M345P11 Galois Theory

Lectured by Prof Alessio Corti at Imperial College London

Comments or corrections should be sent to wz302@cam.ac.uk.

Last updated: April, 2018

Contents

1 Introduction 21.1 What is Galois theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Galois theory when dealing with equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 A first example 62.1 The Eisenstein’s criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Field theory 113.1 Algebraic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Fundamental theorem of Galois theory – first half . . . . . . . . . . . . . . . . . . . . . . . 15

4 Normal and separable extensions 174.1 Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Normal extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 Separable degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.4 Separable extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5 Fundamental theorem of Galois theory 245.1 Galois theory with biquadratic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6 Examples of Galois correspondence 266.1 Galois groups of cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266.2 Galois groups of biquadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.3 Quartics equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286.4 Solubilities in radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.4.1 Soluble polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.4.2 Soluble groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.4.3 Sn not soluble for n ≥ 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.4.4 The theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.4.5 Generic equations of degree n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

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1 Introduction

week 2lecture 1

Galois theory is about fields which we denote by K. A field is a ring where 1 6= 0, and where for allx 6= 0, there exists ywith xy = 1.

Example.

1. Q = rational numbers.

2. the fraction field in n variables: k(t1, . . . , tn) = Frac(k[t1, . . . , tn]).

3. R, C.

4. finite field F = Z/(p) for a prime number p.

5. F = k[X]/(f) for a prime (irreducible) polynomial f ∈ k[X].

The last two examples are quotient rings of the form R/m for some maximal ideal m.

6. quadratic surds like Q(√2) = a+ b

√2 : a,b ∈ Q.

7. cyclotomic field: Q(ζn), where ζn = e2πi/n. It is defined to be the smallest subfield of C thatcontains ζn.

For example, F = Q(ω) = Q(√−3). Why?

Ifω ∈ F, thenω−ω2 =√−3 ∈ F.

If√−3 ∈ F, thenω = (−1+

√−3)/2 ∈ F.

1.1 What is Galois theory

week 2lecture 2

Definition. Let K,L be fields. A field homomorphism f : K → L is a set-theoretic function such thatf(a+ b) = f(a) + f(b), f(ab) = f(a)f(b) for all a,b ∈ K, and f(0) = 0, f(1) = 1.

Lemma. Every field homomorphism is injective.

Proof. If a 6= 0, then a has an inverse b. Now f(ab) = f(a)f(b) = 1, and therefore f(a) 6= 0. Then iff(x1) = f(x2), then f(x1 − x2) = 0, and from above x1 − x2 = 0.

Therefore, given a field homomorphism f : K→ L, we can always identify K as the image of f (subset)in L. We call the field homomorphism a field extension, and write as K ⊂ L.

Notation. If k ⊂ K,k ⊂ L, then Embk(K,L) = field homomorphisms f : K→ L : f|k = id.

Example. Q ⊂ Q(√2), R ⊂ C.

Galois theory studies field extensions, and, in particular, those extensions K ⊂ L where a polynomialf ∈ K[X] has no roots in K but has roots in L. For example, X2 − 2 ∈ Q[X] has roots in Q(

√2). We will

learn the following:

Given a field K, ∀f ∈ K[x], there exists an “optimal” extension K ⊂ L such that f(x) splits completelyin L, called the splitting field.

Remark. (Very important) If K ⊂ L is a field extension, then L is naturally a K-vector space.

Definition. An extension K ⊂ L is finite if L is finite dimensional as a K-vector space. The degree of Lover K is the dimension of L, denoted as [L : K] = dimKL.

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We will prove the tower law: suppose K ⊂ L ⊂M are extensions, then K ⊂M is finite if and only ifK ⊂ L and L ⊂M are finite, and [M : K] = [M : L][L : K].

Lemma. Suppose K ⊂ L is finite. Then every f ∈ Embk(L,L) is an isomorphism.

Proof. L is a finite dimensional vector space. An injective linear map from L to itself must also besurjective. Thus we have an inverse. Verify the inverse is a homomorphism.

It follows that for a finite extension k ⊂ L, Embk(L,L) = Autk(L) is the group field isomorphism f :

L→ L : f|k = id. This group is called the Galois group and it is denoted by Gk(L).

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Fundamental theorem of Galois theory

If k ⊂ L is finite, normal and separable, then there is a 1-to-1 correspondence between

intermediate fields K ⊂M ⊂ L←→ subgroups H ≤ G = AutkL

M 7−→ H = σ ∈ G : σ|M = id

M = a ∈ L : ∀h ∈ H,h(a) = a 7−→HNormal and separable: we will see that

1. K ⊂ L normal ⇐⇒ there exists f ∈ K[X] such that the splitting field of f is L.

2. If ch(K) = 0 (i.e. K contains Q), then all extensions are separable.

Example. We start with Q, and consider the splitting field of x3 − 2 = (x− 3√2)(x−ω 3

√2)(x−ω2 3

√2).

So the splitting field must containω (or√−3 equivalently) and 3

√2, i.e. consider Q ⊂ K = Q( 3

√2,√−3).

Clearly, G = AutQ(K) permutes the three roots of x3 − 2.

Suppose α is a root, so α3 − 2 = 0, and σ ∈ G = EmbQ(K,K).Then, 0 = σ(0) = σ(α)3 − σ(2). But σ ∈ G so it fixes all rationals, and therefore σ(α)3 − 2 = 0.

So we have a group homomorphism ρ : G→ S3. This map, in this case, is actually an isomorphism1.Here we represent S3 by

S3 = 〈σ = (123), τ = (23) | σ3 = τ2 = e, τστ = σ2〉.

We draw a diagram with partial order (subgroup) along the vertical direction, and with the correspon-dence at the top of this page in mind:

K = Q( 3√2,√−3)

F1

Q

F2 F3 F4

〈e〉

〈(23)〉

S3

〈(13)〉 〈(12)〉 〈(123)〉

Here, F1 to F4 are the intermediate fields, and we will use the correspondence above to figure outwhat they are. Recall that for each subgroup H, the intermediate field is defined to be

F = a ∈ Q(3√2,√−3) : ∀h ∈ H,h(a) = a.

Recall that if (23) is in G, it corresponds to switching the roots a2,a3.

1The reason for this will become clear later in the course

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a1 = 3√2

a2 = ω 3√2

a3 = ω2 3√2

Let’s start with H = 〈(23)〉. This corresponds to the complex conjugation, i.e. it fixes 3√2. So

F1 = Q( 3√2).

For σ = (123), notice that σ(ω) = σ(a2a1

) =a3a2

= ω. So σ fixesω, and F4 = Q(ω) = Q(√−3). So the

complete picture is the following:

K = Q( 3√2,√−3)

Q( 3√2)

Q

Q(ω 3√2) Q(ω2 3

√2) Q(

√−3)

〈e〉

〈(23)〉

S3

〈(13)〉 〈(12)〉 〈(123)〉

1.2 Galois theory when dealing with equations1

Suppose we want to solve y3 + 3py+ 2q = 0 over some field K 3 ω, and that it splits as (y− a1)(y−

a2)(y− a3) over L (note we have a1 + a2 + a3 = 0).

Suppose the Galois group G of the extension K ⊂ L contains σ = (123). Consider the two quantities:

u = ω2a1 +ωa2 + a3,

v = ωa1 +ω2a2 + a3.

Then we have σ(u) = ω2a2 +ωa3 + a1 = ωu, and σ(v) = ω2v. We can recover the a ′is from u, v:

a1 =ωu+ω2v

3, a2 =

ω2u+ωv

3, a3 =

u+ v

3.

So technically, if we can solve u, v, we can then solve the cubic equation. Notice even further thatσ(u3) = u3,σ(v3) = v3, i.e. σ fixes u3, v3.

Now suppose that G is actually the symmetric group on the three roots, and G contains τ = (12).Note that τ(u) = v, τ(u3) = v3. It follows that u3 + v3 and u3v3 are G-invariant.

Exercise. Find expressions for u3 + v3, u3v3 in terms of p, q and hence write a quadratic equationwith coefficients in K, of which u3, v3 are the two roots.

1cf. Q1 on Ex1

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2 A first example

Example.week 3lecture 1

Let K = Q( 3√2) ⊂ C be the smallest subfield of C that contains 3

√2. (Note that we already

know Q(√2) = a+ b

√2 : a,b ∈ Q)

Clearly, 3√2 ∈ K, 3

√2 · 3√2 = 3√4 ∈ K, and 3

√2 · 3√2 · 3√2 = 2 ∈ K. So we ask the question:

is it true that K is the same as X = Q[3√2] := a+ b

3√2+ c

3√4 : a,b, c ∈ Q?

This is the same as asking: is X a field?

Remark. Let L be a field, and X ⊂ L be a subset. Then X is a field iff 0, 1 ∈ X, and for all a,b ∈ X,a+ b, a− b, a · b, a/b (if b 6= 0) ∈ X.

Addition, subtraction and multiplication are easy to verify for X. So we need to find, for a givena+ b 3

√2+ c 3

√4 6= 0, some x,y, z ∈ Q with (a+ b 3

√2+ c 3

√4)(x+ y 3

√2+ z 3

√4) = 1.

Question/Exercise. Are 1, 3√2, 3√4 linear independent over Q? Let us first assume they are, and we

will go back to this at the end.

We expand out the terms in the product, and we have

(ax+ 2cy+ 2bz) + (bx+ ay+ 2cz)3√2+ (cx+ by+ az)

3√4 = 1.

Assuming linear independence, this is equivalent to solving the linear equation:a 2c 2b

b a 2c

c b a

xyz

=

100

.

And let’s denote the coefficient matrix by A. Then det(A) = a3 − 6abc+ 2b3 + 4c3.

Question: if a,b, c ∈ Q, does det(A) = 0 implies a = b = c = 0?

Let us denote a possible zero to det(A) by (a,b, c). First, since every term in det(A) is of degree3 (homogeneity), we can assume a solution a,b, c ∈ Z. Now we can further assume they do nothave common factor larger than 1 (we can always divide out).

Immediately we see amust be even, as every other term is even. So suppose a = 2a ′. But thenwe have

8a ′3 − 12a ′bc+ 2b3 + 4c3 = 0⇔ 4a ′3 − 6a ′bc+ b3 + 2c3 = 0,

and now bmust also be even. We write b = 2b ′ and

2a ′3 − 6a ′b ′c+ 4b ′3 + c3 = 0.

Now this forces c to be even as well. This contradicts our assumption that a,b, c have no commonfactor larger than one.

So in our case, a+ b 3√2+ c 3

√4 6= 0 implies that not all of a,b, c are 0, which further implies that

det(A) 6= 0. So there is a unique solution of x,y, z, and we can write down the formulae using Cramer’srule:

x =a2 − 2bc

det(A), y =

2c2 − ab

det(A), z =

b2 − ac

det(A).

The upshot is that X is indeed a field, and K = X.

We can of course generalise this result with the following proposition.

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Proposition. Suppose k ⊂ R ⊂ L where k,L are fields, and R is a ring. If R is finite dimensional as a k-vectorspace, then R is a field.

Example. Apply this to k = Q, R = X = Q[ 3√2], and L = C. Then X is a field.

Proof. Fix any 0 6= a ∈ R. Consider the map T : R→ R defined as T(x) = ax. This is obviously a k-lineartransformation. Since R ⊂ L, ax = 0 iff x = 0, and so T is injective. But an injective linear map from afinite dimensional vector space to itself must also be surjective. Thus there exists x ∈ R with ax = 1.

Now we go back to the question: are 1, 3√2, 3√4 linear independent over Q? Suppose not, i.e. there

exist a,b, c ∈ Q not all 0 such that a+b 3√2+ c( 3

√2)2 = 0. This is to say that 3

√2 is a root to the quadratic

equation a+ bx+ cx2 = 0 where a,b, c ∈ Q. Could this possibly happen? We know 3√2 does satisfy

x3 − 2 = 0 though.

Claim: x3 − 2 ∈ Q[x] is irreducible.

We will be dealing with the proof later. Let’s first assume it is the case.

On the other hand, if k is a field, f ∈ k[x] has a degree, deg(f) ∈ N. We then have the Euclid’salgorithm: for all f,g ∈ k[x], there exist q, r ∈ k[x] such that f = q · g+ r and deg(r) < deg(g). Theexistence of Euclid’s algorithm implies that:

• hcf always exists;

• irreducible and prime elements are the same;

• k[x] is a unique factorisation domain.

Now let f ∈ Q[x] be any polynomial. The polynomial x3 − 2 is irreducible implies that either x3 − 2 | f orhcf(f, x3 − 2) = 1. In the latter case, we know there will exist A(x), B(x) ∈ Q[x] such that

Af+B(x3 − 2) = 1.

This is an identity, so in particular it holds for x = 3√2, i.e. A( 3

√2)f( 3√2) = 1, and therefore, 3

√2 is not a

root of f. Equivalently, for any f ∈ Q[x], if 3√2 is a root of f, then x3 − 2must divide f.

Now we can conclude that 1, 3√2, 3√4 are linear independent over Q, since clearly x3 − 2 does not

divide a+ bx+ cx2.

2.1 The Eisenstein’s criterion

The last example suggests that we need some criteria for irreducibility of polynomials in Q[x].week 3lecture 2 Remark. Let p(x) = c0+c1x+ · · ·+cdxd ∈ Z[x]. Suppose x0 = p/q ∈ Q with p,q ∈ Z and hcf(p,q) = 1,

then x0 is a root of p implies that p | c0 and q | cd.

Proof. clearing up the denominators and we get qdc0 +qd−1c1p+ · · ·+ cdpd = 0. Now clearly p |qdc0,and therefore p | c0. Similarly q | cd.

Example. The polynomial x3 − 2 ∈ Q[x] is irreducible.

By the remark above, x3 − 2 does not have any rational roots. But if x3 − 2 were reducible in Q[x], itwould split as x3 − 2 = (x− x0)(x

2 + ax+ b), and therefore would have a rational root x0.

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off therecord

How about polynomials of degree 4 or more, for example, f = 1+ x+ x2 + x3 + x4?

First notice f = (x5 − 1)/(x− 1), and therefore f splits in C as f =∏4k=1(x− e

2πki/5). So the

only way to factorise f in R would be f = (x2− 2 cos2π

5x+ 1)(x2− 2 cos

4π

5x+ 1). So f is irreducible

in Q[x] iff cos2π

5= (√5− 1)/4 and cos

4π

5are rationals, which is clearly not the case.

In the case I do not know the exact value of cos2π

5, I can consider the field extension

Q ⊂ K = Q(e2πi/5),

and the Galois group acts as automorphisms of the group of 5-th roots of unity. So the Galoisgroup is C4. From this I know the value of cos (2π/5) should be in some quadratic surd.

Theorem. (Eisenstein’s criterion) Let f(x) = a0 + a1x+ · · ·+ adxd ∈ Z[x]. Suppose there exists a primenumber p such that p - ad, p |ai for all i < d, and p2 - a0, then f(x) ∈ Q[x] is irreducible.

Example. For any prime number p, the polynomial f(x) = xp−1 + xp−2 + · · ·+ 1 ∈ Q[x] is irreducible.

Proof. We prove an equivalent statement: f(x+ 1) is irreducible. Now notice that f(x) =xp − 1

x− 1, so

f(x+ 1) =(x+ 1)p − 1

x

= xp−1 +

(p

1

)xp−2 + · · ·+

(p

p− 1

).

Obviously now ad = 1, and p - ad; p |ai for all i < d; and a0 = p, p2 - a0.

off therecord

Consider the roots to x12 − 1, and find an irreducible polynomial f(x) ∈ Q[x] that f(e2πi/12) = 0.

We could try to factorise the polynomial:

x12 − 1 = (x6 + 1)(x6 − 1) = (x4 − x2 + 1)(x2 − x+ 1)(x2 + 1)(x2 + x+ 1)(x+ 1)(x− 1).

Now for every factor c | 12, we look at the primitive c-th roots (i.e. not an n-th root for any n < c) ofunity, and define

Φc =∏

ξ primitivec-th root of unity

(x− ξ) ∈ Q[x].

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Then the factorisation we get corresponds to

x12 − 1 = Φ12 ·Φ6 ·Φ4 ·Φ3 ·Φ2 ·Φ1.

For example, the 4-th primitive root i, −i are characterised by the term x2 + 1. So our final answer isΦ12 = x4 − x2 + 1.

Exercise. Show that x4 − x2 + 1 is irreducible over Q.

Exercise. More generally, for all n, define Φc for all c |n such that xn − 1 =∏c |nΦc; also prove

thatΦn is irreducible for all n, and is of degree φ(n), the Euler’s totient function.

Eisenstein’s criterion turns out to be a consequence of the following Gauss’ lemma.

Lemma. Let f(x) = a0 + · · ·+ adxd ∈ Z[x] with hcf(a0, . . . ,ad) = 1. Suppose f(x) splits non-trivially asf(x) = g(x)h(x) in Q[x], then f splits in Z[x].

To prove this, we need some ring theory:

Lemma. Let p ∈ Z be a prime number. Then p ∈ Z[x] is a prime element.

This says if p |a(x)b(x), then either p |a(x) or p |b(x).

Naively let’s consider a,b to be linear functions and p | (a0 +a1x)(b0 + b1x) = (a0b0 + (a0b1 +

a1b0)x+ a1b1x2). Then we must have p |a0b0, p |a0b1 + a1b0, and p |a1b1.

Without loss of generality, suppose p |a0. Then the second condition becomes p |a1b0. Eitherp |a1 and we are done, or p |b0, in which case from p |a1b1 we are always done.

Of course, alternatively, Z[x]/p = Fp[x] is an integral domain, and therefore the ideal (p) isprime.

Proof. (Gauss’ lemma) We know if p is prime in Z, then p is prime in Z[x].

We first kill the denominators: there exists a smallest c ∈N such that cf = g(x)h(x) with g, h ∈ Z[x].We claim that c = 1, and if so, we have a split in Z[x].

If c 6= 1, then there exists a prime number p | c, and therefore p | cf. By the ring theory above, wlogp | g. Then

c

pf =

g

ph

is another split in Z[x] but with c/p < c, contradicting our assumption that c is minimal.

Same idea shows that if R is a UFD, then R[x] is a UFD. So for a field k, k[x1, . . . , xn] is a UFD.

Proof. (Eisenstein’s criterion) By Gauss, if f is not irreducible in Q[x], then there exist h(x),g(x) ∈ Z[x]

such that f(x) = h(x)g(x). Modulo p, we get

0 6= adxd ≡ f(x) ≡ h(x)g(x) mod p

Since Fp[x] is a UFD, there exist d1,d2 with d1 + d2 = d, and bd1 · cd2 = ad such that

h ≡ bd1xd1 , g ≡ cd2x

d2 mod p

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Then we get

h(x) = bd1xd1 + bd1−1x

d1−1 + · · ·+ b1x+ b0 p |bi ∀i < d1g(x) = cd2x

d2 + cd2−1xd2−1 + · · ·+ c1x+ c0 p | ci ∀i < d2

But then, the constant term of f = hg is b0c0, which is divisible by p2, and this is a contradiction.

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3 Field theory

3.1 Algebraic extensions

Now that we know how to make many irreducible polynomials in Q[x], we go back to field theory.Suppose K ⊂ L, then for all a ∈ L, we write K(a) to be the smallest subfield of L that contains K and a.

Remark. If I is some indexing set, and Ki ⊂ L for all i ∈ I, then⋂i∈I Ki is a subfield of L, and K(a) =⋂

K⊂F⊂La∈F

F.

Definition. A field extension K ⊂ L is primitive if there exists an a ∈ L such that K(a) = L.

Definition. The evalutation homomorphism at a is the map φa : K[x]→ K(a) ⊂ L, f(x) 7→ f(a).

Proposition. Consider a field extension K ⊂ L and a ∈ L, and the evaluation homomorphism φa : K[x]→ L.Then there are two exclusive alternatives:

1. φa is injective, and φa extends to φa : K(x)∼−→ K(a) ⊂ L.

2. φa is not injective, and the kernel is the principal ideal generated by a unique monic irreducible polynomialf(x) ∈ K[x]. In this case, φa passes through the quotient to define a field homomorphism:

φa : K[x]/(f) ∼−→ K(a) ⊂ L.

If (1) holds, we say a is transcendental; if (2), we say a is algebraic over K, and f is the minimal polynomial of a.

The last example we have talked about illustrated all of these with K = Q ,L = C, a = 3√2 as an

algebraic extension. We know that a = e, π are transcendental, and Q(e) = Q(π) are isomorphic to Q(x),so this proposition doesn’t say much about transcendental extensions. In fact, we don’t really care abouttranscendental extensions in this course.

Proof.week 4lecture 1

We know K[x] is a Euclidean domain (ED), and thus a principal ideal domain (PID). Hence everyideal, ker(φa) in particular, is of the form (f) for some f ∈ K[x], and clearly there is a unique one that ismonic. We need to prove f is irreducible. However, this comes automatically. Consider the inclusion:

R = K[x]/(f)φa−−→ L.

Since L is a field, R contains no zero-divisors. In terms of elements in K[x], this translates to for allg,h ∈ K[x], if f | gh, then f | g or f | h, which is precisely to say f is prime (or irreducible, same in a ED).

It remains to showφa is an isomorphism from R to K(a). Clearly it is injective. To show it is surjective,first note that R is actually a field (the ideal f is maximal if you took algebra III, or for all g ∈ K[x], eitherf | g, in which case g is 0 in R, or hcf(f,g) = 1 as f is irreducible, and therefore the image of g in R isinvertible). So R ⊂ L is a subfield containing a and K. On the other hand, K(a) is defined to be thesmallest field in L that contains a and K. Together φa is surjective.

Corollary. Suppose K is a field and f(x) ∈ K[x]. Then there exists a field L and a ∈ L such that f(a) = 0.

Proof. Let g | f be a prime factor of f, and consider L = K[x]/g.

Corollary. If K ⊂ L1, K ⊂ L2 are two field extensions, and a1 ∈ L1, a2 ∈ L2 have the same minimal polynomialf(x) ∈ K[x], then there exists a unique isomorphism φ : K(a1)→ K(a2) such that φ(a1) = a2.

Proof. both are isomorphic to K[x]/f.

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Corollary. Let K ⊂ L = K(a) and a is algebraic with minimal polynomial f(x) ∈ K[x]. Let K ⊂ Ω be anotherfield extension. Then

EmbK(K(a),Ω) = roots of f(x) inΩ.

Proof. Suppose ξ ∈ Ω is a root to f. Then consider the evaluation homomorphism at ξ:

φξ : K[x]→ Ω.

Tautologically, kerφξ = (f) (fmust be in kerφξ, and conversely we know ker must be generated by aprime gwhich divides f, but f is already prime). So the proposition above states that there is a map

φξ : K(a) ∼= K[x]/f→ Ω

a 7→ x 7→ ξ.

Conversely, if we write f(x) = b0 + b1x+ · · ·+ xd, and φ : K(a)→ Ω ∈ EmbK(K(a),Ω), then at x = a,we apply φ to get:

0 = φ(0) = b0 + b1φ(a) + · · ·+φ(a)d.

So ξ = φ(a) is a root of f.

week 4lecture 2

The first corollary suggests we always have an extension field. However, f usually will only have oneroot in there.

Example. Consider x7 − 2 ∈ Q[x], L = Q( 7√2). In L[x], f(x) = (x − 7

√2)(x6 + 7

√2x5 + · · · + 7

√26).

Denote by g ∈ L[x] the second factor. Is g irreducible? (Yes, but you can try) If you are not convinced,x3 − 2will also do.

In C, f factors as f =∏7k=1(x− ε

k 7√2). We also have a 1-to-1 correspondence:

EmbQ(Q(7√2), C)↔ a ∈ C : a7 = 2.

So there are 7 embeddings of Q( 7√2) in C:

σk : Q(7√2)→ C

7√2 7→ εk

7√2

Definition. K ⊂ L is algebraic if for all a ∈ L, a is algebraic over K.

Naturally, we ask the two questions:

Question A: Suppose L = K(a) where a is algebraic over K. Does this imply K ⊂ L is algebraic?

Question B: Given K ⊂ L and a,b ∈ L algebraic. Are a+ b,a− b,a · b,a/b (if b 6= 0) algebraic?

To convince you these are not trivial questions, you can try playing around with these questionsusing only elementary tools. For example, if a satisfies some monic irreducible polynomial f(x), and bsatisfies some g(x), what polynomial does a+ b satisfy?

It turns out these questions can be solved with one simple idea, which we will illustrate below.

Definition. Let K ⊂ L be a field extension. The degree of L over K, degK L = [L : K], is the dimension ofL as a K-vector space (also written as dimK L). K ⊂ L is finite if dimK L <∞.

Remark. Notice that finite implies algebraic (not the converse). If we let n = dimK L <∞, and pick a ∈ L.Consider the set

1,a,a2, . . . ,an.

- 12 -

There are n + 1 elements, and therefore they must be linear dependent. In particular, there existλ0, . . . , λn ∈ K such that a satisfies λ0 + λ1x+ · · ·+ λnxn = 0.

Remark. If K ⊂ L, and a ∈ Lwith minimal polynomial f(x) ∈ K[x] of degree n, then [K(a) : K] = n.

Indeed, the set 1,a, . . . ,an−1 must be a basis of K(a) as a vector space over K. Linear independenceis obvious, as the minimal polynomial is of degree n. To see they generate K(a), first notice that the map

φa : K[x]→ K(a)

is surjective (from the proposition). Thus for any b ∈ K(a), there exists some g(x) ∈ K[x] such thatg(a) = b. We could apply Euclidean’s algorithm to get g = qf+ r where deg r < deg f = n. Thenb = g(a) = q(a)f(a) + r(a) = r(a), and r(a) is in the span of the set.

Theorem. (Tower law) If K ⊂ L ⊂ E, then [E : K] = [E : L][L : K]. In particular, K ⊂ E is finite iff both K ⊂ Land L ⊂ E are finite.

Surprisingly, this simply idea can directly leads to the solutions to both questions we asked.

K

K(a)

K(b)

⊂⊂

⊂

⊂K(a,b) = K(a)(b) = K(b)(a) ⊂ L

Corollary. Both questions have answer yes.

Proof to A. We know K ⊂ L = K(a) has degree n, thus this extension is finite. But finite implies algebraic.

Proof to B. Consider K ⊂ K(a) ⊂ K(a,b). We know K ⊂ K(a) is finite. The fact that b is algebraic over Ksurely implies b is algebraic over K(a), so K(a) ⊂ K(a,b) is finite. From tower law, [K(a,b) : K] is finite,thus algebraic.

Corollary. K ⊂ L. Then the set a ∈ L : a is algebraic over K is a field.

Proof. (Tower law) Suppose a1, . . . ,an ⊂ L is a basis of L as a K-vector space, and b1, . . . ,bm ⊂ E abasis of E as a L-vector space. We claim that aibj is a basis of E as a K-vector space.

First we prove this is a generating set. For any x ∈ E, x can be written as

x = λ1b1 + · · ·+ λmbm λj ∈ L,bi ∈ E

= (µ11a1 + · · ·+ µ1nan)b1 + . . . µij ∈ K,ai ∈ L ⊂ E

=∑

µijajbi.

To show the linear independence, suppose 0 =∑µijajbi =

∑j(∑i µijai)bj. Then

∑i µijai = 0 for

all j, and then µij = 0 for all i, j.

Theorem. Suppose K ⊂ L is finite, and K ⊂ E is finite. Then there exists a finite extension K ⊂ Ω andembeddings:

K

L

E

⊂⊂

⊂

⊂Ω

N.B. this theorem does not sayΩ is unique. In fact, we have

- 13 -

Q

Q( 7√2)

Q( 7√2)

⊂⊂

⊂

⊂Q( 7√2) or Q(e2πi/7, 7

√2)

inclusion

7√2 7→ e2πi/7

7√2

Proof. We prove by induction on [L : K]. If [L : K] = 1, then L = K, and we can simply take Ω = E.Otherwise, let a ∈ L \K algebraic with minimal polynomial f(x) = xn + λ1x

n−1 + · · ·+ λn ∈ K[x]. Wehave two cases: f has a root in E or not.

Case 1: if f has a root in E, by the third corollary to the proposition before, we know there is anembedding K(a) → E. Of course we also have K(a) ⊂ L. In a diagram:

K ⊂ K(a)L

E

⊂⊂

99K

99KΩ

By the tower law, [L : K(a)] < [L : K], and by induction there exists such anΩ.

Case 2: we know there exists E ⊂ E ′ such that f has a root in E ′. Then there also exists an embeddingK(a)

σ−→ E ′. In a diagram:

K ⊂ K(a)L

E ⊂⊂ E ′

⊂→ 99K

99KΩ

Now [L : K(a)] < [L : K], and we can use induction again to makeΩ.

week 5lecture 1

We mentioned in one of our remarks that an extension is finite implies it is algebraic. Now we give anequivalent condition.

Definition. K ⊂ L is finitely generated if there exist a1, . . . ,ar ∈ L such that L = L(a1, . . . ,ar).

Corollary. K ⊂ L is finite iff K ⊂ L is algebraic and finitely generated.

Proof. ⇒: We know finite implies algebraic. Now pick a ∈ L \ K. From the tower law, [L : K] = [L :

K(a)][K(a) : K]. But a 6∈ K implies [K(a) : K] > 1, and therefore [L : K(a)] < [L : K]. By induction K(a) isfinitely generated, and L = K(a)(a1, . . . ,ar) = K(a,a1, . . . ,ar).

The converse should be clear from the tower law, as

[L : K] = [L : K(a1, . . . ,ar−1)] · [K(a1, . . . ,ar−1) : K(a1, . . . ,ar−2)] · · · · · [K(a1) : K],

which is a finite product of finite numbers, and is therefore finite.

Theorem. Let K ⊂ L be finite, and K ⊂ Ω be another extension. Then

|EmbK(L,Ω)| ≤ [L : K].

Example. |EmbQ(Q( 7√2), C)| = 7.

- 14 -

Proof. If L = K(a), and f(x) is the minimal polynomial of awith degree n. Then we know there is a 1-to-1correspondence between EmbK(K,Ω) and roots of f inΩ, and there are at most n roots inΩ.

We do the general case by induction on the degree. Suppose K ⊂ L. Choose a ∈ L \ K. ConsiderK ⊂ K(a) ⊂ L, and restriction defines a function of sets:

EmbK(L,Ω)ρ−→ EmbK(K(a),Ω).

Now we want to estimate |ρ−1(φ)| for a given φ ∈ EmbK(K(a),Ω).

From the tower law, we know [L : K(a)] < [L : K], and we can therefore assume by induction that|ρ−1(φ)| ≤ [L : K(a)]. Then

|EmbK(L,Ω)| ≤ [L : K(a)][K(a) : K] = [L : K],

and we are done.

3.2 Fundamental theorem of Galois theory – first half

week 5lecture 2

We will finish off this section proving the easy half of the fundamental theorem of Galois theory.

Definition. Suppose L is a field, and G ⊂ Aut(L). The fixed subfield of G in L is the set K = a ∈ L :

∀g ∈ G,g(a) = a. It is also written as LG.

Theorem. L is a field, G ⊂ Aut(L) a finite group, and K = LG the fixed field of G. Then EmbK(L,L) = G.

Explanation: we previously mentioned that under certain conditions, a finite extension K0 ⊂ Lwillhave the correspondence

intermediate fields K0 ⊂ E ⊂ L←→ subgroups G of EmbK0(L,L) = AutK0L

E 7−→ EmbE(L,L)

K = LG 7−→GSo we are essentially saying that composing the second map with the first map gives identity with noextra condition at all.

N.B. G ⊂ EmbK(L,L)1, so |G| ≤ |EmbK(L,L)| ≤ [L : K] from the last theorem. It suffices to prove[L : K] ≤ |G|.

Proof. Let n = |G|, and pick any a1, . . . ,an+1 ∈ L. We will show they are K-linear dependent, i.e.non-trivial λi ∈ K such that λ1a1 + · · ·+ λn+1an+1 = 0.

Let G = σ1, . . . ,σn. Consider the matrix:

An,n+1 =

σ1(a1) σ1(a2) · · · σ1(an+1)

σ2(a1) σ2(a2) · · · σ2(an+1)...

.... . .

...σn(a1) σn(a2) · · · σn(an+1)

∈ Mn,n+1(L).

1Fix x ∈ G. For any k ∈ K, by definition, any g ∈ Gwill fix k, and therefore xmust also fix k.

- 15 -

In the L-vector space Ln+1, the linear system

A ·

λ1...

λn+1

= 0

will have non-trivial solutions since rk(A) < n+ 1. But solutions occur in L. We need to show λi’s are inK, and we are done.

By rearranging the columns of A if necessary, we may assume that

1. λ1, . . . , λm 6= 0, and λi = 0 for all i > m;

2. λm = 1;

and such thatm is the smallest number satisfying these conditions.

Under these conditions the solution is unique. Suppose we have another solution λ ′1, . . . , λ ′n+1, thenλ1 − λ

′1, . . . , λn+1 − λ ′n+1 will be another solution, and that them-th term will be 0, thus contradicting

the minimality ofm.

We show λj ∈ K by definition, i.e. for all σ ∈ G, σ(λj) = λj. Now for any σ ∈ G, we apply σ to ouroriginal linear equation to get (σ is a L-automorphism):

σσ1(a1) σσ1(a2) · · · σσ1(an+1)

σσ2(a1) σσ2(a2) · · · σσ2(an+1)...

.... . .

...σσn(a1) σσn(a2) · · · σσn(an+1)

σ(λ1)...

σ(λm)

0...0

= 0.

Notice that the map G→ G, g 7→ σ g is injective. So the rows are the same with the ones in Awith adifferent order. RHS is zero, so we can swap the rows to get a different solution:

σ1(a1) σ1(a2) · · · σ1(an+1)

σ2(a1) σ2(a2) · · · σ2(an+1)...

.... . .

...σn(a1) σn(a2) · · · σn(an+1)

σ(λ1)...

σ(λm)

0...0

= 0.

By uniqueness, σ(λj) = λj for all j, and λj ∈ K.

- 16 -

4 Normal and separable extensions

4.1 Splitting fields

Definition. K ⊂ L is a splitting field for f(x) ∈ K[x] (not necessarily irreducible) if

1. f(x) splits completely in L[x], i.e. f(x) =∏

(x− xi) for some xi ∈ L;

2. If λ1, . . . , λm are the roots of f(x) in L, then L = K(λ1, . . . , λm).

Proposition. Given a field K, and f ∈ K[x], we have the properties:

1. There exists a splitting field.

2. If K ⊂ L is a splitting field for f, and K ⊂ Ω is any field in which f splits completely, then EmbK(L,Ω) 6= ∅.

Assuming these properties, we have the following corollary.

Corollary. Any two splitting fields of f ∈ K[x] are isomorphic.

Proof. We know there exist embeddings σ1 : L1 → L2 and σ2 : L2 → L1.

Claim. σ1 is an isomorphism.

σ1 is injective and K-linear. But dimL1 = dimL2, as the existence of σ1 implies dimL1 ≤ dimL2,and the existence of σ2 implies dimL2 ≤ dimL1. So σ1 is an isomorphism.

Example. Q(√2,√3) is the splitting field of (x2 − 2)(x2 − 3). But it is also the splitting field for f(x) =

x4 − 10x2 + 1. Question1. How to show f splits completely without factorising it?

Proof. (proposition)

1. Suppose f is irreducible. Then consider K ⊂ L = K[x]/(f). Tautologically, a = x ∈ L is a root, asf(a) = f(x) = 0 ∈ L. So f has a linear factor X− a in L[X]. Putting in 1 root at a time, and we get asplitting field for f. If f is not irreducible, then we do the same to each of its prime factor.

Remark. the proof shows [L : K] ≤ n! where n is the degree of f.

2. Suppose a ∈ L is a root with minimal polynomial g | f. In particular, g splits completely in Ω.So there exists some b ∈ Ω that is a root of g, and we have an embedding K(a) → Ω, a 7→ b.Inductively, we could construct an embedding L = k(a1, . . . ,an)→ Ω.

Example.week 6lecture 2

A field with 8 elements. Consider

F = F2[x]/(x3 + x+ 1).

Note: x3 + x+ 1 is irreducible over F2, as neither 0 or 1 is a root. Denote by F2(a) the field F wherea3 + a+ 1 = 0.

Question. Does x3 + x+ 1 splits completely in F2(a) (i.e. F = F2(a))?

We can list all 8 elements in F, and make a multiplication table:

0 1 a a+ 1 a2 a2 + 1 a2 + a a2 + a+ 1

×a 0 a a2 a2 + a a+ 1 1 a2 + a+ 1 a2 + 1

×a2 0 a2 a+ 1 a2 + a+ 1 a2 + a a a2 + 1 1

1cf. Q8 on Ex1

- 17 -

Now make long division in F2(a) to get:

x3 + x+ 1 = (x− a)(x2 + ax+ a2 + 1)

and by noticing further that a2 is another root, we split x3 + x+ 1 as

x3 + x+ 1 = (x+ a)(x+ a2)(x+ a2 + a).

Example. Fix a prime number p. For all integerm ≥ 0, there exists a field F with q = pm elements, andit is unique up to isomorpihsm (denoted by Fq).

Proof. (1) Uniqueness: Suppose |F| = q, then Fmust be the splitting field for f(x) = xq − x ∈ Fp[x] overFp[x]. Indeed, for all a ∈ F, we must have aq = a, since either a = 0, and aq = 0 = a, or if a 6= 0, a is inthe group Fq \ 0 which is of order q− 1, and thus aq−1 = 1. But since F only has q elements, we musthave

xq − x =∏a∈F

(x− a).

Thus uniqueness.

(2) Existence: Let F be the splitting field for xq − x. Claim: |F| = q.

Notice that X := a ∈ F : aq − a = 0 ⊂ F actually satisfies X = F, as X is a field, and F is defined to bethe smallest field containing all the roots. The fact that X is closed under multiplication is obvious. Tosee it is closed under addition, notice that in a characteristic p field, (α+β)p = αp +βp, and then therest follows. Therefore if xq − x has q distinct roots, then the claim is true1.

4.2 Normal extensions

Definition. A finite extension K ⊂ L is normal if for all extensions K ⊂ Ω, and for all σ1 : L → Ω,σ2 : L→ Ω in EmbK(L,Ω), we have σ1(L) ⊂ σ2(L).

Remark. K ⊂ L ⊂ E. If K ⊂ E is normal, then L ⊂ E is normal, as any σ ∈ EmbL(E,Ω) also fixes K.

Theorem. For a finite extension K ⊂ L, the following are equivalent:

(I) K ⊂ L is normal

(II) for all irreducible f ∈ K[x], either f has no root in L, or f splits completely in L.

(III) there exists f ∈ K[x] such that K ⊂ L is a splitting field for f.

Proof. (I)⇒ (II): suppose f ∈ K[x] irreducible and has one root in L, say a ∈ L.

Claim. There exists a field extension K ⊂ Ω such that

1. f splits completely inΩ, i.e. there exist λ1, . . . , λm ∈ Ωwith f(x) =∏mi=1(x− λi).

2. for each i = 1, . . . ,m, the embedding σi : K(a)→ Ω extends to σi:

L

⊂

K(a)

⊂

K

Ω⊂σi

σi

1cf. Jacobian criterion

- 18 -

where σ(a) = λi which we know to exist and to be unique.

If this claim is true, then by normality, σi(L) = σ1(L) for all i. In particular, λi ∈ σ1(L) for all i = 1, . . . ,m.So we can write f(x) =

∏mi=1(x− σ

−11 (λi)), and f splits completely in L.

Now we prove the claim. Let K ⊂ E be a splitting field for f. So we have the following diagram:

Lσi

σi

Ωi

K(a) E

⊂

where σi(a) = λi, and Ωi is some other field that L,E embed in. Now by previous theorem, E ⊂ Ωi willgive E ⊂ Ω. This proves the claim.

(II)⇒ (III): Since K ⊂ L is finite, thus it is finitely generated, say L = K(a1, . . . ,al) for some ai ∈ L \K.Let fi(x) ∈ K[x] be the minimal polynomial for ai. Manifestly L is the splitting field for

∏li=1 fi(x).

(III)⇒ (I): We know L = K(λ1, . . . , λm), where f(x) =∏

(x − λi). Suppose L ⊂ Ω, and σ ∈EmbK(L,Ω).

K

L

L

⊂⊂

⊂

→ΩσFor all i, σ(λi) ∈ Ω is a root for f, i.e. σ sends λi : i = 1, . . . ,m to λi : i = 1, . . . ,m. However, σ(L) isgenerated by σ(λi), and therefore (I) follows.

Remark. For the last part of the proof, think of L = Q( 3√2) embeds in C in two different ways. But x3 − 2

splits inΩ = C, so σ : Q( 3√2)→ C is actually Q( 3

√2)→ Q( 3

√2).

Corollary. For every finite extension K ⊂ L, there exists K ⊂ L ⊂ Ω where K ⊂ Ω is normal.

Idea of proof. There exist ai such that L = K(a1, . . . ,al). Let fi be the minimal polynomial for ai, and takeΩ to be the splitting field of

∏li=1 fi(x).

4.3 Separable degree

Suppose K ⊂ L finite. We define the separable degree as

[L : K]s = |EmbK(L,Ω)|,

where L ⊂ Ω is some extension, and K ⊂ Ω is normal.

Remark. Back in a theorem we proved a while ago, [L : K]s ≤ [L : K].

Theorem.

1. [L : K]s is well defined, i.e. independent of the choice ofΩ.

2. If K ⊂ L ⊂ E, then [E : K]s = [E : L]s[L : K]s.

Lemma. Suppose K ⊂ L ⊂ E with K ⊂ E normal. E ⊂ Ω any extension and σ : L→ Ω embedding.

- 19 -

K ⊂ L Ωσ

E⊂ ⊂

⊂ ⊂

ΩEσ

Then σ(L) ⊂ E.

Proof. There exists a bigger field Ω such that σ : L→ Ω extends to σ : E→ Ω. But K ⊂ E is normal. Soσ(E) ⊂ E, and σ(L) ⊂ E.

Proof. (1) Suppose we pick two normal extension K ⊂ Ω1 and K ⊂ Ω2. We can always make a biggerfieldΩ:

K

Ω1

Ω2

Ω

L

We will argue that the degree measure in Ω1 is the same as in Ω, and same for Ω2. We may as wellassumeΩ1 ⊂ Ω2 = Ω.

Then by the previous lemma, any embedding from L toΩ2 has image insideΩ1, thus is an embeddingfrom L toΩ1. So [L : K]s is well defined.

week 7lecture 1

(2) Choose E ⊂ Ω such that K ⊂ Ω is normal. Then every embedding E→ Ω can be restricted to anembedding L→ Ω.

Claim: the restriction map res : EmbK(E,Ω) → EmbK(L,Ω) is surjective, i.e. every embeddingσ : L→ Ω extends to σ : E→ Ω.

First, we can do this in a bigger field:

Eσ

σ

Ω

L Ω

⊂ ⊂

But recall we chose E ⊂ Ω, so we can apply the previous lemma to

K ⊂ E Ωσ

Ω⊂ ⊂

to get σ(E) ⊂ Ω. QED claim.

Now for all σ, res−1(σ) = EmbL(E,Ω), and (since L ⊂ Ω is normal) therefore:

[E : K]s = [E : L]s[L : K]s.

So we have the tower law for the separable degree.

- 20 -

Question. Could it be that the separable degree is the same as the old degree we defined? N.B. Wedefinitely know [L : K]s ≤ [L : K].

If L = K(a), where a has minimal polynomial f ∈ K[x], then EmbK(L,Ω) = α ∈ Ω : f(α) = 0, i.e.[L : K]s is the number of distinct roots of f.

So our question becomes: is there a field K and f ∈ K[x] irreducible such that in some extensionΩ, fsplits completely but the roots are not distinct?

In a characteristic 0 field, for f ∈ K[x], λ ∈ L is a repeated root iff f ′(λ) = 0. So f has no repeated rootsiff hcf(f, f ′) = 1. If f is irreducible, it implies that f ′ 6= 0, then f has no repeated roots.

However, in a characteristic p field, f 6= 0with f ′ = 0 is possible. So we have some problems to dealwith here.

Example. K = Fp(t), the fraction field in t with coefficients in Fp. Consider f = xp − t irreducible. IfL = K(a) where ap = t, then f(x) = (x− a)p. In this case, [L : K]s = 1, [L : K] = p.

4.4 Separable extensions

Definition.week 7lecture 2

Let K be a field. A degree n polynomial f(x) ∈ K[x] is separable if it has n distinct roots insome extension of K. If K ⊂ L, an element a ∈ L is called separable over K if the minimal polynomial ofa is separable.

Example. K = Fp(t), a = p√t is not separable.

Indeed, the minimal polynomial of a is xp − t, since

1. a 6∈ K. If [P(x)/Q(x)]p = t, then Pp = tQp. This is an identity in the ring Fp[t], which is a UFD.So we use the usual argument to show t | P and t | Q.

2. xp − t = (x− a)p in Fp(a) ⊃ Fp(t). If (x− a)k ∈ Fp(t)[x] for some k < p, then expand this, wehave the second term having coefficient ka ∈ Fp(t), which is impossible.

Let K be a field. We can define a derivative operator D : K[x] → K[x], D(∑anx

n) =∑nanx

n−1.This map has the properties that:

1. D(λf+ µg) = λD(f) + µD(g),

2. D(fg) = D(f)g+ fD(g)

for all f,g ∈ K[x], λ,µ ∈ K.

Proposition.

1. f ∈ K[x] is separable iff hcf(f,Df) = 1.

2. If f is irreducible, then f is not separable iff Df = 0.

Proof. (i) Let f ∈ K[x], and K ⊂ Lwhere L splits: f(x) =∏mi=1(x− λi) ∈ L[x]. Then

Df =

m∑i=1

(

m∏j 6=i

(x− λj)).

If f is separable, we claim hcf(f,Df) = 1, as otherwise there exists an iwith x− λi | Df. This is impossiblebecause x− λi divide all but the i-th summand in the sum above. This shows hcf(f,Df) = 1 in L[x], sosurely also in K[x].

- 21 -

On the other hand, if f is inseparable, then there exists λ ∈ L such that (x− λ)2 | f, and f = (x− λ)g(x).Then x− λ | Df, and f,Df have a common factor x− λ in L[x]. We claim they must have a common factorin L[x] as well. Otherwise, we can write φf+ψDf = 1, but then this is an identity also holds in L[x],which is a contradiction.

(ii) If f is irreducible, then hcf(f,Df) 6= 1 iff hcf= f, and then f | Df, which is true iff Df = 0.

Remark. In a char 0 field, all irreducible polynomials are therefore separable. However, in a char p field,for f ∈ K[x], if Df = 0, then there must exist some g ∈ K[x] such that f = g(xp).

Indeed, if we write f = a0 + a1x+ . . . , then

Df = a1 + 2a2x+ · · ·+ papxp−1 + · · ·+ 2pa2px2p−1 + . . .

Therefore ai = 0 for all p - i.

Definition. A field K of characteristic p is perfect if Kp = K, or equivalently, for all a ∈ K, there existb ∈ K such that bp = a.

Example. For all q = pm, the finite field Fq is perfect.

This is because the group homomorphism F×q → F×q defined by b 7→ bq in injective, thus alsosurjective.

Proposition. If K is perfect, then every irreducibly polynomial is separable.

Proof.week 8lecture 1

If the irreducible polynomial f is not separable, then Df = 0, and by the remark above, f = h(xp).We write h(x) = xn + a1x

n−1 + · · ·+ an.

Since K is perfect, there exist bi’s such that bpi = ai. Then consider:

g(x) = xn + b1xn−1 + · · ·+ bn.

Therefore gp = xnp+bp1 (xp)n−1+ · · ·+bpn = h(xp) = f(x). This contradicts the fact that f is irreducible.

This proposition explains why we picked Fp(t) in our first example, as it is the simplest infinite fieldwith characteristic p, so that there would exist irreducible non-separable polynomial.

We would like to describe the separable extension without referring to elements of the fields.

Definition. A finite extension K ⊂ L is separable if: for all K ⊂ F1 ⊂ F2 ⊂ L, if [F2 : F1]s = 1, thenF1 = F2.

Remark. For K ⊂ F ⊂ L, if K ⊂ L is separable, then K ⊂ F and F ⊂ L are separable.

In fact the converse is also true.

Theorem.

1. K ⊂ L is separable iff [L : K]s = [L : K].

2. For K ⊂ F ⊂ L, K ⊂ L is separable iff K ⊂ F and F ⊂ L are both separable.

Proof. Suppose [L : K]s = [L : K], then by two tower laws, we have

[L : K] = [L : F2][F2 : F1][F1 : K]

[L : K]s = [L : F2]s[F2 : F1]s[F1 : K]s

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Now the left sides are equal, and every term in the top row is at least as large as the corresponding termin the bottom row, so they must equal to each other. In particular, [F2 : F1]s = [F2 : F1].

But [F2 : F1] = [F2 : F1]s = 1, so F1 = F2.

For the converse, we first prove the claim:

Claim: Suppose K ⊂ K(a) is separable, then a is separable over K (i.e. its minimal polynomial isseparable over K), and therefore [K(a) : K]s = [K(a) : K].

Let f(x) be the minimal polynomial of a. Assume for a contradiction that f(x) is not separable.Then by previous remark, f = h(xp), where h(x) ∈ K[x] must also be irreducible (if h = h1h2 thenf = h1(x

p)h2(xp)).

Now b = ap satisfies h, and thus h is the minimal polynomial of b. So we can consider the tower

K ⊂ K(b) ⊂ K(a)

where K ⊂ K(b) has degree the degree of h, and K ⊂ K(a) has degree the degree of f. Thus [K(a) :

K(b)] = p. We now claim the minimal polynomial of a over K(b) must be xp − b. Indeed, a satisfies thispolynomial, and it is of degree p. However, this polynomial has p identical roots, and [K(a) : K(b)]s = 1.This contradicts the fact that K ⊂ K(a) is separable. qed claim.

Now suppose K ⊂ L is separable, then pick a ∈ L \K, and consider K ⊂ K(a) ⊂ L. By the remark,both extensions are separable. By claim, [K(a) : K]s = [K(a) : K]. On the other hand, we know[L : K(a)] < [L : K] and K(a) ⊂ L is separable. So we assume by induction that [L : K(a)]s = [L : K(a)]. Bytwo tower laws, we have [L : K] = [L : K]s.

The second part is a simple corollary of the first part. We leave it as an exercise.

The following corollaries are the usual ways of presenting this topic on a textbook. They are directconsequences of the theorems above.

Corollary. An element a is separable over K iff K ⊂ K(a) is separable.

Proof. We proved⇐ in the previous claim. For⇒, if a is separable, then [K(a) : K]s = [K(a) : K], and bytheorem above, K ⊂ K(a) is separable.

Corollary. K ⊂ L is separable iff for all a ∈ L, a is separable over K.

Proof. ⇒ follows from previous corollary. ⇐: first notice that given K ⊂ F ⊂ L, and if a ∈ L is separableover K, then it is separable over F. Therefore, if we pick a and consider the tower K ⊂ K(a) ⊂ L, weknow both extensions are separable, thus K ⊂ L is separable.

As a fact, [L : K]s | [L : K]s, so we can write [L : K] = [L : K]s[L : K]i, where [L : K]i is called theinseparable degree. Moreover, the inseparable degree is a power of p, the characteristic of field K.

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5 Fundamental theorem of Galois theory

Definition.week 8lecture 2

K ⊂ L is a finite Galois extension if it is finite, normal, and separable.

Normal says there are enough roots in L, and separable says all roots are unique.

Theorem. (Fundamental theorem of Galois theory, aka Galois correspondence)Suppose K ⊂ L is finite Galois, then there is a canonical G-equivariant inclusion-reversing bijection between thefollowing sets:

intermediate fields K ⊂ F ⊂ L←→ subgroups H of G = EmbK(L,L)

F 7−→ F∗ = EmbF(L,L)

a ∈ L : φ(a) = a ∀φ ∈ H = H† = LH 7−→H.

Inclusion reversing: the bigger the field, the smaller the group (vice versa):

F1 ⊂ F2, then F∗2 ⊂ F∗1.

G-equivariant: notice G acts on both sets (on fields by applying the map, and on groups by conjugation), and thecorrespondence is compatible with these actions ∗ and †, e.g.:

(gF)∗ = (gHg−1).

We already showed that if H is a finite subgroup of field automorphisms of L, and H† its fixedfield, then EmbH†(L,L) = H, i.e. (H†)∗ = H, and [L : H†] = |H|. In fact, it is clear that H ⊂ (H†)∗ ⊂EmbH†(L,L).

Note: K ⊂ L is normal, so H† ⊂ L is normal (K ⊂ H† ⊂ L).

Proof. Since F ⊂ (F∗)†, it suffices to prove inclusiong in the other direction.

We know K ⊂ L is finite, normal and separable, therefore F ⊂ L is also finite, normal and separable.Recall that

F∗ = φ : L→ L : φ|F = id = EmbF(L,L).

Now F ⊂ L is normal, by the definition of separable degree, |EmbF(L,L)| = [L : F]s, i.e. |F∗| = [L : F]S.

Note also that1 F∗ ⊂ Emb(F∗)†(L,L), and therefore

|F∗| ≤ |Emb(F∗)†(L,L)| = [L : (F∗)†]s.

Now apply tower law of separable degree to F ⊂ (F∗)† ⊂ L:

[L : F]s = [L : (F∗)†]s[(F∗)† : F]s,

where LHS= |F∗|, and [L : (F∗)†]s ≥ |F∗|. So [(F∗)† : F]s must be 1. But the extension is separable, so[(F∗)† : F] = 1, and F = (F∗)†.

5.1 Galois theory with biquadratic extensions

Definition. Let K be a field with charK 6= 2. L/K is biquadratic if it is a splitting field of a polynomialf(x) = (x2 − a)2 − b for some a,b ∈ K.

1In fact we have equality here, i.e. F∗ = ((F∗)†)∗ (by taking H = F∗), but we only need inclusion in this direction.

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Of course, the roots are ±√a±√b. There are many possible intermediate fields between K and L.

Potentially something like the diagram below:

K(√b,√a+√b,√a−√b)

K(√b,√a−√b) K(

√b,√a2 − b) K(

√b,√a+√b)

K(√b) K(

√a2 − b)

K

Of course, in Kwith char 6= 2, if α is not a square, then [K(√α) : K] = 2. But for β ∈ K not a square, could

β be a square in K(√α)?

Say γ = a+ b√αwith γ2 = β. Then

a2 + b2α+ 2ab√α = β.

So ab = 0, and a = 0 or b = 0. But β not a square in K, so we must have b2α = β. E.g. 18 not a square inQ, but 18 = (3

√2)2.

Now let’s apply Galois theory. Let K be a field with char 6= 2, K 3 α,β 6= 0 not squares, and β/α alsonot a square in K. We just saw x2 −β = 0 has no roots in K(

√α).

Consider the splitting field of (x2 −α)(x2 −β), K(√α,√β).

K(√α,√β)

K(√α)

K

2

4

2

This extension is Galois, and G = EmbK(L,L) has size 4. But obviously we have 4 different subgroupsof G. Since C4 only has 3 subgroup, Gmust be C2 ×C2, which has 5 subgroups. So we have one moresubfield. The complete picture is:

K(√α,√β)

K(√α) K(

√αβ) K(

√β)

K

1

C2 × 1 diagonal subgroup 1×C2

C2 ×C2

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6 Examples of Galois correspondence

week 9lecture 1

For the remaining part of this course, we will be mainly dealing with some specific examples ofGalois correspondence, including the famous insolubility of quintic equations.

6.1 Galois groups of cubics

We study f(x) = x3 − σ1x2 + σ2x− σ3 ∈ K[x].

Question. Assume f(x) is irreducible, separable, and L the splitting field of f, then what is the groupGal(L/K)?

Proposition. Suppose f(x) = xn − σ1xn−1 + σ2x

n−2 ± · · · ± σn ∈ K[x] irreducible, separable, and L thesplitting field. Then G ⊂ Sn = group of permutations on the set of roots of f, and it acts transitively on theroots.

Proof. There exists a group homomorphism ρ : G→ Sn, i.e. if g ∈ G, λ∈L a root, then g(λ) is also a root.Indeed, since g ∈ EmbK(L,L), it fixes all elements of K, and therefore 0 = g(f(λ)) = f(g(λ)).

ρ is injective: suppose λ1, . . . , λn ∈ L are roots of f, and suppose g(λi) = λi for all i, then g is identityon K(λ1, . . . , λn), which is precisely L. Thus ρ has trivial kernel.

G acts transitively, i.e. given λ1, λi ∈ L roots, there exists a g ∈ G such that g(λ1) = λi: here we use fis irreducible.

K[x]/f L

K L

φ1

φ2

φ

There exist two embeddings φ1,φ2 : K[x]/f→ Lwhere φ1([x]) = λ1, and φ2([x]) = λi.

Since L/K is normal, there exists φ : L → L such that φ|K = Id, φ φ1 = φ2, and φ(λ1) = λi (Wecan always make bigger fieldΩ, and K ⊂ L ⊂ Ω is normal, thus the image of φ2 contains the image ofφ1).

Now back to cubics, G can only be S3 or a cyclic group of order 3.

Theorem. Assume char K 6= 2, and consider the discriminant∆ = σ21σ22− 4σ

31σ3− 4σ

32+ 18σ1σ2σ3− 27σ

23 ∈

K. Then ∆ is not a square in K iff G = S3.

Corollary. If char K 6= 3, then we can complete the cube and assume f(x) = x3+ax+b, and ∆ = −4a3− 27b2.

If we make a table of the discriminant of the polynomials x3 −nx− 1 (for n 6= 2), we would find that∆ is very rarely a square, i.e. almost always the case that G = S3.

Definition. A polynomial f(x1, . . . , xn) ∈ K[x1, . . . , xn] is symmetric if for all σ ∈ Sn, f(xσ(1), . . . , xσ(n)) =f(x1, . . . , xn). The elementary symmetric polynomial σi ∈ k[x1, . . . , xn] are defined by the formula:

(x− x1)(x− x2) · · · (x− xn) = xn − σ1xn−1 + σ2x

n−2 − . . .

In particular, σi are homogeneous of degree i.

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Let’s write out a few:

σ1 = x1 + x2 + x3 + · · ·+ xnσ2 =

∑i<j

xixj

σk =∑

i1<i2<···<ik

xi1xi2 . . . xik

Theorem. Every symmetric polynomial is a function of the elementary symmetric polynomials, i.e. an element ofK[σ1,σ2, . . . ,σn].

Now let’s go back to discriminant. Let the roots of f be λ1, λ2, λ3 ∈ L, and consider

d = (λ1 − λ2)(λ2 − λ3)(λ1 − λ3) = 6= 0

If char K 6= 2, then d is not symmetric, as (12) ∈ S3 will give −d. However, D = d2 will be symmetric.Exercise: D = ∆(λ1, λ2, λ3).

Now we can prove the theorem regarding discriminant.

Proof. Observe that G ⊂ A3 = C3 iff d ∈ K.

Hopefully it should all make sense too you if you go back to section 1.2 at this stage.

6.2 Galois groups of biquadratics

week 10lecture 2

Consider f(x) = x4 − 2ax2 + (a2 − b) ∈ K[x] where K is a field with char 6= 2. Let L be the splittingfield. We know that if none of b, a2 − bb(a2 − b) is a square in K, then [L : K] = 8, and the roots are±√a±√b. Fix β ∈ Lwith β2 = b, and α,α ′ ∈ Lwith α2 = a+β,α ′2 = a−β.

L

K(α) K(α ′)

K(β)

K

2

2

2

2

2

Task: determine the Galois group G and all subgroups/intermediate fields.

We know |G| = 8, and G ⊂ group of permutations of the 4 roots. Now take any σ ∈ G. Notice thatσ2(β) = σ(b) = b, and therefore σ(β) = ±β.

If σ(β) = β, then σ2(α) = σ(α2) = a+β, and thus σ(α) = ±α. Similarly σ(α ′) = ±α ′. If σ(β) = −β,then σ2(α) = σ(α2) = a−β, and thus σ(α) = ±α ′, and σ(α ′) = ±α. There are 8 possibilities. So in factall 8 cases occur.

α ′

−α α

−α ′

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By direct verification, G is the symmetry group of this square, i.e. G = D8 = 〈σ, τ |σ4 = τ2 = e, τστ =σ−1〉, where σ is the rotation counterclockwise by 90 degrees, and τ is the reflection along the horizontalaxis.

The lattice of subgroups of D8

Consider γ = α · α ′, δ = α+α ′, δ ′ = α−α ′. Then γ2 = a2 − b, and therefore K(γ) is degree 2 over K.Now σ2(α) = −α, σ2(α ′) = −α ′, and so σ2 fixes γ. Similarly στ fixes γ as well. So the fixed fixed fieldcorresponding to 〈σ2,στ〉 is K(γ).

Also (βγ)2 = b(a2 − b), so degree 2 over K. Same reasoning applies.

e

〈τ〉 〈σ2τ〉 〈σ2〉 〈στ〉 〈σ3τ〉

〈σ2, τ〉 〈σ〉 〈σ2,στ〉

D8

L

K(α) K(α ′) K(β,γ) K(δ) K(δ ′)

K(β) K(βγ) K(γ)

K

6.3 Quartics equations

Let f(x) = x4 + ax3 + bx2 + cx+ d ∈ K[x], and L the splitting field. Let’s just say the Galois groupG = S4.

There is a surjective homomorphism ρ : S4 → S3 with kernel V = C2 ×C2 (here ρ has a section andsplits, and S4 = V o S3).

ρ is defined by the actions on the set A,B,C, where A = 1, 4, 2, 3, B = 1, 3, 2, 4, C =

1, 2, 3, 4, and S4 is the usual actions on the numbers 1234. For example, (12)A = B, (12)B = A, andthus we can identify (12) by (AB) ∈ S3. The kernel is e, (14)(23), (13)(24), (12)(34) = V .

Given groups N,H, and actions of H on N, i.e. a group homomorphism ρ : H → Aut(N), we canmake a new group, called the semidirect product G = NoH, where G as a set is just the Cartesianproducts, and the group structure is by:

(n1,h1)(n2,h2) = (n1 · ρ(h)1(n2),h1h2).

This turns H into a subgroup of G, and N a normal subgroup of G, with N∩ K = e.

Nevertheless, we found a normal subgroup V of S4, and S4/V = S3. By Galois correspondence, we

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have a tower: (K = S†

4

)⊂(F = V†) ⊂ (L = e†

)where F ⊂ L has Galois group V = C2 ×C2 and has degree 4, and K ⊂ F has Galois group G/V and hasdegree 6. Notice that since V is a normal subgroup, K ⊂ F is a normal extension, and therefore F is asplitting field. We expect:

1. There is a cubic equation such that its roots generate F.

2. Go from F to L, we take two more square roots.

We want to understand F. Now inside L, V ⊂ S4 acts on the ring R = K[x1, x2, x3, x4], and F = V†. It islogical to look for V− invariants in the ring R.

The most common two sets of invariants (equivalent) are

Y1 = (x1 + x4)(x2 + x3) Z1 = x1x4 + x2x3

Y2 = (x1 + x3)(x2 + x4) or Z2 = x1x3 + x2x4

Y3 = (x1 + x2)(x3 + x4) Z3 = x1x2 + x3x4

(Exercise: K[Y1, Y2, Y3] = K[Z1,Z2,Z3] as the subring of V− invariants inside R)

Now we have done enough, and try to set F = K(Y1, Y2, Y3). In particular, we need to find a cubicequation with Y the roots. By construction, the S3 symmetric polynomials in the Y’s are, upon substitutingx’s, S4 symmetric polynomials; also the coefficients of the cubic are related to the value of S3 symmetricpolynomials in Y’s (or put it differently, the S3 symmetric functions are polynomials in the coefficients off(x)). After some calculations, we conclude Y’s are the solutions to

Y3 − 2bY2 + (b2 + ac− 4d)Y + (a2d+ c2 − abc) = 0.

This finishes step 1.

Now if we set u = x1 + x2, v = x3 + x4, then u + v = −a, uv = Y3. So u, v are solutions toW2 + aW + Y3 ∈ F[W]. Similarly solve for xi + xj, and eventually we get the roots xi.

6.4 Solubilities in radicals

week 11lecture 1

We aim to show: if K contains all roots of unity, and f ∈ K[x] separable polynomial. Then f(x) = 0soluble in radicals implies the Galois group of f is a soluble group. 1

As a consequence: For n ≥ 5, a generic equations of degree n is not soluble in radicals.

1. Make precise what it means for f(x) = 0 to be solubule.

2. Theory of soluble group.

3. Show Sn is not soluble for n ≥ 5.

4. Prove the theorem.

5. Generic equations.1In fact the converse is also true with the help of Kummer theory.

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6.4.1 Soluble polynomials

Here K is always a field containing all roots of unity, f ∈ K[x] separable, and L the splitting field of f.

Definition. K ⊂ F is soluble if there exists a chain of root extensions:

K = K0 ⊂ K1 ⊂ · · · ⊂ Km = F

where for all i = 1, . . . ,m, Ki = Ki−1(ai) and there exists ri such that arii = bi ∈ Ki−1 (if char(K) = p,then also p - ri). A polynomial f(x) is soluble in radicals if L ⊂ Fwhere K ⊂ F is a soluble extension.

Remark. Since ai is a root of Xri − bi ∈ Ki−1[X], and Ki contains all roots of unity, the polynomial mustsplits:

Xri − bi =∏

(x− εai)

for all εri = 1. So all Ki−1 ⊂ Ki are normal

Lemma. Suppose f is soluble, then L ⊂ F where K ⊂ F is soluble and normal.

Proof. Induction onm. Ifm = 1, then by remark above F over K is normal.

Now by induction, we know there is a F ⊃ Km−1 such that K0 ⊂ F is normal and soluble. We need toconstruct E ⊃ Km such that K0 ⊂ E is normal and soluble.

Ω

F E

K0 K1 · · · Km−1 Km Km−1(ri√b)

⊂⊂

⊂ ⊂ ⊂ ⊂

⊂ ⊂

=

We can of course construct anΩ containing both F and Km. Now the goal is to make E ⊂ Ω such thatK0 ∈ E is normal and soluble.

We’d like to set E = F(a). By remark, F ⊂ E is the splitting field of Xr − b. But K0 ⊂ E isn’t normal.

Instead, consider g(x) =∏σ∈Gal(F/K0)(x

r − σ(b)). For any τ ∈ Gal(F/K0), τg = g, and so thecoefficients are fixed by all τ, and therefore g(x) ∈ K0[x]. Let E be the splitting field of g(x). Then K0 ⊂ Eis normal.

6.4.2 Soluble groups

Definition.week 11lecture 2

A group G is soluble if there exists a chain of subgroups

G = G0 B G1 B · · · B Gm = e

where Gi/Gi+1 are abelian for all i = 0, . . . ,m− 1.

Let G be a group. If a,b ∈ G, define the commutator of a,b to be [a,b] = a−1b−1ab. a,b commuteiff [a,b] = e. The derived subgroup G ′ of G is the subgroup of G generated by all commutators.

Remark.

1. G ′ C G, as G acts on G ′ by conjugation and g sends commutators to commutators:

g−1[a,b]g = [g−1ag,g−1bg].

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2. In fact G ′ is characteristic in G. For any automorphism T of G, TG ′ = G ′. Indeed, T [a,b] = [Ta, Tb].In particular, G ′ is normal if we take T the conjugation map.

3. G/G ′ is abelian.

We can say more about G ′. Suppose there is a group homomorphism φ : G → A where A is abelian.Since φ[a,b] = 0 for all a,b, φ factors through G ′ to define a unique homomorphism φ ′ : G/G ′ → A

such that φ = φ ′ρ, where ρ is the natural projection G→ G/G ′. This universal property characterisesthe group G/G ′, and we call this group G/G ′ = Gab the abelianization of G.

G ′ is the commutator subgroup of G, and write G(2) = (G ′) ′ the commutator subgroup of G ′, and soon. This defines a chain of normal subgroups:

G B G ′ B G(2) B · · ·

Claim: G B G(i) for all i ≥ 1.

Look at G(2) first. For g ∈ G, we want g−1G(2)g ⊂ G(2). Now define T : G ′ → G ′ by T(a) = g−1ag.The image is contained in G ′ since g−1G ′g ⊂ G ′ by remark above. So this is an automorphism of G ′.But G(2) is characteristic in G ′, and therefore TG(2) ⊂ G(2).

Lemma. G is soluble iff there exists anm such that G(m) = e.

Proof. ⇐ is trivial. Now for⇒:

Suppose there is a chain G0 B G1 B · · · B Gm = e. We know G/G1 is abelian, and therefore G1contains all commutators, i.e. G1 ⊃ G ′; But G1/G2 is also abelian, and G2 ⊃ [G1,G1] ⊃ [G ′,G ′] = G(2);· · · · · · e = Gm ⊃ G(m). So G(m) = e.

Corollary. For a group G and a surjective group homomorphism φ : G→ G, if G is soluble, then G is soluble.

Proof. For all k, G(k) is the image of G(k) under φ.

Example. S4 is soluble.

In fact we showed S4 = V o S3 = (V oC3)oC2. And clearly V oC3 is A4. So S4 B A4 B V , and Vis abelian, so we are done.

6.4.3 Sn not soluble for n ≥ 5

Let a = (123),b = (145). Then [a,b] = (135). So if n ≥ 5, every 3-cycle is a commutator. Therefore for allk, (Sn)(k) contains all 3-cycles. In particular, it is never trivial.

In fact, An can be generated by 3-cycles. So (Sn)(k) = An for all k.

6.4.4 The theorem

Now f(x) = 0 is soluble in radicals, and so the splitting field L is contained in some F with K ⊂ F normaland soluble:

K = K0 ⊂ K1 ⊂ · · · ⊂ Km = F

where Ki = Ki−1(ai) = Ki−1( ri√bi) over Ki−1 is a root extension for all i.

Key point is: Ki−1 ⊂ Ki is normal [and separable charK - ri].

Claim: Gi = Gal(Ki/Ki−1) is abelian. This follows from the fact that Gi ⊂ µri = ε : εri = 1.

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Proof. (claim)

For any σ ∈ Gi, σ(ai) is another root to Xri − bi = 0. So σ(ai) = ε(σ)ai. This defines a map:Gi → µri . We claim this is an injective group homomorphism.

For any a such that ari = bi, we know a = η · ai, and thus σ(a) = σ(η)σ(ai) = ηε(σ)ai = ε(σ)a.This immediately implies the last claim.

So now Gi are abelian. By the fundamental theorem of Galois theory, the chain of fields extensioncorresponds to a chain a subgroups: (not the same Gi)

G = G0 ⊃ G1 ⊃ · · · ⊃ Gm = e

Since Ki−1 ⊂ Ki is normal, Gi C Gi−1 is normal. Moreover, Gi/Gi−1 is the Galois group of Ki overKi−1, and is therefore abelian. So we conclude: Gal(F/K) is soluble.

However we need to show Gal(L/K) is soluble. Now we have a tower of extensions:

K ⊂ L ⊂ F

K ⊂ L normal, so L is the fixed field of a normal subgroup H C G, and G/H = Gal(L/K). In particular,we have a soluble group G = Gal(F/K), and a surjective group homomorphim G → G/H. By theprevious corollary, G/H is soluble. This finishes the proof.

6.4.5 Generic equations of degree n

f(x) = (x− x1)(x− x2) · · · (x− xn) = xn − σ1xn−1 + · · · ± σn ∈ K(x1, . . . , xn)[x].

So K(σ1, . . . ,σn) ⊂ K(x1, . . . , xn) is the splitting field of f, and is created separable. We can talk aboutthe Galois group of this extension.

Clearly Sn acts on K(x1, . . . , xn) and K(σ1, . . . ,σn) ⊂ S†n, i.e. we have a tower:

K(σ1, . . . ,σn) ⊂ S†n ⊂ K(x1, . . . , xn).

Recall the first half of the fundamental theorem that if G is a finite subgroup of the Galois group, then[L : LG = G†] = |G|. So we have:

[K(x1, . . . , xn) : S†n] = |Sn| = n! ≤ [K(x1, . . . , xn) : K(σ1, . . . ,σn)] ≤ n!

The last inequality comes from the fact that this extension is the splitting field of a degree n polynomial,and hence has maximal possible degree of extension n!. So in fact this extension K(σ1, . . . ,σn) ⊂K(x1, . . . , xn) has precisely degree n!, and the Galois group is Sn.

End of examinable part of notes.

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