suppose b01F + b1t+ · · ·+ bn−1tn−1 = 0. This is the same

as saying that g = 0 or equivalently that g ∈ K[x]f where g = b0 + b1x+ · · ·+ bn−1xn−1.

But as the degree of g is less than the degree of f this can only happen when g = 0, thatis when b0 = b1 = . . . = bn−1 = 0.

Finally if f = a0 + a1x+ · · ·+ anxn. Then

f(t) = a01F + a1t+ · · ·+ antn

= a01 + a1x+ · · ·+ an(x)n

= a0 + a1x+ · · ·+ anxn

= f = 0 = 0F.

15

This finishes the proof. 2

Remarks. (1) This gives us a field extension K1F ⊆ F where K1F ∼= K (See Exer-cise sheet 4 for some clarifications). We should thus perhaps more naturally consider t asa root of the polynomial

a01F + a11Fx+ · · ·+ an1Fxn

in K1F[x]. We then have concretely constructed a simple field extension K1F ⊆ K1F (t).

This is however an unneccessary formalism. We simply identify a ∈ K with a1F in K1Fand think of K itself as a subfield of F. Then our simple field extension is K ⊆ K(t) andour polynomial can remain exactly what it was.

(2) Let f be any polynomial in K[x] of degree at least 1 and let q be any irreduciblefactor, say f = qg. Proposition 2.8 gives us a larger field where q has a root t. But thenf(t) = q(t)g(t) = 0g(t) = 0 and thus f has a root in a larger field.

Examples (1) Consider Q ⊆ Q(i). Now the minimal polynomial of i over Q is x2 + 1.By Poposition 2.6 we then have

Q(i) = Q⊕Qi

where the field operations are like adding and multiplying polynomial expressions andusing i2 + 1 = 0.

(2) Consider the irreducible polynomial x2 + 1 ∈ Q[x]. Using Proposition 2.8, we get afield extension Q ⊆ Q(t) where

Q(t) = Q⊕Qt

where again the field operations are like adding and multiplying polynomial expressionsand using t2 + 1 = 0. Of course Q(t) and Q(i) are isomorphic.

(2) Consider the irreducible polynomial x2 + x+ 1 in Z2[x]. Using Proposition 2.8 we canthen construct a new field

F = Z2(t) = Z2 ⊕ Z2t,

where t2 = t+ 1. We have thus constructed a field with 4 elements.

III. Algebraic field extensions

Let K ⊆ F be a field extension and let a1, . . . , an ∈ F. Consider the ring homomor-phism

K[x1, . . . , xn]→ F, f 7→ f(a1, . . . , an).

The image, denoted by K[a1, . . . , an], is the smallest subring of F that contains K anda1, . . . , an. The subfield

K(a1, . . . , an) = {a/b : a, b ∈ K[a1, . . . , an], b 6= 0}

is the smallest subfield of F that contains K and a1, . . . , an.

Remark. Clearly K(a1, . . . , an) = K(a1)(a2) · · · (an).

16

Definition. Let K ⊆ F be a field extension.

(1) We say that K ⊆ F is finitely generated if there are finitely many elements a1, . . . , an ∈F such that F = K(a1, . . . , an).

(2) We say that K ⊆ F is finite if [F : K] <∞.

Remark. (2) is generally stronger than (1). For example Q ⊆ Q(π) is clearly finitelygenerated but it is not finite. If

∞∑m=0

amπm = 0

where almost all but not all the coefficients are zero, then this would imply that π isalgebraic. This we know is not the case and hence 1, π, π2, . . . are infinitely many linearlyindependent elements in Q(π) and thus Q(π) is infinitely dimensional over Q. Likewise ifwe consider the field F = Q(x) of rational expressions in the free variable x, then Q ⊆ Fis simple and thus finitely generated. It is however not finite. (In fact we know that Q(π)and Q(x) are isomorphic).

Definition. An extension K ⊆ F is said to be algebraic if every element a ∈ F is al-gebraic over K.

Theorem 2.9 Let K ⊆ F be a field extension. The following are equivalent.

(1) K ⊆ F is finite.

(2) There exist finitely many elements a1, . . . , an ∈ F that are algebraic over K and suchthat F = K(a1, . . . , an).

(3) K ⊆ F is finitely generated and algebraic.

Proof ((1) ⇒ (3)). Suppose [F : K] = n and let let a1, . . . , an be a basis for thevector space F over K. Then clearly F = K(a1, . . . , an) and thus the extension is finitelygenerated. To see that it is algebraic let t ∈ F. Then the n + 1 elements 1, t, . . . , tn arelinearly dependent, say

b0 + b1t+ · · ·+ bntn = 0,

for some b0, . . . , bn ∈ K (not all zero). Thus t is a root of the nonzero polynomialf = b0 + · · ·+ bnx

n over K. This shows that the extension is algebraic.

((3)⇒ (2)). This is obvious.

((2) ⇒ (1)). Let Kr = K(a1, . . . , ar). Then Kk+1 = Kk(ak+1). By our assumptionak+1 is algebraic over K and thus also over the larger field Kk (a nonzero polynomialin K[x], mapping ak+1 to zero is also in Kk[x]). By Proposition 2.6 we have that[Kk+1 : Kk] = [Kk(ak+1) : Kk] is finite (same as the degree of the minimal polynomialmk+1 of ak+1 over Kk. Thus the the tower law gives

[F : K] = [Kn : Kn−1][Kn−1 : Kn−2] · · · [K1 : K0] <∞.

17

Corollary 2.10 Let K ⊆M ⊆ F be an ascending chain of fields. Then K ⊆ F is algebraicif and only if K ⊆M and M ⊆ F are algebraic.

Proof (⇒) Suppose K ⊆ F is algebraic. To see that K ⊆ M is algebraic notice thatevery element in M is in F and thus algebric over K. Now we turn to M ⊆ F. Let a ∈ F.As a is algebraic over K there is some 0 6= f ∈ K[x] such that f(a) = 0. But as K ⊆ M,f is also in M[x] and thus a algebraic over M.

(⇐) Let a ∈ F. As M ⊆ F is algebraic there is a polynomial 0 6= f ∈ M[x] with aas a root. Suppose

f = bnxn + bn−1x

n−1 + · · ·+ b0,

with b0, . . . , bn ∈ M. Then in fact f ∈ K(b0, . . . , bn)[x] and thus a algebraic overK(b0, . . . , bn). Thus the extension K(b0, . . . , bn) ⊆ K(b0, . . . , bn, a) is finite by Proposition2.6 (the degree of the extension is the same as the degree of the minimal polynomial of aover K(b0, . . . , bn)). But as b0, . . . , bn ∈ M, and thus algebraic over K, K ⊆ K(b0, . . . , bn)is finite by Theorem 2.9. Hence, using the tower law,

[K(b0, . . . , bn, a) : K] = [K(b0, . . . , bn, a) : K(b0, . . . , bn)] · [K(b0, . . . , bn) : K] <∞.

In particular it then follows from Theorem 2.9 that a is algebraic over K. 2

Corollary 2.11 Let K ⊆ F be a field extension. Then

AF/K = {a ∈ F : a is algebraic over K}

is a subfield of F containing K.

Proof It is clear that K is contained in AF/K as any element a ∈ K is algebraic over K(it is root of the polynomial x − a ∈ K[x]). It remains to see that AF/K is a subfield ofF. For this we only need to show that it is closed under taking addition, multiplication,taking additive inverses and taking multiplicative inverses of non-zero elements. To seethis notice that if a, b ∈ AF/K then a+ b, ab,−a, as well as a−1, when a 6= 0 are in K(a, b).As a, b are algebraic over K, Theorem 2.9 tells us that K ⊆ K(a, b) is algebraic. Thus inparticular a+ b, ab,−a, a−1 are algebraic over K and thus in AF/K. 2.

Definition. AF/K is called the subfield of algebraic elements in F over K.

IV. Splitting fields.

Suppose that f ∈ K[x] is a polynomial of degree n ≥ 1. We can factorise it into aproduct of irreducible polynomials over K[x], say

f = p1 · · · pr.

By Proposition 2.8, p1 has a root t1 in a larger field K1. Now factorise f over K1 into aproduct of irreducibles over K1

f = (x− t1)q2 · · · qs.

18

Again we can apply Proposition 2.8 to find a root t2 in a larger field K2 and then wefactorise into irreducibles over K2, say

f = (x− t1)(x− t2)h3 · · · kl.

Continuing in this manner we arrive (after n steps) to a larger field F over which f splitsinto linear factors

f = c(x− t1) · · · (x− tn).

with c ∈ K and t1, . . . , tn ∈ F.

Definition. Let K be a field and 0 6= f ∈ K[x]. A field F that contains K is calleda splitting field of f over K if

(a) f splits into linear factors over F, that is

f = c(x− t1) · · · (x− tn)

with c ∈ K and t1, . . . , tn ∈ F.

(b) F = K(t1, . . . , tn).

We have seen that a splitting field for f exists. We will next see that they all lookthe same. In fact we will need something slightly stronger later in the course and so oursetting is going to be a bit more general.

Lemma 2.12 Suppose φ : K1 → K2 is a field isomorphism. Let f1 ∈ K1[x] be anirreducible polynomial over K1 and let f2 = φ∗(f1). Let ai be a root of fi in some largerfield Fi. Then there is an isomorphism

ψ : K1(a1)→ K2(a2)

such that ψ(a1) = a2 and ψ(a) = φ(a) for all a ∈ K1.

Proof We know from Proposition 2.6 that K1(a) consists of polynomial expressions f(a1)where f runs through the polynomial ring K1[x]. Likewise K2(a2) consists of polynomialexpressions in a2 over K2. We define ψ : K1(a1) → K2(a2) by ψ(f(a1)) = φ∗(f)(a2) forany f ∈ K1[x]. Thus

ψ(s0 + s1a1 + · · ·+ snan1 ) = φ(s0) + φ(s1)a2 + · · ·+ φ(sn)an2 .

Clearly ψ(s) = φ(s) for all s ∈ K1 and ψ(a1) = a2. It remains to see that ψ is a welldefined field isomorphism. As φ is bijective it is clear that all polynomial expressions ina2 over K2 are covered by ψ and thus ψ is surjective.

ψ is well defined and injective. By Exercise 1 on sheet 2, we know that f2 is irreducibleover K2 and thus f1, f2 are the minimal polynomials of a1, a2. We have

f(a1) = g(a1) ⇔ f1|(f − g)

⇔ f2 = φ∗(f1)|(φ∗(f)− φ∗(g))

⇔ φ∗(f)(a2) = φ∗(g)(a2)

⇔ ψ(f(a1)) = ψ(g(a1)).

19

ψ is a ring homomorphism. This follows from the already established fact that φ∗ :K1[x]→ K2[x] is a ring homomorphism. 2

Remark. Last lemma is intuitively kind of obvious. Suppose that the minimal poly-nomials have degree m. Then

K1(a1) = K1 ⊕K1a1 ⊕ · · · ⊕K1am−11

andK2(a2) = K2 ⊕K2a2 ⊕ · · · ⊕K2a

m−12 .

The structure of the K1(a1) only depends on K1[x] and f1 and likewise the structure ofK2(a2) only depends on K2[x] and f2 = φ∗(f1). As K2 is just another copy of the field K1

we have that K2[x] is like K1[x] and the polynomial f2 is just like f1 where we have justreplaced the coefficients in f1 with the corresponding coefficients in K2.

Theorem 2.13 (Uniqueness of the splitting field) Suppose φ : K1 → K2 is a field iso-morphism. Let f1 ∈ K1[x] be of degree at least one and let f2 = φ∗(f1). Let Fi be asplitting field of fi over Ki. Then there exists a field isomorphism ψ : F1 → F2 such thatψ(a) = φ(a) for all a ∈ K1.

Proof Supposef1 = c(x− a1) · · · (x− an)

with c ∈ K1 and a1, . . . , an ∈ F1. Let g1 be the minimal polynomial of a1 over K1. Theng1|f1 in K1[x]. Let b1 be a root in F2 of the factor φ∗(g1) of f2 = φ∗(f1). By Lemma 2.12,there is an isomorphism

φ1 : K1(a1)→ K2(b1)

such that φ1(a1) = b1 and φ1(a) = φ(a) for all a ∈ K1. Now let g2 be the minimal polyno-mial of a2 over K1(a1). As before it is clear that g2 divides f1 and thus φ∗1(g2)|φ∗1(f1) = f2.Let b2 be a root of φ∗1(g2) in F2. Applying Lemma 2.12 again, there is a field isomorphism

φ2 : K1(a1, a2)→ K2(b1, b2)

such that φ2(a2) = b2 and φ2(a) = φ1(a) for all a ∈ K1(a1). Continuing in this mannerwe arrive at a isomorphism

ψ : K1(a1, . . . , an)→ K2(b1, . . . , bn)

such that ψ(ai) = bi and ψ(a) = φ(a) for all a ∈ K1.

Now let d = φ(c). Then

d(x− b1) · · · (x− bn) = ψ∗(c(x− a1) · · · (x− an)) = ψ∗(f1) = φ∗(f1) = f2

Hence b1, . . . , bn are the roots of f2 in F2 and F2 = K2(b1, . . . , bn). We have thus shownthat ψ is the isomorphism we were after. 2

Corollary 2.14 Let f ∈ K[x] be a polynomial of degree at least 1. Let F1,F2 be splittingfields of f over K. There exists a field isomorphisms ψ : F1 → F2 such that ψ(a) = a forall a ∈ K.

20

Proof. Let K1 = K2 = K and φ = id in Theorem 2.13. 2

Remark. Let F be a finite field. Then we see on sheet 4 that |F| = pn for some prime pand positive integer n. Our next result is a remarkable application of Corollary 2.14. Itgives us a classification of finite fields.

Theorem 2.15 For each prime p and positive integer n, there exists (up to isomorphism)exactly one field of order pn. This is a splitting field of the polynomial

xpn − x

oer Zp.

Proof Let F be a field of order pn. Then F\{0} is a group of order pn−1. By Lagrange’sTheorem we then know that ap

n−1 = 1 for all a ∈ F \ {0}. Hence all the nontrivial el-ements of F are roots of xp

n − x = x(xpn−1 − 1). Of course the same is true for 0 ∈ F.

It follows that F is a splitting field of xpn − x over Zp. As all splitting fields of a given

polynomial are isomorphic it follows that any two fields of the same order are ismorphic.

We still have some work to do as we want to show that all prime powers pn are cov-ered. In order to do this we take arbitrary prime p and positive integer n and let F bea splitting field of xp

n − x over Zp. On problem sheet 5 we will see that the roots of thepolynomial form a subfield of F of order pn. Hence |F| = pn. 2.

We end this chapter by considering the Galois group of K ⊆ F in the case when F isthe splitting field of some polynomial f ∈ K[x].

Lemma 2.16 Suppose K1 ⊆ F1 and K2 ⊆ F2 are field extensions. Suppose ψ : F1 → F2

is a field isomorphism where ψ(K1) = K2. Let f1 ∈ K1[x] and let f2 = ψ∗(f1). If a is aroot of f ∈ K1[x] in F1 then ψ(a) is a root of ψ∗(f) ∈ K2[x] in F2.

Proof Suppose f = α0 +α1x+ · · ·+αnxn. Then ψ∗(f) = ψ(α0)+ψ(α1)x+ · · ·+ψ(αn)xn.Hence

ψ∗(f)(ψ(a)) = ψ(a0)+ψ(a1)ψ(a)+· · ·+ψ(an)ψ(a)n = ψ(a0+a1a+· · ·+anan) = ψ(0) = 0.2

Remark. In fact we are mostly interested in the special case when K1 = K2 and Ψ fixesall the elements in K.

Corollary 2.17 Suppose K ⊆ F is a field extension and let σ ∈ Aut KF. Let a be a rootof some polynomial f ∈ K[x] of degree at least one. Then σ(a) is also a root of f .

Proof. It is worth rewriting the proof for this special case. As σ is a field automorphismthat fixes the elements in K pointwise it fixes all the coefficients of f ∈ K[x]. Supposef = α0 + α1x+ · · ·+ αnx

n. Then

0 = σ(0) = σ(α0 + α1a+ · · ·+ αnan) = α0 + α1σ(a) + · · ·+ αnσ(a)n

and thus σ(a) is a root of f . 2

21

Proposition 2.18 Suppose f ∈ K[X] has n ≥ 1 distinct roots in a splitting field F overK, then G = Aut KF is isomorphic to a subgroup of Sn.

Proof. Suppose that the roots are t1, t2, . . . , tn. Let I = {t1, . . . , tn}. By the last lemmawe know that the elements in G permute the n roots. We thus get a homomorphism

Ψ : Aut KF→ Sym (I), φ 7→ φ|I .

The map is also injective as F = K(t1, . . . , tn) and thus every σ ∈ Aut KF is determined byits values in t1, . . . , tn. We thus have that Aut KF is isomorphic to Im Ψ that is a subgroupof Sym (I). As |I| = n we have that Sym (I) is isormorphic to Sn and hence the result. 2

22

3 Separability, Normality and Galois Extensions

In this section we will introduce some conditions on field extensions K ⊆ F that willensure that we get a 1-1 correspondence between the collection of intermediate fields ofthe extension and the subgroups of the Galois group.

I. Separability.

Definition.

(1) We say that an irreducible polynomial f over K is separable if it has no multipleroot in a splitting field over K.

(2) An element a ∈ F is said to be separable over K if it is algebraic over K and itsminimal polynomial over K is separable.

(3) An algebraic field extension K ⊆ F is said to be separable if all elements a ∈ Fare separable over K.

Remark. Let F1 and F2 be two splitting fields of f ∈ K[x]. By Theorem 2.13 thereis a field isomorphism φ : F1 → F2 such that φ(a) = a for all a ∈ K. Suppose that f hasno multiple roots in F1. Thus

f = c(x− t1) · · · (x− tn)

with c ∈ K and t1, . . . , tn distinct elements in F1. Then

f = φ∗(f) = c(x− φ(t1)) · · · (x− φ(tn)).

As φ is an isomorphism we also have that φ(t1), . . . , φ(tn) are distinct. Thus the definition(1) above makes sense as it doesn’t depend on what the splitting field is. No surprise asall the splitting fields are isomorphic.

Examples.

(1) f = x2 + x+ 1 is separable over Q. It has the distinct roots −1±i√

32

.

(2) Let K = Z2(t), the field of rational expressions over Z2 in the free variable t. Considerthe polynomial

f = x2 − t ∈ K[X].

(a) f is irreducible over K: otherwise (f/g)2 = t for some polynomials f, g ∈ Z2[t]. Butthen

f 2 = tg2

23

which is absurd as the left hand side is of even degree and the right hand side of odd degree.

(b) We show that f is not separable. Let F be a splitting field of f over K and let abe a root of f in F. Then

(x− a)2 = x2 − a2 = x2 − t.

Thus a is double root of f .

Definition. (Formal derivation). We introduce a map D : K[x]→ K[x]. For

f = a0 + a1x+ · · ·+ anxn ∈ K[x]

we letD(f) = a1 + 2a2x+ · · ·+ nanx

n−1.

We call D(f) the formal derivative of f .

Remark. Notice that if K ⊆ F and f ∈ K[x], then the derivative of f ∈ K[x] is thesame as the derivative of f ∈ F[x].

Lemma 3.1 Let f, g ∈ K[x] and a ∈ K.

(1) D(f + g) = D(f) +D(g)(2) D(af) = aD(f)(3) D(fg) = D(f)g + fD(g)

Proof See Problem Sheet 6.

Remark. For some of the results in this section we need to use the following fact.If 0 6= f, g ∈ K[x] and K ⊆ F then a gcd of f, g in K[x] is also a gcd of f, g ∈ F[x]. Tosee this, let d be the greatest common divisor of f, g in K[x]. If f = dh and g = dk withh, k ∈ K[x], then h, k ∈ F[x] as well and thus d|f and d|g in F[x]. Now we know that

d = af + bg

for some a, b ∈ K[x]. It follows from this that if c divides f, g in F[x] then it divides d inF[x]. Hence d is a gcd of f and g in F[x].

Lemma 3.2 Let f ∈ K[x] be a polynomial of degree at least 1. Then f has no multipleroot in a splitting field F if and only if f and D(f) are coprime in K[x].

Proof (⇐). We argue by contradiction and suppose f = (x−a)2g over F. Then, workingin F[x],

D(f) = 2(x− a)g + (x− a)2D(g)

and f and D(f) have the common divisor (x− a) and thus can’t be coprime in F[x] andthen they can’t be coprime in K[x] either by the remark above. But this contradicts ourassumption that f and g were coprime in K[x].

24

(⇒). Suppose that f = c(x − a1) · · · (x − an) over F where c ∈ K and a1, . . . , an aredistinct elements in F. Then

D(f) = c

n∑k=1

(x− a1) · · · (x− ak−1)(x− ak+1) · · · (x− an).

We want to show that f and D(f) are coprime. The only way this would fail is if theyhad some common irreducible factor that would then have to be some (x− aj). However,

D(f)(aj) = c(aj − a1) · · · (aj − aj−1)(aj − aj+1) · · · (aj − an) 6= 0,

which shows that no (x− aj) is a factor of D(f) and thus that f and D(f) are coprime.2

Corollary 3.3 An irreducible polynomial f ∈ K[x] is separable if and only if D(f) 6= 0.

Proof If f is irreducible then f and D(f) have a common divisor if and only if f di-vides D(f). But as D(f) is of lower degree than f , this happens if and only if D(f) = 0. 2

Theorem 3.4 Let f ∈ K[x] be irreducible.

(a) If charK = 0 then f is separable.(b) If charK = p for a prime p, then f is separable if and only if f 6∈ K[xp].

Proof (a) Suppose the leading coefficient of f is an 6= 0. Notice that n ≥ 1. The leadingcoefficient of D(f) is then nan that is not zero as char K = 0.

(b) We have that

D(a0 + a1x+ · · ·+ anxn) = a1 + 2a2x+ · · ·+ nanx

n−1

is zero iff an = 0 for all n that are coprime to p iff the polynomial is in K[xp]. 2

Corollary 3.5 Let K ⊆ F be an algebraic field extension where char K = 0. Then K ⊆ Fis separable.

Proof Let a ∈ F. The minimal polynomial f of a over K is irreducible and by Theorem3.4 it is thus separable. Hence a is separable over K. 2

Remark. Let K ⊆ M ⊆ F be an ascending chain of fields where K ⊆ F is separa-ble. Then both K ⊆ M and M ⊆ F are separable. To see that K ⊆ M is separable leta ∈M. As a ∈ F we know that a is separable over K. To see that M ⊆ F is separable leta ∈ F. The minimal polynomial mK of a in K[x] is a polynomial in M[x] that maps a tozero and thus it must be divisible by the minimal polynomial mM of a in M[x]. As mK isseparable the same is true its divisor mM.

Remark. Let f, g ∈ K[x] and suppose K ⊆ F. Recall from earlier that if d is a gcdof f, g in K[x] then d is also a gcd of f, g ∈ F[x]. In particular if d is the unique monicgcd of g, f in K[x] then d is also the unique monic gcd of g, f in F[x]. This fact will beimportant in the proof of the following result.

25

Theorem 3.6 Let K ⊆ F be a finite extension where either

(a) K is finite

or

(b) K is infinite and K ⊆ F is separable.

Then F = K(θ) for some θ ∈ F.

Proof Suppose first that K is a finite field. Then F is also finite and we know from GroupTheory (http://people.bath.ac.uk/gt223/MA30237.html, Sheet 5: Question 5) that thegroup of units F∗ = F \ {0} is a cyclic group, say generated by θ. Thus F = K(θ).

We can thus assume that K is an infinite field. Notice that K ⊆ F is finitely gener-ated and arguing by induction, it is sufficient to prove this in the case when there are twogenerators, say F = K(α, β). Let f, g be the minimal polynomials of α, β over K. Take asplitting field L of fg over F. Suppose

f = (x− α1) · · · (x− αn)

g = (x− β1) · · · (x− βm),

where α = α1, . . . , αn ∈ L are distinct (as the extension is separable) and likewise β =β1, . . . , βm ∈ L are distinct. Let

θ = α + cβ

where c ∈ K will be specified later. Let

h = f(θ − cx) ∈ K(θ)[x].

Notice that h(β) = f(θ − cβ) = f(α) = 0 and thus x− β is a common irreducible factorof h, g. The aim is to show that one is able to pick c ∈ K such that the unique gcd of h, gin L[x] is x−β (and therefore by the remarks also the unique gcd of h, g in K(θ)[x] whereboth g, h are). This is going to be the case iff none of the factors x− β2, . . . , x− βm of gare factors of h that happens iff none of β2, . . . , βm is a root of h. To see what could gowrong notice that for 2 ≤ j ≤ m,

h(βj) = 0⇔ f(θ − cβj) = 0⇔ θ − cβj = αi for some 1 ≤ i ≤ n⇔ α + cβ − cβj = αi for some 1 ≤ i ≤ n⇔ c = αi−α

β−βjfor some 1 ≤ i ≤ n.

(Notice that the denominator is nonzero as β 6= βj that we know is the case as the exten-sion in separable). As there are only finitely many expressions of the form (αi−α)/(β−βj),with 1 ≤ i ≤ n and 2 ≤ j ≤ m, and as K is infinite, we can choose c ∈ K so that it isdifferent from all of these expressions. As we have pointed out above this then impliesthat none of β2, . . . , βm is a root of h and hence the unique monic gcd of h, g in L[x]is x − β. This is however also the unique monic gcd of h, g in K(θ)[x]. In particularx − β ∈ K(θ)[x] and thus β ∈ K(θ). Then also α = θ − cβ ∈ K(θ) and we have shownthat F = K(α, β) = K(θ). 2

26

II. Normality

Definition. An algebraic field extension K ⊆ F is said to be normal if, for all a inF, the minimal polynomial of a over K splits into linear factors over F.

Remark. Equivalently K ⊆ F is normal if whenever f ∈ K[x] is irreducible over Kwith a root in F, then all the roots of f must be in F.

Examples. (1) R ⊆ C is normal.

(2) Q ⊆ Q( 3√

2) is not normal as 3√

2 is the only real root of x3 − 2.

Lemma 3.7 Let K ⊆ M ⊆ F be an ascending chain of fields where K ⊆ F is normal.Then M ⊆ F is normal.

Proof Let a ∈ F and let fK, fM be the minimal polynomials of a over K,M. As K ⊆ Fis normal we know that fK splits into linear factors over F, say

fK = (x− a1) · · · (x− an)

where a = a1, a2, . . . , an ∈ F. As fK(a) = 0 and fK ∈M[x] it follows that fM|fK and thusfM also splits into a product of linear factors over F. 2

Remarks. (1) In general if K ⊆ M,M ⊆ F are normal it does not follow that K ⊆ F isnormal. For example Q ⊆ Q(

√2),Q(

√2) ⊆ Q( 4

√2) are normal as they are of degree 2

(see Problem Sheet 6). However Q ⊆ Q( 4√

2) is not normal. The element 4√

2 has minimalpolynomial x4 − 2 over Q but only two of the roots are in Q( 4

√2) (in fact only two roots

are real) namely ± 4√

2.

(2) Let K ⊆ M ⊆ F be an ascending chain of fields where K ⊆ F is normal. In gen-eral it does not follow that K ⊆ M is normal. Consider for example Q ⊆ Q( 3

√2) ⊆ A

where A = AC/Q is the field of the algebraic elements in C over Q. Here Q ⊆ A is normal

but Q ⊆ Q( 3√

2) is not normal.

Remark. Let F be a splitting field of some f ∈ K[x] over K, say F = K(a1, . . . , an)where a1, . . . , an ∈ F are the roots of f in F. If F ⊆ F(b) is any field extension then F(b) isa splitting field of f over K(b). This is because F(b) = K(b, a1, a2, . . . , an) and a1, . . . , anare the roots of f (that is also in K(b)[x]). We will make use of this observation in theproof of the following theorem.

Theorem 3.8 The field extension K ⊆ F is normal and finite if and only if F is a splittingfield of some polynomial f ∈ K[x] over K.

Proof (⇒). Suppose F = K(a1, a2, . . . , an) and let fi be the minimal polynomial of aiover K. Then let

f = f1 · · · fn.The roots of f are a1, a2, . . . , an and some extra roots b1, . . . , bm that are also in F asK ⊆ F is normal. Thus

F = K(a1, . . . , an, b1, . . . , bm)

27

and thus a splitting field of f over K.

(⇐). Suppose F a splitting field of f ∈ K[x] over K. The extension K ⊆ F is clearly finiteas it is generated by finitely many algebraic elements over K (see Theorem 2.9) and itremains to see that it is normal.

Let g ∈ K[x] be an irreducible polynomial with a root θ1 ∈ F and let θ2 be any rootof g in a splitting field L of g over F. We want to show that θ2 ∈ F. First notice thatK(θ1) ∼= K[x]/K[x]g ∼= K(θ2) as θ1, θ2 have the same minimal polynomial g over K. Letφ : K(θ1)→ K(θ2) be an isomorphism between the two fields. As F(θj) is the splitting fieldof f over K(θj) we can apply Theorem 2.13 that tells us that there is a field isomorphism

ψ : F(θ1)→ F(θ2)

such that ψ|K(θ1) = φ. From Exercise 1 on Sheet 3 we know that [F(θ1) : K(θ1)] =[ψ(F(θ1)) : ψ(K(θ1))] = [F(θ2) : K(θ2)]. From this and [K(θ1) : K] = deg g = [K(θ2) : K],we get from the Tower Law that

[F(θ1) : K] = = [F(θ1) : K(θ1)][K(θ1) : K]

= [F(θ2) : K(θ2)][K(θ2) : K]

= [F(θ2) : K].

A second application of the tower law gives then

[F(θ1) : F][F : K] = [F(θ2) : F][F : K]

from which we deduce that [F(θ2) : F] = [F(θ1) : F] = 1 (as θ1 ∈ F) and therefore thatθ2 ∈ F. This finishes the proof. 2

III. Galois extensions

Let K ⊆ F be a finite field extension. We have seen on Problem Sheet 3 that in thecase when the extension is not separable or not normal then the Galois Correspondencebetween intermediate subfields of K ⊆ F and subgroups of the Galois Group can fail to be1-1. We will see in next chapter that if we add the constraints that the extension is bothnormal and separable then the Galois Correspondence turns out to be bijective providingsome close interplay between the Galois Group and the field extension. We will end thischapter by characterising such extensions in terms of splitting fields. We start with alemma that will be useful in understanding the size of the Galois Group of a given fieldextension (see the corollary to the lemma). The Lemma is given in a bit more generalform as this will be needed later.

Lemma 3.9 Let K1 ⊆ F1 and K2 ⊆ F2 be finite extensions where [F1 : K1] = [F2 :K2]. Let φ : K1 → K2 be a field isomorphism. Then there are at most [F1 : K1] fieldisomorphisms ψ : F1 → F2 that extend φ, that is such that ψ|K1 = φ.

Proof We prove this by induction on [F1 : K1]. When [F1 : K1] = 1 then F1 = K1,F2 = K2 and ψ = φ is clearly then only isomorphism extending φ. Now suppose [F1 :K1] > 1 and that the result holds whenever the degree is smaller. Let a ∈ F1 \ K1 and

28

let f ∈ K1[x] be its minimal polynomial over K. If ψ : F1 → F2 extends φ then we knowfrom Lemma 2.16 that b = ψ(a) must be a root of φ∗(f) in F2. In which case ψ inducesan isomorphism φb : K1(a)→ K2(b) such that φb(a) = b and φb|K1 = φ. There are at mostdeg f = [K2(a) : K2] = [K1(a) : K1] such roots b of φ∗(f) (see Proposition 2.6) and byinduction hypothesis each such φb has then at most [F1 : K1(a)] extensions τ : F1 → F2.Thus in total we get at most

[K1(a) : K1][F1 : K1(a)] = [F1 : K1]

extensions of φ. This finishes the proof. 2

Corollary 3.10 If K ⊆ F is any finite field extension, then |Aut KF| ≤ [F : K].

Proof Take K1 = K2, F1 = F2 and φ = id in Lemma 3.9. 2.

Remark. We will see that when we add normality and separability we will get equality inthe last result. This is essentially going to follow from next proposition. We have definedearlier what we mean by an irreducible polynomial to be separable. We now extend thedefinition.

Definition. A polynomial f ∈ K[x] is separable if none of its irreducible factors overK have a multiple root .

Remark. Notice that we only require the irreducible factors to have no multiple roots.For example we have that (x− 1)2 ∈ Z2[x] is separable over Z2. Although f has a mul-tiple root, its irreducible factors do not have a multiple root. Notice also that we knowthat all polynomials in K[x] are separable when char K = 0. This follows from Theorem3.4.

Proposition 3.11 Let φ : K1 → K2 be a field isomorphism and let f ∈ K1[x] be aseparable polynomial of degree at least one. Let F1 be a splitting field of f over K1 andlet F2 be a splitting field of φ∗(f) over K2. Then there are exactly [F1 : K1] isomorphismsψ : F1 → F2 that extend φ.

Proof We prove this by induction on [F1 : K1]. Suppose first that [F1 : K1] = 1. Then fsplits into linear factors over K1, say

f = c(x− a1) · · · (x− an)

with c, a1, . . . , an ∈ K1. Hence φ∗(f) = φ(c)(x−φ(a1)) · · · (x−φ(an)) and φ∗(f) also splitsover K2. Hence F2 = K2 and ψ = φ is clearly the unique extension of φ.

Now suppose that [F1 : K1] > 1 and that the result holds whenever the degree is smaller.As F1 6= K1, f must have an irreducible factor g of degree at least 2. Let a be any rootof g in F1. Let b be any root of φ∗(g) (that is then an irreducible factor of φ∗(f)) in F2.By Lemma 2.12 there exists a field isomorphism φb : K1(a) → K2(b) that maps a to band where φb|K1 = φ. As f (and thus φ∗(f)) is separable we have that there are exactlydeg g = [K1(a) : K1] roots b of φ∗(g) (see Proposition 2.6). Now observe that F1 is thesplitting field of f over K1(a) and that F2 is the splitting field of φ∗b(f) = φ∗(f) over

29

K2(b). Thus by induction hypothesis, there are for each root b exactly [F1 : K1(a)] fieldisomorphisms ψ : F1 → F2 that extend φb. As there are [K1(a) : K1] such roots we get intotal at least

[F1 : K1(a)][K1(a) : K1] = [F1 : K1]

extensions of φ. From Lemma 3.9 we know that there are at most [F1 : K1] of these.Hence we get the equality. 2.

Remark. Taking K1 = K2 = K, F1 = F2 = F and φ = id we get in particular thatif F is the splitting field of a separable polynomial f ∈ K[x], of degree at least 1, then|Aut KF| = [F : K].

Theorem 3.12 Let K ⊆ F be a field extension. The following are equivalent.

(a) K ⊆ F is finite, normal and separable.(b) F is a splitting field of some separable polynomial f ∈ K[x] of degree at least 1.

Proof (a)⇒ (b). By Theorem 3.8 we know that F is a splitting field of some polynomialf ∈ K[x] of degree at least 1. To see that f is separable we need to show that none of itsirreducible factors over K have multiple root. Let g thus be one of the irreducible factorsand let a be one of the roots. Then g is the minimal polynomial of a over K and as theextension is separable we know that g has no multiple root.

(b)⇒(a). From Theorem 3.8 we know that the extension is finite and normal. We argueby contradiction and suppose that there is an element a ∈ F whose minimal polynomialg has a multiple root a. Let ψ ∈ Aut KF and consider the induced field isomorphismφ : K(a) → ψ(K(a)), t 7→ ψ(t). We know from Corollary 2.17 that b = ψ(a) is anotherroot of g and then ψ(K(a)) = K(b) and φ = φb : K(a) → K(b) where φb(a) = b andφb(t) = t for all t ∈ K. As g has multiple roots we get less than deg g = [K(a) : K] ofthese φb. By Lemma 3.9 we know that for each b there are at most [F : K(a)] elementsψ ∈ Aut KF that extend φb. Thus we have less than

[F : K(a)] · deg g = [F : K(a)] · [K(a) : K] = [F : K]

elements in Aut KF. This contradicts however Proposition 3.11 (see the remark after it). 2

Definition. We say that a field extension K ⊆ F is a Galois extension if it satisfiesthe following equivalent conditions.

(a) K ⊆ F is finite, normal and separable.(b) F is the splitting field of some separable f ∈ K[x], of degree at least 1, over K.

The following will be one of the main ingredients for the Fundamental Theorem of GaloisTheory.

Theorem 3.13 Let K ⊆ F be a Galois extension. Then

(a) |Aut KF| = [F : K].(b) For any K ⊆M ⊆ F and any φ ∈ Aut KM, there exists ψ ∈ Aut KF such

that ψ|M = φ.

30

Proof Part (a) we have already dealt with (see the remark after Proposition 3.11). Thepart (b) is also a consequence of Propostion 3.11 (see sheet 7 for the details).

31

4 The Fundamental Theorem of Galois Theory

I. The Fundamental Theorem.

Throughout this section K ⊆ F will be a Galois extension.

Proposition 4.1 Let H a subgroup of G = Aut KF. Then [F : FixH] = |H|.

Proof. By Theorem 3.6 we know that the extension K ⊆ F is simple, say F = K(a). LetB = {σ(a) : σ ∈ H}. Notice that τH = H for all τ ∈ H and therefore τ(B) = B. Nowconsider the polynomial f =

∏b∈B(X − b). As τ(B) = B, we have

τ ∗(f) =∏b∈B

(x− τ(b)) =∏

c∈τ(B)

(x− c) =∏c∈B

(x− c) = f.

Thus all the coefficients of f are in M = FixH and f is a polynomial over M. Notice alsothat f must be irreducible over M. To see this argue by contradiction and suppose thatf = gh with g, h ∈ M[x] both of degree less than deg f . Without loss of generality wecan suppose that a is a root of g. Notice that as all the elements of H fix the elements inM = FixH pointwise we have that τ ∗(g) = g for all τ ∈ H. But this implies that τ(a) isa root of τ ∗(g) = g for all τ ∈ H and thus all the elements in B (that is all the roots of f)are roots of g. But this gives us the contradiction that g = f . Thus f is irreducible overM and is thus the minimal polynomial of a over M. It follows that (see Proposition 2.6)

[F : M] = [M(a) : M] = deg f = |B| ≤ |H|.

On the other hand we know from Lemma 2.5 that H ≤ ΦΨ(H) and as M = Ψ(H) ≤ Fis a Galois extension we know from Theorem 3.13 (see Qn1(b) on Sheet 7 for the details)that

|H| ≤ |Φ(Ψ(H))| = [F : M].

Hence [F : FixH] = |H|. 2

Lemma 4.2 Let σ ∈ G = Aut KF and let M be a field such that K ⊆M ⊆ F. Then

Aut σ(M)F = σ · Aut MF · σ−1.

Proof We have that τ ∈ G fixes the elements in σ(M) pointwise if and only if τσ(a) =σ(a) for all a ∈ M if and only if σ−1τσ(a) = a for all a ∈ M. But this happens if andonly if σ−1τσ ∈ Aut MF that is if and only if τ is in σ · Aut MF · σ−1. 2

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We are now ready for the main result of this course. Let us first recall the Galois Corre-spondence. Let

F = {M : K ≤M ≤ F}, and G = {H : H ≤ G = Aut KF}.

We consider the mapsΦ : F → G,M 7→ Aut MF

andΨ : G → F , H 7→ Fix(H).

Theorem 4.3 (The Fundamental Theorem of Galois Theory). Let K ⊆ F be a Galoisextension with Galois group G = Aut KF. Then

(1) |G| = [F : K].(2) Φ and Ψ are mutual inverses and order reversing.(3) If M ∈ F then

[F : M] = |Φ(M)| and [M : K] = |G|/|Φ(M)|.

(4) For M ∈ F , we have that K ⊆M is normal if and only if Φ(M) is a normal subgroupof G and then the Galois group of K ⊆M is isomorphic to G/Φ(M).

Proof (1) This is Theorem 3.13(a).

(2) That the maps are order reversing follows from Lemma 2.5. Let M ∈ F . By Theorem3.13 (see Qn1(b) on Sheet 7 for the details) we have

[F : M] = |Φ(M)|, and [F : ΨΦ(M)] = |ΦΨΦ(M)|.

By Lemma 2.5, Φ = ΦΨΦ and thus it follows that [F : M] = [F : ΨΦ(M)]. By Lemma 2.5again M ≤ ΨΦ(M) and thus we conclude that ΨΦ(M) = M. We have thus shown thatΨΦ = idF .

Let H ∈ G. From Proposition 4.1, we know that

[F : Ψ(H)] = |H| and [F : ΨΦΨ(H)] = |ΦΨ(H)|.

By Lemma 2.5 we have Ψ = ΨΦΨ and thus it follows that |ΦΨ(H)| = |H|. From Lemma2.5 again we know that H ≤ ΦΨ(H) and thus we conclude that ΦΨ(H) = H. We havethus shown that ΦΨ = idG.

(3) We have already established the first identity. For the latter simply observe that

[M : K] =[F : K]

[F : M]= |G|/|Φ(M)|.

(4) We first show that K ⊆M is normal if and only if σ(M) = M for all σ ∈ G = Aut KF.First suppose that K ⊆M is normal and let σ ∈ G. For any a ∈M we have that the σ(a)must be a root of the minimal polynomial of a over K (see Corollary 2.17). As K ⊆ Mis normal we know however that all the roots are in M. Hence σ(a) ∈ M. This showsthat σ(M) ⊆M. Then also σ−1(M) ⊆M and thus M = σσ−1(M) ⊆ σ(M). We have thusσ(M) = M for all σ ∈ G. Conversely suppose that σ(M) = M for all σ ∈ G. Let a ∈ M

33

and let f ∈ K[x] be the minimal polynomial of a over K. Let b be any other root of f in Fwe want to show that b ∈M. By Lemma 2.12 we know that there is a field isomorphismφb : K(a) → K(b) that maps a to b and fixes K pointwise. By Proposition 3.11 we canextend φb to an element τ ∈ G (that is such that τ |K(a) = φb. But by our assumption weknow that τ(M) = M and hence in particular b = τ(a) ∈ M. We have thus establishedour claim.

To finish the proof of (4) we use Lemma 4.2. We have that K ⊆ M is normal if andonly if σ(M) = M for all σ ∈ G. But as Φ is a bijection this happens if and only if

Φ(M) = Φ(σ(M)).

By Lemma 4.2 we know that the right hand side is equal to σ ·Φ(M) · σ−1. Thus K ⊆Mis normal if and only if Φ(M) is normal in G.

Now suppose that K ⊆ M is normal. We have seen above that σ(M) = M for all σ ∈ G.There is thus a well defined group homormorphism Π : G → Aut KM, σ 7→ σ|M. FromTheorem 3.13(b) we know that every τ ∈ Aut KM can be extended to an automorphismσ ∈ G. This tells us that Π is surjective. We apply the 1st Isomorphism Theorem inGroup Theory. The kernel of Π is clearly Φ(M) and thus

Aut KM = im Π ∼= G/ker Π = G/Φ(M).

This finishes the proof of the Fundamental Theorem. 2

II. Examples

We will consider two general examples that are both interesting for us with respect to thegeneral theory. The first deals with finite fields and the 2nd with the roots of unity infields of characteristic 0.

A. Finite fields.

We have seen that for each prime p and each positive integer n there is (up to iso-morphism) exactly one finite field of order pn that consists of the roots of xp

n − x in asplitting field over Zp (and thus is itself the splitting field). Notice that the there are nomultiple roots of xp

n − x and thus it is separable. We thus have that Zp ⊆ F is a Galoisextension and we can apply the Fundamental Theorem of Galois Theory here.

Theorem 4.4 Let F be a field of order pn. Then the Galois group of Zp ⊆ F is cyclic oforder n.

Proof Let G be the Galois group. By the Fundamental Theorem of Galois Theory weknow that |G| = [F : Zp] = n. Consider the map φ : F → F, a 7→ ap. We have that ifφ(a) = φ(b) then 0 = φ(a) − φ(b) = ap − bp = (a − b)p (where the last identiy holds asthe characteristic is p) and thus a = b. This shows that φ injective and as F is finite, φ isthen a bijection. Also φ(1) = 1, φ(ab) = (ab)p = apbb and φ(a + b) = (a + b)p = ap + bp.We thus have φ is an automorphism and by the little Fermats Theorem we also know thatφ(a) = ap = a for all a ∈ Zp. Thus φ is in the Galois group G = Aut ZpF. In order to

34

finish he proof we show that 〈φ〉 = G, i.e. that o(φ) = n. Notice first that φn(a) = apn

= afor all a ∈ F and thus the o(φ) divides n. However if the order of φ was some positiveinteger m < n then we would have ap

m= a for all a ∈ F and thus all the pn elements of

F roots of xpm − x that is absurd as pn > pm. This finishes the proof. 2

Remark. We know from Group Theory that for each divisor d of n there is exactlyone subgroup of this order. Notice how this matches the situation with the field F (seeExercise 5 on Sheet 5) where we have seen that for each divisor d of n there exists exactlyone subfield of order pd.

Theorem 4.5 We have that xpn − x ∈ Zp[x] is the product of all the monic irreducible

polynomials over Zp whose degree divide n.

Proof Let F be the roots of of xpn −x in a splitting field (and thus F the splitting field).

Let f be any irreducible monic factor of xpn − x and let a be any root of f in a F. Then

n = [F : Zp] = [F : Zp[a]] · [Zp[a] : Zp] = [F : Zp[a]] · deg f

and thus deg f divides n. Conversely let f be any irreducible polynomial over Zp of degreed where d|n. Consider the field Zp[x]/Zp[x]f = Zp[t] where f is a the miniaml polynomialof t over Zp. As K = Zp[t] is a field of order pd we know (Exercise 5 on Sheet 5) that Fhas a copy of K as subfield and thus t is root of xp

n − x. As f is the minimial polynomialof t it follows that f |(xpn − x). 2

B. The polynomial xn − 1 ∈ Q[x].

Let F be a splitting field of xn − 1 over Q. From Group Theory we know that theroots form a cyclic multiplicative subgroup of F \ {0}, say 〈w〉 (as it is a finite subgroupof a field). This can also be seen from the fact that all the roots in C can be generatedby w = ei2π/n.

Notice that wk, 1 ≤ k ≤ n, has order n if and only if k is coprime to n and thusthe number of roots of xn− 1 that have order n (that is the primitive nth roots of unity)is φ(n) where φ is the Euler function. Now let

Φn(x) = (x− t1) · · · (x− tφ(n))

where t1, . . . , tφ are the primitive nth roots of unity. This gives us a poynomial of degreeφ(n). Now every root of xn − 1 has order d dividing n and thus

xn − 1 =∏d|n

Φd(x). (1)

Now consider xn − 1 as a polynomial in Z[x]. If we know all the Φd for d < n and d|nthen we can use (1) to determine Φn by dividing xn− 1 by the product of all the Φd withd < n, d|n. This gives us recusively that Φn is a monic polynomial in Z[x] (See problemsheet 8). In fact it turns out that Φn is irreducible over Q.

Theorem 4.6 The polynomial Φn is irreducible over Q.

35

Proof Let w be a primitive nth root of unity in a splitting field of F of xn−1 over Q andlet f be the minimial polynomial of w over Q. As f is monic and divides xn − 1 we havein fact that f ∈ Z[x] (see sheet 8). The aim is to show that f = Φn. We know that f |Φn

as Φn(w) = 0. In order to show that f = Φn it thus remains to see that all the roots ofΦn are roots of f . Now let wk be any of the roots of φn. Thus 1 ≤ k ≤ n and k is coprimeto n. Suppose k = p1 · · · pr is the prime factorisation of k. As k and n are coprime itfollows that none of the primes p1, . . . , pr divide n. The claim that wk = wp1···pr is a rootof f will thus follow from the following claim.

Claim. If θ is a root of f and p is a prime that does not divide n then θp is also a root of f .

To prove the claim we argue by contradiction and suppose there is a prime p where pdoes not divide n and were f(θ) = 0 whereas f(θp) 6= 0. As f is the minimial polynomialof w it must divide xn − 1 in Q[x], say

xn − 1 = fg (2)

with g ∈ Q[x]. In fact g ∈ Z[x] (again as g is monic and divides xn − 1). But thenf(θp)g(θp) = (θp)n − 1 = 0 and thus g(θp) = 0 and θ is a root of the polynomial g(xp).Thus f |g(xp). Consider the ring homomorphism φp : Z→ Zp, a 7→ a. For each polynomiale ∈ Z[x] we let e = φ∗(e) be the corresponding polynomial in Zp[x]. By Fermat’s littleTheorem we have g[x]p = g[xp]. Let k[x] be any irreducible factor of f . As f |g(xp) wehave that k then divides g[x]p = gp and thus (as k is irreducible) k|g. Thus k divides bothf and g and thus k2 divides

f g = xn − 1 ∈ Zp[x]

But then we have that xn − 1 ∈ Zp[x] has a multiple root. For this to be the case wewould (by Theorem 3.2) need xn−1 and D(xn−1) = nxn−1 to have a common irreduciblefactor. But this is absurd as the only irreducible factor of D(xn − 1) = nxn−1 is x thatdoes not divide xn − 1. (Notice that as p 6 |n we have that nxn 6= 0). This contradictionfinishes the proof of the claim and thus of the Theorem. 2

Remark. It follows in particular that Φn is the minimal polynomial of w oer Q.

36

5 Algebraicially closed fields and The Fundamental

Theorem of Algebra

In this chapter all fields will be of characteristic zero.

Definition. A field K is algebraically closed if every polynomial in K[x], of degree atleast 1, has a root in K.

Remark. Suppose K is algebraically closed and that f ∈ K[x] has degree at least 1.Then f has a root say a1 ∈ K and thus f = (x−a1)g for some g ∈ K[x]. If g has degree atleast 1 we can continue and find a root a2 of g in K. Continuing in this manner we see thatf splits into linear factors over K, i.e. f = c(x−a1) · · · (x−an) for some c, a1, . . . , an ∈ K.

The Fundamental Theorem of Algebra states that C is algebraically closed. This the-orem is a kind of a misnomer as C is a topological construction and its properties dependheavily in particular on the fact that R is a complete topological space. One can’t thusprove the Fundamental Theorem of Algebra using only algebraic methods. What we willdo is that we will prove a related result that is more general and is purely algebraic innature with an algebraic proof. We will call this the General Fundamental Theorem ofAlgebra. We will then see that we can deduce the algebraic closure of C from this resultusing only a basic result from Analysis (the Intermediate Value Theorem). The followinglemma gives us an equivalent description of algebraic closure that we will be using.

Lemma 5.1 A field K is algebraically closed if and only if there is no proper finite ex-tension K ⊂ F.

Proof (⇒). We work with the contrapositive. Suppose there is a finite proper extensionK ⊂ F. Let a ∈ F \ K with minimal polynomial f . Then f is irreducible of degree[K(a) : K] ≥ 2 (as a 6∈ K). In particular f has no root in K. Hence K is not algebraicallyclosed.

(⇐). Again we look at the contrapositive. Suppose that K is not algebraically closedand let f be a polynomial of degree at least 2 in K[x] with no root in K. Then we canfind a root a in a larger field F. Thus K ⊂ K(a) is a finite proper extension (as a 6∈ K). 2

Proposition 5.2 Let K be a field where K2 = K and where every polynomial of odddegree has a root in K. Then K is algebraically closed.

Proof Suppose K ⊆ F is a finite extension. We want to show that F = K. By Theorem3.6 we know that there exists θ ∈ F such that F = K(θ) (notice that K ⊆ F is separable as

37

char K = 0). Let f be the minimal polynomial of θ over K. By replacing F by a splittingfield of f over K we can assume that K ⊆ F is a Galois extension (note that f is separableas char K = 0). Now suppose [F : K] = 2rm where m is odd. We show that [F : K] = 1in two steps both which involve the Fundamental Theorem of Galois Theory and a grouptheory result we proved in MA30237: Group Theory.

Step 1. We show that m = 1.

Let G be the Galois group of K ⊆ F. By the Fundamental Theorem of Galois The-ory we have that |G| = [F : K] = 2rm. Let P be a Sylow 2-subgroup of G and let M bethe corresponding subfield of F under the Galois correespondence

F {id}| |

M P| |K G

By the Fundamental Theorem of Galois Theory we have [M : K] = |G|/|P | = 2rm/2r = m.Now let a ∈M and let fa be the minimal polynomial of a over K. Then

m = [M : K] = [M : K(a)] · [K(a) : K]

and thus deg fa = [K(a) : K] divides m. Thus fa is of odd degree and thus has a rootb ∈ K by our assumptions. As fa is irreducible (and has a as a root) it follows thatfa = x− b = x− a and a ∈ K. This shows that M = K and thus m = 1 and [F : K] = 2r.

Step 2. We show that r = 0 and thus [F : K] = 1.

We argue by contradiction and suppose that r ≥ 1. We now have |G| = [F : K] = 2r. ByGroup Theory there exists an ascending chain of normal subgroups of G

{1} = H0 < H1 . . . < Hr = G

where |Hj| = pj. Let M be the subfield of F that corresponds to Hr−1.

F {id}| |

M Hr−1

| |K G = Hr

By the Fundamental Theorem of Galois Theory we have that [M : K] = |G|/|Hr−1| =2r/2r−1 = 2. Let a ∈M\K. Then M = K(a). Let f = x2 +αx+β ∈ K[x] be the minimalpolynomial of a over K. Then

f = (x+ α/2)2 − (α2/4− β).

By our assumptions there is γ ∈ K such that γ2 = α2/4− β. Then

f = (x+ α/2)2 − γ2 = (x+ α/2− γ)(x+ α + γ)

38

that is a factorization in K[x]. This is absurd as f is irreducible over K. By this contra-diction we know thus that r = 0 and that [F : K] = 1. Hence there is no proper finiteextension K ⊂ F and K is algebraically closed. 2.

Remark. Essentially the Propostition tells us that for a field (of characteristic zero)to be algebraically closed, it suffices that every polynomial in K[x] of odd degree has aroot in K and that every element in K has a square root in K. It will then follow that allpolynomials of even degree in K[x] have a root as well in K.

Theorem 5.3 (The general Fundamental Theorem of Algebra). Let K be a field (of char-acteristic zero) where [K∗ : (K∗)2] = 2 and

(a) K∗ = (K∗)2 ∪ (−1)(K∗)2

(b) K2 + K2 = K2

(c) Every polynomial in K[x] of odd degree has a root in K.

Then F = K[x]/K[x](x2 + 1) = K[t] (where t2 + 1 = 0) is algebraically closed.

Proof Let F ⊆ L be a finite extension. We want to show that L = F. Much of the proofwill go through as it did in the proof of the last proposition. Suppose [L : F] = 2rm wherem is odd. Then

[L : K] = [L : F] · [F : K] = 2rm · 2 = 2r+1m.

As in the proof of last proposition we can assume without loss of generality that theextension K ⊆ L is a Galois extension. The same argument as was used in Step 1 of thatproposition shows that the fact, that all polynomials in K[x] of odd degree have a root inK, implies that m = 1. We thus have that [L : F] = 2r. Now the same argument, as wasused in Step 2 to deduce that r = 0 will work here provided that F2 = F. It thus onlyremains to see that any u = a+ tb in F has a square root in F. As (using (a))

K = (K∗)2 ∪ (−1)(K∗)2 ∪ {0} = (K∗)2 ∪ t2(K∗)2 ∪ {0},

we know that all the elements in K have a square root in F. We can thus assume that b 6= 0.

For any v = c + td in F we let v = c − td. Notice then that v · v = c2 + d2 ∈ K2

by (b), say v · v = α2 for α ∈ K. We will use this fact in the proof.

In particular there is c ∈ K such that c2 = a2 + b2 = uu. Notice that as b 6= 0 wehave u, u 6= 0 and thus their product is non-zero giving c 6= 0. Now c is in K and thusc = d2 for some d ∈ F and then

u = c(a/c+ t(b/d)) = d2(a/c+ t(b/c)).

It thus remains to see that a/c+t(b/c) has a square root in F. Notice that (a/c)2+(b/c)2 =(a2 + b2)/c2 = c2/c2 = 1. We can thus assume without loss of generality that u · u = 1.Now pick γ ∈ K such that γ2 = (1 +u)(1 + u) = (1 + a)2 + b2. Notice that this is zonzeroagain as 1 + u, 1 + u are nonzero. Then

(1 + u

γ)2 =

(1 + u)(1 + u)

(1 + u)(1 + u)=

(1 + u)u

(1 + u)u=

(1 + u)u

(u+ 1)= u.

This finishes the proof. 2

39

Theorem 5.4 (The Fundamental Theorem of Algebra). C is algebraically closed.

Proof It suffices to show that K = R satisfies (a),(b),(c) in Theorem 5.3.

(a) Let a > 0 and consider the polynomial f = x2 − a ∈ R[x]. Now f(0) = −a < 0and f(1 +a/2) = 1 + (a/2)2 > 0 and thus we know from the Intermediate Value Theoremfrom Analysis that f has a root t. Then c = t2. This shows that R∗ = R+ ∪ R− =(R∗)2 ∪ (−1)(R∗)2.

(b) This follows from R+0 + R+

0 = R+0 . Notice that ⊇ follows from a = a/2 + a/2.

(c) Consider any monic polynomial f = xn + an−1xn−1 + · · · + a0 of odd degree n. If

z ∈ R, then

f(z) = zn(1 +an−1

z+ · · ·+ a0

zn).

We have that f(z) tends to +∞ as z tends to +∞ and (as n is odd) f(z) tends to −∞as z tends to −∞. Again it follows from the IVT of Analysis that f has a root. 2.

Definition. Let K be a field. A field F is called an algebraic closure of K if K ⊆ F is al-gebraic and the field F is algebraically closed.

Example. C is an algebraic closure of R.

Remark. We explore algebraic closures a bit further on Problem Sheet 9.

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6 Galois’ Theorem

All fields in this chapter are subfields of C.

Definition. (a) We say that an extension K ⊆ F is radical if there is an ascendingchain of fields

K ⊆ K(a1) ⊆ K(a1, a2) ⊆ . . . ⊆ K(a1, . . . , an) = Fsuch that for each 1 ≤ i ≤ n, there exists a prime pi such that api

i ∈ K(a1, . . . , ai−1).

(b) We say that a polynomial f ∈ K[X] is solvable by radicals over K if for a split-ting field L of f over K, there is a radical extension K ⊆ F such that K ⊆ L ⊆ F.

Definition. We say that a group G is solvable if it has an ascending chain of subgroups

{id} = H0 ≤ H1 ≤ . . . ≤ Hn = G

for which Hi is normal in Hi+1 for all i = 0, . . . n − 1 and all the factors Hi/Hi−1,i = 1, . . . , n, are abelian.

Theorem 6.1 (MA30237: Group Theory). A finite group G is solvable if and only if ithas a an ascending chain of subgroups

{id} = H0 ≤ H1 ≤ . . . ≤ Hn = G

for which Hi is normal in Hi+1 for all i = 0, . . . n − 1 and all the factors Hi+1/Hi haveprime orders.

In this chapter we prove Galois’ Theorem on solvability by radicals. There will be a coupleof hurdles that need to be overcome in order to apply the Fundamental Theorem of GaloisTheory. One that will involve the roots of unity.

Theorem 6.2 Let F ⊆ C be the splitting field of xn − 1 over Q. Then the Galois groupof Q ⊆ F is isomorphic to Z∗n.

Proof Let w be a root of Φn in F. Then the roots of Φn are the elements wk where1 ≤ k ≤ n and (k, n) = 1. In particular F = Q(w) and [F : Q] = deg Φn = φ(n) = |Z∗n|.Now any element in the Galois group must map w to some other root wk, 1 ≤ k ≤ nwith (n, k) = 1. Thus it is clear that the Galois group G consists of σk : F → F whereσk(w) = wk and σk fixes Q pointwise. Finally notice that

Π : (Zn)∗ → G, k 7→ σk

is an isomorphism of groups as σkσr(w) = (wr)k = wrk = σkr(w). 2.

41

Proposition 6.3 Let p be a prime and K be a field containing θ = ei2π/p. Supposea ∈ K \ Kp. If K ⊆ F is normal and [F : K] = p, then there exists t ∈ F such thatF = K(t) and tp ∈ K.

Proof Let G be the Galois group of the extension. By the Fundamental Theorem ofGalois Theory we know that |G| = [F : K] = p. Thus G is cyclic say G = 〈σ〉 for someσ ∈ G. Now id, σ, . . . , σp−1 are linearly independent in the vector space of all functionsF → F (see sheet 10). In particular id + θσ + · · · + θp−1σp−1 6= 0 and thus there existsz ∈ F such that

t =

p−1∑k=0

θkσk(z)

is non-zero. Then

σ(t) =

p−1∑k=0

θkσk+1(z)

=1

θ

p−1∑k=0

θk+1σk+1(z)

=1

θ· t.

As t is not fixed by σ we have that t ∈ F\K and F = K(t). Also σ(tp) = (σ(t))p = (t/θ)p =tp and tp ∈ Fix (G) = K where the last equality holds by the Fundamental Theorem ofGalois Theory. 2

Proposition 6.4 Let G be a solvable group.

(a) If H ≤ G, then H is solvable.(b) If N �G, then G/N is solvable.

Proof See sheet 10.

Definition. Let F be the splitting field of f ∈ K[x] over K. We often call the Ga-lois group of K ⊆ F the Galois group of f over K.

Theorem 6.5 (Galois’ Theorem). The polynomial f ∈ K[x] is solvable by radicals overK if and only if the Galois group of f over K is solvable.

Proof of (⇐). Let F be the splitting field of f over K. The aim is to show that we canfind a radical extension K ⊆ L such that K ⊆ F ⊆ L.

Suppose [F : K] = n. By the Fundamental Theorem of Galois Theory we know that|Aut KF| = [F : K] = n. Let w = ei2π/n. Then F(w) is the splitting field of f · (xn − 1)over K.

As K ⊆ F is normal, we have (see the proof of the FTGT (4)) that σ(F) = F for allσ ∈ Aut KF(w). In particular we get a well defined group homomorphism

G = Aut K(w)F(w)→ Aut KF, σ 7→ σ|F.

42

This homomorphism is injective as any σ ∈ G that fixes the elements in F pointwise mustfix all the elements in F(w) (as it already fixes w). Hence G is isomorphic to a subgroupof Aut KF and thus G is solvable by Proposition 6.4. Notice also that Lagrange’s Theoremtells us that |G| divides n.

As G is solvable we can apply Theorem 6.1 (from MA30237) that tells us that we havean ascending chain of subgroups

{id} = Hn < H1 < · · · < Hm = G

whereHi−1�Hi and |Hi|/|Hi−1| = pi for some primes p1, . . . , pm. Let F(w) = F0, . . . ,Fm =K(w) be the corresponding subfields of F(w) under the Galois Correspondence.

F0 = F(w) H0 = {id}| |...

...| |

Fi−1Ψ←− Hi−1

| |Fi Hi

| |...

...Fm = K(w) G

By Exerecise 1 on Sheet 7 we know that Fi ⊆ Fi−1 is normal as Hi−1 �Hi. We also knowfrom the Galois Correspondence that [Fi−1 : Fi] = |Hi|/|Hi−1| = pi. As pi divides |G|(that we have seen previously that divides n) we have that K(w) (and thus Fi) containsei·2pi/pi = wn/pi . We can thus deduce from Proposition 6.3 that Fi−1 = Fi(ti) for someti ∈ Fi−1 where tpi

i ∈ Fi. We have thus established that we get a radical chain

K(w) = Fm ⊆ · · · ⊆ Fi ⊆ Fi−1 ⊆ · · · ⊆ F0 = F(w).

Now suppose that n = q1 · · · qs. Then we get another radical chain

K ⊆ K(wn/q1) ⊆ K(wn/q1q2) ⊆ · · · ⊆ K(wn/q1···qs) = K(w).

Merging this with the radical chain for K(w) ⊆ F(w) we see that K ⊆ F(w) is radical andthen for the splitting field F of f we have K ⊆ F ⊆ F(w). 2.

Remark. The proof turned out to be quite fiddly as we needed K to contain all thepith roots of unity to be able to use Proposition 6.3. This is a technical hurdle butthe core idea of the proof otherwise is the part of the proof that shows that shows thatK(w) ⊆ F(w) is radical knowing that K(w) contains all the necessary roots of unity.

Lemma 6.6 Let p be a prime and let K be a field that contains θ = ei·2π/p. Let a ∈ Ksuch that a is not a pth power of and element in K and let F be a splitting field of

f = xp − a

over K. Then the Galois group G of f over K is cyclic of order p.

43

Proof Let b be a root of xp − a in L. Then the complete list of roots is

b, bθ, bθ2, . . . , bθp−1

and F = K(b) (as θ ∈ K). By Sheet 6 (question 5), we have that xp − a is irreducible.Hence, by the Fundamental Theorem of Galois Theory,

|G| = [K(b) : K] = p.

Hence G is cyclic of order p. 2

Remark. We now want to prove the (⇒) part of Galois’ Theorem. We start thuswith a splitting field F of some polynomial over a field K. As the polynomial is solvableby radicals, there is a radical extension K ⊆ L such that K ⊆ F ⊆ L. Where F lies exactlydepends on the given polynomial, but there will be cases where F = L. It is thus naturalto work with K ⊆ L instead. The extension K ⊆ F is a normal extension and thus theGalois group G will be a quotient of Aut KL and thus (by Proposition 6.4) G is solvableprovided Aut KL is solvable.

There is however a problem with this. In order to apply the Fundamental Theoremof Galois Theory we need the radical extension K ⊆ L to be normal. However this isnot true in general. For example the radical extension Q ⊆ Q( 3

√2) is not normal. This

problem can though be sorted without much difficulty.

Proposition 6.7 Let K ⊆ L be a radical extension. Then there exists a field N such thatL ⊆ N and K ⊆ N is a radical extension that is normal.

Proof See problem sheet 10.

Proof of Galois’ Theorem (⇒). Let f ∈ K[x] that is solvable by radicals. Let Fbe the splitting field of f over K. Then there is a radical extension K ⊆ L such thatK ⊆ F ⊆ L. By Proposition 6.7 we can assume that K ⊆ L is normal (and thus a Galoisextension). By the remark before Propostion 6.7, it suffices to show that Aut KL is solv-able. We can thus assume without loss of generality that K ⊆ F is a radical extension.We thus have a radical chain

K ⊆ K(t1) ⊆ · · · ⊆ K(t1, . . . , tm) = F

where tpi

i = ai ∈ K(t1, . . . , ti−1) for some primes p1, . . . , pm. We now have similar technicaldifficulty as in the proof of the (⇐) part. Namely in order to use Lemma 6.6 we need Kto contain all the pjth roots of unity. We overcome this hurdle in a similar way as before.Let n = p1 · · · pm and let w = ei2π/n. Then F(w) is the splitting field of f(xn − 1) over K(and thus also K(w)). Consider now the radical chain

K(w) ⊆ K(w, t1) ⊆ · · · ⊆ K(w, t1, . . . , tm) = F(w).

We can assume here that tj 6∈ K(w, t1, . . . , tj−1) (as otherwise the term K(w, t1, . . . , tj)is redundant and can be omitted). Now let Fj = K(w, t1, . . . , tj) and let H0, H1, . . . , Hm

be the subgroups of H = Aut K(w)F(w) that correspond to F0, . . . ,Fm under the Galois

44

correspondence.Fm = F(w) Hm = {id}

| |...

...| |

FiΦ−→ Hi

| |Fi−1 Hi−1

| |...

...F0 = K(w) H0 = H

As Fi−1 contains θ = ei2π/pi we know that Fi = Fi−1(ti) is the splitting field of xpi − aiover Fi−1 and thus Fi−1 ⊆ Fi is a normal extension. By Exercise 1 on Sheet 7 we thenknow that Hi �Hi−1 is normal. By Lemma 6.6 we know also that Hi−1/Hi1, the Galoisgroup of Fi−1 ⊆ Fi, is cyclic of order pi. Consider now the larger lattice for K ⊆ F(w).

Fm = F(w) Hm = {id}| |...

...| |

FiΦ−→ Hi

| |Fi−1 Hi−1

| |...

...F0 = K(w) H0 = H

| |K G = Aut KF(w)

We have that K ⊆ K(w) is normal as the latter field is the splitting field of xn − 1 overK. Thus by Exercise 1 on Sheet 7, we have that H � G. By Theorem 6.2 we know thatG/H, the Galois group of K ⊆ K(w), is isomorphic to Z∗n and thus in particular abelian.We have thus seen that we get an ascending chain of subgroups of G

{id} = Hm ≤ Hm−1 ≤ · · · ≤ H0 ≤ H−1 = G.

where Hj �Hj−1 for j = m, . . . , 0 and where Hj−1/Hj is abelian. Thus G is solvable.

Finally Aut KF is isomorphic to a quotient of G (by the Fundamental Theorem of GaloisTheory) as K ⊆ F is normal. Hence the Galois group of K ⊆ F is solvable. 2

Galois Theory originated from the failed attempt to solve the quintic by radicals andit is thus appropriate to end by giving an example of an unsolvable quintic over Q.

An unsolvable quintic over Q.

Recall from Group Theory that the group Sn is solvable if and only if n ≤ 4. Wehave seen earlier (Proposition 2.18) that if f ∈ K[x] has n distinct roots (that happens

45

for example if f is irreducible), then the Galois group is isomorphic to a subgroup of Sn.Hence all polynomials of degree less than or equal to 4 are solvable by radicals.

We now want to find an example of a polynomial f ∈ Q[x] of degree 5, whose Galoisgroup is S5 and thus f not solvable by radicals.

Lemma 6.8 Let p be a prime and suppose that H is a subgroup of Sp that contains a2-cycle and such that |H| is divisible by p. Then H = Sp.

Proof (See Problem Sheet 10)

Proposition 6.9 Let p be a prime and let f be an irreducible polynomial of degree p overQ that has precisely two non-real roots in C. Then the Galois group G of f over Q isisomorphic to Sp.

Proof Let F be the splitting field of f over Q and let τ : C→ C, z 7→ z be the complexconjugation. Then the restriction τ |F ∈ G fixes all the real roots and swaps the twonon-real roots. It follows that G (thought of as a permutation of the roots) contains a2-cycle. Let a be any root of f in F. Then [Q(a) : Q] = p as f is irreducible. Considerthe chain

Q ⊆ Q(a) ⊆ F.

The tower law gives that p divides [F : Q]. But by the Fundamental Theorem of GaloisTheory we have that the latter is equal to |G|. Thus G is a subgroup of Sp, that containsa 2-cycle and whose order is divisible by p. By Lemma 6.8 it follows that G = Sp.

It is now easy to find an explicit example of quintic over Q that is not solvable by radicals.

Example. The polynomial f = x5 − 6x + 3 is not solvable by radicals. (See Prob-lem Sheet 10)

46