Part II – Galois Theory

Based on lectures by Dr C. BrookesNotes taken by Bhavik Mehta

Michaelmas 2017

Contents

0 Introduction 20.1 Course overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1 Field Extensions 31.1 Motivatory Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Review of GRM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Digression on (Non-)Constructibility . . . . . . . . . . . . . . . . . . . . . . . 71.4 Return to theory development . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Separable, normal and Galois extensions 142.1 Trace and Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Normal extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Fundamental Theorem of Galois Theory 243.1 Artin’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 Galois groups of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.3 Galois Theory of Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Cyclotomic and Kummer extensions 324.1 Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.2 Kummer Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.3 Cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.4 Quartics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.5 Solubility by radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5 Final Thoughts 455.1 Algebraic closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455.2 Symmetric polynomials and invariant theory . . . . . . . . . . . . . . . . . . 47

Index 49

1

0 Introduction

The primary motivation of this course is to study the solutions of polynomial equations inone variable to wander whether there is a formula involving roots, a solution by radicals.Quadratics were typically studied in school, while the solution in radicals for cubics andquartics has been known for a long time and studied in particular in 1770 by Lagrange.

In 1799, Ruffini claimed that there were some quintics that can’t be solved by radicals,that is, there is no general formula, but it took until 1824 before Abel used existing ideasabout permutations to produce the first accepted proof of insolubility, before dying in 1829.Galois’ main contribution was in 1831, when he gave the first explanation as to why somepolynomials are soluble by radicals and others are not. He made use of the group of per-mutations of the roots of a polynomial, and realised in particular the importance of normalsubgroups.

Galois’ work was not known generally in his lifetime - it was only published by Liouvillein 1846, who realised that it tied in well with the work of Cauchy on permutations. Galoishad submitted his work for various competitions and for entry into the Ecole Polytechniquein Paris. Unfortunately Galois died in a duel in 1832, leaving a six and a half page letterindicating his thoughts about the future development of his theory.

0.1 Course overview

Most of this course is Galois Theory, but presented in a more modern fashion - in terms offield extensions. Recall from GRM that if f(t) is an irreducible polynomial in k[t] where kis a field, then k[t]/(f(t)) is a field, where (f(t)) denotes the ideal of k[t] generated by f(t),and this new field contains k. In this way, we can see the field k[t]/(f(t)) as a field extensionof k.

Prerequisites Quite a lot of the Groups, Rings and Modules course, but no modulesexcept in one place where it’s useful to know the structure of finite abelian groups. TheDPMMS website has a Galois Theory page with a long history of example sheets and notes,in particular see Tony Scholl’s 2013-4 course page.

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1 Field Extensions

Definition 1.1 (Field extension). A field extension K ≤ L is the inclusion of a field Kinto another field L with the same 0, 1, and where the restriction of + and · (in L) to Kgives the + and · of K.

Example.

(i) Q ≤ R

(ii) R ≤ C

(iii) Q ≤ Q(√

2) ={λ+ µ

√2∣∣ λ, µ ∈ Q }

(iv) { λ+ µi | λ, µ ∈ Q } = Q(i) ≤ C

Suppose K ≤ L is a field extension. Then L is a K-vector space using the addition fromthe field structure and scalar multiplcation given by the multiplication in the field L.

Definition 1.2 (Degree). The degree of L over K is dimK L, the K-vector space dimensionof L. This may not be finite. We typically denote this by |L : K|. If |L : K| <∞, then theextension is finite, otherwise the extension is infinite.

Example.

(i) |C : R| = 2, with R-basis 1, i

(ii) |Q(i) : Q| = 2, with Q-basis 1, i

(iii) Q ≤ R is an infinite extension.

Theorem 1.3 (Tower law). Suppose K ≤ L ≤ M are field extensions. Then |M : K| =|M : L| |L : K|.

Proof. Assume that |M : L| < ∞, and |L : K| < ∞. Take an L-basis of M , given by

{ f1, . . . , fb }, and a K-basis of L given by { e1, . . . , ea }. Take m ∈ M , so m =∑bi=1 µifi

for some µi ∈ L. Similarly, µi =∑aj=1 λijej for some λij ∈ K, so

m =

b∑i=1

a∑j=1

λijejfi

Thus { ejfi | 1 ≤ j ≤ a, 1 ≤ i ≤ b } span M .Linear independence: It’s enough to show that if 0 = m =

∑∑λijejfi then λij are all

zero. However if m = 0 the linear independence of fi forces each µi = 0. Then the linearindepedence of ej forces λij all to be zero, as required.

The Tower law will not be proved for infinite extensions, but observe that if M is aninfinite extension of L then it is an infinite extension of K, and similarly if L is an infiniteextension of K then the larger field M must also be an infinite extension of K.

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1.1 Motivatory Example

Observe Q ≤ Q(√

2)≤ Q

(√2, i)

(i) Q(√

2)

has basis 1,√

2 over Q.

(ii) Q(√

2, i)

has basis 1, i as a Q(√

2)–vector space.

(iii) Q(√

2, i)

has basis 1,√

2, i, i√

2 over Q.∣∣∣Q(i,√2)

: Q∣∣∣ = 4 = 2 · 2 =

∣∣∣Q(i,√2)

: Q(√

2)∣∣∣ ∣∣∣Q(√2

): Q∣∣∣

Any intermediate field strictly between Q and Q(√

2, i)

must be of degree 2 by the Towerlaw.

What are these intermediate fields? There are Q(√

2), Q(i) and Q

(i√

2), but are these

all?The Galois correspodence arising in the Fundamental Theorem of Galois Theory gives

an order reversing bijection between the lattice of intermediate subfields and the subgroupsof a group of ring automorphisms of the big field (in this case Q

(i,√

2)) that fix the smaller

field elementwise. For instance, consider the ring automorphisms of Q(i,√

2)

that fix Q:

e :√

2 7−→√

2

i 7−→ i

g :√

2 7−→√

2

i 7−→ −i

h :√

2 7−→ −√

2

i 7−→ i

gh :√

2 7−→ −√

2

i 7−→ −i

Notice that i and −i play the same role in the field Q(√

2, i), both roots of t2 + 1 = 0,

similarly√

2 and −√

2 are both roots of t2 − 2 = 0. The automorphism e is seen to beidentity, and g is conjugation. These four form the group of order 4 =

∣∣Q (√2, i)

: Q∣∣.

Q

Q(i) Q(i√

2)

Q(√

2)

Q(i,√

2)

{e, g} {e, h} {e, gh}

{e, g, h, gh}

{e}

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The recipe for producing an intermediate subfield from a subgroup is to take the elementsof Q

(i,√

2)

which are fixed by all elements of the subgroup. For instance, Q(i√

2)

is thefield of elements fixed by both e and gh.

This correspondence doesn’t always work for all finite field extensions. It works for Galoisextensions. In the correspondence, normal extensions correspond to normal subgroups. Inthis example, all subgroups are normal and the extensions are normal. We’ll also prove thePrimitive Element Theorem, which in the context of finite extensions of Q tells us that theyare necessarily of the form Q(α) for some α, for instance Q(i,

√2) = Q(i+

√2).

1.2 Review of GRM

Definition 1.4 (Algebraic). Suppose K ≤ L is a field extension. Take α ∈ L and define

Iα = { f ∈ K[t] | f(α) = 0 }

We say α is algebraic over K if Iα 6= 0. Otherwise α is transcendental. We say L isalgebraic over K if α is algebraic over K for all α ∈ L.

Remark. We can see Iα is an ideal of K[t] since it is the kernel of the ring homomorphismK[t]→ L given by f(t) 7→ f(α).

Example.

(i)√

2 is algebraic over Q

(ii) π is not algebraic over Q

Lemma 1.5. Let K ≤ L be a finite field extension. Then L is algebraic over K.

Proof. Let |L : K| = n, and take α ∈ L. Consider 1, α, α2, . . . , αn, which must be linearlydependent in the n-dimensional K–vector space L. So,

∑ni=0 λiα

i = 0 for some λ ∈ K notall zero, and hence α is a root of f(t) =

∑ni=0 λit

i, so α is algebraic over K. α was arbitrary,so L is algebraic over K.

Definition 1.6 (Minimal polynomial). The non-zero ideal Iα (where α is algebraic over K)is principal since K[t] is a principal ideal domain. In particular, we can say Iα = (fα(t))where fα(t) can be assumed to be monic. Such a monic fα(t) is the minimal polynomialof α over K.

Remark. Multiplication by α within the field L gives a K–linear map L → L, an auto-morphism (if α 6= 0). In GRM, we have seen the minimal polynomial of a linear map isunique.

Example.

(i) The minimal polynomial of√

2 over Q is t2 − 2.

(ii) The minimal polynomial of√

2 over R is t−√

2.

Lemma 1.7. Suppose K ≤ L is a field extension, α ∈ L and α is algebraic over K. Thenthe minimal polynomial fα(t) of α over K is irreducible in K[t] and Iα is a prime ideal.

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Proof. Suppose fα(t) = p(t)q(t). We aim to show p(t) or q(t) is a unit in K[t]. But0 = fα(α) = p(α)q(α), so p(α) = 0 or q(α) = 0, without loss of generality take p(α) = 0,thus p(t) ∈ Iα.

But Iα = (fα(t)), so p(t) = fα(t)r(t), giving fα(t) = fα(t)r(t)q(t) and so r(t)q(t) = 1 inK[t], and q(t) is a unit, as required. Recall from GRM that irreducible elements of K[t] areprime and hence generate prime ideals of K[t]. So Iα is a prime ideal.

Definition 1.8 (Simple extension). Suppose K ≤ L is a field extension and α ∈ L. K(α) isdefined to be the smallest subfield of L containing K and α. It’s called the field generatedby K and α. We say that L is a simple extension if L = K(β) for some β ∈ L.

Given α1, . . . , αn ∈ L, K ≤ L. K(α1, . . . αn) is the smallest field containing α1, . . . , αn.It is the field generated by K and α1, . . . , αn.

On the other hand K[α] is the ring generated by K and α, in particular the image ofK[t] under the map f(t) 7→ f(α).

Theorem 1.9. Suppose K ≤ L is a field extension and α ∈ L is algebraic over K. Then

(i) K(α) = K[α]

(ii) |K(α) : K| = deg fα(t) where fα(t) is the minimal polynomial of α over K.

Proof.

(i) Clearly K[α] ≤ K(α). We aim to show that any non-zero element β of K[α] is a unit,so K[α] is a field.

By definition of K[α], we have β = g(α) for some g(t) ∈ K[t]. Since β = g(α) 6= 0,g(t) /∈ Iα = (fα(t)). Thus fα(t) - g(t).

From Lemma 1.7, fα(t) is irreducible and K[t] is a PID, we know ∃r(t), s(t) ∈ K[t]with

r(t)fα(t) + s(t)g(t) = 1 ∈ K[t].

Hence s(α)g(α) = 1 in K[α], and so β = g(α) is a unit, as required.

(ii) Let n = deg fα(t) We’ll show that T = { 1, α, α2, . . . , αn−1 } is a K–vector space basisof K[α].

Spanning: If fα(t) = tn+an−1tn−1+· · ·+a0 with ai ∈ K, then αn = −an−1αn−1−· · ·−

a0. This implies αn is a linear combination of { 1, α, α2, . . . , αn−1 }, and an easy induc-tion shows that αm for m ≥ n is likewise a linear combination of { 1, α, α2, . . . , αn−1 },so we have spanning.

Linear independence: Suppose λn−1αn−1+. . .+λ0 = 0. Let g(t) = λn−1t

n−1+. . .+λ0.Since g(α) = 0, we have g(t) ∈ Iα = (fα(t)). So g(t) = 0 or fα(t) | g(t). The latteris not possible since deg fα(t) > deg gα(t) so g(t) = 0 in K[t] and all the λi’s arezero.

Corollary 1.10. If K ≤ L is a field extension and α ∈ L, then α is algebraic over K if andonly if K ≤ K(α) is finite.

Proof.

(⇒) By Theorem 1.9, |K(α) : K| = deg fα(t) ≤ ∞.

(⇐) Lemma 1.5

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Corollary 1.11. Let K ≤ L be a field extension with |L : K| = n. Let α ∈ L, thendeg fα(t) | n.

Proof. Use the Tower law on K ≤ K(α) ≤ L. We deduce that |K(α) : K| divides |L : K|.Theorem 1.9(ii) gives deg fα(t) = |K(α) : K|.

1.3 Digression on (Non-)Constructibility

Schedules mention ‘other classical problems’ and we are now in a position to tackle some ofthese using Corollary 1.11.

A classical question from Greek geometry concerns the existence or otherwise of con-structions using ruler and compasses (where a ruler refers to a single unmarked straightedge). If you’re an expert you can divide a line betwen 2 points into arbitrarily many equalsegments, you can bisect an angle, or you can produce parallel lines. Given a polygon youcan produce a square of the same area or double the area. However,

1. You cannot duplicate the cube using ruler and compasses (given a cube you can’tproduce a cube of double the volume)

2. You cannot trisect the angle π/3 using ruler and compasses.

3. The circle cannot be squared using ruler and compasses (given a circle you can’tconstruct a square of the same area)

Assume we’re given a set P0 of points in R2, and we can formalise our operations.

Ruler operation Draw a straight line through any two points in P0.

Compass operation Draw a circle with the centre being a point in P0 and radius thedistance between a pair of points in P0.

Definition 1.12 (Constructible). The points of intersection of any two distinct lines or cir-cles drawn using these operations are constructible in one step from P0. More generally,a point r ∈ R2 is constructible from P0 if there is a finite sequence r1, r2, . . . , rn = r suchthat ri is constructible in one step from P0 ∪ {r1, . . . , ri−1}.

Exercise. Construct the midpoint of a line between two points.

Let K0 be the subfield of R generated by Q and the co-ordinates of the points in P0. Letri = (xi, yi) and set Ki = Ki−1(xi, yi).

Thus K0 ≤ K1 ≤ K2 ≤ · · · ≤ Km ≤ R.

Lemma 1.13. xi, yi are both roots in Ki of quadratic polynomials in Ki−1[t].

Proof. There are three cases for ri: line meets line, line meets circle, circle meets circle. Wedo the second case only here.

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A

B

C

X

Y

The line is defined by two points A = (p, q) and B = (r, s) while the circle is defined witha centre C = (t, u) and radius w. Then, points X and Y satisfy the equation of the linex−pr−p = y−q

s−q , and the equation of the circle (x− t)2 + (y − u)2 = w2. Solving these togethergives coordinates of X and Y satisfying quadratic polynomials over Ki−1. The other twocases are left as an exercise for the reader.

Theorem 1.14. If r = (x, y) is constructible from a set P0 of points in R2 and if K0 isthe subfield of R generated by Q and the coordinates of the points in P0, then the degrees|K0(x) : K0| and |K0(y) : K0| are powers of two.

Proof. Continue with the previous notation of Ki = Ki−1(xi, yi). By the Tower law,

|Ki : Ki−1| = |Ki−1(x, y) : Ki−1(x)| |Ki−1(x) : Ki−1|

But Lemma 1.13 tells us that |Ki−1(x) : Ki−1| must be 1 or 2 depending on whether thequadratic polynomial arising in the lemma is reducible or not, using Theorem 1.9(ii). Sim-ilarly, |Ki−1(x, y) : Ki−1(x)| is 1 or 2.

So |Ki : Ki−1| = 1, 2 or 4, (but in fact 4 cannot happen), hence by the Tower law,|Kn : K0| = |Kn : Kn−1| |Kn−1 : Kn−2| . . . |K1 : K0| is a power of two.

If r = (x, y) is constructible from P0, then

x, y ∈ Kn and K0 ≤ K0(x) ≤ Kn

K0 ≤ K0(y) ≤ Kn

and the Tower Law again gives that |K0(x) : K0| and |K0(y) : K0| are also powers of 2.

To use this for proofs about non-constructibility we need to be reasonably expert atworking out minimal polynomials.

Theorem 1.15. Let f(t) be a primitive integral polynomial. Then f(t) is irreducible inQ[t] if and only if it is irreducible in Z[t].

Proof. A special case of Gauss’ lemma from GRM.

Theorem 1.16 (Eisenstein’s criterion). Let f(t) = antn + an−1t

n−1 + · · · + a0 ∈ Z[t].Suppose there is a prime p such that

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(i) p - an

(ii) p | an−1, p | an−2, . . . , p | a0

(iii) p2 - a0

Then f(t) is irreducible in Z[t]

Proof. Recall from GRM.

Example. For p a prime, consider f(t) = tp−1 + tp−2 + · · ·+ 1. This is irreducible over Qby considering f(t+ 1) and using p as the prime in Eisenstein’s criterion.

Another method is to consider an integral polynomial f(t) (mod p). If f(t) is irreduciblein Z[t] then it is reducible over Z/pZ. So, if we find a prime p such that f(t) (mod p) isirreducible then f(t) is irreducible in Z[t].

Example. t3 + t+ 1 is irreducible mod 2. If it were reducible it would have a linear factorand so the polynomial would have a root mod 2. But 0, 1 are not roots. So, t3 + t + 1 isirreducible in Z[t], hence irreducible in Q[t].

Remark. On a later example sheet you’ll meet an irreducible polynomial in Z[t] which isreducible mod p for all primes p.

Theorem 1.17. The cube cannot be duplicated by ruler and compasses.

Proof. The problem amounts to whether given a unit distance, one can construct pointsdistance α apart, where α satisfies t3 − 2 = 0. Starting with points P0 = {(0, 0), (1, 0)} canwe produce (α, 0)?

No. If we could, Theorem 1.14 would say |Q(α) : Q| is a power of 2. But |Q(α) : Q| = 3since |Q(α) : Q| = deg fα(t) where fα(t) is the minimal polynomial of α over Q. α satisfiest3 − 2, which is irreducible over Z by Eisenstein’s criterion hence irreducible over Q. Sot3 − 2 is the minimal polynomial fα(t).

Theorem 1.18. The circle cannot be squared using ruler and compasses.

Proof. Starting with (0, 0) and (1, 0), we must construct (√π, 0) so that we have a square of

side length√π and hence area π. But π and hence

√π is transcendental over Q (Lindemann

- not proved here). Theorem 1.14 tells us we can’t do this construction.

1.4 Return to theory development

Lemma 1.19. Let K ≤ L be a field extension. Then

(i) α1, . . . , αn ∈ L are algebraic over K if and only if K ≤ K(α1, . . . , αn) is a finite fieldextension.

(ii) If K ≤ M ≤ L such that K ≤ M is finite, then there exist α1, . . . , αn ∈ L such thatK(α1, . . . , αn) = M .

Proof.

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(i) By Corollary 1.10, α is algebraic over K if and only if K ≤ K[α] is a finite fieldextension. αi is algebraic over K and hence algebraic over K(α1, . . . , αi−1) and so

|K(α1, . . . , αi) : K(α1, . . . , αi−1)| <∞.

By the Tower law applied to

K ≤ K(α1) ≤ K(α1, α2) ≤ · · · ≤ K(α1, . . . , αn),

we get |K(α1, . . . , αn) : K| <∞.

Conversely, consider K ≤ K(αi) ≤ K(α1, . . . , αn). Then the tower law says that if|K(α1, . . . , αn) : K| <∞ then |K(αi) : K| <∞ and by Corollary 1.10, αi is algebraicover K.

(ii) If |M : K| = n then M is an n-dimensional K-vector space, so there exists a K-basisα1, . . . , αn over M . Then K(α1, . . . , αn) ≤ M . However, any element of M is a K-linear combination of α1, . . . , αn and so lies in K(α1, . . . , αn), so M = K(α1, . . . , αn).

Definition 1.20 (Homomorphism over a field). Suppose K ≤ L, K ≤ L′ are field exten-sions. A K-homomorphism φ : L→ L′ is a ring homomorphism such that φ|K = id.

A K-homomorphism is a K-isomorphism if it is a ring isomorphism.

Lemma 1.21. Suppose K ≤ L, K ≤ L′ are field extensions. Then

(i) Any K-homomorphism φ : L→ L′ is injective and K ≤ φ(L) is a field extension.

(ii) If |L : K| = |L′ : K| <∞ then any K-homomorphism φ : L→ L′ is a K-isomorphism.

Proof.

(i) L is a field and kerφ is an ideal of L.

Note 1 7→ 1 and so kerφ can’t be the whole of L, hence kerφ = {0}. So φ(L) is a fieldand K ≤ φ(L) is a field extension.

(ii) φ is an injective K-linear map, so |φ(L) : K| = |L : K|. In general, |φ(L) : K| ≤|L′ : K|, but since |L : K| = |L′ : K| by assumption, we have |φ(L) : K| = |L′ : K|,hence φ(L) = L′ and φ is a K-isomorphism L → L′. (If L′ = L then φ would be aK-automorphism also.)

Definition 1.22 (Splitting). Let K ≤ L be a field extension and f(t) ∈ K[t]. We say fsplits over L if

f(t) = a(t− α1)(t− α2) · · · (t− αn)

where a ∈ K and α1, . . . , αn ∈ L.We say L is a splitting field for f over K if L = K(α1, . . . , αn).

Remark. This is equivalent to saying that L is a splitting field for f over K iff

(i) f splits over L

(ii) if K ≤M ≤ L and f splits over M then M = L.

Example.

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1. Consider f(t) = t3 − 2 over Q.

Q( 3√

2) is not a splitting field for f over Q, but Q( 3√

2, ω 3√

2, ω2 3√

2) is a splitting fieldover Q where ω is a primitive cube root of unity ω = e2πi/3

Q( 3√

2) Q(ω)

Q

Q( 3√

2, ω)

≤ 3

23

≤ 2

Starting in the bottom right, going clockwise:

• The minimal polynomial of ω over Q is t2 + t+ 1, so |Q(ω) : Q| = 2.

• t3 − 2 is irreducible over Q by Eisenstein’s criterion, so∣∣Q( 3√

2) : Q∣∣ = 3.

• ω satisfies t2+t+1 over Q and hence over Q( 3√

2) and so∣∣Q( 3√

2, ω) : Q( 3√

2)∣∣ ≤ 2.

• 3√

2 satisfies t3 − 2 over Q and hence over Q(ω), so∣∣Q( 3√

2, ω) : Q(ω)∣∣ ≤ 3.

Then by the Tower law, 2 |∣∣Q( 3√

2, ω) : Q∣∣ and 3 |

∣∣Q( 3√

2, ω) : Q∣∣, so we deduce∣∣∣Q(

3√

2, ω) : Q(ω)∣∣∣ = 3∣∣∣Q(

3√

2, ω) : Q(3√

2)∣∣∣ = 2

2. Take f(t) = (t2 − 3)(t3 − 1). The splitting field for f over Q is

Q(√

3,−√

3, ω, ω2, 1)

=Q(√

3, ω) = Q(√

3, i)

since ω = −1+i√3

2 .

3. t2 − 3 and t2 − 2t− 2 have the same splitting field over Q: Q(√

3).

4. Take f(t) = t2 + t+ 1 in F2[t].

f(t) is irreducible over F2 since it has no roots in F2 and hence no linear factors inF2[t]. Hence, F2[t]/(t2 + t+ 1) is a field.

Set α = t+ (t2 + t+ 1) ∈ F2[t]/(t2 + t+ 1), then

F2[t]

(t2 + t+ 1)= F2[α].

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The elements are 0, 1, α, 1 + α noting that α2 + α+ 1 = 0 and we have characteristic2, so α2 = α+ 1.

f(t) splits over F2[α]:f(t) = (t− α)(t− 1− α)

Thus F2[α] is the splitting field for f over F2.

We can use this final construction to produce splitting fields in general.

Theorem 1.23 (Existence of splitting fields). Let K be a field and f(t) ∈ K[t]. Then thereexists a splitting field for f over K.

Proof. If deg f = 0 then K is the splitting field for f over K.Suppose deg f > 0 and pick an irreducible factor g(t) of f(t) in K[t], noting that K ≤

K[t]/(g(t)) is a field extension.Take

α1 = t+ (g(t)) ∈ K[t]/(g(t)),

then K[t]/(g(t)) = K(α1) and g(α1) = 0 in K(α1). Therefore f(α1) = 0 in K(α1) and wecan write f(t) = (t− α1)h(t) in K(α1)[t].

Repeat, noting that deg h(t) < deg f(t) and so we get

f(t) = a(t− α1)(t− α2) · · · (t− αn)

where a is a constant in K. Thus, we have a factorisation of f(t) in K(α1, . . . , αn)[t], andso K(α1, . . . , αn) is a splitting field for f over K.

Theorem 1.24 (Uniqueness of splitting fields). If K is a field and f(t) ∈ K[t], then thesplitting field for f over K is unique up to K-isomorphism, that is, if there are two suchsplitting fields L and L′, there is a K-isomorphism φ : L→ L′.

Proof. Suppose L and L′ are splitting fields for f(t) ∈ K[t] over K. We need to show thatthere is a K-isomorphism L→ L′.

Suppose K ≤ M ≤ L and there exist M ′ with K ≤ M ′ ≤ L′ and a K-isomorphismψ : M →M ′. Clearly some M exists (we can take M = K), so we pick M so that |M : K|is maximal among all such M,M ′, ψ.

We must show M = L and M ′ = L′. Note that if M = L then f(t) splits over M :

f(t) = a(t− α1) · · · (t− αn) ∈M [t]

Apply ψ, we get an induced map M [t]→M ′[t].

f(t) = ψ(f(t)) = ψ(a)(t− ψ(α1)) · · · (t− ψ(αn))

Thus f(t) splits over ψ(M) = M ′. But L′ is a splitting field and M ′ ≤ L′, so M ′ = L′.So, suppose M 6= L and we’ll get a contradiction of maximality of M . Since M 6= L,

there is a root α of f(t) in L which isn’t in M . Factorise f(t) = g(t)h(t) in M [t] so thatg(t) is irreducible in M [t] while g(α) = 0 in L. Then there exists a K-homomorphismM [t]/(g(t))→ L given by t+ (g(t)) 7→ α which has image M(α).

The K-isomorphism M [t] → M ′[t] induced by ψ maps g(t) ∈ M [t] to γ(t) ∈ M ′[t].f(t) = g(t)h(t) in M [t] yields f(t) = γ(t)δ(t) in M ′[t].

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We have a field extension M ′ ≤ M ′[t]/(γ(t)) and there exists a M ′-homomorphismM ′[t]/(γ(t))→ L′ given by t+ (γ(t)) by picking a root α′ of γ(t) in L′. However γ(t) | f(t)in M ′[t] and hence in L′[t] and so α′ is also a root of f(t) in L′. The M ′-homomorphismgives a K-isomorphism

M ′[t]/(γ(t))→M ′(α′)

and so we have a K-isomorphism M(α)→M ′(α′). This contradicts the maximality of M ,since M �M(α).

Definition 1.25 (Normal extension). A field extension K ≤ L is normal if for every α ∈ Lthe minimal polynomial fα(t) of α over K splits over L.

Theorem 1.26. Let K ≤ L be a finite field extension. Then K ≤ L is normal ⇐⇒ L isthe splitting field for some f(t) ∈ K[t].

Proof. Later.

Example 1.27. Let F be a finite field with |F| = m. F has characteristic p for some p andFp ≤ F, therefore m = pr for some r. The non-zero elements form the multiplicative groupof order m− 1 = n and they satisfy tn − 1: they are roots of tn − 1. So,

tn − 1 = (t− α1) · · · (t− αn)

where α1, . . . , αn are the non-zero elements of F. Thus F is the splitting field for tn− 1 overFp. By Theorem 1.24, any other field with m elements is Fp-isomorphic to F. (We’ll showlater that there is a field of m elements.)

Theorem 1.28. Let G be a finite subgroup of the multiplicative group of a field K. ThenG is cyclic. In particular, the multiplicative group of a finite field is cyclic.

Proof. Let |G| = n. By the structure theorem of finite abelian groups from GRM,

G ∼= Cqm11× Cqm2

2× · · · × Cqmrr

with qi prime, not necessarily distinct. However if q = qi = qj for some i 6= j, there are atleast q2 distinct solutions of tq−1 = 0 in K (since Cq×Cq ∼= subgroup of G). But in a field(or even an integral domain), a polynomial of degree q has at most q roots, a contradiction.So all the qi are distinct and hence G is cyclic, generated by (g1, . . . , gr) where gi generatesCqmii

using the Chinese Remainder Theorem.

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2 Separable, normal and Galois extensions

Definition 2.1 (Separable polynomial). Let K be a field and f(t) ∈ K[t]. Suppose f(t) isirreducible in K[t] and L is a splitting field for f(t) over K. Then f(t) is separable overK if f(t) has no repeated roots in L.

For general f(t) we say f(t) is separable over K if every irreducible factor in K[t] isseparable over K.

All constant polynomials are separable.

Definition 2.2 (Formal differentiation). If K is a field then formal differentiation

D : K[t]→ K[t]

tn 7→ ntn−1

is a K-linear map. We denote this by D(f(t)) = f ′(t).

Lemma 2.3. Let K be a field and f(t), g(t) ∈ K[t]. Then:

(a) D(f(t)g(t)) = f ′(t)g(t) + f(t)g′(t) (Leibniz’ rule)

(b) Assume f(t) 6= 0. Then f(t) has a repeated root in a splitting field L if and only iff(t) and f ′(t) have a common irreducible factor in K[t].

Proof.

(a) D is a K-linear map and so we only need to check for f(t) = tn, g(t) = tm. Left asan exercise.

(b) Let α be a repeated root in a splitting field L, then

f(t) = (t− α)2g(t) ∈ L[t]

f ′(t) = (t− α)2g′(t) + 2(t− α)g(t)

and so f ′(α) = 0. Therefore the minimal polynomial fα(t) of α in K[t] divides bothf(t) and f ′(t) and thus fα(t) is a common irreducible factor of f(t) and f ′(t).

Conversely, let h(t) be a common irreducible factor of f(t) and f ′(t) in K[t]. Pick aroot α in L of h(t).

So f(α) = 0 = f ′(α), thus f(t) = (t − a)g(t) in L[t], and f ′(t) = (t − a)g′(t) + g(t).Since f ′(α) = 0 we have (t− a) | f ′(t). and so (t− a) | g(t). Hence (t− a)2 | f(t) andwe have a repeated root.

Corollary 2.4. If K is a field and f(t) ∈ K[t] is irreducible:

(i) If the characteristic of K is 0, then f(t) is separable over K.

(ii) If the characteristic of K is p > 0, then f(t) is not separable if and only if f(t) ∈ K[tp].

Proof. By Lemma 2.3, f(t) is not separable over K if and only if f(t) and f ′(t) have acommon irreducible factor. Since we’re assuming f(t) is irreducible, this is equivalent tosaying f ′(t) = 0.

f(t) = antn + an−1t

n−1 + · · ·+ a0

f ′(t) = nantn−1 + · · ·+ a1

Thus f ′(t) = 0 ⇐⇒ iai = 0 for all i > 0.

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(i) If charK = 0 then f ′(t) 6= 0 for any non-constant polynomial, so f(t) is separableover K.

(ii) If charK = p > 0 then if f ′(t) = 0 we have iai = 0 for all i > 0, so f(t) is notseparable ⇐⇒ f(t) ∈ K[tp].

Definition 2.5 (Separable extension). We say α ∈ L is separable over K if its minimalpolynomial is separable over K.

L is separable over K if all α ∈ L are separable over K.If fα(t) = (t−α)n = tn−αn where n is a power of p(= charK), we say that α is purely

inseparable over K.

Example.

(1) Let Q ⊆ L be an algebraic field extension. Then L is separable over Q.

(2) Let L = Fp(X) be the rational functions in X over Fp, and K = Fp(Xp). ThenK ≤ L is not separable. Observe that if f(t) = tp −Xp ∈ K[t] then f ′(t) = 0. Buttp−Xp = (t−X)p in L[t] since the binomial expansion gives other terms divisible byp and hence 0 since the characteristic is p.

However f(t) is irreducible in K[t]: Suppose f(t) = g(t)h(t) in K[t] and hence in L[t].So we get g(t) = (t −X)r for some 0 < r < p if the factorisation is non-trivial. Butthis would mean Xr was in K. However, ∃integers a, b such that ar + bp = 1. So

(Xr)a(Xp)b ∈ K and so X ∈ K, thus we’d have X = u(Xp)v(Xp) , a contradiction.

Thus f(t) = tp − Xp is the minimal polynomial of X over K, thus X is purelyinseparable over K and K ≤ L is not separable.

(3) Take F a finite field with order m, a power of p where charF = p. Consider f(t) = tn−1where n = m− 1. This is separable over Fp since we saw that f(t) has distinct linearfactors in F[t].

Remark. It’s useful to have an alternative approach to separability of field extensionswithout having to check separability of minimal polynomials for all elements of the largerfield. This is where we start thinking about K-homomorphisms.

Lemma 2.6. Let M = K(α), where α is algebraic over K and let fα(t) be the minimalpolynomial of α over K.

Then, for any field extension K ≤ L, the number of K-homomorphisms of M to L isequal to the number of distinct roots of fα(t) in L. Thus this number is ≤ deg fα(t) =|K(α) : K| = |M : K|.

K

L

M = K(α)

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Proof. We saw in Lemma 1.21 that any K-homomorphism M → L is injective, and we have

K(α) ∼=K[t]

(fα(t)).

For any root β of fα(t) in L we can define a K-homomorphism

K[t]

(fα(t))→ L

t+ (fα(t)) 7→ β

Thus we get a K-homomorphism M → L.Conversely, for any K-homomorphism φ : M → L the image φ(α) must satisfy

fα(φ(α)) = 0.

These processes are inverse to each other, giving a 1-1 correspondence

{K homomorphisms M → L} ←→ {roots of fα(t) ∈ L}.

Example. Take K = Q, L = Q( 3√

2), and use α = 3√

2, with minimal polynomial fα(t) =t3 − 2.

Then, there is only one K-homomorphism M = K( 3√

2) = L→ L, the identity map.

Corollary 2.7. The number of K-homomorphisms K(α)→ L = deg fα(t) ⇐⇒ L is largeenough, in particular L contains a splitting field for fα(t) and α is separable over K.

Proof. Immediate from Lemma 2.6.

Lemma 2.8. Let K ≤M be a field extension and M1 = M(α1) (where α1 is algebraic overM). Let f(t) be the minimal polynomial of α1 over M and let K ≤ L. Let φ : M → L bea K-homomorphism. Then there is a correspondence

{Extensions φ1 : M1 → L of φ} ←→ {roots of φ(f(t)) ∈ L}.

K

M

M1

L

Remark. Lemma 2.6 is the special case M = K and φ = inclusion of K in L.

Proof. f(t) is irreducible in M [t], so φ(f(t)) is irreducible in φ(M)[t]. Any extension φ1 :M1 → L of φ produces a root φ1(α) of φ(f(t)).

Conversely, given a root γ of φ(f(t)) in L,

M1 = M(α1) ∼=M [t]

(f(t))∼=

φ(M)[t]

(φ(f(t)))∼= φ(M)(φ) ≤ L.

Thus we get an extension φ1 of φ as required.

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Corollary 2.9. If L is large enough, the number of φ1 which extend φ is equal to thenumber of distinct roots of f(t) in L. This is equal to |M1 : M | ⇐⇒ α is separable overM .

Proof. Immediate from Lemma 2.8.

Corollary 2.10. Let K ≤ M ≤ N be finite field extensions, K ≤ L. Let φ : M → L be aK-homomorphism. Then the number of extensions of φ to maps θ : N → L is ≤ |N : M |.Moreover, such a θ exists if L is large enough.

Proof. Pick α1, . . . , αr so that N = M(α1, . . . , αr) and set Mi = M(α1, . . . , αi). Then we’vegot

M ≤M1 ≤M2 ≤ · · · ≤Mr = N.

Using Lemma 2.8, there are

≤ |M1 : M | extensions φ1 : M1 → L of φ

≤ |M2 : M1| extensions φ2 : M2 → L of φ1

...

≤ |Mr : Mr−1| extensions φr : Mr → L of φr−1

By the Tower law, the number of extensions θ : N → L (recall N = Mr) of φ : M → L is

≤ |Mr : Mr−1| |Mr−1 : Mr−2| · · · |M1 : M | = |N : M |

where the last part comes from the proof of Lemma 2.8 - we need L to contain roots.

Remark 2.11. This proof together with Corollary 2.9 shows that the number of extensionsθ of φ is equal to |N : M | iff L is large enough (so that everything splits) and αi is separableover M(α1, . . . , αi−1).

Lemma 2.12. Let K ≤ N be a field extension with |N : K| = n and N = K(α1, . . . , αr)say. Then the following are equivalent:

(i) N is separable over K.

(ii) Each αi is separable over K(α1, . . . , αi−1).

(iii) If K ≤ L is large enough there are exactly n distinct K-homomorphisms N → L.

Proof. (i) ⇒ (ii). N is separable over K =⇒ αi is separable over K. The minimalpolynomial of αi over K(α1, . . . , αi−1) divides the minimal polynomial of αi over K (inK(α1, . . . , αi−1)[t]).

So if the latter has distinct roots in a splitting field then the former does. So αi separableover K =⇒ αi separable over K(α1, . . . , αi−1).

(ii) ⇒ (iii) follows from Remark 2.11.(iii) ⇒ (i). Assume (iii) is true and (i) false, aiming for a contradiction. So, ∃β ∈ N

that is not separable over K, so there are � |K(β) : K| K-homomorphisms φ : K(β) → Lby Corollary 2.7.

By Corollary 2.10, φ extends to ≤ |N : K(β)| extensions θ : N → L, and so there are� |N : K(β)| |K(β) : K| K-homomorphisms N → L, contradiction.

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Definition 2.13 (Separably generated). We say M = K(α1, . . . , αr) is separably gener-ated by α1, . . . , αr over K if each αi is separable over K.

Corollary 2.14. A finite extension is separable ⇐⇒ it is separably generated.

Proof. Lemma 2.12.

Lemma 2.15. If K ≤M ≤ L finite field extensions, M ≤ L, then

K ≤M, M ≤ L are both separable ⇐⇒ K ≤ L is separable

Proof. Example sheet.

Example 2.16. Take F a finite field, |F| = m. Multiplicative group of order n = m − 1is cyclic. Take a generator α, then F = Fp(α), and so the minimal polynomial of α dividestn−1. Observe this has distinct roots, which are all of F\{0} since αn = 1. So the minimalpolynomial of α is separable, and F = Fp(α) is separable over Fp.

Theorem 2.17 (Primitive Element Theorem). Any finite separable extension K ≤M is asimple extension, that is, M = K(α) for some α, called a primitive element.

Proof. First deal with the case where K is a finite field. Then M is also finite and we cantake α to be a generator of the multiplicative group of M , which is cyclic.

Now assume K is an infinite field.Since K ≤ M is a finite extension, M = K(α1, α2, . . . , αn) for some αi. It is enough to

show that any field M = K(α, β) with β separable over K is of the form K(γ).Take f(t) and g(t) to be the minimal polynomials of α and β over K and let L be the

splitting field for f(t)g(t) over K(α, β). Say the distinct zeros of f(t) in L are α = α1, . . . , αaand of g(t) are β = β1, . . . , βb.

By separability, b = deg g(t). Choose λ ∈ K such that all αi + λβj are distinct, which ispossible since K is infinite. Set γ = α+ λβ.

Let F (t) = f(γ − λt) ∈ K(γ)[t]. We have g(β) = 0 and F (β) = f(α) = 0. Thus F (t)and g(t) have a common zero.

Any other common zero would have to be βj for some j > 1. But then F (βj) =f(α+λ(β−βj)). By assumption, α+λ(β−βj) is never an αi and so F (βj) 6= 0. Separabilityof g(t) says its linear factors are all distinct, so (t − β) is a highest common factor of F (t)and g(t) in L[t].

However the minimal polynomial h(t) of β overK(γ) then divides F (t) and g(t) inK(γ)[t]and hence in L[t]. This implies h(t) = t−β and so β ∈ K(γ). Therefore α = γ−λβ ∈ K(γ)and so K(α, β) ⊂ K(γ) and equality holds since γ ∈ K(α, β).

Example. From our example in Section 1.1, Q ≤ Q(√

2, i), we had intermediate subfields.

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Q

Q(i) Q(i√

2)

Q(√

2)

Q(i,√

2)

If we follow the procedure of the proof of Theorem 2.17, α =√

2, β = i, f(t) = t2 − 2,g(t) = t2 + 1. Consider

√2 + λi where ±

√2 ± λi are distinct. For instance λ = 1, so

Q(√

2, i) = Q(√

2 + i).

2.1 Trace and Norm

Definition 2.18 (Trace and norm). Let K ≤ M be a finite field extension, and α ∈ M .Multiplication by α gives a K-linear map θα : M →M .

Then we define

Trace of α over K is given by TrM/K(α) = trace of θα ∈ K.

Norm of α over K is given by NM/K(α) = determinant of θα ∈ K.

Note these are dependent on the field extension.

Theorem 2.19. With the above notation, suppose fα(t) = ts + as−1ts−1 + · · ·+ a0 is the

minimal polynomial for α over K. Let r = |M : K(α)|, then the characteristic polynomialof θα is (fα(t))r.

Note|M : K| = |M : K(α)| |K(α) : K| = rs.

Then TrM/K(α) = −ras−1 and NM/K = ((−1)sa0)r.

Proof. Regard M as a K(α)-vector space with basis 1 = β1, . . . , βr. Now take the K-vectorspace basis 1, α, α2, . . . , αs−1 of K(α). So, 1, α, α2, . . . , αs−1, β2, β2α, . . . , β2α

s−1, β3, . . . isa K-vector space basis for M . Multiplication by α in K(α) is represented by matrix

A =

0 0 0 . . . 0 −a01 0 0 . . . 0 −a10 1 0 . . . 0 −a20 0 1 . . . 0 −a3...

......

. . ....

...0 0 0 . . . 1 −as−1

an s× s matrix whose characteristic polynomial is fα(t).

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Multiplication by α in M is represented by the rs× rs matrixA 0 0 . . . 00 A 0 . . . 00 0 A . . . 0...

......

. . ....

0 0 0 . . . A

whose characteristic polynomial is (fα(t))r.

Look at the terms of this characteristic polynomial to get the trace and norm.

Theorem 2.20. Let K ≤M be a finite separable field extension and |M : K| = n, α ∈M .Let K ≤ L be large enough so that there are n distinct K-homomorphisms

σ1, σ2, . . . , σn : M −→ L.

Then the characteristic polynomial of θα : M →M (the multiplication map) is

n∏i=1

(t− σi(α))

hence

TrM/K(α) =

n∑i=1

σi(α) and NM/K(α) =

n∏i=1

σi(α).

Proof. Write

fα(t) = (t− α1) . . . (t− αs) ∈ L[t]

= ts + as−1ts−1 + . . .+ a0

the minimal polynomial of α over K (where L large enough implies fα(t) splits in L). Thereare s K-homomorphisms K[α]→ L corresponding to maps sending α to αi.

Each of these extends in |M : K(α)| ways to give K-homomorphisms M → L (by sepa-rability and Corollary 2.9).

However each of these extensions of a map sending α → αi still sends α → αi. Setr = |L : K(α)|. Thus there are r maps sending α → αi for each i. Thus if the n(= rs)distinct K-homomorphisms M → L are σ1, . . . , σn, then

n∑i=1

σi(α) = r(α1 + α2 + · · ·+ αs) = −ras−1 = TrM/K(α)

n∏i=1

σi(α) = ((−1)sa0)n = NM/K(α).

Theorem 2.21. Let K ≤ M be a finite separable extension. Then we define a K-bilinearform

T : M ×M → K

(x, y) 7−→ TrM/K(xy).

Then this is non-degenerate and in particular the K-linear map TrM/K : M → K is non-zero,and hence surjective.

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Remark. If K ≤ M is a finite extension which is not separable then TrM/K : M → K isalways zero and so T : M ×M → K is degenerate (see example sheet).

Proof. Separability and finiteness give M = K(α) for some α, by Theorem 2.17. We have aK-basis 1, α, α2, . . . , αn−1 of K(α) where n = |M : K|. The K-bilinear form is representedby

A =

TrM/K(1) TrM/K(α) . . .TrM/K(α) TrM/K(α2) . . .

......

. . .

.

Let L be the splitting field of the minimal polynomial fα(t) of α over K.Thus fα(t) = (t− α1) · · · (t− αn) with α1, . . . , αn ∈ L. The entries in A are of the form

TrM/K(αe) which is αe1 + · · ·+ αen using Theorem 2.20.Now consider ∆ =

∏i<j(αi − αj), the discriminant of V :

V =

1 1 . . . 1a1 a2 . . . ana21 a22 . . . a2n...

.... . .

...an−11 an−12 . . . an−1n

.

Observe that V V T = A, and 0 6= ∆2 =∣∣V V T ∣∣ = |A|, so A is non-singular and therefore the

bilinear form T is non-degenerate.

Remark. We’ll meet ∆ again shortly, as it is the discriminant of the polynomial fα(t).

2.2 Normal extensions

We restate Definition 1.25 and Theorem 1.26:

Definition 1.25 (Normal extension). An extension K ≤ L is normal if for every α ∈ Lthe minimal polynomial fα(t) of α over K splits over L.

Theorem 1.26. Let K ≤ L be a finite field extension. Then K ≤ L is normal ⇐⇒ L isthe splitting field for some f(t) ∈ K[t].

Proof. Assume K ≤ M is normal. Pick α1, . . . , αr ∈ M so that M = K(α1, . . . , αr). Letfαi(t) be the minimal polynomial for αi over K.

Letf(t) = fα1

(t)fα2(t) . . . fαr (t).

By normality, each fαi(t) splits over M and therefore f(t) splits over M . M is the splittingfield of f(t) over K since if β1, . . . , βm are the roots of f(t) then M = K(β1, . . . , βm).

Conversely, suppose M is a splitting field for f(t) over K. Thus M = K(β1, . . . , βm)where the βj are the roots of f(t) in M .

Take α ∈ M . Let fα(t) be the minimal polynomial of α over K. Let M ≤ L largeenough so that fα(t) splits in L and consider K-homomorphisms φ : M → L. φ(βj) is also aroot of f(t) and is therefore one of the βjs. Injectivity of K-homomorphisms (Lemma 1.21)implies that φ(βj) generate the βj .

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M = K(β1, . . . , βm) and so φ is determined by the images of the βj and thus φ(M) = M .However if αi is a root of fα(t) in L, there is a K-homomorphism

K(α) −→ K(αi) ≤ Lα 7−→ αi.

This extends by Corollary 2.10 to a K-homomorphism φ : M → L with φ(α) = αi. Butφ(M) = M , so αi ∈M . Thus M is normal over K.

Remark. As for separability, the property of ‘normality’ is equivalent to ‘normally gener-ated’. In particular, a finite extension K ≤ L is normal if and only if L = K(α1, . . . , αr),with each fαi(t) splitting over L.

Definition 2.22 (Automorphism group). Let K ≤ M be a finite field extension. Its K-automorphism group is AutK(M) = {φ | φ a K-homomorphism M →M }.

From Lemma 1.21 we know that such K-homomorphisms are isomorphisms and thushave inverses. Composition gives the group operation.

Lemma 2.23.AutK(M) ≤ |M : K| .

Proof. Corollary 2.10.

Theorem 2.24. Let K ≤M be a finite field extension. Then |AutK(M)| = |M : K| iff theextension is both normal and separable.

Definition 2.25 (Galois extension). A finite field extension that is normal and separableis a Galois extension.

Definition 2.26 (Galois group). Let K ≤ M be a Galois extension. Then, the K-automorphism group of M is the Galois group of M over K. Write this as Gal(M/K).

Remark. Some authors use ‘Galois group’ for the automorphism group even when theextension is not Galois.

Proof of Theorem 2.24. (⇒). Suppose |AutK(M)| = |M : K| = n. Let L be large enoughcontaining M .

The n distinct K-homomorphisms φ : M → M ≤ L give us n K-homomorphismsφ : M → L and Lemma 2.12 says that M is separable over K. For normality, pick α ∈ Mwith minimal polynomial fα(t) over K.

Take M = K(α1, . . . , αm) as in the proof of Corollary 2.10 with α = α1 and L = M .We only get |M : K| extensions of the inclusion K ↪→ M if each inequality in the proof isan equality. In particular we need the number of K-homomorphisms K(α1) → M to be|K(α1) : K|.

But then Lemma 2.6 says we have |K(α) : K| distinct roots of fα(t) in M . Thus fα(t)splits over M .

Conversely, suppose K ≤ M is separable and normal. Then for K ≤ M ≤ L withL large enough, separability implies there are |M : K| K-homomorphisms φ : M → L byLemma 2.12. However since K ≤M is normal, it is the splitting field for some polynomialf(t) ∈ K[t] (Theorem 1.26) and thus M = K(α1, . . . , αn), where f(t) = (t−α1) · · · (t−αn).Note that φ(aj) is also a root of φ(f(t)) = f(t) and is therefore one of the αjs. Thusφ(M) = M . Thus we have |M : K| K-homomorphisms φ : M →M .

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Remark 2.27. In the proof we have shown that if K ≤ M ≤ L and φ : M → L is aK-homomorphism and K ≤M is normal, then φ(M) = M .

Example.

(1) Q ≤ Q(√

2, i) is Galois. Gal(Q(√

2, i)/Q) has 4 elements:

σ :√

2 7−→ ±√

2

i 7−→ ±i

In particular, Gal(Q(√

2, i)/Q) ∼= C2 × C2, all non-identity elements have order 2.

(2) Consider f(t) = t3 − 2. The splitting field over Q is Q( 3√

2, ω), where ω is a primitivecube root of 1.

Thus Q ≤ Q( 3√

2, ω) is Galois and∣∣Q( 3√

2, ω) : Q∣∣ = 6 (we saw this in the example

before Theorem 1.23).

σ1 :

{3√

2 7−→ 3√

2

ω 7−→ ωidentity

σ2 :

{3√

2 7−→ ω 3√

2

ω 7−→ ωorder 3

σ3 :

{3√

2 7−→ ω2 3√

2

ω 7−→ ωorder 3

σ4 :

{3√

2 7−→ 3√

2

ω 7−→ ω2

complex conjugationorder 2

σ5 :

{3√

2 7−→ ω2 3√

2

ω 7−→ ω2order 2

σ6 :

{3√

2 7−→ ω 3√

2

ω 7−→ ω2order 2

Observe also σ4σ2σ−14 = σ3, so this is the dihedral group.

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3 Fundamental Theorem of Galois Theory

3.1 Artin’s Theorem

Definition 3.1 (Fixed field). Let K ≤ L be a field extension and H ≤ AutK(L). Thefixed field of H is,

LH := {α ∈ L | σ(α) = α for all σ ∈ H }

Exercise. Check it is a field and K ≤ LH ≤ L.

Theorem 3.2 (Fundamental Theorem of Galois Theory). Let K ≤ L be a finite Galoisextension. Then

(i) there is a 1 to 1 correspondence

{intermediate subfields K ≤M ≤ L} ←→ {subgroups H of Gal(L/K)}M 7−→ AutM (L)

LH ←− [ H

This is called the Galois correspondence.

(ii) H is a normal subgroup of Gal(L/K) iff K ≤ LH is normal iff K ≤ LH is Galois.

(iii) If H C Gal(L/K) then the map

θ : Gal(L/K) −→ Gal(LH/K)

given by restriction to LH is a surjective group homomorphism with kernel H.

Remark.

(i) Observe that M ≤ L is Galois and so we could have written Gal(L/M) instead ofAutM (L) in (i). To see this: separability follows from Lemma 2.15. Normality: Ifα ∈ L, the minimal polynomial of α over M divides the minimal polynomial of α overK. But the latter splits over L.

(ii) If K ≤M is normal then Remark 2.27 says that if σ : L→ L then σ(M) = M and sowe can talk about the restriction of α to M giving an automorphism of M .

Example.

(i) Q ≤ Q(√

2, i). We saw in Section 1 the lattice of intermediate fields and subgroupsGal(Q(

√2, i)/Q) ∼= C2 × C2 abelian. All subgroups are normal and intermediate

subfields are normal extensions of Q.

(ii) Q ≤ Q( 3√

2, ω):

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Q( 3√

2, ω)

Q(ω) Q( 3√

2) Q(ω 3√

2) Q(ω2 3√

2)

Q

{e}

{σ1, σ2, σ3} {σ1, σ4} {σ1, σ5} {σ1, σ6}

D6∼= {σ1, σ2, σ3, σ4, σ5, σ6}

The subgroup H of order 3 is normal but those of order 2 are not.

Q ≤ Q(ω) is normal←→ H of order 3

Gal(Q(3√

2, ω)/Q) 7−→ Gal(Q(ω)/Q)

D6 7−→ C2

This final mapping has kernel H, and the image is generated by conjugation.

Theorem 3.3 (Artin’s Theorem). Let K ≤ L be a field extension and H a finite subgroupof AutK(L). Let M = LH . Then M ≤ L is a finite Galois extension, and H = Gal(L/M).

Remark. This implies some of the Galois correspondence

H LH Gal(L/LH)

we get back to H.

Proof of Artin’s Theorem. Take α ∈ L.First step: Show that |M(α) : M | ≤ |H|. Let

{α1, . . . , αn}︸ ︷︷ ︸all distinct

= {φ(α) | φ ∈ H } .

Define g(t) =∏ni=1(t−αi). Each φ induces a homomorphism L[t]→ L[t] that sends g(t)

to itself, since φ is permuting the αi. So the coefficients of g(t) are fixed by all φ ∈ H andthus they all lie in LH = M . Thus g(t) ∈M [t].

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By definition, g(α) = 0 since α is one of the αi. Hence the minimal polynomial fα(t) ofα over M divides g(t). Thus |M(α) : M | = deg fα(t) ≤ deg g(t) ≤ |H|. We’ve shown that αis algebraic over M . Moreover, fα(t) is separable since g(t) is. Thus M ≤ L is a separableextension.

Next step: Show that M ≤ L is a simple extension. Pick α ∈ L with |M(α) : M |maximal. We’ll show that L = M(α) for this choice of α. Suppose β ∈ L. Then M ≤M(α, β) is finite and is separably generated and hence is a finite separable extension byLemma 2.12.

By the Primitive Element Theorem, M(α, β) = M(γ) for some γ. But M ≤ M(α) ≤M(γ). The maximality of |M(α) : M | forces M(α) = M(γ). Thus β ∈ M(γ) = M(α) andso L = M(α) so |L : M | ≤ |H|.

Finally,|L : M | = |M(α) : M | ≤ |H| ≤ |AutM (L)| ≤

Lemma 2.23

|L : M |

We must have equality throughout, and so |L : M | = |AutM (L)| = |H|. Hence by Theo-rem 2.24 we have M ≤ L is a finite Galois extension and H = Gal(L/M).

Theorem 3.4. Let K ≤ L be a finite field extension. Then the following are equivalent:

(i) K ≤ L is Galois

(ii) LH = K when H = AutK(L)

Remark. The theorem allows some authors yet another alternative definition of a ‘Galoisextension’.

Proof. (i) ⇒ (ii): Let M = LH where H = AutK(L). By Artin’s Theorem, M ≤ L is aGalois extension, and |L : M | = |Gal(L/M)| and H = Gal(L/M).

However if K ≤ L is Galois then |H| = |AutK(L)| = |L : K| by Theorem 2.24. Thus|L : M | = |L : K| and so M = K.

(ii) ⇐ (i): Use Theorem 3.3.

Proof of Fundamental Theorem of Galois Theory.

(i) Composing the maps H → LH and M → Gal(L/M) gives H → H by Theorem 3.3.Also M −→ Gal(L/M) −→ LH where H = Gal(L/M) yields M since M ≤ LH whereH = Gal(L/M) and∣∣L : LH

∣∣ =(2.24)(3.3)

|H| = |Gal(L/M)| =(2.24)

|L : M |

So M = LH .

(ii) Take H ≤ Gal(L/K), then LφHφ−1

= φ(LH) when φ ∈ Gal(L/K). So by (i), H isnormal iff φ(LH) = LH . Set M = LH .

We’ll show that K ≤M is normal iff φ(M) = M ∀φ ∈ Gal(L/K). K ≤M is normal=⇒ φ(M) = M by remark 2 after the statement of Fundamental Theorem of GaloisTheory.

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Conversely if φ(M) = M ∀φ ∈ Gal(L/K), pick α ∈ M and let fα(t) be its minimalpolynomial over K. Take β to be a root of fα(t) in L (possible by normality). Thenthere is a K-homomorphism

K(α) ∼= K[t](fα(t))

K(β) ∼= K[t](fα(t))

≤ L

α β.

This extends to a K-homomorphism φ : L→ L.

However we are assuming φ(M) = M and so φ(α) = β ∈M . Thus K ≤M is normal.Note that K ≤ LH is separable since K ≤ LH ≤ L and K ≤ L separable.

(iii) By remark 2 after statement of Theorem 3.2, the restriction map

θ : Gal(L/K)→ Gal(LH/K)

is defined. Surjectivity follows from being able to extend a K-homomorphism LH →LH ≤ L to aK-homomorphism L→ L by Corollary 2.10. ClearlyH ≤ Ker θ. However

|L : K||Ker θ|

=Gal(L/K)

|Ker θ|=∣∣Gal(LH/K)

∣∣ by surjectivity of θ

=∣∣LH : K

∣∣ since K ≤ LH is Galois

=|L : K||L : LH |

by Tower law

So |Ker θ| =∣∣L : LH

∣∣ =∣∣Gal(L/LH)

∣∣ = |H| by Theorem 3.3, so H = Ker θ.

3.2 Galois groups of polynomials

Definition 3.5 (Galois group of polynomial). Let f(t) be a separable polynomial ∈ K[t]and let K ≤ L with L a splitting field for f(t). Then the Galois group of f(t) over K is

Gal(f) := Gal(L/K).

Since L is a splitting field for f(t), L = K(α1, . . . , αn) where α1, . . . , αn are the roots off(t) in L. Observe that if φ ∈ Gal(L/K) it maps

{roots of f(t)} −→ {roots of f(t)}

Thus φ permutes the αi. Moreover, if φ fixes each αi, it also fixes all elements of L, and sois the identity map. Thus Gal(f) may be regarded as a permutation group of the n roots,so Gal(f) ≤ Sn.

Lemma 3.6. Suppose f(t) is separable, f(t) = g1(t) · · · gs(t) with gi(t) irreducible in K[t]is a factorisation in K[t]. Then the orbits of Gal(f) on the roots of f(t) correspond to thefactors gj(t).

Two roots are in the same orbit ⇐⇒ they are roots of the same gj(t).

In particular, if f(t) is irreducible in K[t] there is one orbit, i.e., Gal(f) acts transitively onthe roots of f(t).

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Proof. Let αk, αl be in the same orbit under Gal(f). Thus there is φ ∈ Gal(f) with αl =φ(αk). But if αk is a root of gj(t) then φ(αk) = αl is also a root of gj(t).

Conversely, if αk, αl are roots of gj(t) then

K(αk) K[t](gj(t))

K(αl) L∼=

φ0

∼= ≤

with φ0(αk) = αl. φ0 extends to a φ : L → L ∈ Gal(L/K), thus αk, αl are in the sameorbit.

Since the Galois group of an irreducible polynomial acts transitively, and is a subgroupof Sn, it is useful to know what the transitive subgroups of Sn are.

Lemma 3.7. The transitive subgroups of Sn for n ≤ 5 are

n = 2: S2 (∼= C2)n = 3: A3 (∼= C3), S3

n = 4: C4, V4, D8, A4, S4

n = 5: C5, D10, H20, A5, S5

where H20 is generated by a 5-cycle and a 4-cycle.

Proof. Exercise.

Theorem 3.8. Let p be a prime, and f(t) irreducible ∈ Q[t] of degree p. Suppose f(t) hasexactly 2 non-real roots in C. Then Gal(f) over Q ∼= Sp.

Proof. Gal(f) acts on the p distinct roots of f(t) in a splitting field L of f(t) (in C). ByLemma 3.6, the irreducibility of f(t) implies that Gal(f) is acting transitively on the p roots.By the orbit-stabiliser theorem, p | |Gal(f)| but |Gal(f)| ≤ |Sp| = p! and so Gal(f) has aSylow p-subgroup of order p, necessarily cyclic. Thus, Gal(f) contains a p-cycle.

The supposition that we have precisely 2 non-real roots gives that complex conjugationyields a transposition in Gal(f). The p-cycle and transposition generate the whole of Sp.

Example 3.9. Consider f(t) = t5 − 6t+ 3 ∈ Q[t]. Then Gal(f) ∼= S5.

Proof. f(t) is irreducible by Eisenstein’s criterion with p = 3. We want to show that f(t)has three real roots, two non-real ones and apply Theorem 3.8.

f(−2) = −17, f(−1) = 8, f(1) = −2, f(2) = 23

and f ′(t) = 5t4 − 6 which has two real roots. From the intermediate value theorem, f hasat least three real roots, and by Rolle’s theorem there are at most three real roots, so weare done.

Definition 3.10 (Discriminant). Let f(t) ∈ K[t] with distinct roots α1, . . . , αn in a splittingfield (with f(t) not necessarily irreducible). Let

∆ =∏i<j

(αi − αj).

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Then the discriminant D = D(f) of f is

D = ∆2 =∏i<j

(αi − αj)2

= (−1)n(n−1)

2

∏i 6=j

(αi − αj).

Remark. We’ve already met this in the proof of Theorem 2.21.

Lemma 3.11. Let f(t) be separable ∈ K[t] of degree n with charK 6= 2. Then

Gal(f) ≤ An ⇐⇒ D(f) is a square in K.

Proof. Let L be a splitting field of f(t) over K. Then D(f) 6= 0 and is fixed by all elementsof G = Gal(L/K) as the latter permutes the roots. Thus D ∈ K, since LG = K (by Galoiscorrespondence).

On the other hand, if σ ∈ G then σ(∆) = (sgnσ)∆ where we’re regarding G as a subgroupof Sn and the signature of σ:

sgnσ =

{+1 if σ even

−1 if σ odd

(This is where we need charK 6= 2).Thus if G ≤ An we get that ∆ is fixed by all σ ∈ G. Thus ∆ ∈ K = LG. Otherwise if

G � An, we get σ(A) = −∆ if σ is an odd permutation, and so ∆ /∈ K = LG. Note that ifD does have square roots, they must be ±∆.

Example 3.12. For n = 2, take f(t) = t2 + bt+ c. Then D(f) = b2 − 4c.For n = 3, consider f(t) = t3 + ct + d. Then we can check that D(f) = −4c3 − 27d2.

Any general monic cubic can be written in the above form by suitable substitutions.Take f(t) = t3 − t − 1 ∈ Q[t]. f is irreducible since it is irreducible in Z[t] since it is

irreducible mod 2. D(f) = −23, which is not a square in Q, so Gal(f) ∼= S3.f(t) = t3 − 3t− 1 ∈ Q[t] is irreducible since it is irreducible mod 2. Now D(f) = 81, so

Gal(f) ∼= A3.

For irreducible quartics, we saw that the possible Galois groups are C4, V4, D8, A4, S4.Those that are subgroups of A4 are V4 and A4. From looking at the discriminant, onegets information as to whether the group is one of V4, A4 or is C4, D8, S4. We need furthermethods to pin down which group we are dealing with.

Theorem 3.13 (Mod p reduction). Let f(t) ∈ Z[t] be monic of degree n with n distinctroots in a splitting field. Let p be a prime such that f(t), the reduction of f(t) mod p alsohas n distinct roots in a splitting field. Let f(t) = g1(t) · · · gs(t) be the factorisation intoirreducibles in Fp[t] with nj = deg gj(t). Then Gal(f) ↪→ Gal(f) and has an element ofcycle type (n1, n2, . . . , ns).

Proof. We will talk about the last sentence after thinking about Galois groups of finite fields.The fact that Gal(f) ↪→ Gal(f) is from Number Fields - see Tony Scholl’s teaching page onGalois.

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Example 3.14. For f(t) = t4 + dt + e, we have D(f) = −27d4 + 256e3, not provedhere. Consider f(t) = t4 − t − 1, which is irreducible since it is irreducible mod 2. ThenD(f) = −283, which is not a square in Q.

Now take p = 7.

f(t) = t4 − t− 1

= (t+ 4)(t3 + 3t2 + 2t+ 5) (mod 7)

and the latter factor is irreducible over F7 as it is a cubic which has no roots in F7.By Theorem 3.13, Gal(f) contains an element of cycle type (1, 3), i.e. a 3-cycle. We

deduce Gal(f) ∼= S4, as the other possibilities that contain an odd permutation do notcontain 3-cycles.

3.3 Galois Theory of Finite Fields

Recall what we already know from Section 1. From Example 1.27, a finite field F is ofcharacteristic p > 0, p necessarily prime, and |F| = pr for some r. The multiplicative groupof F is cyclic (Theorem 1.28). It is a splitting field for tn − 1 over Fp where n = pr − 1. Bythe Uniqueness of splitting fields, this is unique. Observe we could also describe F as thesplitting field of tp

r − t over Fp. What we haven’t shown yet is that for any pr, there is afield F with F = pr.

Definition 3.15 (Frobenius automorphism). Let F be a finite field of characteristic p. Thenthe Frobenius automorphism of F is

φ : F −→ Fα 7−→ αp.

Remark.

(i) (α+ β)p = αp + βp since all other terms in binomial expansion are divisible by p.

(ii) Fp is fixed under this, so it is an Fp-automorphism. Since tpr

splits as a product ofdistinct linear factors (t − α) in F, we have that Fp ≤ F is a Galois extension and sowe consider Gal(F/Fp) = G. It is of order r since |F : Fp| = r.

Theorem 3.16 (Galois groups of finite fields). Let F be a finite field with |F| = pr. ThenFp ≤ F is a Galois extension with Gal(F/Fp) = G, a cyclic group with the Frobeniusautomorphism as generator.

Proof. It remains to show that the order of the Frobenius automorphism is r. Supposeφs = id. Then αp

s

= α ∀α ∈ F. But tps − t has at most ps roots in F, so we deduce that

s ≥ r. Observe that φr = id since αpr

= α, ∀α ∈ F.

Now apply the Fundamental Theorem of Galois Theory:

{Fp ≤M ≤ F intermediate fields M} ←→ {subgroups H ≤ G}

where G = Gal(F/Fp) is cyclic.But we know all about subgroups of a cyclic group with generator φ of order r. There

is exactly one subgroup of order s for each s | r generated by φrs . The corresponding

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intermediate subfields are the fixed fields F〈φrs 〉, and

∣∣∣F : F〈φrs 〉∣∣∣ = s. By the Tower Law,∣∣∣F〈φ rs 〉 : Fp

∣∣∣ = rs . Observe that all subgroups of cyclic groups are normal and therefore all

our intermediate fields are normal extensions of Fp.By Theorem 3.2 part (iii), Gal(F〈φ

rs 〉/Fp) ∼= Gal(F/Fp)/H where H = 〈φ rs 〉.

Corollary 3.17. Let Fp ≤ M ≤ F be finite fields. Then Gal(F/M) is cyclic, generated byφu, where φ is the Frobenius automorphism and |M | = pu and M is the fixed field of 〈φu〉.

Proof. Set n = rs .

Theorem 3.18 (Existence of finite fields). Let p be a prime and u ≥ 1. Then there is afield of order pu, unique up to isomorphism.

Proof. Consider the splitting field L of f(t) = tpu − t over Fp. It is a finite Galois extension

Fp ≤ L. However the roots of f(t) form a field, the fixed field of φu. Set L = F and|F : Fp| = u.

Remark (About Theorem 3.13). We’ll discover in Number Fields that Gal(f) ↪→ Gal(f)if f(t) ∈ Z[t]. We factorised f(t) = g1(t) · · · gs(t), a product of irreducibles. We know fromLemma 3.6 that the orbits of Gal(f) correspond to the factorsiation.

We now know Gal(f) is cyclic generated by the Frobenius map, which must have cycletype (n1, . . . , ns) where nj = deg gj(t)

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4 Cyclotomic and Kummer extensions

4.1 Cyclotomic extensions

Definition 4.1 (Cyclotomic extension). Suppose charK = 0 or p prime where p - m. Themth cyclotomic extension of K is the splitting field L of tm − 1.

Remark. f(t) = tm − 1 and f ′(t) = mtm−1 have no common roots, and so the roots off(t) are distinct, the mth roots of unity. They form a finite subgroup µm of L×, and henceby Theorem 1.28 a cyclic group 〈ξ〉. Thus, L = K(ξ) is a simple extension.

Definition 4.2 (Primitive mth root of unity). An element ξ ∈ µm is a primitive mthroot of unity if µm = 〈ξ〉.

Choosing a primitive mth root of unity determines µm → Z/mZ an isomorphism. Notethat ξi is a generator of µm iff (i,m) = 1 and so the primitive mth roots of unity correspondto elements of (Z/mZ)×.

Now consider Galois groups of cyclotomic extensions: we’ll see they must be abelian.Observe f(t) = tm−1 is separable and so the extension K ≤ L is Galois. Let G = Gal(L/K).An element σ ∈ G sends a primitive mth root of unity ξ to a primitive mth root of unityξi where (i,m) = 1. Then ξ 7→ ξi determines a K-homomorphism KL(ξ) → KL(ξ) withξ 7→ ξi. Thus we get an injective map:

Definition 4.3 (Group of roots of unity).

θ : G −→ (Z/mZ)×

This is a group homomorphism: If σ(ξ) = ξc, φ(ξ) = ξj then (σφ)(ξ) = σ(ξj) = ξij . HenceG is abelian.

Thus we regard G as a subgroup of (Z/mZ)×.

Definition 4.4 (Cyclotomic polynomial). The mth cyclotomic polynomial is

Φm(t) =∏

i∈(Z/mZ)×(t− ξi),

the product of the linear factors of tm−1 corresponding to the primitive mth roots of unity.

Remark.f(t) = tm − 1 =

∏i∈Z/mZ

(t− ξi) =∏d|m

Φd(t).

Example. Take K = Q.

Φ1(t) = t− 1, Φ2(t) = t+ 1, Φ3(t) = t2 + t+ 1, Φ4(t) = t2 + 1, Φ8(t) = t4 + 1

since t8 − 1 = (t− 1)(t+ 1)(t2 + 1)(t4 + 1).

Lemma 4.5. Φm(t) ∈ Z[t] if charK = 0 (with Q ↪→ K, prime subfield). Φm(t) ∈ Fp[t] ifcharK = p (with Fp ↪→ K, prime subfield).

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Proof. Induct on m. m = 1 is clearly true.For m > 1, consider

f(t) = tm − 1 = Φm(t)

∏d|md6=m

Φd(t)

.

Note that∏

d|md 6=m

Φd(t) is monic and is defined in Z[t] or Fp[t] by induction.

If charK = 0, we deduce Φm(t) ∈ Q[t] by division of polynomials and by Gauss’ Lemmait is in Z[t]. If charK = p > 0, we deduce by division that Φm(t) ∈ Fp[t].

Lemma 4.6. The homomorphism θ : G→ (Z/mZ)× defined in Definition 4.3 is an isomor-phism iff Φm(t) is irreducible.

Proof. We know from Lemma 3.6 that the orbits of G = Gal(L/K) correspond to thefactorisation of f(t) in K[t]. In particular, the primitive mth roots of unity form one orbitiff Φm(t) is irreducible. Then θ is surjective iff Φm(t) is irreducible.

Theorem 4.7. Let L be the mth cyclotomic extension of finite field F = Fq where q = pn.Then the Galois group G = Gal(L/F) is isomorphic to the cyclic subgroup of (Z/mZ)×

generated by q.

Proof. We know from Corollary 3.17 that G is generated by α 7→ αpn

= αq so θ(G) = 〈q〉 ≤(Z/mZ)×.

Remark. Thus if (Z/mZ)× is not cyclic then θ is not surjective for any finite field F andΦm(t) is reducible over F.

Example. Take F = F3, then Φ8(t) = t4 + 1 = (t2 + t − 1)(t2 − t − 1). t8 − 1 factorisesas a product of linear and quadratic polynomials mod 3, so the splitting field L = F9 is theunique field of order 9 whose multiplicative group is cyclic C8.

(Z/8Z)× = {1, 3, 5, 7} ∼= C2 × C2

We have |L : F3| = 2, so |Gal(L/F3)| = 2 hence Gal(L/F3) cyclic order 2, so θ is notsurjective.

What happens when K = Q?

Theorem 4.8. For all m > 0, Φm(t) is irreducible in Z[t] and hence in Q[t]. Thus θ inDefinition 4.3 is an isomorphism and thus Gal(Q(ξ)/Q) ∼= (Z/mZ)× where ξ = primitivemth root of unity.

Remark. We already know this when m = p prime by substitution and Eisenstein’s crite-rion.

Proof of Theorem 4.8. Gauss’ Lemma gives us that irreducibility in Z[t] implies irreducibil-ity in Q[t]. From Lemma 4.5, irreducibility corresponds to surjectivity of θ. It’s left to showthat Φm(t) is irreducible in Z[t].

Suppose not, and Φm(t) = g(t)h(t) in Z[t] with g(t) irreducible. monic and deg g(t) �deg Φm(t). Let Q ≤ L be the mth cyclotomic extension and ξ be a root of g(t), ξ primitivemth root of unity.

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Claim: if p - m, p prime, then ξp is also a root of g(t) in L. Suppose not. Then ξp isalso a primitive mth root of 1, since p - m, as a root of Φm(t). By the supposition, ξp is aroot of h(t). Define r(t) = h(tp). Then r(ξ) = 0 but g(t) is the minimal polynomial of ξover Q. So g(t) | r(t) in Q[t].

By Gauss’ Lemma, r(t) = g(t)s(t) with s(t) ∈ Z[t]. Now reduce mod p. r(t) = g(t)s(t).But r(t) = h(tp) = (h(t))p. If a(t) is any irreducible factor of g(t) in Fp[t] then a(t) | (h(t))p

and so a(t) | h(t). But then (a(t))2 | g(t)h(t) = Φm(t). Hence Φm(t) has a repeated rootand thus tm − 1 has repeated root mod p. Contradiction, since p - m, so claim is true.

Now consider a root γ of h(t). Then it is also a primitive root of 1 and so γ = ξi for somei with (i,m) = 1. Write i = p1 · · · pk factorisation with pj prime, not necessarily distinct,pj - m. Applying the claim repeatedly we get that γ is a root of g(t), and so Φm(t) has arepeated root.

Hence Φm(t) is irreducible over Q.

Definition 4.9 (Cyclic, abelian extension). An extension K ≤ L is cyclic if the extensionis Galois and Gal(L/K) is cyclic. Similarly, it is called abelian if Gal(L/K) is abelian.

Example. In Corollary 3.17 we saw that for finite fields F ≤ L is cyclic and from Defini-tion 4.3 cyclotomic extensions are abelian. However Theorem 4.8 says that Gal(Q(ξ)/Q) ∼=(Z/mZ)× and so if m = 8, Q ≤ Q(ξ) is an abelian, non-cyclic extension, where ξ primitive8th root of unity.

4.2 Kummer Theory

We consider Galois extensions K ≤ L where L is a splitting field of a polynomial of theform tm − λ with λ ∈ K.

Theorem 4.10. Let f(t) = tm − λ ∈ K[t] and charK - m. Then the splitting field L off(t) over K contains a primitive mth root of unity ξ and Gal(L/K(ξ)) is cyclic of orderdividing m. Moreover f(t) is irreducible over K(ξ) iff |L : K(ξ)| = m.

Remark.L {e}

K(ξ) Gal(L/K(ξ))

K Gal(L/K)

cyclic

abelian

As Gal(L/K(ξ)) is cyclic, Gal(L/K(ξ)) C Gal(L/K).By Theorem 3.2 part (iii), Gal(L/K)/Gal(L/K(ξ)) ∼= Gal(K(ξ)/K) is abelian.

Proof of Theorem 4.10. Since tm − λ and mtm−1 are coprime, we know that tm − λ hasdistinct roots α1, . . . , αm in the splitting field L. Since (αiα

−1j )m = λλ−1 = 1, the elements

1 = α1α−11 , α2α

−11 , . . . , αmα

−11 are m distinct mth roots of unity in L and so

tm − λ = (t− β)(t− ξβ)(t− ξ2β) · · · (t− ξm−1β) ∈ L[t]

where β = α1 and ξ primitive mth root of unity.

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So L = K(ξ, β). Let σ ∈ Gal(L/K(ξ)), which is determined by its action on β. Notethat σ(β) is another root of tm − λ and so σ(β) = ξj(σ)β, where 0 ≤ j(σ) < m. Also, ifσ, τ ∈ Gal(L/K(ξ)) then

τσ(β) = τ(ξj(σ)β) = ξj(σ)τ(β) = ξj(σ)ξj(τ)β

since ξ is fixed by τ . Thus σ → j(σ) gives a group homomorphism

θ : Gal(L/K(ξ))→ Z/mZ.

Note that j(σ) = 1, only if σ is the identity and so θ is injective. Hence Gal(L/K(ξ)) ∼=subgroup of Z/mZ. Finally |L : K(ξ)| = |Gal(L/K(ξ))| ≤ m with equality exactly when theaction of Gal(L/K(ξ)) is transitive on the roots, i.e. when tm − 1 is irreducible over K(ξ)by Lemma 3.6.

Example. Consider f(t) = t6 + 3. If ξ is a primitive 6th root of unity, ξ = −ω, with ω aprimitive cube root of 1.

Q(ξ) = Q(ω) = Q(

1

2(1 +

√−3)

)= Q(

√−3)

f(t) is irreducible over Q by Eisenstein’s criterion with p = 3. However over Q(ξ) = Q(√−3)

it factorises asf(t) = (t3 −

√−3)(t3 +

√−3)

Let L be the splitting field of f(t). From Theorem 4.10 since f(t) factorises over K(ξ),Gal(L/K(ξ)) � Z/6Z.

L {e}

Q(ξ) Gal(L/Q(ξ))

Q Gal(L/Q)

3 3

2 2

Gal(L/Q) is dihedral of order 6, and Gal(L/Q(ξ)) ↪→ Z/6Z is cyclic of order 3. The elementof order 2 is given by complex conjugation. Take β a root of f(t). Then, the roots are

β ξβ ξ2β ξ3β ξ4β ξ5β

β ω2β ωβ

−ωβ −β ω2β

A 3-cycle is given by permuting β, ω2β, ωβ. Conjugating by complex conjugation yields theinverse of the 3-cycle - a dihedral relation.

We have a converse of Theorem 4.10:

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Theorem 4.11. Suppose K ≤M is a cyclic extension with |L : K| = m, where charK - mand that K contains a primitive mth root of unity. Then ∃λ ∈ K such that tm − λ isirreducible over K and K is the splitting field of tm − λ over K. If β is a root of tm − λ inL, then L = K(β).

Definition 4.12 (Kummer extension). A cyclic extension K ≤ L with |L : K| = m, wherecharK - m and K contains a primitive mth root of unity is a Kummer extension.

The proof of Theorem 4.11 uses

Lemma 4.13. Let φ1, . . . , φn be embeddings of a field K into a field L. Then there do notexist λ1, . . . , λn not all zero such that λ1φ1(x) + · · ·+ λnφn(x) = 0 ∀x ∈ K.

Proof. Example sheet 2, question 10.

Proof of Theorem 4.11. Let Gal(L/K) = 〈σ〉 of order m. Observe that 1, σ, σ2, . . . , σm−1

are distinct maps L→ L, and we can apply Lemma 4.13. There exists α ∈ L such that

β = α+ ξσ(α) + · · ·+ ξm−1σm−1(α) 6= 0

where ξ is a primitive mth root of unity. Observe that σ(β) = ξ−1β 6= β and so β /∈ K, thefixed field of Gal(L/K).

σ(βm) = (σ(β))m = βm. Let λ = βm ∈ K. But tm − λ = (t− β)(t− ξβ) · · · (t− ξm−1β)in L[t], and so K(β) is the splitting field of tm − λ over K (recall ξ ∈ K). Observe that1, σ, . . . , σm−1 are distinct K-automorphisms of K(β) and so |K(β) : K)| ≥ m.

So L = K(β) = K(ξβ) since ξ ∈ K. tm − λ is the minimal polynomial of β over K andhence is irreducible.

Definition 4.14 (Extension by radicals). A field extension K ≤ L is an extension byradicals if ∃K = L0 ≤ L1 ≤ · · · ≤ Ln = L such that each Li ≤ Li+1 is either cyclotomic orKummer extension. A polynomial f(t) ∈ K[t] is soluble by radicals if its splitting fieldlies in an extension by radicals.

4.3 Cubics

We’ve already seen that if f(t) is a monic irreducible cubic in K[t] with L its splitting fieldover K,

Gal(f) = Gal(L/K) = G is A3 or S3

(since irreducibility =⇒ action on roots is transitive, and transitive subgroups of S3 areA3 or S3).

L

K(∆)

K

{e}

G ∩A3

G

1 or 2 1 or 2

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where ∆2 = D(f) the discriminant of f . But to see that we can solve f by radicals we wantto make use of Theorem 4.11, and so we need to adjoin the appropriate roots of unity.

Now we get a bigger picture, denoting ω a primitive cube root of 1.

L(ω)

K(∆, ω)

K(∆)

L

K

1 or 2

3

1 or 2

1 or 2

From the Tower law, |L(ω) : K(∆, ω)| = 3. Hence Gal(L(ω)/K(∆, ω)) ∼= C3. We can applyTheorem 4.11 to see that L(ω) = K(∆, ω)(β), where β is a root of an irreducible polynomialt3−λ ∈ K(δ, ω)[t]. In fact, from the proof of Theorem 4.11 we see that β = α1 +ωα2 +ω2α3

where α1, α2, α3 roots of f(t).Now all the extensions K ≤ K(∆) ≤ K(∆, ω) ≤ L(ω) are cyclotomic or Kummer. So

f(t) is soluble by radicals.

In practice

Given an irreducible cubic

f(t) = t3 + at2 + bt+ c

= (t− α1)(t− α2)(t− α3)

we have α1 + α2 + α3 = −a.Replace the αis by α′i = αi+

a3 so that α′1 +α′2 +α′3 = 0. They are roots of a polynomial

t3 + pt+ q and K(α1, α2, α3) = K(α′1, α′2, α′3). We have D(g) = −4p3 − 27q2.

Set β = α′1 + ωα′2 + ω2α′3, γ = α′1 + ω2α′2 + ωα′3. Then

βγ = α′1 + α′2 + α′3 + (ω + ω2)(α′1α′2 + α′1α

′3 + α′2α

′3)

= (α′1 + α′2 + α′3)2 − 3(α′1α′2 + α′1α

′3 + α′2α

′3)

= −3p

and so β3γ3 = −27p3.

β3 + γ3 = (α1 + ωα′2 + ω2α′3)3 + (α′1 + ω2α′2 + ωα′3)3 + (α′1 + α′2 + α′3)3

= 3(α′31 + α′32 + α′33 ) + 18α′1α′2α′3

= −27q

since α′3i = −pα′i − q and so α′31 + α′32 + α′33 = −3q.

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So β3 and γ3 are roots of a quadratic t2 + 27qt− 27p3, hence they are

−27

2q ± 3

√−3

2

√−27q3 − 4p3 = −27

2q ± 3

√3

2

√D.

We can solve for β3 and γ3 in K(√−3,√D) = K(ω,∆). We can get β by adjoining a cube

root of β3, and then set γ = −3pβ . Finally we solve in L(ω) for α′1, α

′2, α′3 (remembering

α′1 + α′2 + α′3 = 0), using

α′1 =1

3(β + γ), α′2 =

1

3(ω2β + ωγ), α′3 =

1

3(ωβ + ω2γ).

4.4 Quartics

As with the cubics by making a substitution of the form α′i = αi + a4 we may assume the

sum of the roots to be zero, and so the t3 term is zero.

f(t) = t4 + bt2 + ct+ d monic irreducible

= (t− α1)(t− α2)(t− α3)(t− α4)

Let L = K(α1, α2, α3, α4) be the splitting field for f(t) over K and set M = LG∩V4 . TheFundamental Theorem of Galois Theory gives Gal(M/K) ∼= G

G∩V4.

L

M = LG∩V4

K(∆)

K

normalextension

of K

{e}

G ∩ V4 CG

G ∩A4

G = Gal(L/K) ≤ S4

Take θ : S4 → S3 with ker θ = V4, so S4/V4 ∼= S3. Consider θ|G : G → S3, soker θ|G = G ∩ V4, and G

G∩V4

∼= im θ|G ≤ S3.We therefore go looking for a cubic for which M is the splitting field, called the resolvent

cubic.Set x = α1 + α2, γ = α1 + α3, z = α1 + α4 and so

α1 =1

2(x+ y + z), α2 =

1

2(x− y − z), α3 =

1

2(−x+ y − z), α4 =

1

2(−x− y + z).

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Thus K(α1, α2, α3, α4) = K(x, y, z).

x2 = (α1 + α2)2 = −(α1 + α2)(α3 + α4)

y2 = (α1 + α3)2 = −(α1 + α3)(α2 + α4)

z2 = (α1 + α4)2 = −(α1 + α4)(α2 + α3)

These are distinct, e.g. if y2 = z2 then y = ±z and so either α3 = α4 or α1 = α2. Thex2, y2, z2 are permuted by G and are fixed by G ∩ V4 by observation. So K(x2, y2, z2) ≤M = LG∩V4 .

Claim: We have equality M = K(x2, y2, z2). Consider the resolvent cubic g(t) =(t− x2)(t− y2)(t− z2) ∈ K[t]. Note that its coefficients are fixed by G and so lie in K.

Observe D(f) = D(g), (check on example sheet). So K(∆) ≤ K(x2, y2, z2). Now observethat Gal(L/K(x2, y2, z2)) = Gal(K(x, y, z)/K(x2, y2, z2)).

K(x2, y2, z2) ≤ K(x, y2, z2) ≤ K(x, y, z2) ≤ K(x, y, z)

where each extension is degree 1 or 2. So∣∣K(x, y, z) : K(x2, y2, z2)

∣∣ divides 8. So, theelements of Gal(L/K(x2, y2, z2)) have order dividing 8. But Gal(L/K(x2, y2, z2)) ≤ G∩A4,so Gal(L/K(x2, y2, z2)) ≤ G ∩ V4. By the Fundamental Theorem of Galois Theory, M =K(x2, y2, z2).

Consider the coefficients of g(t): x2, y2, z2 are permuted by G and so the coefficients ofg(t) are fixed by G and therefore in K.

x2 + y2 + z2 = −2b,

x2y2 + x2z2 + y2z2 = b2 − 4d,

xyz = −c, x2y2z2 = c2,

So g(t) = t3 + 2bt2 + (b2 − 4d)t− c2

We know how to solve cubics and so we can solve for x2, y2, z2. Then we can solve forx, y, z. Then use the formulae from earlier to find α1, α2, α3, α4.

L

M

K(∆)

K

normalextension

of K

{e}

G ∩ V4 CG

G ∩A4

G = Gal(L/K) ≤ S4

1 or 3 1 or 3

Example. For f(t) = t4 + 4t2 + 2, we have g(t) = t3 + 8t2 + 8t reducible.

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4.5 Solubility by radicals

Now suppose we have a Galois extension K ≤ L with

K = L0 ≤ L1 ≤ · · · ≤ Lm = L

such that Li ≤ Li+1 is either cyclotomic or Kummer extension.Let G = Gal(L/K). There is a corresponding chain of subgroups of G

G = G0 ≥ G1 ≥ · · · ≥ Gm = {e}

with Gi = Gal(L/Li), Li = LGi from Theorem 3.2.However each extension Li ≤ Li+1 is Galois and we know

Gi+1 = Gal(L/Li+1) C Gal(L/Li) = Gi

hence Gi/Gi+1∼= Gal(Li+1/Li), which is abelian if Li ≤ Li+1 is cyclotomic and cyclic if

Li ≤ Li+1 is Kummer.

Definition 4.15 (Soluble group). A group is soluble if there is a chain of subgroups

{e} = Gm CGm−1 C · · ·CG1 CG0 = G

with Gi/Gi+1 abelian.

Example.

• S3 is soluble: {e}C 〈(123)〉C S3.

• S4 is soluble: {e}C V4 CA4 C S4 and A4/V4 ∼= C3, S4/A4∼= C2.

• A5 is simple, and therefore any normal subgroup is A5 and so any chain would havea non-abelian quotient. So, A5 is not soluble.

Lemma 4.16. A finite group G is soluble if and only if we have

{e} = Gm CGm−1 C · · ·CG1 CG0 = G

with Gi/Gi+1 cyclic.

Proof. (⇐) is immediate. (⇒). We know about the structure of finite abelian groups. If Aabelian then there is a chain

{e} = Ar CAr−1 C · · ·CA0 = A

with Ar/Ar+1 cyclic. Thus if we have a chain with abelian factors Gi/Gi+1 we can refine itto have cyclic factors.

Definition 4.17 (Derived subgroup). The derived subgroup G′ of a group G is thesubgroup generated by all the commutators g1g2g

−11 g−12 for g1, g2 ∈ G.

Lemma 4.18. Let K CG. Then G/K abelian ⇐⇒ G′ ≤ K.

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Proof.

G/K abelian ⇐⇒ Kg1Kg2Kg−11 Kg−12 = K ∀g1, g2 ∈ G

⇐⇒ g1g2g−11 g−12 ∈ K

⇐⇒ G′ ≤ K.

Remark.

(1) Next term in Representation Theory, we will prove Burnside’s theorem: If |G| = paqb

for p, q distinct primes then G is soluble.

(2) The Feit-Thompson theorem says that if |G| is odd, then G is soluble.

(3) There’s an analogue of Sylow’s theorems due to Philip Hall. For all |G| = mn withm,n coprime there is a subgroup of order m iff G is soluble.

Definition 4.19 (Derived series). The derived series {G(m)} of G is defined inductively:

G(0) = G

G(1) = G′

G(2) = (G′)′

G(j+1) = (G(j))′

Thus G = G(0) BG(1) BG(2) B · · · with G(j)/G(j+1) abelian.

Lemma 4.20. For G finite, G is soluble ⇐⇒ G(m) = {e} for some m.

Proof. If G(m) = {e} then the derived series gives a chain in the definition of solubility.Conversely if there is such a chain

GBG1 BG2 B · · ·BGm = {e}

with Gi/Gi+1 abelian then an easy induction shows that G(j) ≤ Gj and so G(m) = {e}.

Lemma 4.21.

(i) Let H ≤ G, G soluble. Then H soluble.

(ii) Let H CG, then G soluble ⇐⇒ H and G/H both soluble.

Proof.

(i) G soluble =⇒ G(m) = {e} by Lemma 4.20. But H(m) ≤ G(m) and so H soluble byLemma 4.20.

(ii) Let H C G, then G soluble =⇒ H soluble by (i). G soluble =⇒ G(m) = {e}, say.Observe that (

G

H

)′=G′H

H≤ G

H.

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Similarly,

(G

H)(j) =

G(j)H

H≤ G

H.

Thus (G/H)(m) = H/H, a trivial subgroup of G/H and so G/H soluble.

Now consider the converse. Suppose that H and G/H are soluble. H(r) = {e} and(G/H)(s) = H/H. But (

G

H

)(s)

=G(s)H

H

so G(s)H = H thus G(s) ≤ H. Hence G(r+s) ≤ H(r) = {e}. Thus G is soluble byLemma 4.20.

Example. S5 is not soluble since its subgroup A5 is not soluble.

Theorem 4.22. Let K be a field and f(t) ∈ K[t]. Assume charK = 0. Then f(t) is solubleby radicals over K ⇐⇒ Gal f over K is soluble.

Remark. We don’t need to restrict to charK = 0. What we need to do for a particularf(t) is to avoid a finite number of bad characteristics (avoid characteristics ≤ deg f(t)).

Corollary 4.23. If f(t) is a monic irreducible polynomial ∈ K[t] with Gal(f) ∼= A5 or S5

then f(t) is not soluble by radicals (with charK = 0).

Example. In Example 3.9, we considered f(t) = t5 − 6t + 3 ∈ Q[t]. Gal(f) over Q is S5,and so f is not soluble by radicals.

Lemma 4.24. If K ≤ N is an extension by radicals then ∃N ′ with N ≤ N ′ with K ≤ N ′

is an extension by radicals, with K ≤ N ′ a Galois extension.

Proof of Theorem 4.22. Suppose f(t) is soluble by radicals. Thus if L is the splitting fieldof f(t) over K then L lies in an extension of K by radicals

K = L0 ≤ L1 ≤ · · · ≤ Lm

with each Li ≤ Li+1 cyclotomic or Kummer.With Lemma 4.24, we may assume Lm is Galois over K. By Fundamental Theorem of

Galois Theory there is a corresponding chain of subgroups of Gal(Lm/K). Our previousdiscussion at the beginning of this section (before Lemma 4.13) we know that Gal(Lm/K)is soluble.

But F ≤ L ≤ Lm with K ≤ L Galois. By the Fundamental Theorem of Galois Theory,Gal(L/K) ∼= Gal(Lm/K)/Gal(Lm/L).

But quotients of soluble groups are soluble, so Gal(L/K) is soluble.

Proof of Lemma 4.24. We have K = L0 ≤ L1 ≤ · · · ≤ Lm with each Li ≤ Li+1 cyclotomicor Kummer, and we want to embed this into a Galois extension of the same form.

Assume charK = 0. By the Primitive Element Theorem, Lm = K(α1) for some α1.Let g(t) be the minimal polynomial of α1 over K with splitting field M . Thus M =K(α1, α2, . . . , αn) where α1, . . . , αn are roots of g(t).

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There are K-homomorphisms

φi : M −→M

α1 7−→ αi

extending the K-homs K(α1)→ K(αi) ≤M .The tower K ≤ φi(K) ≤ φi(L1) ≤ · · · ≤ φi(Lm) = K(αi) with cyclotomic or Kum-

mer extensions as before, Consider Lm = K(α1) ≤ φ2(L1)(α1) ≤ φ2(L2)(α1) ≤ · · · ≤φ2(Lm)(α1) = K(α1, α2).

Consider the extension φ2(Lj)(α1) ≤ φ2(Lj+1)(α1):

if Lj ≤ Lj+1 is cyclotomic then all the roots of unity adjoined are now in Lm =K(α1) and so φ2(Lj)(α1) = φ2(Lj+1)(α1).

if Lj ≤ Lj+1 is Kummer then we obtain Lj+1 by adjoining roots of an element ofLj and so we obtain φ2(Lj+1) by adjoining roots of an element in φ2(Lj). Hence weget from φ2(Lj)(α1) to φ2(Lj+1)(α1) by adjoining roots of an element of φ2(Lj). Soit’s a Kummer extension.

Now continue to get suitable chain K(α1, α2) ≤ · · · ≤ K(α1, α2, α3).Thus we get a suitable chain from K to K(α1, . . . , αn) = M . Observe that K ≤ M is

Galois.

Converse of Theorem 4.22. Suppose G = Gal(f) over K is soluble (and charK = 0). LetL be the splitting field of f(t) over K and so |G| = |L : K| = n. Set m = n! and let ξ be aprimitive root of unity and consider L(ξ).

L(ξ)

K(ξ) L

K

≤ n

n

Our proof is similar to that used for cubics. Observe that |L(ξ) : K(ξ)| ≤ n. By thePrimitive Element Theorem L = K(α) for some α with minimal polynomial g(t) say ofdegree n. Then L(ξ) = K(ξ)(α) and the minimal polynomial of α over K(ξ) divides g(t)and so is of degree ≤ n.

Then Gal(L(ξ)/K) is soluble since Gal(L(ξ)/L) is soluble and Gal(L/K) ∼= Gal(L(ξ)/K)Gal(L(ξ)/L)

soluble by Fundamental Theorem of Galois Theory and Lemma 4.21. Then the subgroupGal(L(ξ)/K(ξ)) ≤ Gal(L(ξ)/K) is soluble by Lemma 4.21.

Thus there is a chain of subgroups

Gal(L(ξ)/K(ξ)) = G0 BG1 B · · ·BGm = {e},

with Gi/Gi+1 cyclic (using Lemma 4.16).

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Now use the Fundamental Theorem of Galois Theory to get a corresponding chain offields K(ξ) ≤ K1 ≤ · · · ≤ Km = L(ξ), with each Ki ≤ Ki+1 Galois, with cyclic Galoisgroup. By Theorem 4.11, all these extensions are Kummer (not all the extensions are ofdegree ≤ n and so we have the appropriate roots of unity). Thus we’ve embedded L in anextension of K by radicals.

Example.

(1) Take f(t) = t4 + 4t2 + 2, irreducible over Q by Eisenstein’s criterion, and it has roots

±√−2±

√2 (two conjugate pairs). Complex conjugation gives double transposition.

The resolvent cubic is g(t) = t3 + 8t2 + 8t which is reducible with roots 0,−4± 2√

2.

We have L = K(√−4 + 2

√2,√−4− 2

√2), and M = K(−4 + 2

√2,−4 − 2

√2) =

K(√

2). So |L : K| = 8, and the Galois group is dihedral.

(2) f(t) = t4 + 2t + 2 is irreducible over Q by Eisenstein’s criterion. Its discriminant is101 · 42, not a square, and the resolvent cubic is g(t) = t3 − 8t− 4, irreducible since itis t3 + 2t+ 1 mod 5, irreducible. So there is a 3-cycle in the Galois group.

Gal(f) is a transitive subgroup of S4 and not of A4 containing a 3-cycle, so Gal(f) isS4.

(3) f(t) = t5 − t− 1 is irreducible over Q since it is irreducible mod 5, and so the Galoisgroup contains a 5-cycle and is a transitive subgroup of S5. Mod 2, it factorises as aproduct of an irreducible cubic and quadratic: (t3 + t2 + 1)(t2 + t + 1). So Gal(f̄) isgenerated by an element of cycle type 3, 2, so Gal(f) contains an element of cycle type3, 2, so g3 is a transposition. A subgroup of S5 containing a 5-cycle and a transpositionmust be S5 itself. So Gal(f) = S5 over Q.

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5 Final Thoughts

5.1 Algebraic closure

Definition 5.1 (Algebraically closed). A field L is algebraically closed if any f(t) ∈ L[t]splits into a product of linear factors in L[t].

Remark. This is equivalent to saying that any f(t) ∈ L[t] has a root in L, or that anyalgebraic extension of L is L itself.

Definition 5.2 (Algebraic closure). An extension K ≤ L is an algebraic closure of K ifK ≤ L is algebraic and L is algebraically closed.

Lemma 5.3. If K ≤ L is algebraic and every polynomial in K[t] splits completely over L,then L is an algebraic closure of K.

Proof. We need to show L is algebraically closed. Suppose L ≤ L(α) is a finite extension,and fα(t) = tn + an−1t

n−1 + · · · + a0 is the minimal polynomial of α over L. Let M =K(a0, a1, . . . , an−1). Then M ≤ M(α) is a finite extension. But each ai is algebraic overK and so |M : K| < ∞. Hence |M(α) : K| < ∞ by Tower law and so α is algebraic overK. The minimal polynomial over K must split over L, and so α ∈ L. Thus any algebraicextension of L is L itself.

Example. A = {α ∈ C | α algebraic over Q } It is a subfield of C: If α, β are algebraic overQ then |Q(α, β) : Q| <∞ so if γ = α+β, α−β, αβ or α/β (if β 6= 0) we get Q(γ) ≤ Q(α, β)and so |Q(γ) : Q <∞. So γ is algebraic over Q so γ ∈ A. Therefore A = Q is an algebraicclosure of Q.

If we want to prove existence and uniqueness of algebraic closures in general then weneed to appeal to Zorn’s Lemma. This appears in Logic and Set Theory, and it’s equivalentto the Axiom of Choice and to the Well Ordering Principle.

Definition 5.4 (Partial order). (S,≤) is a partial order on S if

(i) ∀x ∈ S x ≤ x

(ii) x ≤ y and y ≤ z =⇒ x ≤ z.

(iii) if x ≤ y and y ≤ x then x = y.

S is totally ordered if for any x, y ∈ S either x ≤ y or y ≤ x. A chain is a partiallyordered set (S,≤) that is a totally ordered subset.

Lemma 5.5 (Zorn’s Lemma). Let (S,≤) be a non-empty partially ordered set. Supposethat any chain has an upper bound in S. Then S has a maximal element.

Lemma 5.6. Let R be a ring. Then R has a maximal ideal.

Proof. Let S be the set of proper ideals of R. This is is non-empty, since (0) is proper.Partially order S by inclusion. Any ideal I is proper ⇐⇒ 1 /∈ I. Any chain of proper idealshas an upper bound in S, namely the union of the chain. Zorn’s Lemma gives that S has amaximal element, i.e. a maximal ideal of R.

Theorem 5.7 (Existence of algebraic closures). For any field K there is an algebraic closure.

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Proof. Let

S = { (f(t), j) | f(t) irreducible, monic in K[t], 1 ≤ j ≤ deg f }

For each pair s = (f(t), j) ∈ S we introduce an indeterminate Xs = Xf,j . Consider thepolynomial ring K[Xs : s ∈ S] and set

f̃(t) = f(t)−deg g∏j=1

(t−Xf,j) ∈ K[Xs : s ∈ S][t].

Let I C K[Xs : s ∈ S] generated by all the coefficients of all the f̃(t). Denote thecoefficents of f̃(t) by af,l for 0 ≤ l ≤ deg f .

Claim: I 6= K[Xs : s ∈ S]. Proof: Suppose 1 ∈ I and aim for a contradiction.

b1af1,l1 + · · ·+ bNafN ,lN = 1 in K[Xs : s ∈ S]. (+)

Let L be a splitting field for f1(t) · · · fN (t).

For each i, fi splits over L. fi(t) =∏deg fij=1 (t − aij). Define a K-linear ring homomor-

phism, identity on K,

θ : K[Xs : s ∈ S] L

Xfi,j αij

Xs 0 otherwise.

This induces a map K[Xs : s ∈ S]→ L[t]. Then

θ(f̃i(t)) = θ(fi(t))−deg fi∏j=1

θ(t−Xfi,j)

= fi(t)−deg fi∏j=1

(t− αi,j) = 0.

But then θ(afi,j) = 0 since afi,j are the coefficients of f̃i(t). But applying θ to (+) we get0 = 1.

Then I is a proper ideal of K[Xs : s ∈ S]. By Zorn’s Lemma there is a maximal idealP of K[Xs : s ∈ S] containing I. Set L1 = K[Xs : s ∈ S]/P , a field. Thus we have a fieldextension K ≤ L1.

Claim: L1 is an algebraic closure of K. First show K ≤ L1 is algebraic: L1 is generatedby the maps xf,j of the Xf,j . However f̃(t) has coefficients in I and so its image L1[t] is thezero polynomial. Thus in L1[t],

f(t) =∏

(t− xf,j) (∗)

and so f(xf,j) = 0. Thus the xf,j are algebraic.Any element of L1 involves only finitely many of the xi,j and so is algebraic over K.

Moreover from (∗) any f(t) ∈ K[t] splits completely over L1.The result follows from Lemma 5.3.

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Theorem 5.8. Suppose θ : K → L is a ring homomorphism and L is algebraically closed.Suppose K ≤ M is an algebraic extension. Then θ can be extended to a homomorphismθ : M → L (i.e. φ|K = θ).

Proof. Let

ξ = { (N,φ) | K ≤ N ≤M,φ a homomorphism N → L extending θ } .

Partially order ξ with (N1, φ1) ≤ (N2, φ2) if N1 ≤ N2 and φ2|N2= φ1. ξ is non-empty since

(K, θ) ∈ ξ.If there is a chain (N1, φ1) ≤ · · · then set N =

⋃Nλ. This is a subfield of M , and we can

define ψ : N → L as follows: if α ∈ N then α ∈ Nλ for some λ and we set ψ(α) = φλ(α).This is well defined.

Then (N,ψ) is an upper bound for our chain ξ.Zorn’s Lemma applies and gives a maximal element of ξ, (N,φ). We now show N = M .

Given α ∈M , it is algebraic over K, and hence over N . Let fα(t) be its minimal polynomialover N . But φf(t) is in L[t] and so splits completely over L, since L is algebraically closed.

So φf(t) = (t− β1) · · · (t− βr), say. Since φf(Bγ) = 0 then there is a map

N(α) ∼= N [t](fα(t)) L

α β1 extendingφ

Maximality of (N,φ) implies that N(α) = N . So α ∈ N , so N = M .

Theorem 5.9 (Uniquness of algebraic closures). If K ≤ L1, L ≤ L2 are two algebraicclosures of K then there exists an isomorphism φ : L1 → L2.

Proof. By Theorem 5.8 there is a homomorphism φ : L1 → L2 extending the embedding ofK into L2. Since K ≤ L2 is algebraic, so too is φ(L1). But L1 is algebraically closed andso φ(L1) is algebraically closed. So L2 = φ(L1) and φ is an isomorphism.

5.2 Symmetric polynomials and invariant theory

In the buildup of the Fundamental Theorem of Galois Theory we met Artin’s Theorem: LetK ≤ L and H a finite group of AutK(L). Let M = LH . Then M ≤ L is a Galois extensionand H = Gal(L/M).

Example. L = K(X1, . . . , Xn). Let Sn permute the variables. These permutations induceK-automorphisms of L. By Artin’s Theorem, if M = LSn ≤ L then it is Galois andGal(L/M) = Sn. Thus Sn is a Galois group of some field extension.

We know that we can regard any finite group G as a subgroup of some Sn, and so we seethat any finite group is a Galois group of some field extension (using Fundamental Theoremof Galois Theory). Let

f(t) = (t−X1)(t−X2) · · · (t−Xn) ∈M [t]

= tn − s1tn−1 + s2tn−2 + · · ·+ (−1)nsn.

Thus s1 = X1 + · · ·+Xn, s2 =∑i<j XiXj , and sn = X1X2 · · ·Xn.

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Definition 5.10 (Elementary symmetric polynomials). These si are the elementary sym-metric polynomials.

Theorem 5.11. The fixed field M = Lsn = K(s1, . . . , sn) and the s1, . . . , sn are alge-braically independent over K (in L).

Definition 5.12. α1, . . . , αn are algebraically independent over K if the ring homo-morphism K[Y1, . . . , Yn]→ K[α1, . . . , αn] ≤ L is an isomorphism where K[Y1, . . . , Yn] is thepolynomial ring in Y1, . . . , Yn.

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Index

algebraic, 5algebraic closure, 45

existence, 45uniqueness, 47

algebraically closed, 45Artin’s theorem, 25automorphism group, 22

constructible, 7cyclotomic polynomial, 32

degree, 3derived series, 41derived subgroup, 40discriminant, 28

Eisenstein’s criterion, 8elementary symmetric polynomials, 48

field extension, 3by radicals, 36cyclic, 34cyclotomic, 32Kummer, 36normal, 13normally generated, 22separable, 15separably generated, 18simple, 6

finite field

existence, 31uniqueness, 30

fixed field, 24formal derivative, 14frobenius automorphism, 30fundamental theorem, 24

Galois extension, 22Galois group, 22

of polynomial, 27

K-homomorphism, 10

minimal polynomial, 5

norm, 19

poset, 45Primitive element theorem, 18primitive root, 32

separable polynomial, 14soluble, 40splitting field, 10

existence, 12uniqueness, 12

tower law, 3trace, 19

Zorn’s lemma, 45

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