Galois Theory

Willem A. de Graaf

Contents

Preface 1

1 Galois Theory 31.1 How it all began . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Cardano and the others . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.3 Ruffini and Abel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.4 Galois . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Separable polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Separable, normal and Galois extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 The Galois correspondence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Cyclotomic fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5.1 Cyclotomic polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.2 The Galois group of a cyclotomic extension . . . . . . . . . . . . . . . . . . . . 20

1.6 Solvability of polynomial equations by radicals . . . . . . . . . . . . . . . . . . . . . . 231.6.1 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.6.2 Radical extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.6.3 A polynomial not solvable by radicals . . . . . . . . . . . . . . . . . . . . . . . 311.6.4 The discriminant of a polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . 321.6.5 Casus irreducibilis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2 Applications of Galois theory 352.1 The fundamental theorem of algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.2 Proving that certain primitive functions are not elementary . . . . . . . . . . . . . . . 352.3 Galois descent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.4 Galois cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Index 53

Bibliography 54

Preface

Galois theory started, now about two hundred years ago, with the work of Galois (1811-1832) onroot of polynomials. His goal was to give a criterion for deciding whether a given polynomial wassolvable by radicals or not. The criterion he came up with involved a (for his time) new algebraicstructure, namely groups. Galois introduced them to study the symmetries that exist between theroots of a polynomial. For this reason Galois theory was a very important step in the development ofalgebra as we know it today. Moreover the theory of Galois provided a striking example of the powerof group actions. After the publication of Galois’ work (in 1846) his theory has seeen a tremendousdevelopment and is now key to understanding various areas of algebra and geometry.

The theory found in Galois’ work is rather different from the way it is done nowadays. One ofthe reasons for this is the development that mathematics and in particular algebra has undergonesince the times of Galois. When Galois was alive there was no group theory (indeed, the origin ofthis concept lies in his work) and almost no field or ring theory. To mention just one main differencebetween Galois’ and the modern approach: Galois used permutations of the roots of a polynomial,whereas nowadays we use automorphisms of a splitting field. For a thorough account of the historyof the study of the roots of a polynomial we refer to the book by Tignol, [Tig01].

In these notes we describe some of the basic concepts and constructions of modern Galois theory,focusing on field extensions of finite degree. (The infinite case is very important and interesting aswell, but requires a lot more technical tools.) We will need many basic results from algebra, and forthese I will refer to my course notes on algebra, available at

http://www.science.unitn.it/~degraaf/algnotes/algebranotes.pdf

A reference like 3(Algebra refers to Chapter 3 of the algebra notes. The main concepts that weneed from basic algebra are groups (3(Algebra) and field extensions (Chapter 4(Algebra)). From thelatter especially the concept of splitting field (Section 4.6(Algebra)) will play a major role.

There are a great many great books on Galois theory (most of them with the same title as thesenotes). We cannot list all of them, so we just mention [Bos18], [Cox12].

2 CONTENTS

Chapter 1

Galois Theory

We start with a very brief description of the historical roots of Galois theory; here we see how theconcept and use of symmetry has been developed.

1.1 How it all began

Galois theory is concerned with roots of polynomials. The main problem that led Galois to invent histheory was the one of finding radical expressions for the roots of a polynomial (or proving that nosuch expressions can be found). This problem has a rather long history as demonstrated by certainclay tablet dating back to at least 1500 BC, and found in current Iraq:

This shows that the ancient Babylonians knew how to solve an equation of degree two, in modernnotation: x2 + bx+ c = 0. Given this knowledge the question comes up whether we can find formulasfor the roots of polynomials of higher degree, for example of degree three: x3 + ax2 + bx+ c = 0. Thehistory regarding those equations starts rather spectacularly in the italian renaissance.

1.1.1 Cardano and the others

In the sixteenth century mathematicians from northern Italy found magnificent formulas for findingthe roots of polynomial equations of degree three and four. The main players in this story are Scipionedel Ferro (1465-1526, from Bologna), Nicolo Fontana, called il Tartaglia (1499-1557, from Brescia) andGirolamo Cardano (1501-1576, from Milano).

Scipione del Ferro Nicolo Tartaglia Girolamo Cardano

4 Galois Theory

Scipione del Ferro was the first to find a formula for the roots of polynomials of degree 3. However,he did not publish it in any form, preferring to keep his discovery to himself, written down in a littlenotebook. Apparently the mathematicians of those days had a bit of a habit to keep their discoveriesto themselves. The reason for this tendency to secrecy was that they often tried to impress otherpeople with their mathematical skill, and even fought mathematical duels between each other. It isclear that the person with more secrets was more likely to excell at such events. However, in thecenturies after del Ferro’s death his notebook was lost, so we do not know exactly what he discovered.

Before he passed away del Ferro revealed the secret of his formula for the cubic equation to a studentof his, Antonio Maria del Fiore. Antionio del Fiore went round in mathematical circles braggingabout his knowledge. Tartaglia, who was interested in the problem of solving cubic equations himself,heard about it, and accepted a cartello di matematica disfida from Antonio del Fiore. This was amathematical duel where each contestant gave thirty mathematical problems to his opponent. Thewinner would be the one who managed to solve more problems. Tartaglia proposed thirty problemsof various nature, whereas del Fiore wrote down thirty cubic equations. However, in the meantimeTartaglia had on his own discovered the formula for solving the cubic, so he easily won the contest.

At around that time Girolamo Cardano was writing a book, called the Ars Magma (or in full ArtisMagnae, Sive de Regulis Algebraicis Liber Unus), on algebra. He learned about the contest won byTartaglia and was very keen on including the formula for the cubic in his book. However, Tartaglia,like Scipione del Ferro, wanted to keep the secret to himself, and only revealed the formula to Cardanowhen the latter promised not to publish it.

Lodovico Ferrari (1522-1565), who was a collaborator of Cardano, then also found a way to solveequations of degree four. One step of his method consisted in solving an equation of degree 3. Sowithout Tartaglia’s formula Cardano and Ferrari could not publish their new discovery either. At thispoint Cardano and Ferrari recalled the contest with Antionio del Fiore, who had obtained the solutionto the cubic from his bolgnese master Scipione del Ferro. They went to Bologna where they found delFerro’s notebook that had been kept by Annibale della Nave, who was his successor at the universityof Bologna. Because he had now found the formula also from a different source, Cardano found thathe was no longer bound by the promise made to Tartaglia and published the formulas for the cubicand quartic equations in his Ars Magna. Cardano did write that the formulas for the cubic were dueto del Ferro and Tartaglia. But the latter was extremely annoyed just the same.

1.1.2 Lagrange

After the publication of the Ars Magna the obvious question arose whether a formula could be foundto solve equations of degree 5. The first to systematically investigate this question was Joseph-LouisLagrange (1736-1813)

Joseph-Louis Lagrange

who in his great book Reflexions sur la resolution algebrique des equations devised a method forsolving polynomial equations of arbitrary degree. For degrees 2,3,4 his method yielded the knownformulas, but for degree 5 the method failed.

Here we have a short look at Lagrange’s method. Consider, for example, an equation of degree 3,x3 + ax2 + bx+ c = 0. Let α1, α2, α3 be its solutions. Lagrange considered symmetric expressions inthese αi; these are polynomials in the αi that remain invariant when we permute the αi. For example

α1α2α23 + α2

1α2α3

1.1 How it all began 5

is not symmetric because if we interchange α1, α2 the expression becomes

α2α1α23 + α2

2α1α3 = α1α2α23 + α1α

22α3

which is not equal to the original expression. On the other hand,

α41α2α3 + α1α

42α3 + α1α2α

43

is symmetric. A fundamental fact concerning these symmetric expressions (known before Lagrange)is the following: a symmetric expression in α1, α2, α3 can be computed as a polynomial in a, b, c. Forexample,

α41α2α3 + α1α

42α3 + α1α2α

43 = a3c− 3abc+ 3c2.

Now let ω = −1+√−3

2 , then ω ∈ C has the property ω 6= 1 but ω3 = 1. Set

z1 = 13(α1 + ωα2 + ω2α3)

z2 = 13(α1 + ωα3 + ω2α2)

z3 = 13(α2 + ωα1 + ω2α3)

z4 = 13(α2 + ωα3 + ω2α3)

z5 = 13(α3 + ωα1 + ω2α2)

z6 = 13(α3 + ωα2 + ω2α1),

and define the so called resolvent polynomial r = (x − z1)(x − z2)(x − z3)(x − z4)(x − z5)(x − z6).Because permuting the αi means that we permute the zi it follows that the coefficients of r aresymmetric expressions in the αi. In particular, they can be expressed as polynomials in a, b, c. In fact,

a small calculation shows that r = x6 + qx3 − p3

27 where q = 2a3

27 −ab3 + c and p = −a2

3 + b. We seethat by some miracle r has a form that it makes it easy to find its roots: it is a quadratic polynomialin x3. So by finding its roots we find the zj and from there we can find the αi. This then yields thesame formulas as found by del Ferro and Tartaglia.

One of the main reasons why Lagrange’s method is important is his use of symmetric expressions.It is one of the the first occasions that symmetry, understood as invariance under some transformations(in this case they are permutations of the roots) is used in a systematic way.

1.1.3 Ruffini and Abel

Paolo Ruffini Niels Henrik Abel

Paolo Ruffini (1765-1822) was an italian mathematician and medical doctor. In 1799 he publisheda book, entitled Teoria generale delle equazioni, with a proof of the impossibility of resolving equationsof degree five by radicals. He sent a copy of the book to Lagrange but never received an answer. Hisargument was a valid attempt at proving the impossibility of resolving the quintic, but in the end itwas not quite correct.

Niels Henrik Abel (1802-1829) was a norwegian mathematician who in 1824 gave the first fullyaccepted proof that it is not possible to solve the quintic by radicals. He used a similar strategy asRuffini and essentially closed the gap in Ruffini’s proof. Abel, like Lagrange, used permutations of the

6 Galois Theory

roots of a general equation of degree 5. One essential step in his proof was the following fact (statedin modern language): let the symmetric group S5 act on the set X, then the orbit of an x ∈ X cannothave 3 or 4 elements.

1.1.4 Galois

Evariste Galois

Evariste Galois (1811-1832) was a french mathematician and revolutionary. He died in a duel,of which the circumstances are not exactly known, at the age of 20. By that time he had written apaper entitled Memoire sur les conditions de resolubilite des equations par radicaux on the problem ofsolving polynomial equations by radicals. He submitted several versions of it to the french academyof sciences, but some were lost and the final one, submitted in 1831, was rejected on the grounds thatit was hard to read and did not have a criterion that could easily be applied to a given polynomial.(In fact, in order to apply the theory of Galois to a polynomial one first has to compute its splittingfield, which can be rather difficult.)

When Galois started to work on solving polynomial equations he probably considered the questionas to the solvability of the general equation of degree 5 as closed (and settled by Abel). However, thereobviously are polynomials of degree 5 that can be solved by radicals; for a trivial example considerx5− 2. So Galois considered a different question: when is a given polynomial solvable by radicals? Inhis solution he went far beyond the mathematics of people such as Lagrange and Abel. For example,Lagrange had considered all permutations of the roots of a polynomial, but one of the main resultsof Galois was that there is a group Gf of permutations of the roots of a polynomial f such that1) every expression in the roots that is invariant under these permutations is rational (that is, liesin Q), 2) every expression in the roots that is rational is invariant under these permutations. It isimportant to remark here that Galois used a notion of invariance different from Lagrange’s: whileLagrange was only considered with the form of an expression (that is, he saw them as expressions informal indeterminates), Galois was concerned with the value of the expression (i.e., the exact numericalvalue).

In 1842, so ten years after Galois’ death, Joseph Liouville somehow received Galois’ papers andstarted reading them. A year later he communicated to the french academy that he would publishthe work of Galois, which he did in 1846.

1.2 Separable polynomials

Let F be a field and f ∈ F [x] an irreducible polynomial. Then we say that f is separable if the rootsof f in a splitting field of f all have multiplicity one (in other words, when f has exactly deg f distinctroots in its splitting field). Because two different splitting field of f are isomorphic via an isomorphismwhose restriction to F is the identity (Corollary 4.6.8(Algebra)), this notion does not depend on thechoice of splitting field. More generally we say that a not necessarily irreducible f ∈ F [x] is separableif its irreducible factors are separable.

Remark 1.2.1 Opinion seems to be divided as to what the best definition of separability for reduciblepolynomials should be. Many authors use a simpler definition, and define a polynomial to be separableif it has roots of multiplicity one in its splitting field. With this definition the polynomial x4 + 2x3 +3x2 + 2x + 1 ∈ Q[x] is not separable as it is equal to (x2 + x + 1)2, whereas with our definition it

1.2 Separable polynomials 7

is separable. This difference usually does not lead to confusion, because the only polynomials whoseseparability is of interest are the irreducible ones.

There is a simple criterion for deciding whether a given polynomial is separable or not based onLemma 4.7.2(Algebra). Let f ∈ F [x] be irreducible and suppose that it is not separable. Then thementioned lemma says that f and f ′ have a common root. That means that gcd(f, f ′) (which lies inF [x]!) has degree at least 1. As f is irreducible that implies that f = gcd(f, f ′). Hence f divides f ′

which, because of their respective degrees, is only possible when f ′ = 0. Write f = a0+a1x+· · ·+amxm;then f ′ = a1 + 2a2x+ · · ·+mamx

m−1. So f ′ = 0 if and only if iai = 0 for i ≥ 1. That implies that thecharacteristic of F is a prime p > 0 and ai = 0 whenever p does not divide i. In other words, we have

f = a0 + apxp + · · ·+ akpx

kp,

that is f is a polynomial in xp. This shows one implication of the following lemma.

Lemma 1.2.2 Let F be a field and let f ∈ F [x] be irreducible. Then f is not separable if and only ifthe characteristic of F is a prime p and f is a polynomial in xp.

Proof. We still need to prove the other direction. Suppose that the characteristic of F is a primep and f is a polynomial in xp. Then f ′ = 0. Consider a root α of f in a splitting field E. Thenf = (x−α)g for some g ∈ E[x]. Hence f ′ = g+(x−α)g′ and as f ′ = 0 it follows that f = −(x−α)2g′,so that f is not separable. 2

Remark 1.2.3 We expand a little on remark 1.2.1. Consider f = x4 + x2 + 1 ∈ F2[x]. This is apolynomial in x2, but f = (x2 + x + 1)2 so f is reducible. Now x2 + x + 1 is not a polynomial in x2

so it is separable. Therefore, according to our definition, f is separable as well.

The following rather technical theorem will play an important role for us. In flavour it is similarto the theorem that says that two splitting fields are isomorphic (Theorem 4.6.7(Algebra)). Also inthis case we start with two base fields F , F that are isomorphic via the map a 7→ a. But now, on theleft we do not take a splitting field but an extension of F of the form E = F (α1, . . . , αn). On the rightwe take a field L such that the minimal polynomial of each αi splits into linear factors over L (or,equivalently, such that L contains a splitting field of the product of these polynomials). The theoremcounts the number of injective field homomorphisms σ : E → L that extend a 7→ a.

From Section 4.6(Algebra) we recall that the isomorphism F → F extends to an isomorphismF [x]→ F [x], f 7→ f .

Theorem 1.2.4 Let F ⊂ M be fields and let α1, . . . , αn ∈ M be algebraic over F . Let fi ∈ F [x] bethe minimal polynomial of αi. We suppose that each fi is separable. Let E = F (α1, . . . , αn) and letL ⊃ F be an extension such that each fi splits into linear factors over L. Then the number of injectivefield homomorphisms σ : E → L with σ(a) = a for a ∈ F is exactly [E : F ].

Proof. We prove this theorem by induction on n. If n = 0 then E = F so there exists only oneσ (given by σ(a) = a for a ∈ F ) and [E : F ] = 1. So the statement is trivial in this case. Nowlet α1, . . . , αn+1 ∈ M be algebraic over F with separable minimal polynomials f1, . . . , fn+1. LetE = F (α1, . . . , αn+1) and let L ⊃ F be an extension containing a splitting field of f1 · · · fn+1.

First note that f1 ∈ F [x] is separable as well (this follows immediately from Lemma 1.2.2). So ithas t = deg f1 distinct roots β1, . . . , βt ∈ L. By Lemma 4.6.6(Algebra) for 1 ≤ i ≤ t we have a uniqueinjective homomorphism ψi : F (α1)→ L with ψi(a) = a for a ∈ F and ψi(α1) = βi.

Let

A = σ : E → L | σ is an injective field homomorphism and σ(a) = a for all a ∈ F.

Let σ ∈ A and write f1 = a0 + a1x+ · · ·+ atxt with ai ∈ F . Then

f1(σ(α1)) = a0 + a1σ(α1) + · · ·+ atσ(α1)t

= σ(a0) + σ(a1)σ(α1) + · · ·+ σ(at)σ(αt1)

= σ(a0 + a1α1 + · · ·+ atαt1) = σ(f1(α1)) = σ(0) = 0.

8 Galois Theory

It follows that σ(α1) = βi for a certain i. Therefore the restriction of σ to F (α1) is equal to a ψi. SoA = A1 ∪ · · · ∪At (disjoint union) where

Ai = σ ∈ A | σ|F (α1) = ψi.

Let g2, . . . , gn+1 ∈ F (α1)[x] be the minimal polynomials of respectively α2, . . . , αn+1 over F (α1). Thengj divides fj (4.3.4(Algebra)). Hence if we let gj be the image of gj under ψi, then gj divides fj . Wesee that L is an extension of F (βi) such that the polynomials g2, . . . , gn+1 split into linear factors overL. So by induction we have |Ai| = [E : F (α1)]. Hence

|A| = |A1|+ · · ·+ |At|= [E : F (α1)] + · · ·+ [E : F (α1)]

= t[E : F (α1)] = [E : F (α1)][F (α1) : F ] = [E : F ]

(by the degree formula, Theorem 4.3.3(Algebra)). 2

Example 1.2.5 Let i =√−1 ∈ C and ω = 4

√2 ∈ R. Using the degree formula (Theorem 4.3.3(Alge-

bra)) one sees that [Q(i, ω) : Q] = 8. So by the previous theorem it follows that there are 8 injectivefield homomorphisms σ : Q(i, ω) → Q(i, ω) with σ(a) = a for all a ∈ Q. Let σ be such a homomor-phism. Viewing Q(i, ω) as an 8-dimensional vector space over Q we have that σ is a linear map. Butas σ is injective it has to be surjective as well. It follows that σ is an automorphism.

We can actually describe all these automorphisms. For that note that an automorphism σ :Q(i, ω)→ Q(i, ω) with σ(a) = a for a ∈ Q is uniquely determined by the values σ(i), σ(ω). But σ(i)has to be a root of x2 + 1 and σ(ω) is a root of x4−2. In other words, σ(i) can be ±i and σ(ω) can be±ω, ±iω. So we have 8 choices for the pair (σ(i), σ(ω)). Because we need 8 automorphisms it followsthat each of these choices yields an automorphism.

Note that without using Theorem 1.2.4 we would only be able to conclude that we have at most 8automorphisms. To prove that we have exactly 8 of them (without using the theorem) would involvechecking that each choice in fact defines an automorphism, a rather laborious task.

Lemma 1.2.6 Let E ⊃ F be a field extension of finite degree. Then E is not equal to the union of afinite number of fields E1, . . . , Es with F ⊂ Ei ( E.

Proof. If E is finite then E∗ is a cyclic group (Proposition 3.7.9(Algebra)). Also each E∗i is a cyclicgroup. Because these are strictly smaller, they cannot contain an element of order |E∗|. So the lemmafollows in this case.

Now suppose that E is infinite and write [E : F ] = n. Then E is an n-dimensional vector spaceover F and the Ei are finite dimensional subspaces. Let α1, . . . , αn be a basis of E over F . Each Ei iscontained in a (generally not unique) (n− 1)-dimensional subspace Vi ⊂ E. Since Vi has codimension1, there are ei,1, . . . , ei,n ∈ F such that

Vi = a1α1 + · · ·+ anαn | ai ∈ F, ei,1a1 + · · ·+ ei,nan = 0.

In other words, if hi = ei,1x1 + · · · + ei,nxn ∈ F [x1, . . . , xn], then Vi consists of those∑

i aiαi suchthat hi(a1, . . . , an) = 0. let h = h1 · · ·hs. Then h(a1, . . . , an) = 0 if and only if

∑i aiαi lies in the

union of the Vi. But h is not the zero polynomial. So since F is infinite, there are b1, . . . , bn ∈ F withh(b1, . . . , bn) 6= 0. But then

∑i biαi lies outside the union of the Vi, hence outside the union of the Ei. 2

Theorem 1.2.7 (Primitive Element Theorem) Let F ⊂ M be fields and α1, . . . , αn ∈ M alge-braic over F . Set E = F (α1, . . . , αn). Suppose that the minimal polynomial of each αi is separable.Then there is an α ∈ E such that E = F (α).

1.3 Separable, normal and Galois extensions 9

Proof. Let fi be the minimal polynomial of αi over F . Let L be an extension of F containing asplitting field of each fi over F . According to Theorem 1.2.4 there are t = [E : F ] injective fieldhomomorphisms σi : E → L with σi(a) = a for a ∈ F (for 1 ≤ i ≤ t). For 1 ≤ i < j ≤ t set

Ei,j = α ∈ E | σi(α) = σj(α).

These are proper subfields of E. So by Lemma 1.2.6 there is an α ∈ E lying outside all Ei,j , i.e., suchthat σi(α) 6= σj(α) for i 6= j. In other words, σ1(α), . . . , σt(α) are all distinct.

Let f be the minimal polynomial of α. Then a small calculation (very similar to the one performedin the proof of Theorem 1.2.4) shows that σi(α) is a root of f . Hence deg(f) ≥ t. So [F (α) : F ] ≥ t.But as [E : F ] = t and F (α) ⊂ E it follows that F (α) = E. 2

An α ∈ E as in the theorem is called a primitive element of E. Note that in the case of a finitefield this term is used for elements with a substantially stronger property (they have to generate themultiplicative group E∗, Section 4.7.3(Algebra)).

Example 1.2.8 Let E = Q(√

2,√

3). Then also E = Q(√

2 +√

3), so√

2 +√

3 is a primitive elementof E.

Remark 1.2.9 From the proof of the primitive element theorem we see that if α ∈ E is such thatσ1(α), . . . , σt(α) are distinct, then α is a primitive element. Consider for example the field Q(i, ω)from Example 1.2.5. Let α = i+ ω. Then the images of α are ±i± ω, ±i± iω. These are all distinct,hence α is a primitive element.

1.3 Separable, normal and Galois extensions

A field extension E ⊃ F is said to be algebraic if every element of E is algebraic over F . We leave itas an exercise to show that extensions of finite degree are automatically algebraic.

Definition 1.3.1 Let E ⊃ F be an algebraic field extension. This extension is called

separable if the minimal polynomial of every element of E over F is separable,

normal if every irreducible polynomial in F [x] that has one root in E splits as a product of linearfactors over E (that is, has all its roots in E),

Galois if it is both separable and normal.

Let E be a field then we let Aut(E) be the set of all field automorphisms of E. This set is calledthe automorphism group of E. We note that with respect to the composition of functions Aut(E) isindeed a group because

composition of functions is associative,

the identity map is an automorphism, and serves as neutral element of Aut(E),

if σ, τ are automorphisms of E then σ τ is an automorphism as well (so Aut(E) is closed undercomposition),

if σ is an automorphism of E then so is its inverse σ−1 (indeed, let α, β ∈ E; then thereare α′, β′ ∈ E with α = σ(α′), β = σ(β′) and hence σ−1(α + β) = σ−1(σ(α′) + σ(β′)) =σ−1(σ(α′ + β′)) = α′ + β′ = σ−1(α) + σ−1(β) and in the same way, σ−1(αβ) = σ−1(α)σ−1(β)).

Example 1.3.2 Consider the field Q(i, ω) as in Example 1.2.5. Let σ be an automorphism of Q(i, ω).It is straightforward to see that σ(a) = a for a ∈ Q (this essentially follows from σ(1) = 1). Sofrom Example 1.2.5 it follows that Aut(Q(i, ω)) has 8 elements. These are described in the mentioned

10 Galois Theory

example. More in detail, we have Aut(Q(i, ω)) = σ1, . . . , σ8 where σj(i), σj(ω) are as in the followingtable

σ1 σ2 σ3 σ4 σ5 σ6 σ7 σ8i i i i i −i −i −i −iω ω iω −ω −iω ω iω −ω −iω

We can use this to compute product relations between the σj . For example, σ4σ7(i) = σ4(σ7(i)) =σ4(−i) = −i and σ4σ7(ω) = σ4(σ7(ω)) = σ4(−ω) = iω; it follows that σ4σ7 = σ6. Also σ7σ4(i) =σ7(i) = −i and σ7σ4(ω) = σ7(−iω) = σ7(−i)σ7(ω) = i · −ω = −iω; so that σ7σ4 = σ8. Continuinglike this we can complete the multiplication table:

σ1 σ2 σ3 σ4 σ5 σ6 σ7 σ8

σ1 σ1 σ2 σ3 σ4 σ5 σ6 σ7 σ8σ2 σ2 σ3 σ4 σ1 σ6 σ7 σ8 σ5σ3 σ3 σ4 σ1 σ2 σ7 σ8 σ5 σ6σ4 σ4 σ1 σ2 σ3 σ8 σ5 σ6 σ7σ5 σ5 σ8 σ7 σ6 σ1 σ4 σ3 σ2σ6 σ6 σ5 σ8 σ7 σ2 σ1 σ4 σ3σ7 σ7 σ6 σ5 σ8 σ3 σ2 σ1 σ4σ8 σ8 σ7 σ6 σ5 σ4 σ3 σ2 σ1

Let E be a field and let G ⊂ Aut(E) be a subgroup. Then we set

EG = α ∈ E | σ(α) = α for all σ ∈ G.

It is straightforward to see that this is a subfield of E. So we have an extension E ⊃ EG.

Lemma 1.3.3 Let E be a field, G ⊂ Aut(G) a finite subgroup and F = EG. Furthermore, for α ∈ Eset Aα = σ(α) | σ ∈ G. Then the minimal polynomial of α over F is∏

β∈Aα

(x− β).

In particular, E/F is algebraic.

Proof. For τ ∈ G we define τ : E[x]→ E[x] by

τ(a0 + · · ·+ amxm) = τ(a0) + · · ·+ τ(am)xm.

It is straightforward to see that τ is a ring automorphism (that is, it is bijective and respects multi-plication and addition). Let

g =∏β∈Aα

(x− β).

Then τ(g) =∏β(x − τ(β)). As G is a group we have that τσ | σ ∈ G = G. Hence τ(β) | β ∈

Aα = Aα. So τ(g) = g, and writing g = a0 + · · ·+ amxm it follows that τ(ai) = ai for all i. As this

holds for all τ ∈ G we get that ai ∈ EG = F . In other words, g ∈ F [x].

Since α ∈ Aα we see that g(α) = 0. Hence α is algebraic over F . Let f ∈ F [x] be its minimalpolynomial. By Proposition 4.3.4(Algebra), f divides g. Writing f = b0 + · · ·+ bmx

m with bi ∈ F , wecompute for σ ∈ G,

f(σ(α)) = b0 + b1σ(α) + · · ·+ bmσ(α)m = σ(b0 + · · ·+ bmαm) = σ(f(α)) = 0.

(Where we use that σ(a) = a for all a ∈ F .) Hence all β ∈ Aα are roots of f and therefore g dividesf . Because both polynomials are monic we have f = g. 2

1.3 Separable, normal and Galois extensions 11

Example 1.3.4 Consider the field Q(i, ω) as in Examples 1.2.5, 1.3.2. Let G = Aut(E), then as wewill see later we have EG = Q.

Let α = i + ω. By using the description of the automorphisms in Example 1.2.5 we see that Aαconsits of

±i± ω, ± i± iω.

So Aα has 8 elements and therefore the minimal polynomial of α has degree 8. In particular α is aprimitive element.

Now let β = ω+ iω. Then Aβ consists of the elements ±ω± iω. It has four elements and thereforethe minimal polynomial of ω + iω has degree 4.

Proposition 1.3.5 Let E be a field, G ⊂ Aut(E) a finite subgroup and F = EG. Then E/F is aGalois extension and [E : F ] = |G|.

Proof. Let α ∈ E. By Lemma 1.3.3 we see that its minimal polynomial splits into linear factors overE and that its roots all have multiplicity one. It follows that E/F is normal and separable, and henceGalois.

For the second statement, we note that [E : F ] = 1 if and only if G consists just of the identity(as F = EG). So we may assume that [E : F ] > 1 and |G| > 1.

Now define a sequence α1, α2, . . . ∈ E in the following way. We let α1 ∈ E \ F . Suppose thatα1, . . . , αm are defined. If F (α1, . . . , αm) = E then the sequence stops, otherwise we let αm+1 be anyelement of E \ F (α1, . . . , αm).

We note that [F (α1, . . . , αn) : F ] is finite as all αi are algebraic over F . Secondly, by the degreeformula,

[F (α1, . . . , αn+1) : F ] = [F (α1, . . . , αn+1) : F (α1, . . . , αn)][F (α1, . . . , αn) : F ].

Because αn+1 6∈ F (α1, . . . , αn) we have that the first factor is greater than 1. Hence we see that

[F (α1, . . . , αn+1) : F ] > [F (α1, . . . , αn) : F ].

On the other hand, by the theorem of the primitive element (Theorem 1.2.7) there are βn ∈ Esuch that F (α1, . . . , αn) = F (βn). However, by Lemma 1.3.3 the minimal polynomial of βn hasdegree at most |G|. Hence [F (βn) : F ] ≤ |G| and we see that the sequence is finite. In particular,E = F (α1, . . . , αk) = F (βk), for a certain k. As just noted, this also implies that [E : F ] ≤ |G|.

In the same way as in the proof of Theorem 1.2.7 we see that there is a β ∈ F such that σ(β) 6= τ(β)for σ, τ ∈ G, σ 6= τ . By Lemma 1.3.3 the minimal polynomial of β has degree |G|. It follows that also[E : F ] ≥ |G|. 2

For a field extension E/F we define its Galois group as

Gal(E/F ) = σ ∈ Aut(E) | σ(a) = a for all a ∈ F.

Let σ, τ ∈ Gal(E/F ). It is obvious that σ(τ(a)) = a for a ∈ F , so that στ ∈ Gal(E/F ) as well.Applying σ−1 to the equality σ(a) = a we see that also σ−1 ∈ Gal(E/F ). The identity obviously liesin Gal(E/F ). Hence the latter is a subgroup of Aut(E).

Remark 1.3.6 Let E/F be an extension of finite degree. Then Gal(E/F ) is a finite group. Indeed,let α1, . . . , αn be a basis of E over F . Then a σ ∈ Gal(E/F ) is completely determined by the valuesσ(αi) for 1 ≤ i ≤ n. But each αi is algebraic over F and σ(αi) is a root of the minimal polynomialof αi over F . Therefore σ(αi) can only take a finite number of values. Hence there are only a finitenumber of possible σ.

Theorem 1.3.7 Let E/F be an extension of finite degree and set G = Gal(E/F ). The following areequivalent:

(i) E/F is a Galois extension,

12 Galois Theory

(ii) E is the splitting field over F of a separable f ∈ F [x],

(iii) [E : F ] = |G|,

(iv) F = EG.

Proof. Suppose that E/F is Galois, that is, separable and normal. Because E/F is of finite degreethere are α1, . . . , αn ∈ E with E = F (α1, . . . , αn). But then by Theorem 1.2.7 there is an α ∈ E withE = F (α). Let f ∈ F [x] be its minimal polynomial. As E/F is normal, f splits into linear factorsover E. So E contains a splitting field of f over F . Obviously this splitting field must contain α andtherefore is equal to F (α) = E.

Next suppose that (ii) holds. We use Theorem 1.2.4 with F = F , a = a for all a ∈ F , L = E.Note that an injective field homomorphism E → E is automatically an isomorphism (indeed: we onlyneed to show surjectivity, but the image of the map is a subspace of E of the same dimension as E,therefore it is equal to E). Hence the conclusion of the theorem translates to |G| = [E : F ].

Suppose that (iii) holds. Set K = EG. From the definition of Galois group we see that F ⊂ K.By Proposition 1.3.5 we see that [E : K] = |G|. So by applying the degree formula (Theorem4.3.3(Algebra)) we get [K : F ] = 1 implying K = F .

Finally, by Remark 1.3.6 we have that G is finite. So it is immediate from Proposition 1.3.5 that(iv) implies (i). 2

Example 1.3.8 Consider the field Q(i, ω) as in Examples 1.2.5, 1.3.2, 1.3.4. From these examples itimmediately follows that |Gal(Q(i, ω)/Q| = 8 = [Q(i, ω) : Q]. Hence it is a Galois extension. In fact,Q(i, ω) is a splitting field of x4 − 2 over Q.

1.4 The Galois correspondence

The Galois correspondence is the main technical tool for applications of Galois groups, linking, as itdoes, the intermediate fields of a field extension to the subgroup structure of its Galois group.

Let E/F be an extension of fields. Let K be a field with F ⊆ K ⊆ E. Then we say that K is anintermediate field of the extension E/F . If it is clear with which extension we are working then wealso simply say that K is an intermediate field.

Remark 1.4.1 Let E/F be a Galois extension of finite degree, and let K be an intermediate field.Then E/K is a Galois extension as well. This can be seen in several ways. We do it just by using thedefinition of Galois extension. Let f ∈ K[x] be irreducible with a root α ∈ E. Because E is of finitedegree over F we have that α is algebraic over F . Let g be its minimal polynomial ovcer F . Thenf divides g by Proposition 4.3.4(Algebra) But E/F is normal, hence g splits as a product of linearfactors over E. Hence the same is true for f . Furthermore, g is separable, hence f is separable as well.

Lemma 1.4.2 Let E/F be a Galois extension of finite degree and set G = Gal(E/F ).

(i) Let H ⊂ G be a subgroup and set K = EH . Then K is an intermediate field and Gal(E/K) = H.

(ii) Let K be an intermediate field and set H = Gal(E/K). Then H is a subgroup of G and K = EH .

Proof. By definition of Galois group we have F ⊂ K so thatK is an intermediate field. By Proposition1.3.5 we see that [E : K] = |H|. By Theorem 1.3.7(iii) we see that [E : K] = |Gal(E/K)|. From thedefinition of Galois group it is immediate that H ⊂ Gal(E/K). Hence |H| = |Gal(E/K)| shows thatH = Gal(E/K).

It is obvious that H in (ii) is a subgroup of G. By part (iv) of Theorem 1.3.7 we conclude thatK = EH . 2

1.4 The Galois correspondence 13

Theorem 1.4.3 (Galois correspondence) Let E/F be a Galois extension of finite degree and setG = Gal(E/F ). Define

A = subgroups H of G B = fields K with F ⊆ K ⊆ E .

Then we have bijective maps α : A → B, β : B → A given by α(H) = EH , β(K) = Gal(E/K). Wehave that α, β are each other’s inverses and they invert the inclusion relation, that is

H1 ⊂ H2 =⇒ EH1 ⊃ EH2

K1 ⊂ K2 =⇒ Gal(E/K1) ⊃ Gal(E/K2).

Proof. From Lemma 1.4.2(i) we see that β(α(H)) = H and Lemma 1.4.2(ii) states that α(β(K)) = K.This also immediately implies that α, β are bijective.

For the last statement, suppose that H1 ⊂ H2 and let α ∈ EH2 . Then σ(α) = α for all σ ∈ H2.Hence σ(α) = α for all σ ∈ H1, so that α ∈ EH1 . The second implication is shown in an analogousmanner. 2

The bijective maps of the theorem are together called the Galois correspondence. With thiscorrespondence it is possible to translate field theoretical problems to group theoretical problems.It has proved to be an extremely potent tool in a number of contexts.

An immediate question is the following. Let’s fix a property P of (sub-) groups, and take the setof all subgroups of a given Galois group that have P ; is it possible to characterize the intermediatefields corresponding to these subgroups? (An analogous question can be formulated by starting witha property of intermediate fields.) The next proposition has the answer to this question (and a bitmore) when P is the property to be a normal subgroup. Subgroups of this kind turn out to correspondto intermediate fields that are stable.

Definition 1.4.4 Let E/F be a Galois extension and G = Gal(E/F ). An intermediate field K (sowe have F ⊆ K ⊆ E) is said to be stable if for all σ ∈ G and α ∈ K we have σ(α) ∈ K (in otherwords, if σ sends K to K).

Proposition 1.4.5 Let E/F be a Galois extension of finite degree and G = Gal(E/F ). Let H ⊂ Gbe a subgroup and K = EH the corresponding intermediate field. The following are equivalent

(i) H is a normal subgroup,

(ii) K is a stable intermediate field,

(iii) K/F is a Galois extension.

Furthermore, if one of these (and hence all three) conditions is satisfied then Gal(K/F ) ∼= G/H.

Proof. Suppose that H is normal and let α ∈ K, σ ∈ G. Then for τ ∈ H we have σ−1τσ ∈ H so thatσ−1τσ(α) = α, which is the same as τ(σ(α)) = σ(α). Since this holds for all τ ∈ H it follows thatσ(α) ∈ K. We conclude that K is stable.

We show that (ii) implies (iii). So suppose that K is stable and set G = Gal(K/F ). We willshow that KG = F . Because K is stable we can restrict a σ ∈ G to K and immediately get thatσ|K : K → K is an element of G. Suppose that KG ) F and let a ∈ KG , a 6∈ F . Then because E/Fis a Galois extension it follows that there is a σ ∈ G with σ(a) 6= a. But then σ|K is an element of Gnot leaving invariant a. This is a contradiction and it follows that KG = F . By Theorem 1.3.7 we seethat K/F is Galois.

Now suppose that (iii) holds, i.e., K/F is Galois. We want to show (i), but in order to do that wefirst show (ii). Let α ∈ K and σ ∈ G. Let f ∈ F [x] be the minimal polynomial of α. Then σ(α) is aroot of f as well. Now f is an irreducible polynomial in F [x] having a root in K, so as K/F is normal

14 Galois Theory

it follows that f splits into linear factors over K. In particular σ(α) ∈ K as well, and we see that Kis stable. Because of this we can define a group homomorphism

Ψ : G→ Gal(K/F ), Ψ(σ) = σ|K .

(Note that it is obvious that this is a group homomorphism.) We consider its kernel. A σ ∈ G liesin ker Ψ if and only if σ(α) = α for all α ∈ K. But that is the same as saying that σ ∈ Gal(E/K).By the Galois correspondence the latter is equal to H. We see that ker Ψ = H and therefore H is anormal subgroup of G (Theorem 3.4.7(Algebra)).

Finally, suppose that the three conditions are satisfied. Again we consider the homomorphismΨ. Again using Theorem 3.4.7(Algebra) we see that Ψ(G) ∼= G/N so that |Ψ(G)| = |G|

|H| . On the

other hand, by the degree formula (Theorem 4.3.3(Algebra)) we have [E : F ] = [E : K][K : F ].Furthermore, [E : F ] = |G|, [E : K] = |H|, [K : F ] = |Gal(K/F )| by Theorem 1.3.7. Hence|Ψ(G)| = |Gal(K/F )| from which it follows that Ψ(G) = Gal(K/F ). By Theorem 3.4.7(Algebra) weconclude that G/H ∼= Gal(K/F ). 2

Example 1.4.6 Consider the field Q(i, ω) as in Example 1.3.8. In that example we have observedthat it is a Galois extension of degree 8 of Q. In Example 1.3.2 we determined the multiplication tableof Gal(Q(i, ω)). Now we first find all its subgroups and then the corresponding intermediate fields.For finding the subgroups of a group we use the following observations:

The order of a subgroup of a finite group G divides the order of G (Corollary 3.2.7(Algebra)).

Let g be an element of order m of a group. Then 1, g, g2, . . . , gm−1 is the smallest subgroupcontaining g (Lemma 3.7.2(Algebra)).

Let H be a group of prime order p. Let h ∈ H, h 6= 1. Then the order of h is p and H =1, h, . . . , hp−1 (this follows immediately from Proposition 3.7.4(Algebra)).

Write G = Gal(Q(i, ω)/Q); then |G| = 8 so that a subgroup of G can have order 1, 2, 4, 8. There isonly one subgroup of order 1, namely σ1. Similarly, there is only one subgroup of order 8, namelyG itself. For the other subgroups we need to work a bit harder.

First consider the subgroups of order 2. By the observation above they have the form σ1, τwhere τ is of order 2, i.e., τ2 = σ1. From the table in Example 1.3.2 we immediately see that we getthe subgroups σ1, σi for i = 3, 5, 6, 7, 8.

Now we look at the subgroups of order 4. We first try to find the cyclic subgroups of order 4. Forthat we need elements of order 4. From the multiplication table of G we see that σ2, σ4 have order 4(whereas σ1 has order 1 and all other elements have order 2). But the subgroups generated by thesetwo elements are both equal to σ1, σ2, σ3, σ4. So we find only one cyclic subgroup of order 4. If asubgroup of order 4 is not cyclic then it is of the form σ1, τ1, τ2, τ3 where the elements τi have order2. (Indeed, a τi cannot have order 1 as then τi = σ1, it cannot have order 3 as 3 does not divide4, and it cannot have order 4 as then the subgroup would be cyclic.) So we have to find all sets ofthree elements of order 2 lying in a subgroup of order 4. If we take τ1 = σ3, τ2 = σ5 then we see thatin fact we get the subgroup σ1, σ3, σ5, σ7. If we take τ1 = σ3, τ2 = σ6 then we get the subgroupσ1, σ3, σ6, σ8. The choices τ1 = σ3, τ2 = σ7 and τ1 = σ3, τ2 = σ8 yield the subgroups that we haveseen already. If we take τ1 = σ5, τ2 = σ6 then because σ5σ6 = σ4 we do not get a subgroup of order4. Continuing like this we see that either we get a subgroup that we have already seen, or we get nosubgroup at all.

Concluding, we found one subgroup of order 1, five subgroups of order 2, three subgroups of order4 and one subgroup of order 8. They are displayed, along with their inclusion relations in Figure 1.3.

Now we find the intermediate field corresponding to each of these subgroups. Because σ1 isthe identity we obviously have Q(i, ω)σ1 = Q(i, ω). Because Q(i, ω) is a Galois extension of Q itimmediately follows that Q(i, ω)G = Q (Theorem 1.3.7). For the other subgroups we can alwayscompute the intermediate field by solving a set of linear equations. This is also illustrated below.However, we also have some tricks that sometimes help to find the corresponding intermediate fieldwith much less effort:

1.4 The Galois correspondence 15

Figure 1.3: Subgroups of Gal(Q(i, ω)/Q).

Gal(Q(i, ω)/Q)

σ1, σ3, σ5, σ7 σ1, σ2, σ3, σ4 σ1, σ3, σ6, σ8

PPPPPPPPPPPP

σ1, σ3

PPPPPPPPPPPP

σ1, σ5 σ1, σ7 σ1, σ6 σ1, σ8

AAAA

AAAA

σ1

XXXXXXXXXXXXXXXX

PPPPPPPPPPP

We note the following fact, whose proof we leave as an exercise. Let E/F be a Galois extension,G = Gal(E/F ) and let H ⊂ G be a subgroup. Let K ⊂ E be an intermediate field such that

K ⊂ EH and [K : F ] = |G||H| . Then K = EH . So if we guess a candidate K for EH and it

happens to be of the correct degree over F then we have our intermediate field.

We can start from the “field side” and list some obvious intermediate fields, and try and identifythe subgroups to which they correspond.

Now we look at the subgroups of order 2, starting with σ1, σ3. By looking at the action of σ3 itis obvious that i is fixed. But also ω2 is fixed (as ω is sent to −ω). So a candidate for the intermediatefield is K = Q(i, ω2). For its degree we have [K : Q] = [K : Q(ω2)] · [Q(ω2) : Q]. Now ω2 =

√2 so

[Q(ω2) : Q] = 2. Furthermore, Q(ω2) ⊂ R so that x2+1 has no roots in Q(ω2) and hence is irreducible

in Q(ω2)[x]. It follows that [K : Q(ω2)] = 2 and [K : Q] = 4 which is equal to |G||H| . We conclude that

K = Q(i, ω)σ1,σ3.Also the other subgroups of order 2 correspond to intermediate fields of degree 4 over Q. But we

also immediately see some intermediate fields of degree 4, namely the fields Q(α) where α runs overthe roots of x4 − 2. However, here ω,−ω and iω,−iω give the same fields. So we immediately havetwo fields, namely Q(ω) and Q(iω). By checking which order 2 subgroups leave these fields fixed weimmediately get Q(ω) = Q(i, ω)σ1,σ5, Q(iω) = Q(i, ω)σ1,σ7.

For the remaining two groups of order 2 it is perhaps not immediately clear to which field theybelong, so we resort to linear equations. For this we first describe a basis of Q(i, ω) over Q. We havethat 1, ω, ω2, ω3 is a basis of Q(ω) over Q. Since x2 + 1 is irreducible over Q(ω) we have that 1, iis a basis of Q(i, ω) over Q(ω). Using the argument in the proof of the degree formula (Proposition4.3.3(Algebra)) we see that 1, ω, ω2, ω3, i, iω, iω2, iω3 is a basis of Q(i, ω) over Q. Now consider thesubgroup σ1, σ6 and let a = a1 + a2ω+ a3ω

2 + a4ω3 + a5i+ a6iω+ a7iω

2 + a8iω3 with ai ∈ Q be an

element of Q(i, ω). Then

σ6(a) = a1 + a2iω − a3ω2 − a4iω3 − a5i+ a6ω + a7iω2 − a8ω3.

Hence σ6(a) = a if and only if a3 = a5 = 0, a2 = a6 and a4 = −a8. It follows that

Q(i, ω)σ1,σ6 = a1 + a2ω + a4ω3 + a2iω + a7iω

2 − a4iω3 | ai ∈ Q.

Since the minimal polynomial of ω+ iω has degree 4 (Example 1.3.4) we see that this field is equal toQ(ω + iω).

16 Galois Theory

The subgroup σ1, σ8 is treated similarly. Now we compute

σ8(a) = a1 − a2iω − a3ω2 + a4iω3 − a5i− a6ω + a7iω

2 + a8ω3

and σ8(a) = a if and only if −a2 = a6, a3 = a5 = 0, a4 = a8. Therefore

Q(i, ω)σ1,σ8 = a1 + a2ω + a4ω3 − a2iω + a7iω

2 + a4iω3 | ai ∈ Q.

Moreover, in the same way as for ω+ iω it is seen that the minimal polynomal of ω− iω has degree 4and the above field is equal to Q(ω − iω).

The subgroups of order 4 correspond to intermediate fields of degree 2. For the subgroup σ1, σ2, σ3, σ4we immediately see that its fixed field is Q(i) (it is fixed, and has degree 2 over Q).

The intermediate fields corresponding to the other two groups of order 4 can be determined bysolving a set of linear equations. Consider the subgroup σ1, σ3, σ6, σ8, write a as above and compute

σ3(a) = a1 − a2ω + a3ω2 − a4ω3 + a5i− a6iω + a7iω

2 − a8iω3

σ6(a) = a1 + a2iω − a3ω2 − a4iω3 − a5i+ a6ω + a7iω2 − a8ω3.

Because σ8 = σ3σ6 we have that σ3(a) = σ6(a) = a implies that σ8(a) = a. So

Q(i, ω)σ1,σ3,σ6,σ8 = a ∈ Q(i, ω) | σ3(a) = σ6(a) = a.

Now from the above equations we see that σ3(a) = a if and only if a2 = a4 = a6 = a8 = 0. Secondly,σ6(a) = a if and only if a3 = a5 = 0, a2 = a6 and a4 = −a8. We see that all coefficients are 0 excepta7 which can be chosen freely. Hence the intermediate field we are after is a1 + a7iω

2 | a1, a7 ∈ Q =Q(iω2). In the same way we see that the field corresponding to the last remaining group is Q(ω2).Of course, if we had seen that Q(iω2), Q(ω2) are intermediate fields of degree 2, then we could haveassociated them directly to these groups.

We can put the intermediate fields in a diagram, see Figure 1.4, obtaing the “same” picture as theone with the subgroups, with the difference that the inclusion relations go the other way. Note thatthe Galois correspondence implies that this figure contains all subfields of Q(i, ω), a result which isnot easily obtained otherwise.

Figure 1.4: Intermediate fields of Q(i, ω)/Q.

Q

Q(ω2) Q(i) Q(iω2)

PPPPPPPPPPPP

Q(i, ω2)

PPPPPPPPPPPP

Q(ω) Q(iω) Q(ω + iω) Q(ω − iω)

AAAA

AAAA

Q(i, ω)

XXXXXXXXXXXXXXXX

PPPPPPPPPPP

1.5 Cyclotomic fields 17

Example 1.4.7 Let f = x4 − 4x2 + 2. By Eisenstein’s criterion (Theorem 2.5.14(Algebra)) f isirreducible. Hence E = Q[x]/〈f〉 is a field (Proposition 4.4.1(Algebra)). Writing α = [x] we have thatthe elements 1, α, α2, α3 form a basis of E over Q so that E = a0 + a1α+ a2α

2 + a3α3 | ai ∈ Q.

We have f(α) = 0 or α4 = 4α2−2. We claim that α1 = α, α2 = −α, α3 = α3−3α, α4 = −(α3−3α)are roots of f in E. For α1 we have already seen this. For α2 it is also obvious since f involves onlyeven powers of x. For α3 we first compute α6 = α2 · α4 = 4α4 − 2α2 = 14α2 − 8 and

(α3 − 3α)2 = α6 − 6α4 + 9α2 = −α2 + 4

and(α3 − 3α)4 = (−α2 + 4)2 = −4α2 + 14.

So we see that indeed α43 − 4α2

3 + 2 = 0. Then immediately α4 = −α3 is also a root of f .It follows that E contains a splitting field of f . But since the smallest subfield of E containing Q

and the αi obviously contains E itself, we see that E is a splitting field of f . Hence E/Q is a Galoisextension (Theorem 1.3.7).

Now we determine the Galois group Gal(E/Q). First note that a σ ∈ Gal(E/Q) is completelydetermined by the value of σ(α). Secondly, σ(α) must be a root of f , so we have at most fourpossibilities: σ(α) = αi, 1 ≤ i ≤ 4. On the other hand, because E/Q is Galois, we have |Gal(E/Q)| =[E : Q] = 4 (Theorem 1.3.7). It follows that each choice σ(α) = αi must yield an element of Gal(E/Q).We conclude that Gal(E/Q) = σ1, σ2, σ3, σ4, where σi(α) = αi.

We can compute products between the σi. For example, σ3σ3(α) = σ3(σ3(α)) = σ3(α3 − 3α) =

(α3 − 3α)3 − 3(α3 − 3α). As seen above we have (α3 − 3α)2 = −α2 + 4, and hence

(α3 − 3α)3 = (α3 − 3α)(−α2 + 4) = −α5 + 7α3 − 12α = 3α3 − 10α.

It follows that σ3σ3(α) = −α so that σ3σ3 = σ2. In an analogous manner we find the other products,and compute the multiplication table:

σ1 σ2 σ3 σ4

σ1 σ1 σ2 σ3 σ4σ2 σ2 σ1 σ4 σ3σ3 σ3 σ4 σ2 σ1σ4 σ4 σ3 σ1 σ2

In this case a proper nontrivial subgroup can only have order 2, and must be of the form σ1, τwith τ2 = σ1. We see that we only have σ1, σ2. So together with the obvious subgroups σ1 andGal(E/Q) we have three subgroups.

The intermediate fields corresponding to σ1, Gal(E/Q) are E and Q respectively (the latterbecause E/Q is a Galois extension). It remains to find the intermediate field corresponding to σ1, σ2.Let a = a0 + a1α+ a2α

2 + a3α3 ∈ E; then

σ2(a) = a0 − a1α+ a2α2 − a3α3.

Hence σ2(a) = a if and only if a1 = a3 = 0. It follows that Eσ1,σ2 = a0 + a2α2 | ai ∈ Q = Q(α2).

1.5 Cyclotomic fields

Let F be a field and n ≥ 1 an integer. An ω ∈ F with ωn = 1 is called an n-th root of unity in F .It is primitive if ωk 6= 1 for 1 ≤ k ≤ n − 1. We can reformulate these concepts in group theoreticallanguage by considering the group F ∗ = F \ 0. A root of unity is an element of finite order. It isan n-th root of unity if its order divides n. It is a primitive n-th root of unity if its order is exactly n.

Example 1.5.1 Let ω = 1√2(1 + i) ∈ C. Then

ω2 = i, ω3 = 1√2(−1 + i), ω4 = −1, ω5 = 1√

2(−1− i), ω6 = −i, ω7 = 1√

2(1− i), ω8 = 1.

Hence ω is a primitive 8-th root of unity.

18 Galois Theory

It is clear that not all fields have primitive n-th roots of unity. For example, the only roots of unityin Q are ±1. In order to make a meaningful study of roots of unity we must work with fields thatcontain enough of them, that is, fields containing a splitting field of xn − 1. Now we have a lemmacollecting some immediate properties of the set of n-th roots of unity; in order to prove it we need alittle lemma.

Lemma 1.5.2 Let G be a group and g ∈ G of finite order n. Let i be an integer, i ≥ 1. Then theorder of gi is n

gcd(n,i) .

Proof. Let k be the order of gi. Set d = gcd(n, i) and write i = i′d. Then (gi)n/d = (gn)i′

= 1.Hence k divides n

d (Lemma 3.7.6(Algebra)). Also we have gik = (gi)k = 1. Hence again by Lemma3.7.6(Algebra), n divides ik. But that implies that n

d divides k. It follows that k = nd . 2

Lemma 1.5.3 Let F be a field and let E ⊃ F be an extension containing a splitting field of xn − 1.Set

Un(E) = ω ∈ E | ωn = 1.

Suppose that the characteristic of F is 0, or a prime not dividing n.

(i) Un(E) is a cyclic subgroup of E∗ of order n.

(ii) A ζ ∈ Un(E) generates Un(E) (that is, every element of Un(E) can be written as a power of ζ)if and only if it is a primitive n-th root of unity.

(iii) Let ζ ∈ Un(E) be a primitive n-th root of unity. For i ≥ 1 we have that ζi is also a primitiven-th root of unity if and only if gcd(n, i) = 1.

Proof. Set f = xn − 1. Then f ′ = nxn−1 which is nonzero because of our hypothesis on thecharacteristic. The only root of f ′ is 0, which is not a root of f . Therefore f only has roots ofmultiplicity 1 in its splitting field (Lemma 4.7.2(Algebra)). Hence |Un(E)| = n. It is obvious thatUn(E) is a subgroup of E∗ and it is cyclic by Proposition 3.7.9(Algebra).

Let ζ ∈ Un(E) and denote its order by |ζ|. The cyclic subgroup of Un(E) generated by ζ has orderexactly |ζ|. So ζ generates Un(E) if and only if |ζ| = n if and only if ζ is a primitive n-th root of unity.

The last item follows by (ii) together with Lemma 1.5.2. 2

Remark 1.5.4 If the characteristic p divides n then we write n = psn0 where p does not dividen0. Let α be an n-th root of unity in some extension of F . Then by Lemma 4.7.1(Algebra) we have0 = αn− 1 = (αn0)p

s − 1 = (αn0 − 1)ps. It follows that α is an n0-th root of unity. So we lose nothing

if we restrict our study to n-th roots of unity where the characteristic of the ground field does notdivide n.

1.5.1 Cyclotomic polynomials

Let Pn be the set of primitive n-th roots of unity in C. Define

Φn =∏ζ∈Pn

(x− ζ),

which a-priori is a polynomial in C[x]. It is called the n-th cyclotomic polynomial. In this section wewill describe a method for computing Φn without first computing all elements of Pn and evaluatingthe product. Then we will show that Φn ∈ Z[x] and that Φn is irreducible in Q[x].

Lemma 1.5.5 Let D(n) be the set of all positive divisors of n (including 1 and n). Then

xn − 1 =∏

d∈D(n)

Φd.

1.5 Cyclotomic fields 19

Proof. Consider the group Un(C) ⊂ C∗ of order n (Lemma 1.5.3). For d ∈ D(n) we have that Pd isexactly the set of elements of Un(C) that have order d. So Un(C) is the disjoint union of the sets Pdwhere d runs through D(n). Hence

xn − 1 =∏

ζ∈Un(C)

(x− ζ) =∏

d∈D(n)

∏ζ∈Pd

(x− ζ) =∏

d∈D(n)

Φd.

2

We can use this lemma inductively to compute Φn for n ≥ 1. Indeed, we have Φ1 = x−1, Φ2 = x+1and Φ3 = x3−1

x−1 = x2 + x+ 1. Also Φ1Φ2Φ4 = x4 − 1, and Φ1Φ2 = x2 − 1 so that Φ4 = x4−1x2−1 = x2 + 1.

For another example, Φ1Φ3Φ9 = x9 − 1 and Φ1Φ3 = x3 − 1 and therefore Φ9 = x9−1x3−1 = x6 + x3 + 1.

Proposition 1.5.6 Φn ∈ Z[x].

Proof. We prove this by induction on n. The induction hypothesis is that Φm ∈ Z[x] for all m < n.Let g be the product of all Φd for d a divisor of n, d 6= n. By the induction hypothesis g ∈ Z[x]. Bythe previous lemma gΦn ∈ Z[x]. Suppose that Φn 6∈ Z[x] and write Φn = a0 + · · ·+ asx

s, ai ∈ C. Letu be maximal with au 6∈ Z. Write g = b0 + · · ·+ btx

t where bi ∈ Z. Let c be the coefficient of xu+t ingΦn. Then

c = btau + (bt−1au+1 + bt−2au+2 + · · · ).

The term in brackets lies in Z. So because c ∈ Z and bt = 1 it follows au ∈ Z which is a contradiction.We conclude that Φn ∈ Z[x]. 2

Theorem 1.5.7 Φn is irreducible in Q[x].

Proof. Fix a primitive n-th root of unity ζ0 in C and let f ∈ Q[x] be its minimal polynomial. Thenf divides Φn by Proposition 4.3.4(Algebra). So there is a g ∈ Q[x] with Φn = fg. Because Φn, f aremonic the same holds for g. By Gauss’ lemma (Lemma 2.5.13(Algebra)) there are rational numbersα, β such that αf , βg lie in Z[x] and Φn = (αf)(βg). But f is monic, so αf ∈ Z[x] implies α = ±1.Similarly we get β = ±1. We conclude that f, g ∈ Z[x].

Now let ζ ∈ Pn be a root of f (there is at least one such ζ). Let p be a prime not dividing n. Thenby Lemma 1.5.3(iii) ζp ∈ Pn. We claim that ζp is also a root of f . Suppose that this is not the case,then ζp is a root of g and hence ζ is a root of h = g(xp). Therefore f and h have a nontrivial commonfactor. Consider the homomorphism ψp : Z[x]→ Fp[y] given by

ψp(a0 + · · ·+ asxs) = [a0]p + [a1]py + · · ·+ [as]py

s.

Since f and h have a nontrivial common factor the same holds for ψp(f), ψp(h). Also ψp(h) =ψp(g)(yp) = (ψp(g))p (where the last equality follows from Lemma 4.7.1(Algebra), along with thefact that [a]pp = [a]p for all a ∈ Z, see Theorem 2.5.22(Algebra)). Hence ψp(f), ψp(g) must have anontrivial common factor. But fg divides xn − 1 so ψp(fg) = ψp(f)ψp(g) implies that ψp(f)ψp(g)divides yn − [1]p. Thus the latter polynomial has a root of multiplicity at least 2 in some extensionof Fp. But that is excluded by Lemma 4.7.2(Algebra) as the derivative of yn − [1]p is [n]py

n−1 whichonly has [0]p as a root because [n]p 6= [0]p. We conclude that ζp must be a root of f .

Finally, let ω ∈ Pn and write ω = ζi0 with gcd(n, i) = 1 (Lemma 1.5.3(iii)). Then i = p1 · · · pk,where the pj are primes not dividing n. By repeatedly applying the claim above we get that

ζp10 , ζp1p20 = (ζp10 )p2 , . . . , ζi0 = ζp1···pk0

are roots of f . In other words, all elements of Pn are roots of f and hence Φn divides f . It followsthat f = Φn and Φn is irreducible. 2

20 Galois Theory

1.5.2 The Galois group of a cyclotomic extension

Let F be a field and let E ⊃ F be a splitting field of xn − 1. Then we say that E is a cyclotomicextension of F . By Remark 1.5.4 we may assume that the characteristic of F does not divide n. Thenwe have that E = F (ζ), where ζ is any primitive n-th root of unity in E. Indeed, by Lemma 1.5.3(ii)F (ζ) contains all roots of xn− 1. Furthermore, as xn− 1 has roots of multiplicity 1 in E we have thatE/F is Galois (Theorem 1.3.7).

Now we briefly introduce a class of groups that will play an important role in the sequel. For n ≥ 1we set

(Z/nZ)∗ = [a]n ∈ Z/nZ | gcd(a, n) = 1.

If gcd(a, n) = gcd(b, n) = 1 then also gcd(ab, n) = 1 so that [a]n[b]n = [ab]n lies in (Z/nZ)∗. Further-more, if gcd(a, n) = 1 then there exist u, v ∈ Z with ua + vn = 1. Hence [u]n[a]n = [1]n. It followsthat (Z/nZ)∗ is a multiplicative group. Its order is denoted ϕ(n); the function ϕ : N → N is calledEuler’s ϕ-function.

Theorem 1.5.8 Let n ≥ 1. Let F be a field of characteristic 0 or of characteristic p > 0 where p isa prime not dividing n. Let E be a splitting field of xn − 1 over F . Then

(i) For σ ∈ Gal(E/F ) there is a unique iσ ∈ Z with 1 ≤ iσ ≤ n − 1, gcd(iσ, n) = 1 such thatσ(ω) = ωiσ for all ω ∈ Un(E).

(ii) The map

ψ : Gal(E/F )→ (Z/nZ)∗ defined by ψ(σ) = [iσ]n

is an injective group homomorphism.

(iii) ψ is surjective if and only if Φn is irreducible in F [x].

Proof. Fix a primitive n-th root of unity ζ ∈ Un(E). Let σ ∈ Gal(E/F ); then σ(ζ) is also a primitiven-th root of unity. Hence by Lemma 1.5.3 σ(ζ) = ζiσ for some iσ with gcd(iσ, n) = 1. Because ζn = 1we may assume that 1 ≤ iσ ≤ n − 1. Then iσ is uniquely determined by σ for ζj = ζiσ implies thatj ≡ iσ mod n (because ζ has order n). Finally, if ω ∈ Un(E) then ω = ζk for a certain k. Henceσ(ω) = σ(ζk) = (ζiσ)k = ωiσ .

Let σ, τ ∈ Gal(E/F ). Then σ(τ(ζ)) = στ(ζ) implies iσiτ ≡ iστ mod n. So [iσ]n[iτ ]n = [iσiτ ]n =[iστ ]n, which is another way of saying that ψ(σ)ψ(τ) = ψ(στ). We see that ψ is a group homorphism.Since E = F (ζ) we have that a σ ∈ Gal(E/F ) is uniquely determined by the value of σ(ζ). Thereforeψ is injective.

By Lemma 1.5.3(iii) it follows that |Pn| = ϕ(n). Thus deg Φn = ϕ(n). Secondly, ζ is a root ofΦn. Indeed: it is a root of xn − 1 so by Lemma 1.5.5 it is a root of a Φd for a divisor d of n. Butwe cannot have d < n as otherwise ζ would be a root of xd − 1 as well, and it therefore would not beprimitive. Now ψ is surjective if and only if |Gal(E/F )| = ϕ(n) if and only if [E : F ] = ϕ(n) (Theo-rem 1.3.7(iii)) if and only if the minimal polynomial of ζ is Φn if and only if Φn is irreducible in F [x]. 2

For the base field Q we summarize our findings in the following corollary.

Corollary 1.5.9 Let E be the splitting field of xn− 1 over Q. Then E = Q(ζ) where ζ is a primitiven-th root of unity. The minimal polynomial of ζ is Φn. Furthermore, Gal(E/Q) is isomorphic to(Z/nZ)∗ and the isomorphism sends [i]n ∈ (Z/nZ)∗ to σi ∈ Gal(E/Q) where σ(ζ) = ζi.

Remark 1.5.10 Let ζ = e2πin ∈ C, then ζ is a primitive n-th root of unity. Hence Q(ζ) is a splitting

field of xn − 1 over Q. By the previous corollary the Galois group Gal(Q(ζ)/Q) contains an elementσ with σ(ζ) = ζn−1. We have σ2(ζ) = ζ so that σ2 = 1 and H = 1, σ is a subgroup of order

2. Furthermore, since σ(ζ) = e−2πin we see that σ is the restriction of complex conjugation to Q(ζ).

Therefore Q(ζ)H = Q(ζ) ∩ R. So for instance it follows that [Q(ζ) ∩ R : Q] = ϕ(n)2 .

1.5 Cyclotomic fields 21

Example 1.5.11 Let ζ ∈ C be a primitive 15-th root of unity and consider the extension Q(ζ)/Q.Here we determine its Galois group and explicitly describe its Galois correspondence.

From Theorem 1.5.7 it follows that Φ15 is the minimal polynomial of ζ. By Lemma 1.5.5 we seethat x15−1 = Φ1Φ3Φ5Φ15. Now Φ1Φ5 = x5−1 and Φ3 = x2+x+1. Furthermore, x15−1 = (x5)3−1 =

(x5− 1)((x5)2 + x5 + 1), so that Φ15 = x10+x5+1x2+x+1

= x8− x7 + x5− x4 + x3− x+ 1. From this it follows

that [Q(ζ) : Q] = 8, that 1, ζ, ζ2, . . . , ζ7 is a basis of Q(ζ) over Q and ζ8 = ζ7 − ζ5 + ζ4 − ζ3 + ζ − 1.Theorem 1.5.8 says that

Gal(Q(ζ)/Q) = σ1, σ2, σ4, σ7, σ8, σ11, σ13, σ14

where σi(ζ) = ζi and σiσj = σij mod 15. This yields the following multiplication table

σ1 σ2 σ4 σ7 σ8 σ11 σ13 σ14

σ1 σ1 σ2 σ4 σ7 σ8 σ11 σ13 σ14σ2 σ2 σ4 σ8 σ14 σ1 σ7 σ11 σ13σ4 σ4 σ8 σ1 σ13 σ2 σ14 σ7 σ11σ7 σ7 σ14 σ13 σ4 σ11 σ2 σ1 σ8σ8 σ8 σ1 σ2 σ11 σ4 σ13 σ14 σ7σ11 σ11 σ7 σ14 σ2 σ13 σ1 σ8 σ4σ13 σ13 σ11 σ7 σ1 σ14 σ8 σ4 σ2σ14 σ14 σ13 σ11 σ8 σ7 σ4 σ2 σ1

We determine the subgroups of Gal(Q(ζ)/Q). The orders of the elements σ1, σ2, . . . , σ14 are respec-tively 1, 4, 2, 4, 4, 2, 4, 2. Hence we have three subgroups of order 2: H1 = σ1, σ4, H2 = σ1, σ11,H3 = σ1, σ14. A subgroup of order 4 is either cyclic or consists of three elements of order 2 apartfrom σ1. We immediately see that we get H4 = σ1, σ2, σ4, σ8, H5 = σ1, σ4, σ7, σ13 (the cyclicsubgroups), and H6 = σ1, σ4, σ11, σ14.

Figure 1.5: Subgroups of Gal(Q(ζ)/Q).

Gal(Q(ζ)/Q)

H4 H5 H6

PPPPPPPPPPPP

H1 H2 H3

PPPPPPPPPPPP

AAAA

σ1

!!!!!!!!!!!!!

Now we determine the intermediate fields corresponding to these subgroups. First we look aroundfor some obvious intermediate fields. We see that ζ3 is a primitive fifth root of unity so Q(ζ3) is afield extension of degree 4 of Q. Hence it corresponds to a subgroup of order 2. By looking at theavailable subgroups of that order we see that it must be H2, that is, Q(ζ3) = Q(ζ)H2 . Secondly, ζ5

is a primitive third root of unity, so that [Q(ζ5) : Q] = 2 and this field corresponds to a subgroupof order 4. By looking at which of those groups leave ζ5 invariant we see that it must be H5, i.e.,Q(ζ5) = Q(ζ)H5 .

22 Galois Theory

We deal with the other groups by solving sets of linear equations. First consider H1. Let a =a0 + a1ζ + · · ·+ a7ζ

7. Then

σ4(a) =(a0 − a2 − a6 + a7) + (a2 + a4 − a7)ζ + (−a3 + a6)ζ2 + (−a2 − a6)ζ3

+ (a1 + a2 − a7)ζ4 + (−a2 + a5 + a7)ζ5 − a6ζ6 + (a2 − a3 + a6 − a7)ζ7.

Hence σ4(a) = a if and only if

−a2 − a6 + a7 = 0

−a1 + a2 + a4 − a7 = 0

−a2 − a3 + a6 = 0

−a2 − a3 − a6 = 0

a1 + a2 − a4 − a7 = 0

−a2 + a7 = 0

−2a6 = 0

a2 − a3 + a6 − 2a7 = 0,

which are equivalent to a6 = 0, a2 = −a3 = a7, a1 = a4. Hence

Q(ζ)H1 = a0 + a1ζ + a2ζ2 − a2ζ3 + a1ζ

4 + a5ζ5 + a2ζ

7 | ai ∈ Q.

The element ζ + ζ4 has exactly four different images under Gal(Q(ζ)/Q). Therefore its minimalpolynomial over Q has degree 4 (Lemma 1.3.3) and consequently it is a primitive element of Q(ζ)H1 ,that is, Q(ζ)H1 = Q(ζ + ζ4).

The computation for H3 is analogous, therefore we omit the details. Here the linear equations onthe ai are equivalent to

a1 + a2 − a5 − a6 − a7 = 0

a4 + a5 + a6 = 0

a2 + a3 − a5 = 0

a1 + 2a2 + 2a3 + a4 = 0,

so that

Q(ζ)H3 = a0+a1ζ+a2ζ2+a3ζ

3+(−a1−2a2−2a3)ζ4+(a2+a3)ζ

5+(a1+a2+a3)ζ6+(−a2−2a3)ζ

7 | ai ∈ Q.

Again using Lemma 1.3.3 we see that the minimal polynomial of ζ − ζ4 + ζ6 has degree 4, so thatQ(ζ)H3 = Q(ζ − ζ4 + ζ6).

Note that H4 is cyclic and generated by σ2. Hence a ∈ Q(ζ)H4 if and only if σ2(a) = a. The linearequations equivalent to that reduce to a1 = a4 = −2a5, a2 = −a3 = a7 = −a5, a6 = 0, so that

Q(ζ)H4 = a0 − 2a5ζ − a5ζ2 + a5ζ3 − 2a5ζ

4 + a5ζ5 − a5ζ7 | a5 ∈ Q.

In this case it is obvious that Q(ζ)H4 = Q(θ) with θ = 2ζ + ζ2 − ζ3 + 2ζ4 − ζ5 + ζ7 (in fact, it can beshown the the minimal polynomial of θ is x2 − 3x+ 6).

Finally, σ4σ14 = σ11 and hence a ∈ Q(ζ)H6 if and only if σ4(a) = σ14(a) = a. But the linearequations corresponding to σ4(a) = a and σ14(a) = a have already been studied above. Together theyamount to a1 = a4 = a5 = a6 = 0, a2 = −a3 = a7. So

Q(ζ)H6 = a0 + a2ζ2 − a2ζ3 + a2ζ

7 | a2 ∈ Q = Q(ζ2 − ζ3 + ζ7).

(The minimal polynomial of ζ2− ζ3 + ζ7 is x2− x− 1.) Figure 1.6 has the subfields together with theinclusions between them.

1.6 Solvability of polynomial equations by radicals 23

Figure 1.6: Subfields of Q(ζ).

Q

Q(θ) Q(ζ5) Q(ζ2 − ζ3 + ζ7)

PPPPPPPPPPPP

Q(ζ + ζ4) Q(ζ3) Q(ζ − ζ4 + ζ6)

PPPPPPPPPPPP

AAAA

Q(ζ)

!!!!!!!!!!!!!

1.6 Solvability of polynomial equations by radicals

In this section we look at Galois’ main application of his theory: a criterion for deciding whether theroots of a given polynomial can be expressed by radicals. The criterion says that this happens preciselywhen the Galois group of the splitting field of the polynomial is solvable. The latter is a property ofgroups which we study first. Then we derive Galois’ criterion and use it to find a polynomial of degree5 whose roots cannot be expressed by radicals. Then after an intermezzo on discriminants we cometo the final topic of this chapter, the so-called irreducible case, which explains why, when expressingthe roots of a polynomial of degree 3 by radicals we always need square roots of negative numbers,also if all roots lie in R.

1.6.1 Solvable groups

Definition 1.6.1 A group G is called solvable if there is a series of subgroups G = G1 ⊃ G2 ⊃ · · · ⊃Gm = 1 such that Gi+1 is a normal subgroup of Gi and Gi/Gi+1 is abelian for 1 ≤ i ≤ m− 1.

For example abelian groups are solvable. Slightly less trivially, the dihedral groups Dn (Section3.1.2(Algebra)) are solvable. Indeed, let G2 ⊂ Dn be the subgroup consisting of the σ[i]. As seen inExample 3.3.2(Algebra) this is a normal subgroup. As Dn/G2 has order 2 it is abelian. Secondly, G2

is abelian as well, so our series is Dn ⊃ G2 ⊃ G3 = σ[0].In order to show that a given group is solvable it suffices to find a series of subgroups as in the

definition. Proving that a given group is not solvable seems much more difficult. It is our next objectiveto derive a criterion for doing that.

Let G be a group and A ⊂ G. Let H be the intersection of all subgroups of G that contain A. Itis obvious that H is a subgroup of G, and that every subgroup of G that contains A must also containH. We say that H is the subgroup of G generated by A.

Definition 1.6.2 Let G be a group. For g, h ∈ G we set [g, h] = g−1h−1gh, which is called thecommutator of g, h. Let A be the set of all commutators of all elements of G then by [G,G] wedenote the subgroup of G generated by A. It is called the commutator subgroup of G.

We note that [g, h] = 1 if and only if gh = hg, i.e., if and only if g, h commute. It follows that Gis abelian if and only if [G,G] is the trivial subgroup.

24 Galois Theory

Lemma 1.6.3 Let G be a group. Then [G,G] is a normal subgroup and the quotient G/[G,G] isabelian. Moreover, for a normal subgroup N ⊂ G we have that G/N is abelian if and only if Ncontains [G,G].

Proof. Let h ∈ [G,G] and g ∈ G then ghg−1 = hh−1ghg−1 = h[h, g−1]. We see that ghg−1 ∈ [G,G]and therefore it is a normal subgroup.

Let N be a normal subgroup of G and for g ∈ G write g for the coset gN . The definition of themultiplication in quotient groups implies that [g, h] = [g, h]. Furthermore, G/N is abelian if and onlyif [g, h] = 1 for all g, h ∈ G. But this is the same as [g, h] = 1, which in turn is equivalent to [g, h] ∈ N .We conclude that G/N is abelian if and only if [g, h] ∈ N for all g, h ∈ G. The latter is obviouslyequivalent to [G,G] ⊂ N . 2

Definition 1.6.4 Let G be a group and define G(1) = G and for k ≥ 1, G(k+1) = [G(k), G(k)]. Theseries G = G(1) ⊃ G(2) ⊃ · · · is called the derived series of G.

Proposition 1.6.5 Let G be a group. Then G is solvable if and only if there is an s ≥ 1 withG(s) = 1.

Proof. First of all, if G(s) = 1 then using Lemma 1.6.3 we see that the derived series is a seriessatisfying Definition 1.6.1. Hence G is solvable.

If G is solvable then let G = G1 ⊃ G2 ⊃ · · · ⊃ Gm = 1 be a series as in Definition 1.6.1. Byinduction we show that G(k) ⊂ Gk. This is certainly true for k = 1. Suppose it holds for a certaink ≥ 1. Because Gk/Gk+1 is abelian we have [Gk, Gk] ⊂ Gk+1 by Lemma 1.6.3. Hence

G(k+1) = [G(k), G(k)] ⊂ [Gk, Gk] ⊂ Gk+1.

The conclusion is that G(m) = 1. 2

Corollary 1.6.6 Subgroups and quotients of solvable groups are solvable.

Proof. Let G be a solvable group and H a subgroup. By induction it is straightforward to see thatH(k) ⊂ G(k). So by the previous proposition, H is solvable as well.

Let N ⊂ G be a normal subgroup. Let π : G → G/N be the surjective homomorphism withπ(g) = gN . We claim that π(G(k)) ⊃ (G/N)(k) for k ≥ 1. For k = 1 this is obvious, so sup-pose k ≥ 1 and π(G(k)) ⊃ (G/N)(k). As in the proof of Lemma 1.6.3 we write g for the coset gN .Let g, h ∈ (G/N)(k); then we may assume that g, h ∈ G(k). Then [g, h] = [g, h] = π([g, h]) andhence [g, h] ∈ π(G(k+1)). It follows that π(G(k+1)) is a subgroup of G/N containing all [g, h] forg, h ∈ (G/N)(k). Hence it contains (G/N)(k+1). Our claim is proved, and it immediately implies thatG/N is solvable. 2

Now we investigate the solvability of the symmetric groups Sn introduced in Section 3.1.1(Algebra).As shown there we can write every element of Sn as a product of disjoint cycles. Because

(i1, . . . , ik) = (i1, i2)(i2, i3) · · · (ik−1, ik)

we can also write every element as a product of 2-cycles. However, this way of writing is far fromunique. For example (1, 2)(2, 3)(1, 2) = (1, 3). We see that not even the number of 2-cycles that appearin different expressions for a given π ∈ Sn is always the same. But we can prove that the parity ofthe number of 2-cycles is constant. For this we define an inversion of a π ∈ Sn to be a pair (i, j) withi < j but π(i) > π(j). We let inv(π) be the number of inversions of π. For example, π = (1, 2, 3) ∈ S3has inversions (1, 3), (2, 3) so that inv(π) = 2.

Lemma 1.6.7 Let π, σ ∈ Sn. Then inv(πσ) ≡ inv(π) + inv(σ) mod 2.

1.6 Solvability of polynomial equations by radicals 25

Proof. Set

A = (i, j) | 1 ≤ i < j ≤ n, σ(i) < σ(j), πσ(i) < πσ(j)B = (i, j) | 1 ≤ i < j ≤ n, σ(i) < σ(j), πσ(i) > πσ(j)C = (i, j) | 1 ≤ i < j ≤ n, σ(i) > σ(j), πσ(i) < πσ(j)D = (i, j) | 1 ≤ i < j ≤ n, σ(i) > σ(j), πσ(i) > πσ(j).

Then inv(πσ) = |B|+ |D|, inv(π) = |B|+ |C|, inv(σ) = |C|+ |D|. The lemma follows. 2

Corollary 1.6.8 Let π ∈ Sn and write π = π1 · · ·πk = σ1 · · ·σl, where the πi, σj are 2-cycles. Thenk ≡ l mod 2.

Proof. Note that the number of inversions of a 2-cycle is odd. So using the previous lemma we seethat inv(π) ≡ k mod 2 and inv(π) ≡ l mod 2. 2

The elements of Sn that can be written as a product of an even number of 2-cycles are called even.The others are called odd. It is obvious that the set of even permutations forms a subgroup of Sn. Itis called the alternating group, and denoted An. Consider the 2-cycle (1, 2). It is immediate that Snis the disjoint union of An and (1, 2)An. Hence |An| = 1

2 |Sn| =12n!. Now we show that [Sn, Sn] = An.

For that we first need a little lemma.

Lemma 1.6.9 Let A ⊂ Sn be the set consisting of all 3-cycles. Then An is generated by A.

Proof. Let H ⊂ Sn denote the group generated by A. We have (i, j, k) = (i, j)(j, k) so that 3-cyclesare even and hence H is contained in An.

Now consider two 2-cycles, π = (i, j), σ = (k, l). If i, j = k, l then πσ = 1. If i, j∩k, l con-sists of one element, say j = k, then πσ = (i, j, l). If i, j, k, l are disjoint then πσ = (i, k, j)(k, l, i).We conclude that πσ lies in H. Hence every even permutation lies in H, in other words, An ⊂ H. 2

Proposition 1.6.10 [Sn, Sn] = An.

Proof. For π ∈ Sn we have that π and π−1 have the same parity. Hence [π, σ] = π−1σ−1πσ is even.It follows that [Sn, Sn] ⊂ An. On the other hand,

[(i, j), (i, k)] = (i, j)(i, k)(i, j)(i, k) = (i, j, k).

Hence [Sn, Sn] contains every 3-cycle and therefore it contains An be the previous lemma. 2

Theorem 1.6.11 The group Sn is solvable for n = 2, 3, 4 and not solvable otherwise.

Proof. We have that S2 is abelian and hence solvable. The group A3 is of order 3 and abelian, hence[A3, A3] = 1. Using Propositions 1.6.10, 1.6.5 we conclude that S3 is solvable. Let

V4 = 1, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3).

For π ∈ Sn we have π(i, j)(k, l)π−1 = (π(i), π(j))(π(k), π(l)). It follows that V4 is a normal subgroupof A4 (even of S4). The quotient A4/V4 has order 3 and hence is abelian. So S4 ⊃ A4 ⊃ V4 ⊃ 1 is aseries as in Definition 1.6.1. So S4 is solvable as well.

Now suppose that n ≥ 5. Let i, j, k, s, t be distinct elements of 1, . . . , n. Then

[(i, s, j), (i, t, k)] = (j, s, i)(k, t, i)(i, s, j)(i, t, k) = (i, j, k).

Hence [An, An] contains every 3-cycle and by Proposition 1.6.10 we see that [An, An] = An. By Propo-sition 1.6.5 Sn is not solvable. 2

Now we want to show that a finite solvable group has a descending series of subgroups such thatthe successive quotients have prime order. For this we need a theorem due to Cauchy.

26 Galois Theory

Theorem 1.6.12 (Cauchy) Let G be a finite group and p a prime dividing |G|. Then G has anelement of order p.

Proof. Let T be the set of all p-tuples (g1, . . . , gp) where gi ∈ G and g1 · · · gp = 1. Note that the firstp−1 elements of such a p-tuple can be chosen arbitrarily in G, and then the last element is determinedby gp = (g1 · · · gp−1)−1. Hence |T | = |G|p−1 and in particular we see that p divides |T |.

Let T0 be the subset of T consisting of the tuples (g, . . . , g) with gp = 1. Of course T0 contains(1, . . . , 1). Hence the theorem is equivalent to the statement |T0| > 1. (Indeed, if gp = 1 and g 6= 1then the order of g is p by Lemma 3.7.6(Algebra).)

For (g1, . . . , gp) ∈ T set ϕ(g1, . . . , gp) = (gp, g1, . . . , gp−1). Because

gpg1 · · · gp−1 = gpg1 · · · gp−1gpg−1p = gpg−1p = 1

we see that ϕ maps T to T . Note that it has an obvious inverse and hence ϕ is a bijection. Furthermore,ϕ maps T0 and T1 = T \ T0 to themselves.

Let t ∈ T1. Observe that ϕp(t) = t. Let k ≥ 1 be minimal with ϕk(t) = t. Then k ≤ p and writep = qk + r with 0 ≤ r < k. Then

t = ϕp(t) = ϕr((ϕk)q(t)) = ϕr(t),

and it follows that r = 0 and k = p because p is prime. It follows that the elements t, ϕ(t), . . . , ϕp−1(t)are all distinct. So T1 is the disjoint union of subsets t, ϕ(t), . . . , ϕp−1(t). It follows that p divides|T1|. As p divides |T | we see that it must divide |T0| as well and wqe are done. 2

Remark 1.6.13 We can reformulate this proof using the language of group actions (see Section3.5(Algebra)). Let G = 1, ϕ, . . . , ϕp−1. Then G is a group of order p (it is a subgroup of the groupST1 of all permutations of T1). It naturally acts on T1. The stabilizer of a t ∈ T1 is a subgroup ofG. However, because this group has p elements, its only subgroups are 1 and G itself. Hence thestabilizer of each t ∈ T1 is trivial. So by the orbit-stabilizer theorem (Corollary 3.5.13(Algebra)) itfollows that each orbit of G on T1 has p elements.

Theorem 1.6.14 Let G be a finite solvable group. Then G has a series of subgroups G = G1 ⊃ G2 ⊃· · · ⊃ Gm = 1 such that Gi+1 is a normal subgroup of Gi and Gi/Gi+1 is abelian and of prime order.

Proof. The proof is by induction on |G|. The induction hypothesis is that all solvable groups ofcardinality less than |G| have a series as in the theorem. Because G is solvable it has a normalsubgroup N such that G 6= N and G/N is abelian. Write H = G/N . If the order of H is not primethen write |H| = ps, where p is a prime and s > 1 an integer. By Theorem 1.6.12 H has an elementof order p, and thus it has a subgroup M of order p. Since H is abelian, M is automatically normal.Now consider the homomorphisms π1 : G → H, π1(g) = gN , and π2 : H → H/M , π2(h) = hM andlet π = π2 π1. Then π is a group homomorphism and let N ′ = ker(π) be its kernel. Then N ′ is notall of G because π1 is surjective and the kernel of π2 is not all of H. We have that N ′ strictly containsN and G/N ′ is abelian as well (this follows for example from Lemma 1.6.3). We can continue thisprocess and eventually we find a normal subgroup N ′′ of G such that N ′′ 6= G, G/N ′′ is abelian andG/N ′′ is of prime order. Set G2 = N ′′.

By Corollary 1.6.6 we have that G2 is solvable. So by the induction hypothesis there is a series ofsubgroups G2 ⊃ G3 ⊃ · · · ⊃ Gm = 1 with the properties stated in the theorem. It follows that Ghas such a series as well. 2

1.6.2 Radical extensions

In this section we prove an amazing theorem by Galois: there are expressions for the roots of a givenpolynomial, involving arithmetic operations and taking n-th roots (for n ≥ 2) if and only if the Galoisgroup of the spliting field of the polynomial is solvable. In order to be able to do that we first translatethe existence of the mentioned expressions into a property of the splitting field.

In this section all fields are of characteristic 0.

1.6 Solvability of polynomial equations by radicals 27

Definition 1.6.15 A field extension E/F is said to be radical if there is a tower of fields F = F1 ⊂F2 ⊂ · · · ⊂ Fm+1 = E such that for 1 ≤ i ≤ m we have Fi+1 = Fi(αi) where αi ∈ Fi+1 satisfiesαnii ∈ Fi for some ni > 0.

With the notation of this definition write βi = αmii . Then βi ∈ Fi and αi can be thought of as anni-th root of βi.

Definition 1.6.16 Let F be a field and f ∈ F [x]. Then f is called solvable by radicals if there is aradical extension E/F such that E contains a splitting field of f .

Note that solvability by radicals exactly means that there are expressions for the roots of f involvingthe arithmetic operations, elements of F and n-th roots n

√. Also we remark that it is not enough

to require that the splitting field itself is a radical extension, as there are examples of splitting fieldsthat are contained in radical extensions but are not radical extensions themselves.

Proposition 1.6.17 Let E/F be a radical extension. Then there is an extension K/E such that K/Fis radical and Galois.

Proof. Because E/F is radical there is a tower as in Definition 1.6.15. In particular E = F (α1, . . . , αm).Let fi be the minimal polynomial of αi over F . Set f = f1 . . . fm and let K ⊃ E be a splitting field off over E. Since the αi are roots of f we have that K is also a splitting field of f over F . The extensionK/F is Galois by Theorem 1.3.7 (note that f is automatically separable because the characteristic is0).

Write Gal(K/F ) = σ1, . . . , σn, where σ1 is the identity. Using Lemma 1.3.3 we see that for eachroot β of fi there is a σj with β = σj(αi). Hence K = F (σj(αi) | 1 ≤ i ≤ m, 1 ≤ j ≤ n). Set

Fi,j = F (σ1(α1), . . . , σ1(αm), σ2(α1), . . . , σ2(αm), σ3(α1), . . . , σj(α1), . . . , σj(αi)).

Then for 1 ≤ i ≤ m − 1 we have Fi+1,j = Fi,j(σj(αi+1)). Now σj(αi+1)ni+1 = σj(α

ni+1

i+1 ). Butαni+1

i+1 ∈ Fi+1 = F (α1, . . . , αi). Hence αni+1

i+1 can be written in terms of elements of F and α1, . . . , αi(and arithmetic operations). It follows that σj(α

ni+1

i+1 ) lies in F (σj(α1), . . . , σj(αi)) ⊂ Fi,j .Secondly, for 1 ≤ j ≤ n− 1 we have F1,j+1 = Fm,j(σj+1(α1)) and σj+1(α1)

n1 = σj+1(αn11 ) = αn1

1 ∈F ⊂ Fm,j . We conclude that we have the tower

F ⊂ F1,1 ⊂ F2,1 ⊂ · · · ⊂ Fm,1 ⊂ F1,2 ⊂ F2,2 ⊂ · · · ⊂ Fm,2 ⊂ · · · ⊂ Fm,n = K

and therefore K/F is radical. 2

Lemma 1.6.18 Let K/F be an extension such that there is an α ∈ K with K = F (α) and αn ∈ Ffor some n > 0. Suppose that F contains a primitive n-th root of unity. Then K/F is Galois andGal(K/F ) is abelian.

Proof. Let ω ∈ F be a primitive m-th root of unity. Consider the polynomial f = xn − αn ∈ F [x].The roots of f in K are α, ωα, . . . , ωn−1α and they are all distinct as ω is primitive. Therefore f splitsinto linear factors in K[x] and we see that K is a splitting field of f over F . So by Theorem 1.3.7K/F is Galois.

A σ ∈ Gal(K/F ) is uniquely determined by the value of σ(α), which has to be a root of f . Henceσ(α) = ωiα for a certain i with 0 ≤ i < n. Let τ be a second element of Gal(K/F ) and writeτ(α) = ωjα. Because ω ∈ F we have σ(ω) = τ(ω) = ω implying that

στ(α) = ωi+jα = τσ(α).

Hence Gal(K/F ) is abelian. 2

Proposition 1.6.19 Let F be a field and f ∈ F [x]. Let E be a splitting field of f over F . If f issolvable by radicals then Gal(E/F ) is solvable.

28 Galois Theory

Proof. By Definition 1.6.16 E is contained in a field K such that K/F is radical. By Proposition1.6.17 we may assume that K/F is Galois as well. Let F = F1 ⊂ · · · ⊂ Fm+1 = K be a tower ofextensions such that Fi+1 = Fi(αi), α

nii ∈ Fi. We now would like to apply the Galois correspondence

and obtain a series of subgroups of the Galois group. However, in order to conclude anything on thesesubgroups we need the previous lemma that assumes that the base field contains certain roots of unity.The trick is to add all needed roots of unity and to put them at the bottom of the tower.

Set n = n1 · · ·nm and let L be the splitting field of xn − 1 over K. Since K/F is Galois it is thesplitting field of some h ∈ F [x]. Hence L is the splitting field of h(xn − 1) over F and we see thatL/F is Galois. Furthermore, as remarked in Section 1.5.2 we have L = K(ω) where ω is a primitiven-th root of unity. So L = F (α1, . . . , αm, ω) and we consider the tower

F ⊂ F (ω) ⊂ F (ω, α1) ⊂ F (ω, α1, α2) ⊂ · · · ⊂ F (ω, α1, . . . , αm) = L.

Set G = Gal(L/F ). To the fields in the tower we apply the Galois correspondence, that is, we setGi = Gal(L/F (ω, α1, . . . , αi)) (so that G0 = Gal(L/F (ω))).

Note that all fields, from F (ω) onwards, contain a primitive ni-th root of unity for all i. (Indeed, ifwe set si = n/ni then ωsi is a primitive ni-th root of unity.) For 0 ≤ i ≤ m−1 consider the extensions

F (ω, α1, . . . , αi) ⊂ F (ω, α1, . . . , αi, αi+1) ⊂ L.

By Lemma 1.6.18 the first of these is Galois with abelian Galois group. So by Proposition 1.4.5 wehave that Gi+1 is normal in Gi and the quotient is abelian. Furthermore, F (ω) is the splitting field ofxn − 1 over F so that F (ω)/F is Galois with abelian Galois group (the latter by Theorem 1.5.8(ii)).Again using Proposition 1.4.5 we see that G0 is normal in G with abelian quotient. So we have a seriesof subgroups

G ⊃ G0 ⊃ · · · ⊃ Gm = 1

satisfying the requirements of Definition 1.6.1. Hence G is solvable.But that is not the group that we were interested in! Consider the extensions F ⊂ E ⊂ L. Here

E/F is Galois, so that N = Gal(L/E) is a normal subgroup and Gal(E/F ) ∼= G/N (again by Propo-sition 1.4.5). Now with Corollary 1.6.6 we conclude that Gal(E/F ) is solvable. 2

Lemma 1.6.20 Let K/F be a Galois extension of prime degree p. Suppose that F contains a primitivep-th root of unity. Then there is an α ∈ K such that K = F (α) and αp ∈ F .

Proof. Write H = Gal(K/F ) then |H| = p by Theorem 1.3.7. Let σ ∈ H be different from theidentity. Then σ has order p by Proposition 3.7.4(Algebra). Hence σ generates H, that is

H = 1, σ, σ2, . . . , σp−1.

Now we view K as a vector space over F and σ : K → K as a linear map. Let λ be an eigenvalueof σ (this λ could lie in an extension of F ). Then σp = 1 implies that λp = 1 and because of theassumption on F we see that F contains all eigenvalues of σ. Now suppose that the only eigenvalueof σ is 1. Then the characteristic polynomial of σ is f = (x− 1)p. The Cayley-Hamilton theorem nowimplies that f(σ) = 0, that is, (σ − IK)p = 0, where IK is the identity map on K. Because σp = 1 wealso have that g(σ) = 0 with g = xp − 1. Furthermore gcd(f, g) = x− 1 so there are u, v ∈ F [x] withuf +vg = x−1 (Theorem 2.3.11(Algebra)). But that implies that σ = 1, which is not the case. Henceσ has an eigenvalue λ not equal to 1. Let α ∈ E be a corresponding eigenvector, i.e., σ(α) = λα.Then α 6∈ F (as otherwise σ(α) = α). Because [K : F ] = p there are no intermediate fields other thanF and K. Hence K = F (α). Finally, σ(αp) = (λα)p = λpαp = αp and as σ generates H we have thatαp ∈ KH = F . 2

Proposition 1.6.21 Let F be a field and f ∈ F [x]. Let E be a splitting field of f over F . Supposethat Gal(E/F ) is solvable. Then f is solvable by radicals.

1.6 Solvability of polynomial equations by radicals 29

Proof. Write G = Gal(E/F ). By Theorem 1.6.14 G has a normal subgroup N such that G/N isabelian of prime order p. Let E′ ⊃ E be a splitting field of xp − 1. In the same way as in the proofof Proposition 1.6.19 we see that E′ = E(ω) where ω is a primitive p-th root of unity and that E′/Fis Galois. Since also E/F is Galois we have that σ(E) = E for all σ ∈ Gal(E′/F ) (Proposition 1.4.5).So we can define the homomorphism

ψ : Gal(E′/F (ω))→ G with ψ(σ) = σ|E .

(Note that Gal(E′/F (ω)) is a subgroup of Gal(E′/F ).)

We have that ψ is injective. Indeed, let σ ∈ Gal(E′/F (ω)) be such that ψ(σ) = 1. Then σ(α) = αfor α ∈ E. Furthermore, σ(ω) = ω is automatic from σ ∈ Gal(E′/F (ω)). Hence σ is the identity onE′. So the kernel of ψ is trivial and hence ψ is injective. It follows that Gal(E′/F (ω)) is isomorphicto a subgroup of G.

We prove the theorem by induction on |G|. The induction hypothesis is that for all Galois exten-sions L/K with Gal(L/K) solvable and of order less than |G| we have that L is contained in a radicalextension of K. We consider two cases.

In the first case we have |Gal(E′/F (ω))| < |G|. By the induction hypothesis E′ is contained in aradical extension M of F (ω). But since F ⊂ F (ω) is radical, M is also a radical extension of F .

In the second case we have |Gal(E′/F (ω))| = |G|. Then Gal(E′/F (ω)) is isomorphic to G and itfollows that Gal(E′/F (ω)) also has a normal subgroup H ′ such that the quotient is abelian of orderp. Let M be the corresponding intermediate field, that is,

M = E′H′

= α ∈ E′ | σ(α) = α for all σ ∈ H ′.

By the Galois correspondence (more precisely Lemma 1.4.2(i)) we have that H ′ = Gal(E′/M). Nowwe apply Proposition 1.4.5 and find that M/F (ω) is a Galois extension with Galois group isomorphicto Gal(E′/F (ω))/H ′ which has p elements. Hence [M : F (ω)] = p and from Lemma 1.6.20 we seethat there is an α ∈M with M = F (ω, α) and αp ∈ F (ω).

As |Gal(E′/M)| = |H ′| < |G| we infer by induction that there is a radical extension R of M withE′ ⊂ R. But we have just seen that M is a radical extension of F , so that R is also a radical extensionof F . Because it contains E the theorem is proved. 2

We unite the two previous propositions in the following theorem due to Galois.

Theorem 1.6.22 (Galois) Let F be a field and let f ∈ F [x] have splitting field E over F . Then fis solvable by radicals if and only if Gal(E/F ) is solvable.

Example 1.6.23 We can use the strategy of the proof of Proposition 1.6.21 to find radical expressionsfor the roots of a given f ∈ F [x] such that the Galois group G of its splitting field E is solvable. Thisworks especially well when we have a series of subgroups G = G1 ⊃ G2 ⊃ · · · ⊃ Gs = 1 with Gi+1

normal in Gi and Gi/Gi+1 abelian of order 2. Ideed, let Fi = EGi ; then E = Fs ⊃ Fs−1 ⊃ · · · ⊃ F1 = Fwhere [Fi : Fi−1] = 2. Write Gal(Fi/Fi−1) = τ0, τ1 where τ0 is the identity and τ21 = τ0. Note thatFi−1 contains a primitive second root of unity, namely -1. So by Lemma 1.6.20, Fi = Fi−1(αi−1) withα2i−1 ∈ Fi−1. Finally, by the proof of the mentioned lemma, finding such an αi−1 amounts to solving

the equation τ1(αi−1) = −αi−1. So we can construct a radical tower and in the end find an expressionfor a root of f using only square roots.

We illustrate this in the case of f = x15 − 1 ∈ Q[x]. The Galois correspondence for its splittingfield is described in Example 1.5.11. Consider the subgroups

G ⊃ H5 = σ1, σ4, σ7, σ13 ⊃ H1 = σ1, σ4 ⊃ σ1

corresponding to the chain of subfields Q ⊂ Q(ζ5) ⊂ Q(ζ + ζ4) ⊂ Q(ζ). In order to execute the aboveprocedure we need the groups

Gal(Q(ζ5)/Q), Gal(Q(ζ + ζ4)/Q(ζ5)), Gal(Q(ζ)/Q(ζ + ζ4)).

30 Galois Theory

By the Galois correspondence (Lemma 1.4.2(i)) we have that Gal(Q(ζ)/Q(ζ+ζ4)) = H1 and Gal(Q(ζ)/Q(ζ5)) =H5.

Since G is abelian, all subgroups are normal and in particular H1 is a normal subgroup of H5. Soby Proposition 1.4.5 we see that Q(ζ + ζ4)/Q(ζ5) is Galois and Gal(Q(ζ + ζ4)/Q(ζ5)) is isomorphicto H1/H5. The latter group consists of the cosets σ1H5, σ7H5. The proof of Proposition 1.4.5 nowshows that Gal(Q(ζ + ζ4)/Q(ζ5)) = σ1, σ7, where σi denotes the restriction of σi to Q(ζ + ζ4).

In the same way we see that Gal(Q(ζ5)/Q) is isomorphic to G/H5. This group consists of thecosets σ1H5 and σ2H5. Hence Gal(Q(ζ5)/Q) = σ1, σ2, where σi denotes the restriction of σi toQ(ζ5).

Now we write the chain of subfields as a radical tower. We start with Q(ζ5) ⊃ Q. Findingα ∈ Q(ζ5) with Q(ζ5) = Q(α) and α2 ∈ Q amounts to solving σ2(α) = −α. Write α = a + bζ5 witha, b ∈ Q. Then σ2(α) = a + bζ10 = a − b − bζ5. So σ2(α) = −α is the same as 2a = b. We choosea = 1, b = 2 and hence α = 1 + 2ζ5. (Then α2 = −3.)

Next we consider Q(ζ + ζ4) ⊃ Q(ζ5). Write θ = ζ + ζ4 then σ7(θ) = ζ7 + ζ28 = 1− ζ − ζ4 + ζ5 =1 + ζ5 − θ. To find a β ∈ Q(θ) with Q(θ) = Q(β) and β2 ∈ Q(ζ5) amounts to finding a β ∈ Q(θ) withσ7(β) = −β. Write β = a + bθ with a, b ∈ Q(ζ5). Then σ7(β) = a + (1 + ζ5)b − bθ. So σ7(β) = −βamounts to 2a + (ζ5 + 1)b = 0. We choose a = ζ5 + 1 and b = −2 so that β = 1 + ζ5 − 2(ζ + ζ4).(Then β2 = 5ζ5.)

Finally we look at Q(ζ) ⊃ Q(ζ+ζ4). We first compute the minimal polynomial of ζ over the secondfield. According to Lemma 1.3.3 the minimal polynomial of ζ is (x− ζ)(x− σ4(ζ)), which is equal tox2 − (ζ + ζ4)z + ζ5. In particular we have the relation ζ2 = (ζ + ζ4)ζ − ζ5. (Which, of course, is alsoeasy to see directly.) Write γ = a+bζ with a, b ∈ Q(ζ+ζ5). Then σ4(γ) = a+bζ4 = a+b(ζ+ζ4)−ζ).So σ4(γ) = −γ amounts to 2a+ (ζ+ ζ4)b = 0. We choose a = ζ+ ζ4, b = −2 so that γ = (ζ+ ζ4)−2ζ.(Then γ2 = (ζ + ζ4)2 − 4ζ(ζ + ζ4) + 4ζ2 = (ζ + ζ4)2 − 4ζ5.)

Now we construct an injective homomorphism Q(ζ) → C in the following way. We start withthe identity homomorphism ψ0 : Q → C. By Lemma 4.6.6(Algebra) there exists a unique injectivehomomorphism ψ1 : Q(α)→ C with ψ1(a) = a for a ∈ Q and ψ1(α) =

√−3. Note that ζ5 = α−1

2 , so

that ψ1(ζ5) =

√−3−12 . So by the same lemma there exists an injective homomorphism ψ2 : Q(α, β)→ C

such that ψ2(a) = ψ1(a) for all a ∈ Q(α) and ψ2(β) =√

52(√−3− 1). We have ζ+ζ4 = 1

2(ζ5+1−β) =α+1−2β

4 and therefore

ψ2(ζ + ζ4) =1 +√−3− 2

√52(√−3− 1)

4.

Again by Lemma 4.6.6(Algebra) we get an injective homomorphism ψ3 : Q(α, β, γ)→ C with γ2(a) = afor all a ∈ Q(α, β) and

ψ2(γ) =

√√√√√1 +√−3− 2

√52(√−3− 1)

4

2

− 2(√−3− 1).

We have ζ = 12(ζ + ζ4 − γ). So we also get a radical expression for ψ3(ζ). After some manipulation it

is seen to be

1

8

(1 +√−3−

√10(√−3− 1)−

√20− 20

√−3− 2

√−3

√10(√−3− 1)− 2

√10(√−3− 1)

).

(Here we have to be a bit careful as something like√u can indicate two complex numbers. With ψ1

we fix a choice for√−3; with ψ2 we fix

√10(√−3− 1). Then

√−3√

10(√−3− 1) is the product of

these two. If we would write this as√−30(

√−3− 1) then we would introduce another choice, not

necessarily compatible with the first two.)

1.6 Solvability of polynomial equations by radicals 31

1.6.3 A polynomial not solvable by radicals

Here we give a polynomial of degree five whose roots are not expressible by radicals. By Theorem1.6.22 along with Theorem 1.6.11 a polynomial whose splitting field has Galois group isomorphic toS5 has this property. The main problem now is to show that the splitting field of a given polynomialhas this Galois group. Indeed, because |S5| = 120 we cannot simply list its elements and compute amultiplication table. We need to resort to some tricks to show that the Galois group is isomorphic toS5. For this we first show how to identify a Galois group with a permutation group, a constructionthat is of wider interest.

Let F be a field, f ∈ F [x] a separable polynomial and E ⊃ F its splitting field. Let α1, . . . , αn ∈ Ebe the roots of f without repetitions (that is, we just take the roots of the irreducible factors of f).Then by definition E = F (α1, . . . , αn) (Definition 4.6.1(Algebra)). Let σ ∈ Gal(E/F ). For 1 ≤ i ≤ nwe have that σ(αi) is a root of f , hence equal to an αj . Moreover, for i1 6= i2 we cannot haveσ(αi1) = σ(αi2) because σ is injective. It follows that there is a πσ ∈ Sn with

σ(αi) = απσ(i) for 1 ≤ i ≤ n.

Define a map ψ : Gal(E/F )→ Sn by ψ(σ) = πσ.

Lemma 1.6.24 ψ is an injective group homomorphism.

Proof. Let σ, τ ∈ Gal(E/F ). Then στ(αi) = σ(τ(αi)) translates to

απστ (i) = απσπτ (i)

so that πστ (i) = πσπτ (i) for all i. In other words, πστ = πσπτ , which just means that ψ is a grouphomomorphism.

If ψ(σ) = ψ(τ) then σ(αi) = τ(αi) for all i, and because E = F (α1, . . . , αn) this implies thatσ = τ . We see that ψ is injective. 2

Let H = ψ(Gal(E/F )). Then H is a subgroup of Sn and ψ : Gal(E/F ) → H is an isomorphism.So Gal(E/F ) is isomorphic to a subgroup of Sn.

Now let f = x5 − 6x + 3 ∈ Q[x] with splitting field E ⊂ C. By Eisenstein’s criterion (Theorem2.5.14(Algebra)) f is irreducible. So it has five distinct roots in C and by the above constructionwe have that Gal(E/Q) is isomorphic to a subgroup H of S5. With a few steps we prove that thissubgroup is equal to S5.

H contains a 5-cycle. Let α ∈ E be a root of f then [Q(α) : Q] = 5, so since [E : Q] =[E : Q(α)][Q(α) : Q] (Theorem 4.3.3(Algebra)) we have that [E : Q] is divisible by 5. Because|Gal(E/Q)| = [E : Q] we have the same for the order of the Galois group. Hence by Theorem 1.6.12H contains an element of order 5. Now the order of a k-cycle in Sn is k. By listing the various possibledecompositions of an element of S5 as a product of disjoint cycles and using Proposition 3.7.7(Algebra)it is seen that an element of order 5 must be a 5-cycle, denote it π5.

H contains a 2-cycle. By computing the derivative of f we see that the graph of f (seen as afunction R → R) has one positive maximum and one negative minimum. Therefore f has exactlythree real roots, denoted α3, α4, α5, and hence exactly two non-real roots, denoted α1, α2. The latterare complex conjugates. Let σ : C→ C be complex conjugation, i.e., σ(x+ iy) = x− iy for x, y ∈ R.Then σ permutes the roots of f and therefore maps E to itself. Furthermore, σ is an automorphism ofC and hance its restriction to E is as well. We have σ(α1) = α2, σ(α2) = α1, σ(αi) = αi for i = 3, 4, 5.Hence πσ = (1, 2).

Write π5 = (1, i2, i3, i4, i5). Then π25 = (1, i3, i5, i2, i4), π35 = (1, i4, . . .), π

45 = (1, i5, . . .). We see

that H contains the 5-cycle (1, 2, j3, j4, j5). By relabelling the αi for i = 3, 4, 5 we now may assumethat jk = k and that H contains (1, 2, 3, 4, 5).

Finally, (1, 2) and (1, 2, 3, 4, 5) generate S5 (we leave this verification as an exercise). The conclusionis that H = S5.

32 Galois Theory

1.6.4 The discriminant of a polynomial

Let F be a field and f ∈ F [x]. Let E be a splitting field of f and write f = γ(x − α1) · · · (x − αn)where γ ∈ F and αi ∈ E. Then

D(f) = γ2n−2∏

1≤i<j≤n(αi − αj)2

is the discriminant of f . (The γ2n−2 is a normalizing factor to make some formulas work out better.)As an immediate property we mention that D(f) = 0 if and only if f has a root of multiplicity at

least 2.

Example 1.6.25 Let f = ax2 + bx+ c = a(x− α1)(x− α2). Then α1 + α2 = − ba , α1α2 = c

a . Hence

(α1 − α2)2 = (α1 + α2)

2 − 4α1α2 =b2

a2− 4

c

a.

It follows that D(f) = a2(α1 − α2)2 = b2 − 4ac.

Example 1.6.26 The discriminant of a polynomial is a symmetric polynomial in the roots of f (thismeans that it is unchanged when we permute the roots in any way). It can be shown that this impliesthat the discriminant of a monic polynomial can be expressed as a polynomial in the coefficients of f .For example, if f = x3 + ax2 + bx+ c then

D(f) = a2b2 − 4a3c+ 18abc− 4b3 − 27c2.

So it is possible to decide whether f has a root of multiplicity at least 2 without computing any ofthe roots explicitly!

Proposition 1.6.27 Let f be a field and f ∈ F [x]. Suppose that f is separable and square-free. Letα1, . . . , αn be the (necessarily distict) roots of f in its splitting field E. Consider the injective grouphomomorphism ψ : Gal(E/F )→ Sn constructed in Section 1.6.3. Then

(i) D(f) ∈ F ,

(ii) ψ(Gal(E/F )) is contained in the alternating group An if and only if D(f) is a square in F .

Proof. Let P = (i, j) | 1 ≤ i < j ≤ n. For π ∈ Sn define the map π : P → P by

π(i, j) =

(π(i), π(j)) if π(i) < π(j)

(π(j), π(i)) if π(i) > π(j).

The inverse of π is τ , where τ = π−1. In particular it follows that π is bijective. For (i, j) ∈ P weset δ(i,j) = αi − αj . Then δπ(i,j) = απ(i) − απ(j) if (i, j) is not an inversion of π and it is equal to−(απ(i) − απ(j)) if (i, j) is an inversion of π.

Now let σ ∈ Gal(E/F ) and write ψ(σ) = πσ. Let ∆ =∏

(i,j)∈P δ(i,j). Then

σ(∆) =∏

(i,j)∈P

(απσ(i) − απσ(j)) = (−1)inv(πσ)∏

(i,j)∈P

δπσ(i,j) = (−1)inv(πσ)∆.

(Where inv(π) is the number of inversions of π, see Section 1.6.1.)Hence

σ(∆2) = σ(∆)2 = ((−1)inv(πσ)∆)2 = ∆2.

We conclude that ∆2 is fixed under all σ ∈ Gal(E/F ) so that ∆2 ∈ F . So the same holds forD(f) = γ2n−2∆2 where γ is the leading coefficient of f .

Secondly it follows that σ(∆) = ∆ if and only if inv(πσ) is even. By definition that is the sameas πσ ∈ An. It follows that ψ(Gal(E/F )) ⊂ An if and only if ∆ ∈ F . But the latter is obviouslyequivalent to D(f) being a square in F . 2

Remark 1.6.28 Note that when f has roots of multiplicity at least 2 then D(f) = 0 and in particular,D(f) ∈ F .

1.6 Solvability of polynomial equations by radicals 33

1.6.5 Casus irreducibilis

Let f ∈ Q[x] be of degree 3 with real roots. Then when using the known formulas for finding aroot of f one invariably finds square roots of negative numbers along the way. This phenomenon wasextensively studied by the bolognese mathematician Rafael Bombelli, who in the 1572 edition of hisbook L’Algebra looks at the following example. Let f = x3 − 15x − 4. The roots of f are 4 and−2±

√3. However, the formulas give as one of the solutions

3√

2 + 11i+ 3√

2− 11i.

Bombelli noticed that (2 + i)3 = 2 + 11i, (2 − i)3 = 2 − 11i so that the sum above reduces to(2 + i) + (2− i) = 4. As Bombelli put it: “Somma 4; e tanto vale la Cosa.” (The sum is 4 and that isthe value of the thing.)

Here we prove the “irreducible case”: if f is irreducible then there always must appear squareroots of negative numbers in a radical expression for a root of f .

Lemma 1.6.29 Let F be a field of characteristic 0 and f ∈ F [x] be monic and irreducible of degree3. Let E ⊃ F be a splitting field and write D = D(f). Let α ∈ E be one of the roots of f . ThenE = F (α,

√D).

Proof. Let β, γ ∈ E be the other two roots of f . Of course, α ∈ E and√D = ±(α− β)(α− γ)(β − γ)

also lies in E. Hence F (α,√D) ⊂ E.

Let K = F (α,√D). We now show that β, γ ∈ K. First of all, in K[x] we have a factorization

f = (x − α)g, where g ∈ K[x] is of degree 2. But g = (x − β)(x − γ). Hence β + γ ∈ K. Next weobserve that g(α) = (α− β)(α− γ) also lies in K. So K contains

√D

g(α)=±(α− β)(α− γ)(β − γ)

(α− β)(α− γ)= ±(β − γ).

Since β + γ and β − γ both lie in K, so do β, γ.So K is a subfield of E containing F and the roots of f . It follows that K = E. 2

Lemma 1.6.30 Let F be a field and p a prime. Let a ∈ F . If the polynomial xp − a is reducible inF [x] then it has a root in F .

Proof. We may assume a 6= 0.Let E ⊃ F be a splitting field of f = xp − a. Let α ∈ E be a root of f . Let β ∈ E be a second

root of f . Then(βα

)p= 1 so that β = αω where ω ∈ E satisfies ωp = 1. It follows that αω1, . . . , αωp

are the roots of f where ωpi = 1.Suppose that f = gh with g, h ∈ F [x] and deg(g),deg(h) ≥ 1. Write k = deg(g); hence

g = (x− ωi1α) · · · (x− ωikα).

Let ζ = ωi1 · · ·ωik then ζp = 1 and the constant coefficient of g is ±ζαk. So ζαk ∈ F .Moreover, gcd(p, k) = 1 and hence there are integers u, v with up + vk = 1 (Theorem 2.2.3(Alge-

bra)). But (ζαk)p = (αp)k = ak hence

a = a1 = (au)p(ak)v = (au)p(ζαk)p = (auζαk)p.

We conclude that a = βp for a β ∈ F and f has a root in F . 2

Theorem 1.6.31 (Casus irreducibilis) Let f ∈ Q[x] be of degree 3 and irreducible. Suppose thatf has three real roots. Let E ⊂ R be a splitting field of f . Let K ⊃ E be a radical extension of Qcontained in C. Then K is not contained in R.

34 Galois Theory

Note that the radical extension K exists by Theorem 1.6.22. Indeed, by the construction in Section1.6.3 we see that the Galois group Gal(E/Q) is isomorphic to a subgroup of S3 which is solvable.Proof. Suppose that K ⊂ R. We will derive a contradiction from that.

By definition we have a tower Q = F1 ⊂ F2 ⊂ · · · ⊂ Fm+1 = K with Fi+1 = Fi(αi) and αmii ∈ Fifor 1 ≤ i ≤ m. We observe that we may assume that each mi is prime. Indeed, suppose that mi = pswhere p is a prime and s > 1. Then we replace Fi ⊂ Fi+1 in the tower by Fi ⊂ Fi(β) ⊂ Fi+1 whereβ = αsi : we have βp ∈ Fi and Fi+1 = Fi(β, αi) with αsi ∈ Fi(β). Continuing this process we arrive ata longer tower where all exponents are prime.

Let D be the discriminant of f . Since√D ∈ K (by Lemma 1.6.29) we can also consider the

intermediate fields Li = Q(√D,α1, . . . , αi) for 0 ≤ i ≤ m (so L0 = Q(

√D)). This yields the tower

Q ⊂ L0 ⊂ L1 ⊂ · · · ⊂ Lm = K.

We note that f is irreducible in L0[x]. Indeed, otherwise f has a root β in L0 because f is of degree3. But [Q(β) : Q] = 3 because f is irreducible. Since D ∈ Q we have that [Q(

√D) : Q] ≤ 2, and that

is a contradiction (a field extension of degree 2 cannot contain an extension of degree 3).Now f is reducible in Lm[x]. Hence there is an i ≥ 0 such that f is irreducible in Li[x] but reducible

in Li+1[x]. Writing α = αi+1, p = mi+1 we have Li+1 = Li(α) with αp ∈ Li where p is prime.Set h = xp − αp ∈ Li[x]. The roots of h in C are α, ζα, ζ2α, . . . , ζp−1α, where ζ ∈ C is a primitive

p-th root of unity. If p = 2 then these are α,−α which are not contained in Li. If p > 2 then exceptα none of the roots of h are real, so that again we conclude that Li contains no root of h. By Lemma1.6.30 we see that h is irreducible in Li[x]. Hence h is the minimal polynomial of α and [Li+1 : Li] = p.In particular, there are no intermediate fields M with Li (M ( Li+1 (by the degree formula).

Because f is reducible in Li+1[x] we have that Li+1 contains a root γ of f . But since E = Q(√D, γ)

(Lemma 1.6.29) we see that E is contained in Li+1. In particular, Li+1 contains all three roots of f ,and therefore Li+1 contains a splitting field of f over Li. From the remark on the intermediate fieldsit now follows that Li+1 is a splitting field of f over Li.

Also Li(γ) is an intermediate field and as γ 6∈ Li it follows that Li+1 = Li(γ). As f is irreduciblein Li[x] it is the minimal polynomial of γ over Li. Hence [Li+1 : Li] = 3 and we have that p = 3.

Now we turn back to h. Its roots in C are α, ζα, ζ2α. By hypothesis α is real, also implying thatthe other two are not real. But Li+1 is the splitting field of f over Li, and in particular the extensionLi+1/Li is normal. Because h is irreducible and has a root in Li+1 it must have all its roots in Li+1.But that contradicts our assumption that K is contained in R.

So we have obtained a contradiction, and it follows that K cannot be contained in R. 2

Chapter 2

Applications of Galois theory

2.1 The fundamental theorem of algebra

The so-called fundamental theorem of algebra states that every polynomial in C[x] has a root in C(or, in other words, that C is algebraically closed). This theorem has a long history and many proofsexist. Here we describe a proof using Galois theory.

First we show that it is enough to prove that every polynomial in R[x] has a root in C. For z =x+iy ∈ C (with x, y ∈ R) let z = x−iy denote its complex conjugate. Then z 7→ z is an automorphismof C. We extend this to an autmorphism of C[x] by mapping a polynomial a0 + a1x+ · · ·+ amx

m to

a0 + a1x+ · · ·+ amxm. Let f ∈ C[x]; then ff = ff = ff so that ff ∈ R[x]. But if α ∈ C is a root of

ff then it has to be a root of f or of f . In the second case α is a root of f . So if we know that everypolynomial in R[x] has a root in C then the same follows for every polynomial in C[x].

Secondly we remark that every polynomial of degree 2 in C[x] has a root in C. Because ofthe quadratic formula this is equivalent to each element of C having a square root in C. So letz = x + iy ∈ C, where x, y ∈ R. Furthermore (u + iv)2 = (u2 − v2) + 2uvi; hence z = (u + iv)2 withu, v ∈ R if and only if u2 − v2 = x, 2uv = y over R. Because

√x2 + y2 ≥ |x| there are u, v ∈ R with

u2 =1

2x+

1

2

√x2 + y2, v2 = −1

2x+

1

2

√x2 + y2.

If we choose u to be positive and the sign of v equal to the sign of y then these u, v are solutions tothe original equations, and hence z = (u+ iv)2.

Now let f ∈ R[x] and suppose that it has no roots in C. Let E/C be its splitting field over C.Hence [E : C] > 1. Then E/R is the splitting field of the polynomial f(x2 + 1). Let G = Gal(E/R)and write |G| = 2sm with m odd. Now by Sylow’s theorem in group theory (see for example [DF04],§4.5, Theorem 18) there is a subgroup H of G of order 2s. Set K = EH. Then [K : R] = m which isodd. Let α ∈ K, then the degree of the minimal polynomial of α over R is [R(α) : R] which must beodd as well by the degree formula. It is an elementary fact of analysis that every polynomial of odddegree in R[x] has a root in R. Hence the minimal polynomial of α over R can only be irreducible ifits degree is 1. Therefore we must have m = 1 and |G| = 2s.

Write G = Gal(E/C). Then G is a subgroup of G so that |G| = 2t for some t > 0. It can be shownthat groups of prime power order are solvable (let P be such a group; one first shows that the centre Zof P is non-trivial ([DF04], §6.1, Theorem 1); then P/Z is solvable by induction, and as Z is solvablethe same follows for P ). Hence by Theorem 1.6.14 G has a normal subgroup H with [G : H] = 2. SetL = EH . Then [L : C] = 2. Let α ∈ L \ C. Then its minimal polynomial over C has degree 2. Aswe have seen above, each such a polynomial has a root in C. Hence we have a contradiction and itfollows that f has a root in C.

2.2 Proving that certain primitive functions are not elementary

We know, for example, that∫x3 dx = 1

4x4,∫

sin(x) dx = − cos(x),∫xex

2dx = 1

2ex2 . However, we

have never seen a nice expression for∫ex

2dx. Of course, we can define the function p : R≥0 → R by

36 Applications of Galois theory

p(x) =∫ x0 e

t2 dt with which we have∫ex

2dx = p(x) (or, in other words, p′(x) = ex

2). But by doing it

this way we do not feel that we have really solved the problem because we define the solution in termsof an integral. Indeed, by writing

∫f(x) dx we ask for a primitive function of f(x), or, in other words,

a function whose derivative is f(x). By using the function p we say nothing other than “a primitveof ex

2exists and it is called p”. In fact, we could use this “method” also in other cases, for example

defining q(x) =∫ x0 t

3 dt we also have that q′(x) = x3. Here it is obvious that the expression q(x) = 14x

4

gives much more information on the function q.Here we will show that for

∫ex

2dx there does not exist a “nice” expression. In order to prove this,

and to arrive at a satisfactory definition of what a “nice” expression is, we will consider functions asformal objects, that is as elements of a field to which we can apply arithmetical operations. The onlyextra property, not generally satisfied by field elements, is that we have a notion of derivative, whichalso is a formal operation.

For convenience all fields that we consider in this section are of characteristic 0.

Definition 2.2.1 A differential field is a field F together with an operation ′ : F → F (called thederivation of F ) such that (a+ b)′ = a′ + b′ and (ab)′ = a′b+ ab′ for all a, b ∈ F .

Example 2.2.2 Let z be an indeterminate over C and consider the field

C(z) =

p

q| p, q ∈ C[z], q 6= 0

.

For ′ we can take the “normal” ddz , i.e., (

p

q

)′=p′q − pq′

q2,

where p′, q′ are defined as in Section 4.7.1(Algebra).We remark that it is possible to consider also other derivations, such as z d

dz .

Lemma 2.2.3 Let F be a differential field with derivation ′. Then we have the following (for a, b, a1, . . . , an ∈F ∗)

(i) 1′ = 0,

(ii) ( 1a)′ = − a′

a2,

(iii) (ab )′ = a′b−ab′b2

,

(iv) (an)′ = na′an−1 for n ∈ Z,

(v)(aν11 ···a

νnn )′

aν11 ···a

νnn

= ν1a′1a1

+ · · ·+ νna′nan

for all ν1, . . . , νn ∈ Z.

Proof. Using Definition 2.2.1 we see that 1′ = (1 · 1)′ = 1′ · 1 + 1 · 1′ = 1′ + 1′ implying that 1′ = 0.Therefore 0 = 1′ = (a · 1a)′ = a′ · 1a + a · ( 1a)′ which implies (ii).

Continuing, (ab )′ = (a · 1b )′ = a′ · 1b + a · (− b′

b2) from which we get (iii).

The proof of (iv) for n > 0 is a straightforward induction. For n < 0 we write an = ( 1a)−n. Since

−n > 0 we see that (an)′ = −n( 1a)−n−1(− a′

a2) = na′an−1.

The last statement is perhaps best proved first for n = 2. Then we have (aν11 aν22 )′ = ν1a

′1aν1−11 aν22 +

ν2aν11 a′2aν2−12 immediately implying the formula. The general case works in the same way. Alterna-

tively it can be shown by induction on n. 2

Keep the notation of the previous lemma and set

C(F ) = a ∈ F | a′ = 0.

The lemma immediately implies that this is a subfield of F . It is called the field of constants of F .

2.2 Proving that certain primitive functions are not elementary 37

Let E,F be differential fields with F ⊂ E. Then we say that E is a differential field extension of Fif E/F is a field extension and the restriction of the derivation of E to F is equal to the derivation ofF (so, denoting the derivations of F and E by ′ and ∂ respectively, we have ∂(a) = a′ for all a ∈ F ).

Let F be a differential field and E a differential field extension of F . Let t ∈ E. Then we saythat t is the logarithm of a ∈ F if t′ = a′

a and we say that t is the exponential of a ∈ F if t′ = a′t.Furthermore, E is said to be an elementary extension of F if there is a tower of differential fieldextensions

F = F1 ⊂ F2 ⊂ · · · ⊂ Fm+1 = E (2.2.1)

with Fi+1 = Fi(ti) where ti ∈ Fi+1 is algebraic over Fi or the logarithm of an element of Fi or theexponential of an element of Fi (note the similarity with Definition 1.6.15).

Let F = C(z) with derivation ′ = ddz (see Example 2.2.2). Let Ω ⊂ C be open and connected. A

function f : Ω→ C is said to be meromorphic if its singularities in Ω are poles, and every singularity isisolated. If f and g are two meromorphic functions on Ω then the sum f + g, product fg and quotientf/g are meromorphic again, the latter of course only if g 6= 0. Hence the set of meromorphic functionson Ω forms a field denoted M(Ω) (see [J93], §4.2). If f is holomorphic on Ω then so is its derivativef ′. Hence M(Ω) is a differential field. Moreover, the elements of F = C(z) are meromorphic on C sothat M(Ω) is a differential extension of F .

An elementary function is a meromorphic function δ : Ω→ C where Ω is a connected open subsetof C, such that F (δ) is an elementary extension of F (note that F (δ) is a well-defined subfield ofM(Ω)).

Example 2.2.4 Here are some examples of elementary functions

e

√1

z2+1 , log(z3 − z − 1), sin(z).

Here the first function is meromorphic on C \ ±i. For the last one note that sin(z) = eiz−e−iz2i .

Now we have some technical facts on differential extensions.

Lemma 2.2.5 Let F be a differential field of characteristic 0 with derivation ′. Let E be an extensionof F of finite degree. Then there exists a unique derivation ∂ of E with ∂(a) = a′ for all a ∈ F .

Proof. Let x be an indeterminate and consider the polynomial ring F [x]. Define two maps D0, D1 :F [x]→ F [x] by

D0(

m∑i=0

aixi) =

m∑i=0

a′ixi

D1(

m∑i=0

aixi) =

m∑i=1

iaixi−1.

First suppose that a derivation ∂ as in the lemma exists. Then for f ∈ F [x] and α ∈ E we have∂(f(α)) = D0(f)(α) +D1(f)(α)∂(α). Now let f be the minimal polynomial of α, then D1(f)(α) 6= 0(as D1(f) has smaller degree) and hence

∂(α) = −D0(f)(α)

D1(f)(α).

We see that ∂ is uniquely determined.

Now we prove the existence of ∂. By the theorem of the primitive element (Theorem 1.2.7) we haveE = F (α) for a certain α ∈ E. Let g ∈ F [x] and define Dg : F [x]→ F [x] by Dg(f) = D0(f)+gD1(f).Then for f1, f2 ∈ F [x] we have

Dg(f1 + f2) = Dg(f1) +Dg(f2), Dg(f1f2) = Dg(f1)f2 + f1Dg(f2),

38 Applications of Galois theory

because these relations hold with D0 and D1 in place of Dg. Now let f ∈ F [x] be the minimal

polynomial of α and set β = −D0(f)(α)D1(f)(α)

. Because E = F (α) there is a g0 ∈ F [x] with g0(α) = β. But

then Dg0(f)(α) = 0.Let γ ∈ E then γ = h(α) for a certain h ∈ F [x]. Then we set ∂(γ) = Dg0(h)(α). We need to check

that this is independent of the choice of h. So let h ∈ F [x] be a second polynomial with h(α) = γ.Then (h − h)(α) = 0 so that f divides h − h (Proposition 4.3.4(Algebra)), or h = h + qf for someq ∈ F [x]. But then Dg0(h) = Dg0(h) +Dg0(q)f + qDg0(f) and it follows that Dg0(h)(α) = Dg0(h)(α).So ∂ is well-defined. The relations satisfied by Dg0 show that ∂ is a derivation of E. It obviouslysatisfies the required property. 2

Lemma 2.2.6 Let F be a differential field and E a differential extension of F of finite degree. Denotethe derivation of both fields by ′. Let σ ∈ Gal(E/F ). Then σ(α′) = σ(α)′ for all α ∈ E.

Proof. Define ∂ : E → E by ∂(α) = σ−1(σ(α)′). It is straightforward to check that ∂ is a derivationof E. For a ∈ F we have ∂(a) = a′, so that by the uniqueness part of Lemma 2.2.5 it follows that∂ =′. Hence α′ = σ−1(σ(α)′) for all α ∈ E. By applying σ to both sides we get the result. 2

Proposition 2.2.7 Let E be a differential extension of the differential field F , and let the derivationof both fields be denoted ′. We suppose that the field of constants of E is equal to the field of constantsof F . Let α ∈ E be transcendental over F .

(i) Suppose that α′ ∈ F . Let f = anxn + · · ·+ a0 ∈ F [x] be a polynomial of degree n > 0. There is

a g ∈ F [x] with f(α)′ = g(α). Moreover deg(g) = n if an is not a constant and deg(g) = n − 1if an is a constant.

(ii) Suppose that α′

α ∈ F . Let f ∈ F [x] be of degree > 0. Then there is a g ∈ F [x] with deg(g) =deg(f) and f(α)′ = g(α). Moreover, g = µf for a certain µ ∈ F if and only if f = anx

n, an ∈ F .

Proof. We start with (i). Using Lemma 2.2.3 we see that f(α)′ = a′nαn + (nanα

′ + a′n−1)αn−1 + · · · .

Hence g exists. If a′n 6= 0 (that is, an is not a constant) then deg(g) = n. Suppose that a′n = 0. Ifnanα

′ + a′n−1 = 0 as well then (nanα + an−1)′ = 0 and by the hypothesis on the fields of constants

it follows that nanα + an−1 ∈ F . But that implies α ∈ F contrary to the hypothesis that α istranscendental over F . Hence nanα

′ + a′n−1 6= 0 and we have deg(g) = n− 1.

For (ii) write α′

α = γ ∈ F . Then for a ∈ F , a 6= 0 we have

(aαn)′ = (a′ + naγ)αn.

If a′+naγ = 0 then (aαn)′ = 0 and aαn ∈ F contrary to the hypothesis on α. Hence (aαn)′ = bαn fora certain b ∈ F . Hence g exists. It is also obvious that if f = anx

n then g = µf . Conversely, supposethat g = µf and that f has more than one monomial. Then we can write f = akx

k + alxl + · · ·

where ak, al 6= 0. Also from the above it follows that g = (a′k + kakγ)xk + (a′l + lalγ)xl + · · · . So

a′k + kakγ = µak and a′l + lalγ = µal, so thata′kak

+ kα′

α =a′lal

+ lα′

α . A small computation shows thatthis is entails (

akαk

alαl

)′= 0.

But that implies that the term in brackets lies in F , contrary to the hypothesis on γ. We concludethat f = anx

n. 2

Now we have a short intermezzo on partial fraction decompositions of rational functions. We startwith a lemma treating a special case, followed by the main theorem.

Lemma 2.2.8 Let F be a field and f, p ∈ F [x] with deg(p) > 0 and deg(f) < sdeg(p) for some s > 0.Then there are unique ai ∈ F [x] with deg(ai) < deg(p) and f

ps = a1p + a2

p2+ · · ·+ as

ps .

2.2 Proving that certain primitive functions are not elementary 39

Proof. The existence of the ai follows from the division with remainder theorem (Proposition 2.3.8(Al-gebra)). Indeed, define q0 = f , and for i ≥ 1, supposing that q0, . . . , qi−1 have been defined, letqi, as+1−i be the unique elements of F [x] with qi−1 = qip + as+1−i and deg(as+1−i) < deg(p). Thesequence stops when i = m with qm = 0. Then for j ≥ 1 we have

f = qjpj +

j−1∑k=0

as−kpk.

Let 0 ≤ t < s be such that t deg(p) ≤ deg(f) < (t + 1) deg(p). Then qt+1 = 0 and f = as + as−1p +· · ·+ as−tp

t. Dividing by ps we see that the ai exist.Secondly, if we have ai as in the lemma, then it is clear that as+1−i is the remainder of qi−1 upon

division by p. Therefore it is uniquely determined. 2

Theorem 2.2.9 Let f, g ∈ F [x] with deg(g) > 0. Write g = pn11 · · · pnrr , where the pi are irreducible

and distinct. Then there are unique h, ai,j ∈ F [x] with deg(ai,j) < deg(pi) and

f

g= h+

r∑i=1

ni∑j=1

ai,j

pji.

Proof. Let g1, g2 ∈ F [x] be such that gcd(g1, g2) = 1. Then there are a, b ∈ F [x] with ag1 + bg2 = 1(Theorem 2.3.11(Algebra)). Hence f

g1g2= fb

g1+ fa

g2. Applying this repeatedly we see that there are

fi ∈ F [x] with fg = f1

pn11

+ · · ·+ frpnrr

. Write fi = hipnii + vi, where hi, vi ∈ F [x] and deg(vi) < ni deg(pi).

Setting h = h1 + · · · + hr and applying the previous lemma to the quotients vipnii

we see that the

polynomials of the theorem exist.Now we come to the uniqueness part. Note that g

ai,j

pjiis a polynomial of degree < deg(g). So

multiplying by g we see that h is a quotient of f upon division by g. Hence h is uniquely determined.By taking all terms with denominator equal to a power of pi together and putting them in a singlefraction we get bi ∈ F [x] with deg(bi) < deg(pi) and

f

g− h =

r∑i=1

bipnii

.

We first show that these bi are uniquely determined. Suppose that

r∑i=1

bipnii

=r∑i=1

cipnii

.

Fix j with 1 ≤ j ≤ r and set q = g

pnjj

. Then

r∑i=1

bipnii

=bj

pnjj

+∑i 6=j

bipnii

=bj

pnjj

+ uq

for some u ∈ F [x]. Similarly,∑r

i=1cipnii

=cj

pnjj

+ vq for some v ∈ F [x]. Then (bj − cj)q = (v − u)p

njj .

But since gcd(pnjj , q) = 1 this implies that p

njj divides bj − cj which, because of the degrees, implies

that bj − cj = 0.It follows that the bi are uniquely determined. We now get the uniqueness of the ai,j by applying

the previous lemma to the quotient bipnii

. 2

Theorem 2.2.10 Let F be a differential field with derivation ′. Let a ∈ F and suppose that thereexists an elementary extension E of F , with the same field of constants, and such that there exists a

40 Applications of Galois theory

y ∈ E with y′ = a. Then there are c1, . . . , cn, u1, . . . , un, v in F , where c1, . . . , cn are constants andsuch that

a =n∑i=1

ciu′iui

+ v′.

Proof. By definition there exists a tower of differential extensions F = F1 ⊂ · · · ⊂ Fm+1 = E asin (2.2.1). We use induction on m ≥ 0. If m = 0 then E = F and we can take ci = 0, ui = 1,v = y. Now suppose m ≥ 1 and and that the theorem holds when the tower consists of m fields.Applying the induction hypothesis to the tower F2 ⊂ · · · ⊂ Fm+1 = E we see that there are c1, . . . , cn,

u1, . . . , un, v ∈ F2 such that a =∑n

i=1 ciu′iui

+ v′. Moreover F2 = F (α) where α is algebraic over F orit is the logarithm or the exponential of an element of F . Accordingly we distinguish three cases.

Case I: α is algebraic over F . Then there exist f1, . . . , fn, g ∈ F [x] with ui = fi(α), v = g(α).(Note that ci ∈ F because they are constants.) Let K ⊃ F2 be the splitting field of the minimalpolynomial of α over F . By Lemma 2.2.5 there exists a unique derivation (which we also denote ′)of K extending the derivation of F . Write G = Gal(K/F ) and let β1, . . . , βs be the distinct elementsof σ(α) | σ ∈ G. Fix a βj and let σ ∈ G be such that σ(α) = βj . Using Lemma 2.2.6 we see thatσ(fi(α)′) = σ(fi(α))′ = fi(βj)

′. So because σ(a) = a we get

a =

n∑i=1

cifi(βj)

′

fi(βj)+ g(βj)

′.

Applying 1s

∑sj=1 to both sides and using Lemma 2.2.3(v) we obtain

a =

n∑i=1

cis

(fi(β1) · · · fi(βs))′

fi(β1) · · · fi(βs)+

(g(β1) + · · ·+ g(βs)

s

)′.

As fi(β1) · · · fi(βs) and g(β1) + · · ·+ g(βs) are fixed by all elements of G they lie in F . Hence we havefound an expression of the required form.

Now we have a small intermezzo concerning some generalities on the case where α is transcendentalover F . In that case for each w ∈ F (α) there are f, g ∈ F [x] with gcd(f, g) = 1 and w = f(α)

g(α) . A

small calculation shows that w′

w = f(α)′

f(α) −g(α)′

g(α) . So we may assume that there exist polynomials

fi ∈ F [x] such that ui = fi(α). Furthermore, because of Lemma 2.2.3(v) we may assume that the fiare irreducible and monic. By the partial fractions theorem (Theorem 2.2.9) there are h, pj , qj ∈ F [x]with qj monic and irreducible, deg(pj) < deg(qj) and

v = h(α) +t∑

j=1

pj(α)

qrjj (α)

.

We say that this is the partial fractions expansion of v.Case II: α is transcendental and the logarithm of b ∈ F . The latter means that α′ = b′

b . Inparticular α′ ∈ F . So by Proposition 2.2.7(i) we have that there is a ki ∈ F [x] with deg(ki) = deg(fi)−1

and fi(α)′ = ki(α). Henceu′iui

= ki(α)fi(α)

. Moreover, the partial fractions expansion of kifi

is simply kifi

.

Now ifpj(α)

qrjj (α)

appears in the partial fractions expansion of v, then the partial fractions expansion of

v′ contains a fraction with denominator qrj+1j (α). But rj + 1 ≥ 2 so such a fraction can never cancel

against a a ki(α)fi(α)

. Furthermore∑

iu′iui

+ v′ ∈ F . Hence the partial fractions expansion of v cannot have

such terms, whence v = h(α). But then we cannot have deg(fi) > 0 as otherwise the fraction ki(α)fi(α)

does not cancel. It follows that ui ∈ F for all i. But then also v′ ∈ F so that v = dα + e for certain

d, e ∈ F . Hence α =∑n

i=1 ciu′iui

+ d b′

b + e′ is an expression of the required form.

Case III: α is transcendental and the exponential of b ∈ F . Then α′

α = b′. From Proposition2.2.7(ii) we get that if f ∈ F [x] is monic, irreducible, and f 6= x then f(α)′ is a polynomial in α

of the same degree, and it is not a scalar multiple of f(α). Henceu′iui

= ki(α)fi(α)

where in this case

2.3 Galois descent 41

deg(ki) = deg(fi). So if fi 6= x then the partial fractions expansion of kifi

is of the form mi + lifi

where

mi ∈ F and li ∈ F [x], deg(li) < deg(f). On the other hand, if fi = x thenu′iui

= b′ ∈ F . Again, if in

the partial fractions decomposition of v we have a termpj(α)

qrjj (α)

with qj 6= x, then in the partial fractions

decomposition of v′ we will have a term with denominator qrj+1j (α), which cannot cancel against any

of theu′iui

. It follows that we can only have qj = x and therefore v =∑

j∈Z ajtj where aj ∈ F . But

then we also must have ui ∈ F if fi 6= x. In particularu′iui∈ F for all i so that v′ ∈ F as well. Note

that (ajαj)′ = bjα

j for a certain bj ∈ F which cannot be zero because α is transcendental. Thereforewe must have v ∈ F . If every ui lies in F then we are done. But if one ui, say u1, is equal to α then

we have α = c1α′

α +∑n

i=2 ciu′iui

+v′ =∑n

i=2 ciu′iui

+(c1b+v)′ and again we have the required expression. 2

Theorem 2.2.11 (Liouville) Let f, g ∈ C(z). Then∫feg dz is elementary if and only if there is an

a ∈ C(z) with f = a′ + ag′.

Proof. Firstly, if such an a exists then∫feg dz = aeg.

For the other implication let E = C(z, t) where t = eg. Suppose that∫feg dz is elementary. By

definition that means that there exists an elementary extension of E containing an element y withy′ = ft. By the previous theorem it follows that there are c1, . . . , cn ∈ C, u1, . . . , un, v ∈ E with

tf =n∑i=1

ciu′iui

+ v′.

Set F = C(z). Then E = F (t) and t′ = gt. We now perform exactly the same analysis as in the thirdcase of the proof of Theorem 2.2.10. Again we obtain that v =

∑j∈Z bjt

j and ui ∈ F or ui = t for

precisely one i. Then v′ =∑

j∈Z(b′j + jg′bj)tj and

∑ni=1 ci

u′iui∈ F . Hence

tf = (

n∑i=1

ciu′iui

+ b′0) +∑j 6=0

(b′j + jg′bj)tj .

It follows that the first term has to be zero, whereas each term in the second summation is zero exceptfor j = 1. Hence tf = (b′1 + g′b1)t from which it follows f = b′1 + b1g

′ and we see that we can takea = b1. 2

Corollary 2.2.12∫ez

2dz is not elementary.

Proof. We apply the previous theorem with f = 1, g = z2. According to the theorem the integral iselementary if and only if there is an a ∈ C(z) with 1 = a′ + 2az. Write a = p

q with p, q ∈ C[z] and

gcd(p, q) = 1. Then a′+2az = 1 amounts to p′q− q′p+2zpq = q2. It follows that q|q′p, which becausep, q are coprime, implies that q|q′ and hence q = 1. So a = p is a polynomial. But then we cannothave a′ + 2az = 1 because deg(2az) = deg(a′) + 2. 2

2.3 Galois descent

In this section we let E/F be a Galois extension of finite degree. Let V be a vector space over E.Then V is also a vector space over F . A U ⊂ V is said to be an F -subspace if u+v ∈ U for all u, v ∈ Uand αu ∈ U for all α ∈ F , u ∈ U (that is, if U is a subspace of V when the latter is considered as avector space over F ).

Example 2.3.1 Let E = C, F = R, V = C2, whose elements we write as row vectors. Let U1 bethe R-subspace spanned by (1, i), (1,−i). Let U2 be the R-subspace spanned by (1, 1), (i, i). Both

42 Applications of Galois theory

spaces (as vector spaces over R) have dimension 2. For i = 1, 2 let CUi denote the vector space over Cspanned by the same vectors as Ui. Then we see that dimCU1 = 2 whereas dimCU2 = 1. We expressthis by saying that U1 is an R-form of V , whereas U2 is not.

Definition 2.3.2 The F -subspace U of V is called an F -form of V if there is a basis of U that is alsoa basis of V .

By the next lemma, if there is a basis of U that is not a basis of V then U is not an F -form of V .

Lemma 2.3.3 Let V be a vector space over E and U an F -form of V . Then every basis of U is alsoa basis of V .

Proof. Let u1, u2, . . . be a basis of U that is also a basis of V . Let v1, v2, . . . be a second basis of U .Let v ∈ V then v is a linear combination (with coefficients in E) of some of the ui. Writing these aslinear combinations of the vj we see that v is a linear combination of the vj .

Consider v1, . . . , vm for some m ≥ 1. There is an n such that each vi lies in the space (over F )spanned by u1, . . . , un. Write vi =

∑nj=1 cijuj with cij ∈ F . Set ci = (ci1, . . . , cin). Then the vectors

ci are linearly independent over F . By elementary linear algebra that implies that they are linearlyindependent over E. But that means that v1, . . . , vm is linearly indepedent over E. It follows thatv1, v2, . . . is a basis of V . 2

Constructing F -forms of V is called descent. The main question is which F -forms a given V canhave. Of course, this depends very strongly on the given data (E, F , V ). This question is mostinteresting if the vector spaces have an extra structure, for example that of an algebra. But for themoment we will just focus on vector spaces.

Let V be a vector space over E and let U be an F -form of V . Let u1, u2, . . . be a basis of U overF . Then by Lemma 2.3.3 this is a basis of V as well. Write G = Gal(E/F ). Then for σ ∈ G wedefine a map Tσ : V → V by Tσ(

∑i αiui) =

∑i σ(αi)ui (where, if the set of ui is infinite, all but a

finite number of the αi are zero). This map does not depend on the choice of the basis. Indeed, letv1, v2, . . . be a second basis of U over F and let T ′σ be the corresponding map. There are δij ∈ F withui =

∑j δijvj . Hence

T ′σ(∑i

αiui) = T ′σ(∑j

(∑i

αiδij)vj) =∑j

∑i

σ(αiδij)vj =∑j

∑i

σ(αi)δijvj =∑i

σ(αi)ui = Tσ(∑i

αiui).

The maps Tσ have the following properties:

Tσ(u+ v) = Tσ(u) + Tσ(v) for σ ∈ G, u, v ∈ VTσ(αu) = σ(α)Tσ(u) for σ ∈ G, u ∈ V, α ∈ ETσ Tτ = Tστ for σ, τ ∈ G

T1 = IV .

(2.3.1)

(Here IV denotes the identity map on V .) Moreover, we obviously have U = u ∈ V | Tσ(u) =u for all σ ∈ G. Hence we have the following lemma.

Lemma 2.3.4 Let V be a vector space over E and let U be an F -form of V . Then there is a set ofmaps Tσ : V → V | σ ∈ G with (2.3.1) such that U = u ∈ V | Tσ(u) = u for all σ ∈ G.

We also have the converse of this. In order to prove it we need a lemma on Galois extensions thatis of independent interest.

Lemma 2.3.5 Let E/F be a Galois extension of finite degree n. Write Gal(E/F ) = σ1, . . . , σn andlet α1, . . . , αn ∈ E be a basis of E over F . Then the matrix (σi(αj))1≤i,j≤n is invertible.

2.3 Galois descent 43

Proof. Consider the system of linear equationsσ1(α1)x1 + σ1(α2)x2 + · · ·+ σ1(αn)xn = 0

σ2(α1)x1 + σ2(α2)x2 + · · ·+ σ2(αn)xn = 0

...

σn(α1)x1 + σn(α2)x2 + · · ·+ σn(αn)xn = 0

The given matrix is invertible if and only if this system only has the trivial solution. So suppose thatthere are nontrivial solutions, and let (λ1, . . . , λn) be such a solution with the minimal number ofnonzero coordinates. We may assume that λ1, . . . , λr 6= 0, λr+1, . . . , λn = 0. After dividing we mayassume that λ1 = 1. Not all λi are in F as otherwise the αi are linearly dependent over F (note thatthere is a j such that σj is the identity). So we may assume that λ2 6∈ F . As E/F is Galois there is aσj with σj(λ2) 6= λ2. We have the equations

σi(α1) + σi(α2)λ2 + · · ·+ σi(αr)λr = 0 for 1 ≤ i ≤ n. (2.3.2)

To thse equations we apply σj and use the fact that σjσi = σk and σk runs over Gal(E/F ) if σi doesso. Therefore we also have

σk(α1) + σk(α2)σj(λ2) + · · ·+ σk(αr)σj(λr) = 0 for 1 ≤ k ≤ n. (2.3.3)

Now we subtract (2.3.3) for k = i from (2.3.2) and we see that we get a nontrivial solution with fewernonzero coordinates. This is a contradiction and the lemma is proved. 2

Lemma 2.3.6 Let V be a vector space over E. Let Tσ | σ ∈ G be a set of maps Tσ : V → Vsatisfying (2.3.1). Set U = u ∈ V | Tσ(u) = u for all σ ∈ G. Then U is an F -form of V .

Proof. It is immediate that U is an F -subspace of V .Let v ∈ V . Then for α ∈ E we set vα =

∑σ∈G σ(α)Tσ(v). Then vα ∈ U because for τ ∈ G we

have Tτ (vα) =∑

σ∈G τσ(α)Tτσ(v), which is equal to vα. Let α1, . . . , αn ∈ E be a basis of E over F .Write G = σ1, . . . , σn. By Lemma 2.3.5 the matrix A = (σj(αi))1≤i,j≤n is invertible. Furthermorewe have vαi =

∑nj=1 σj(αi)Tσj (v), so the inverse of the matrix A allows us to express the Tσi(v) as

E-linear combinations of the vαi . Since there is an i with σi = 1 we see in particular that v itself isan E-linear combination of the vαi . But the latter lie in U . Hence V is spanned by the elements of U .

Let u1, u2, . . . be a basis of U and suppose that it is linearly dependent over E. Let v1, . . . , vrbe a E-linearly dependent subset of the ui of smallest cardinality. (In other words, there are no setsof ui of cardinality < r that are E-linearly dependent.) By hypothesis there are λi ∈ E such thatλ1v1 + · · ·+λrvr = 0. By the minimality of r no λi can be zero, and after dividing we have λ1 = 1. Atleast one λi does not lie in F ; we may assume that λ2 6∈ F . Then there is a σ ∈ G with σ(λ2) 6= λ2.After applying Tσ to the above relation we get

v1 + σ(λ2)v2 + σ(λ3)v3 + · · ·+ σ(λr)vr = 0.

So after subtracting we get a nontrivial linear dependency (λ2 − σ(λ2))v2 + · · ·+ (λr − σ(λr))vr = 0.But this involves fewer than r elements, and hence we have a contradiction. 2

Theorem 2.3.7 Let V be a vector space over E. There is a bijection between the set of F -forms ofV and the set of sets of maps Tσ : V → V | σ ∈ G satisfying (2.3.1).

Proof. The maps between these sets are given by Lemma’s 2.3.4, 2.3.6. Lemma 2.3.4 also says thatif we start with an F -form U and consider the corresponding set of transformations Tσ, then theF -form corresponding to that set is precisely U . Secondly, let Tσ be a set of transformations with(2.3.1) and let U be the corresponding F -form. Let u1, u2, . . . be a basis of U . Then the transfor-mations corresponding to U are defined by T ′σ(

∑i αiui) =

∑i σ(αi)ui. But then obviously T ′σ = Tσ. 2

44 Applications of Galois theory

As an application we sketch the a theorem from algebraic geometry. Let R = E[x1, . . . , xn],V = En. Then for a subset A ⊂ R we consider the set of the common zeros

V(A) = v ∈ V | f(v) = 0 for all f ∈ A.

This is called an affine variety. It is clear that V(A) = V(I), where I is the ideal of R generated byA. Secondly, given a set X ⊂ V we define its vanishing ideal

I(X) = f ∈ R | f(v) = 0 for all v ∈ X.

An affine variety X = V(A) is said to be defined over F if I(X) is generated (as ideal) by polynomialsin F [x1, . . . , xn]. Note that G acts on V by σ · (v1, . . . , vn) = (σ(v1), . . . , σ(vn)).

Theorem 2.3.8 X = V(A) is defined over F if and only if σ(X) ⊂ X for all σ ∈ G.

Proof. Suppose that X is defined over F . Set I = I(X); then also X = V(I). Let f ∈ F [x1, . . . , xn]be a generator of I. Because f has coefficients in F we see that for σ ∈ G and v ∈ V we havef(σ · v) = σ(f(v)). Hence if v ∈ X we get f(σ(v)) = σ(f(v)) = σ(0) = 0. Because this holds for allf ∈ I we obtain σ(v) ∈ X.

Now suppose that σ(X) ⊂ X for all σ ∈ G. As σ ∈ G is invertible this implies that σ(X) = Xfor all σ ∈ G. For σ ∈ G and f ∈ R we let Tσ(f) be the polynomial obtained from f by letting σ acton the coefficients of f . Then the maps Tσ satisfy (2.3.1). Furthermore, for v ∈ V and f ∈ R wehave σ(f(v)) = Tσ(f)(σ · v). Let v ∈ X and σ ∈ G and let w ∈ X be such that w = σ(v). Then forf ∈ I we have 0 = σ(f(v)) = Tσ(f)(w). So since w runs through all of X if v does so, we see thatTσ(f) ∈ I. Because the Tσ are invertible we get that Tσ(I) = I for all σ ∈ G. Hence by Theorem2.3.7, f ∈ I | Tσ(f) = f for σ ∈ G is an F -form of I. On the other hand it is clear that this isI ∩F [x1, . . . , xn]. In particular it follows that I ∩F [x1, . . . , xn] generates I, and therefore X is definedover F . 2

Now we have a look at F -forms of algebras. An algebra over a field K is a vector space A over Ktogether with a bilinear map m : A × A → A. The map m is called the multiplication and often wewrite a · b (or simply ab) instead of m(a, b). The algebra A is called associative if a · (b · c) = (a · b) · cfor all a, b, c ∈ A. One of the main examples of associative algebras are the matrix algebras Mn(K)consisting of the n × n-matrices with coefficients in K. Here the product is the ordinary matrixproduct.

Let A be an algebra over E. An F -form of A is an F -form of the vector space A that is closedunder taking products.

Example 2.3.9 Consider the algebra M2(C). Then M2(R) is a R-form of it. Let H be the R-span ofthe following matrices (

1 00 1

),

(i 00 −i

),

(0 −11 0

),

(0 −i−i 0

).

It is straightforward to show that H is closed under taking products (it is enough to check that theproduct of two basis elements again lies in H). Moreover, the given matrices obviously form a basisof M2(C). So H is another R-form of M2(C). We have that H is the algebra of quaternions. It isnot isomorphic to M2(R) (that is, there is no bijective linear map ψ : M2(R) → H with ψ(XY ) =ψ(X)ψ(Y ) for all X,Y ∈ M2(R)). This can be proved by showing that every nonzero element in Hhas a multiplicative inverse, whereas M2(R) obviously does not have that property.

Let U ⊂ A be an F -form of A. Let u1, u2, . . . be a basis of U , then there are γkij ∈ F with

ui · uj =∑

k γkijuk (where, as always, only a finite number of the γkij are non-zero). For σ ∈ G define,

as before, Tσ : A→ A by Tσ(∑

i ciui) =∑

i σ(ci)ui. Let a =∑

i αiui, b =∑

j βjuj be two elements of

A (so αi, βj ∈ E); then a · b =∑

k

(∑i,j αiβjγ

kij

)uk and therefore

Tσ(a · b) =∑k

∑i,j

σ(αi)σ(βj)γkij

uk =

(∑i

σ(αi)ui

)·

∑j

σ(βj)uj

= Tσ(a) · Tσ(b).

2.4 Galois cohomology 45

So the Tσ corresponding to U are multiplicative, that is, they satisfy Tσ(a · b) = Tσ(a) · Tσ(b) for alla, b ∈ A.

Conversely, let Tσ be a set of multiplicative maps A → A with (2.3.1). Let U = a ∈ A |Tσ(a) = a for all σ ∈ G (i.e., U is the vector space F -form of A corresponding to Tσ). From themultiplicativity of the Tσ it now immediately follows that U is closed under taking products, andtherefore is an F -form of A. We conclude that the F -forms of A correspond bijectively to the sets ofmultiplicative maps Tσ satisfying (2.3.1).

Now let U , V be two F -forms of A corresponding to the maps Tσ, Sσ respectively. Letψ : U → V be an isomorphism; that is ψ is a bijective F -linear map respecting the multiplication.Extend ψ to a map ψ : A → A by ψ(

∑i γiui) =

∑i γiψ(ui), where γi ∈ E. Set S′σ = ψTσψ

−1. Thenthe S′σ satisfy (2.3.1) and are multiplicative. Hence V ′ = a ∈ A | S′σ(a) = a for all σ ∈ G is anF -form of A. But S′σ(a) = a (for all σ) is equivalent to Tσ(ψ−1(a)) = ψ−1(a) (for all σ), which isequivalent to ψ−1(a) ∈ U , which in turn is equivalent to a ∈ V . We see that V ′ = V and thereforeS′σ = Sσ by Theorem 2.3.7. Conversely, if ψ is an automorphism of A with Sσ = ψTσψ

−1 for all σ ∈ Gthen it is immediate that ψ maps U onto V and therefore restricts to an isomorphism ψ : U → V .The conclusion is that the F -forms U , V are isomorphic if and only if there is an automorphism ψ ofA such that Sσ = ψTσψ

−1 for all σ ∈ G.

2.4 Galois cohomology

Galois cohomology offers another perspective on classification problems like the one considered at theend of the previous section. There it is shown that classfying all F -forms of a given algebra A over Eup to isomorphism is equivalent to classifying, up to conjugacy by the group Aut(A), all sets of mapsTσ | σ ∈ G that are multiplicative and satisfy (2.3.1). However, dealing with these maps is not soeasy because they are not linear, indeed, we have Tσ(αa) = σ(α)Tσ(a) for σ ∈ G, α ∈ E, a ∈ A.Galois cohomology provides a way of dealing with sets like the ones above, with the difference thatthe elements are automorphisms.

Example 2.4.1 Consider the set up of Example 2.3.9. Let G = Gal(C/R) and write G = 1, γ,where γ denotes complex conjugation, γ(z) = z for z ∈ C. Let Γ : M2(C) → M2(C) be the mapdefined by

Γ

(a bc d

)=

(a bc d

).

Then G acts on Aut(M2(C)) by γ ·ψ = ΓψΓ (indeed, we have γ ·(γ ·ψ) = ΓΓψΓΓ = ψ). Let T1, Tγ bea set of multiplicative maps with (2.3.1) and define α : G → Aut(M2(C)) by α(1) = Id, α(γ) = TγΓ.Then it is straightforward to see that

α(στ) = α(σ)(σ · α(τ)) for all σ, τ ∈ G. (2.4.1)

Conversely, let α : G → Aut(M2(C)) satisfy (2.4.1). Then α(1) = α(1 · 1) = α(1)1 · α(1) = α(1)2,implying that α(1) = Id. Secondly, from α(γγ) = α(γ)γ ·α(γ) it follows that (α(σ)Γ)2 = Id. Thereforeputting T1 = Id, Tγ = α(γ)Γ we obtain a set of multiplicative maps Tσ : M2(C) → M2(C) | σ ∈ Gwith (2.3.1).

Furthermore, if α′ : G → Aut(M2(C)) is the map corresponding to the set ψT1ψ−1, ψTγψ−1(where ψ ∈ Aut(M2(C))) then

α′(σ) = ψα(σ)(σ · ψ−1) for all σ ∈ G. (2.4.2)

Conversely, let α, α′ : G→ Aut(M2(C)) be two maps with (2.4.1) and suppose that the relation (2.4.2)holds. Set T1 = T ′1 = Id and Tγ = α(γ)Γ, T ′γ = α′(γ)Γ. Then T ′σ = ψTσψ

−1 for all σ ∈ G.

We conclude that the R-forms of M2(C) correspond to maps α : G → Aut(M2(C)) with (2.4.1).Furthermore, two R-forms, corresponding to the maps α, α′, are isomorphic if and only if the relation(2.4.2) holds.

46 Applications of Galois theory

Now we take a step back and consider a more general situation. Let G be a finite group. LetA be a group on which G acts by homomorphisms. That means that we have a map G × A → A,(σ, a) 7→ σ · a as in Section 3.5(Algebra), and for each σ ∈ G the map A → A, a 7→ σ · a is a grouphomomorphsim (i.e., σ · (ab) = (σ ·a)(σ · b) for all a, b ∈ A). Then a 1-cocyle with values in A is a mapα : G→ A with

α(στ) = α(σ)(σ · α(τ)) for all σ, τ ∈ G. (2.4.3)

By Z1(G,A) we denote the set of all 1-cocycles with values in A.

We remark the following:

If α is a 1-cocycle then α(1) = α(1 · 1) = α(1)(1 · α(1)) = α(1)2 implying α(1) = 1 (where thesecond 1 denotes the neutral element of A).

The map α : G→ A with α(σ) = 1 for all σ ∈ G is a 1-cocycle, called the trvial cocycle. In thesequel we denote it by 1.

Let α ∈ Z1(G,A) and a ∈ A. Define αa : G → A by αa(σ) = aα(σ)(σ · a−1) (compare (2.4.2)).Then a short calculation shows that αa is a 1-cocycle as well.

Two 1-cocycles α, α′ ∈ Z1(G,A) are said to be cohomologous if there is an a ∈ A with α′ = αa.In this situation we write α ∼ α′. This is an equivalence relation. (Indeed: α = α1. If α′ = αa

then α = (α′)a−1

. If α′ = αa, α′′ = (α′)b, then α′′ = αba.) The set of equivalence classes is denotedH1(G,A) and called the first cohomology set of G with coefficients in A.

One problem with these cohomolgy sets is that they are difficult to compute. A generalization ofa famous theorem of Hilbert gives these sets for an important class of groups.

Theorem 2.4.2 (Hilbert 90) Let E/F be a Galois extension of finite degree, G = Gal(E/F ). Notethat G acts naturally on the matrix group GL(n,E) (by letting a σ ∈ G act on the coefficients of amatrix in GL(n,E)). We have that H1(G,GL(n,E)) = 1 (where the latter denotes the set consistingof the class of the trivial coclycle only).

Proof. Set U = En. Let u ∈ U and write u = (u1, . . . , un). Then for a σ ∈ G we set σ · u =(σ(u1), . . . , σ(un)). Let α ∈ Z1(G,GL(n,E)) and for σ ∈ G define the map Tασ : U → U by Tασ (u) =α(σ)(σ · u). It is straightforward to see that the set Tασ | σ ∈ G satisfies (2.3.1). Hence

Uα = u ∈ U | Tασ (u) = u for all σ ∈ G

is an F -form of U (Lemma 2.3.6). Let v1, . . . , vn be an F -basis of Uα. Let P ∈ GL(n,E) be the matrixwith columns v1, . . . , vn. For σ ∈ G the matrix σ · P has columns σ · v1, . . . , σ · vn. But vi ∈ Uα, sothat for σ ∈ G we have

vi = Tασ (vi) = α(σ)(σ · vi),

which implies that P = α(σ)(σ · P ) (a product of two elements in GL(n,E)). But then

αP−1

(σ) = P−1α(σ)(σ · P ) = P−1P = IU .

We conclude that α is cohomologous to the trivial cocycle. 2

Definition 2.4.3 A pointed set is a set S with a fixed distinguished element s ∈ S, which is calledthe base point of S. Let S, T be pointed sets with base points s, t respectively. A map of pointed setsis a map f : S → T with f(s) = t. Its kernel is ker(f) = x ∈ S | f(x) = t.

Our main example of a pointed set is the first cohomology set H1(G,A) whose base point is theclass of the trivial cocycle. In the next bit we describe a construction of maps of pointed sets betweencohomology sets.

2.4 Galois cohomology 47

Definition 2.4.4 Let G,G′ be finite groups acting by homomorphisms on the groups A,A′ respectively.Let ϕ : G′ → G, f : A→ A′ be group homomorphisms. Then we say that ϕ and f are compatible if

f(ϕ(σ′) · a) = σ′ · f(a) for all a ∈ A, σ′ ∈ G′.

Remark 2.4.5 We use the notation from the previous definition and suppose ϕ, f are compatible.Write

AG = a ∈ A | σ · a = a for all σ ∈ G.

The obviously f(AG) ⊂ (A′)G.

Proposition 2.4.6 We use the notation from Definition 2.4.4 and suppose that ϕ, f are compatible.Let α, α′ ∈ Z1(G,A) and define β : G′ → A′ by β(σ′) = f(α(ϕ(σ′))) and define β′ similarly using α′.Then β, β′ ∈ Z1(G′, A′). Moreover, if α ∼ α′ then β ∼ β′.

Proof. We have

β(σ′τ ′) = f(α(ϕ(σ′)ϕ(τ ′))) = f(α(ϕ(σ′))(ϕ(σ′) · α(ϕ(τ ′))))

= f(α(ϕ(σ′)))f(ϕ(σ′) · α(ϕ(τ ′)))

= f(α(ϕ(σ′)))(σ′ · f(α(ϕ(τ ′)))

)= β(σ′)(σ′ · β(τ ′)).

Suppose that there is an a ∈ A such that α′ = αa, i.e., α′(σ) = aα(σ)(σ · a−1). We write this withϕ(σ′) in place of σ and apply f to obtain

β′(σ′) = f(a)β(σ′)(σ′ · f(a)−1).

So we see that β′ = βf(a). 2

By the previous proposition we get a map

f∗ : H1(G,A)→ H1(G′, A′)

by f∗([α]) = [β]. The proposition states that this map is well-defined. It obviously maps the class ofthe trivial cocycle to the class of the trivial cocycle. Hence it is a map of pointed sets.

Consider a sequence of maps of pointed sets

. . .fi−1−−−→ Ai

fi−→ Ai+1fi+1−−−→ Ai+2

fi+2−−−→ . . .

The sequence is said to be exact at Ai if im(fi−1) = ker(fi). The sequence is exact if it is exact ateach Ai.

In the next part we work with three pointed sets A,B,C on which G acts. The distinguishedelement will in all cases be denoted 1. We suppose further

A, B are groups on which G acts by homomorphisms,

we have an exact sequence of pointed sets

1→ Af−→ B

g−→ C → 1 (2.4.4)

(that is, ker(f) = 1, im(f) = ker(g), g is surjective),

f is a group homomorphism (so that it is injective),

f, g commute with the action of G, that is f(σ · a) = σ · f(a) for all a ∈ A, σ ∈ G, and similarlyfor g,

48 Applications of Galois theory

for all b, b′ ∈ B we have g(b) = g(b′) if and only if b′ = bf(a) for a certain a ∈ A (that is, g isconstant on the cosets bf(A) - note that f(A) is a subgroup of B- and takes different values ondifferent cosets).

We write AG, BG, CG for the subsets of A,B,C respectively, consisting of all elements that arefixed under all σ ∈ G. So AG consists of all a ∈ A with σ · a = a for all σ ∈ G. Because f, g commutewith the action of G we have f(AG) ⊂ BG and f(BG) ⊂ CG.

Now let c ∈ CG. As g is surjective there exist b ∈ B with g(b) = c; fix such a b. Then g(σ ·b) = g(b)for all σ ∈ G. By the assumption on g this implies that for each σ ∈ G there exists an aσ ∈ A withσ(b) = bf(aσ). But this is the same as f(aσ) = b−1σ(b). Note that aσ is uniquely determined by bbecause f is injective.

Lemma 2.4.7 Define α : G → A by α(σ) = aσ. Then α ∈ Z1(G,A) and its class in H1(G,A) doesnot depend on the choice of b.

Proof. Let σ, τ ∈ G. On the one hand we have f(aστ ) = b−1(στ · b). On the other hand,

f(aσ(σ · aτ )) = f(aσ)(σ · f(aτ )) = b−1(σ · b)(σ · (b−1(τ · b))) = b−1(σ · (τ · b)).

As f is injective it follows that aστ = aσ(στ · aτ ) and α ∈ Z1(G,A).Let b′ ∈ B be a second element with g(b′) = c. This yields an a′σ ∈ A with f(a′σ) = (b′)−1σ · b′.

Since g(b′) = g(b), the assumption on g provides an a′ ∈ A with b′ = bf(a′). But then

(b′)−1σ · b′ = f(a′)−1b−1(σ · b)(σ · f(a′)) = f((a′)−1aσ(σ · a′)).

Because f is injective it follows that a′σ = (a′)−1aσ(σ · a′). But that means that the cocycles definedby b and b′ are cohomologous. 2

By the previous lemma we get a well-defined map

δ0 : CG → H1(G,A)

with δ0(c) = [α], where α ∈ Z1(G,A) is defined by f(α(σ)) = b−1σ · b, where b ∈ B is any elementwith g(b) = c.

Theorem 2.4.8 The following sequence is exact

1→ AGf−→ BG g−→ CG

δ0−→ H1(G,A)f∗−→ H1(G,B).

Proof. The exactness at AG immediately follows from the fact that the kernel of the homomorphismf : A→ B is 1.

We have f(AG) ⊂ ker(g) by the exactness of (2.4.4). Let now b ∈ BG be such that g(b) = 1.Then again by the exactness of (2.4.4) there is an a ∈ A with b = f(a). For σ ∈ G we havef(a) = b = σ · b = σ · f(a) = g(σ · a), so that a ∈ AG as f is injective. Hence b ∈ f(AG).

Now we consider the exactness of the sequence at CG. Let b ∈ BG and set c = g(b). For σ ∈ Glet aσ ∈ A be the element determined by b as above. Then f(aσ) = b−1σ · b = b−1b = 1. Henceδ0(c) is the class of the trivial cocycle and we have shown that g(BG) ⊂ ker(δ0). Now let c ∈ CG,set α = δ0(c) and suppose that α ∼ 1. Let b ∈ B be such that g(b) = c; then f(α(σ)) = b−1σ · bfor all σ ∈ G. Furthermore, α ∼ 1 means that there is an a ∈ A with α(σ) = aσ · a−1. Thenb−1σ · b = f(α(σ)) = f(aσ · a−1) = f(a)σ · f(a)−1, which is equivalent to f(a)−1b−1 = σ · (f(a)−1b−1).We see that f(a)−1b−1 and therefore also bf(a) lie in BG. But g(f(a)) = 1 by exactness of (2.4.4), sog(bf(a)) = g(b) = c and we have shown that ker(δ0) ⊂ f(BG).

Finally we deal with exactness at H1(G,A). Let c ∈ CG and set α = δ0(c). Let b ∈ B be suchthat g(b) = c; then f(α(σ)) = b−1σ · b for all σ ∈ G. Set β = f∗(α); then β(σ) = f(α(σ)) = b−1σ · b.It follows that β ∼ 1. Now let α ∈ Z1(G,A) and set β = f∗(α), i.e., β(σ) = f(α(σ) for σ ∈ G.Suppose that β ∼ 1, that is, there is a b ∈ B with β(σ) = b−1σ · b for all σ ∈ G. Let σ ∈ G. Then

2.4 Galois cohomology 49

b−1σ · b = f(α(σ)) whence σ · b = bf(α(σ)). Therefore σ · g(b) = g(σ · b) = g(bf(α(σ))) = g(b) (againby exactness of (2.4.4)). We see that g(b) ∈ CG. Moreover, because f(α(σ)) = b−1σ · b and g(b) = cwe have by definition of δ0 that δ0(c) = [α]. 2

As an application of this theorem we have the following result.

Proposition 2.4.9 Let the notation be as in Theorem 2.4.2. Let SL(n,E) be the subgroup of GL(n,E)consisting of the matrices of determinant 1. Then H1(G, SL(n,E)) = 1.

Proof. We have the following exact sequence

1→ SL(n,E)i−→ GL(n,E)

det−−→ E∗ → 1

(here i simply maps a matrix to itself) and the maps satisfy all our hypotheses. Then by the previoustheorem

GL(n, F )det−−→ F ∗

δ0−→ H1(G, SL(n,E))f∗−→ H1(G,GL(n,E))

is exact. But by Theorem 2.4.2 the last set in this sequence is trivial. Hence δ0(F ∗) = H1(G, SL(n,E)).Furthermore, det : GL(n, F )→ F ∗ is surjective, so that ker(δ0) = F ∗. In other words, δ0(F ∗) = [1].Putting the two things together proves the proposition. 2

We have already seen how classification of real forms of a complex algebra is equivalent to com-puting a Galois cohomology set (Example 2.4.1). Now we show how Galois cohomology can also beof interest in certain situations for the classification of orbits.

We define an action of BG on CG. Let b ∈ BG and c ∈ CG. There is a b′ ∈ B with g(b′) = c. Wedefine

b · c = g(bb′).

Lemma 2.4.10 We have that g(bb′) does not depend on the choice of b′. This indeed defines an actionof BG on CG.

Proof. Let b′′ ∈ B be such that g(b′′) = c. Then by hypothesis b′′ = b′f(a) for a certain a ∈ A. Henceg(bb′′) = g(bb′f(a)) = g(bb′).

We show that b · c ∈ CG. Let σ ∈ G, then σ · g(bb′) = g((σ · b)(σ · b′)) = g(bσ · b′). Butg(σ · b′) = σ · g(b′) = σ · c = c, so by the first part of the proof, g(bσ · b′) = g(bb′). It follows thatb · c ∈ CG.

Let b1, b2 ∈ B. We show that b1 · (b2 · c) = (b1b2) · c. Let b′2 ∈ B be such that g(b′2) = c, so thatb2 · c = g(b2b

′2). Let b′1 ∈ B be such that g(b′1) = g(b2b

′2), which is equivalent to b′1 = b2b

′2f(a) for a

certain a ∈ A. We now infer b1 · (b2 · c) = g(b1b′1) = g(b1b2b

′2f(a)) = g(b1b2b

′2) = (b1b2) · c. 2

Theorem 2.4.11 Let CG/BG denote the set of orbits of BG on CG. Sending BG · c to δ0(c) yields awell-defined map from CG/BG to ker(f∗) ⊂ H1(G,A). Moreover, this map is a bijection.

Proof. We first show that the map is well-defined. Suppose that c, c′ ∈ CG lie in the same BG-orbit,that is, there is a b ∈ BG with c′ = b · c = g(bb′), where b′ ∈ B is such that g(b′) = c. Then δ0(c) = [α]where f(α(σ)) = (b′)−1σ · b′ and δ0(c′) = [α′] where f(α′(σ)) = (bb′)−1σ · (bb′). But σ · b = b andtherefore (bb′)−1σ · (bb′) = (b′)−1b−1σ · bσ · b′ = (b′)−1σ · b′ = f(α(σ)). As f is injective it follows thatα = α′.

Theorem 2.4.8 directly shows that the map is surjective. So it remains to show injectivity. Letc, c′ ∈ CG, write δ0(c) = α, δ0(c′) = α′ and suppose that α ∼ α′, that is, there is an a ∈ Awith α′(σ) = aα(σ)σ · a−1 for all σ ∈ G. Let b, b′ ∈ B be such that g(b) = c, g(b′) = c′. Thenf(α(σ)) = b−1σ · b, f(α′(σ)) = (b′)−1σ · b′ for all σ ∈ G. Let σ ∈ G, then

(b′)−1σ · b′ = f(α′(σ)) = f(a)f(α(σ))f(σ · a−1) = f(a)b−1(σ · b)(σ · f(a)).

50 Applications of Galois theory

But this is equivalent to b′f(a)b−1 = σ · (b′f(a)b−1). So if we set b = b′f(a)b−1 then we have b ∈ BG.Furthermore b′f(a) = bb, so that c′ = g(b′) = g(b′f(a)) = g(bb) = b · c. We conclude that c and c′ liein the same BG-orbit. 2

Example 2.4.12 Let F be a field. We consider the space M2(F ) consisting of the 2 × 2-matriceswith coefficients in F . Let GL(2, F ) be the group consisting of the invertible elements of M2(F ).Furthermore, SL(2, F ) is the group consisting of the elements of M2(F ) of determinant 1. Both thesegroups act on M2(F ) by conjugation, that is, for T ∈ GL(2, F ) (respectively SL(2, F )) and M ∈M2(F )we set T ·M = TMT−1.

An N ∈M2(F ) is said to be nilpotent if there is an m > 0 with Nm = 0; it can be shown that thisis the case if and only if N2 = 0. Let N ∈M2(F ) be nilpotent and nonzero. Then there is a nonzerov ∈ F 2 such that Nv 6= 0. We have that v,Nv is a basis of F 2 with respect to which the linear mapcorresponding to N has matrix

N0 =

(0 01 0

).

It follows that N is GL(2, F )-conjugate to N0. In other words, the set of nonzero nilpotent elementsof M2(F ) is a single GL(2, F )-orbit.

When we consider the group SL(2, F ) the situation is a bot different. First let F = C and letT ∈ GL(2,C) be such that TNT−1 = N0. Set λ = 1√

det(T )and S = λT . Then S ∈ SL(2,C) and

SNS−1 = N0. So also in this case there is a single orbit of nonzero nilpotent matrices. Second,let F = R. Then if det(T ) > 0 we can do the same thing. However, if det(T ) < 0 then we setλ = 1√

− det(T )and

S = λ

(1 00 −1

)T.

Then again S ∈ SL(2,R), but this time SNS−1 = −N0. Furthermore it is straightforward to see thatN0 and −N0 are not SL(2,R)-conjugate. So here we get two orbits of nilpotent nonzero matrices.

Now we reinterpret this in terms of Galois cohomology. Let G = Gal(C/R) = 1, γ. Hereγ is complex conjugation and acts on M2(C) by the map Γ from Example 2.4.1. B = SL(2,C),A = T ∈ B | T ·N0 = N0 (that is, A is the stabilizer of N0 in SL(2,C)), C = B ·N0 = T ·N0 | T ∈ B(that is, C is the orbit of N0). Then a small calculation shows that

A =

(ε 0c ε

)| ε = ±1, c ∈ C

.

Moreover, if we let f : A → B be the inclusion map (so simply f(T ) = T for T ∈ A) and defineg : B → C by g(T ) = T ·N0 then we get an exact sequence as in (2.4.4). Furthermore, all hypotheseson the sequence (2.4.4) are satisfied. For the last one suppose that g(S) = g(T ). This is equivalent toS ·N0 = T ·N0, which amounts to S−1T ∈ A, or T = SU for a U ∈ A.

We have BG = SL(2,R) and CG = SL(2,C) · N0 ∩M2(R), so CG is the set of nonzero nilpotentelements of M2(R). We now look at the action of BG on CG defined above. Let T ∈ BG and M ∈ CGthen we denote this action by T M . Let T ′ ∈ B be such that T ′ · N0 = M . Then by definitionT M = g(TT ′). But the latter is equal to g(TT ′) = TT ′ · N0 = T · (T ′ · N0) = T ·M (which bydefinition equals TMT−1). So we see that the action of BG on CG is nothing other than the actionwe had already defined. We see that the orbits of BG on CG are the orbits of SL(2,R) on the set ofnonzero nilpotent elements of M2(R).

By Proposition 2.4.9 we have that H1(G,B) = 1. Therefore by Theorem 2.4.8, ker f∗ = H1(G,A).We conclude by Theorem 2.4.11 that the orbits of SL(2,R) on the set of nonzero nilpotent elementsof M2(R) are in bijection with H1(G,A).

Now we compute H1(G,A). Let α : G → A be a cocycle. Then α(1) = I2 (the 2 × 2-identitymatrix) and α(γ) = T with

T =

(ε 0c ε

)

2.4 Galois cohomology 51

with ε = ±1. We have that α is a cocycle if and only if TT = I2. A straightforward computationshows that this happens if and only if c+ c = 0, that is, when c = iu for some u ∈ R. Now let S ∈ Aand write

S

(δ 0c0 δ

)where δ = ±1. Then αS(γ) = STS

−1, which is equal to(

ε 0δε(c0 − c0) + iu ε

).

We see that by choosing c0 = −12δεui the term in the (2, 1) position vanishes. So we have two

cocycles, the trivial one, and the one sending γ to −I2. Therefore we recover the fact that there aretwo SL(2,R)-orbits in the set of nonzero nilpotent elements of M2(R).

Remark 2.4.13 In the previous example it may seem that Galois cohomology is a way to transformeasy problems into more difficult ones. However, there are also situations, analogous to the oneconsidered in the example, where it is not possible to list the real orbits directly. In those cases theapproach via Galois cohomology can be very useful. We refer to [Djo83] for an example.

52 Applications of Galois theory

Index

(Z/nZ)∗, 20An, 25F -form, 42[G,G], 23[g, h], 23Aut(E), 9Φn, 181-cocycle, 46

algebra, 44alternating group, 25automorphism group

of a field, 9

commutator, 23commutator subgroup, 23cyclotomic polynomial, 18

derived series, 24discriminant of a polynomial, 32

Euler’s ϕ-function, 20extension

cyclotomic, 20Galois, 9normal, 9radical, 27separable, 9

fielddifferential, 36of constants, 36

Galois group, 11group

solvable, 23

intermediate field, 12stable, 13

inversion, 24

permutationeven, odd, 25

primitive element, 9

root of unity, 17primitive, 17

separable

polynomial, 6solvable by radicals, 27subgroup

generated by subset, 23

54 Bibliography

Bibliography

[Bos18] Siegfried Bosch. Algebra. Birkhauser, 2018.

[Cox12] David A. Cox. Galois Theory. John Wiley & Sons, 2012.

[DF04] David S. Dummit and Richard M. Foote. Abstract algebra. John Wiley & Sons, Inc., Hoboken,NJ, third edition, 2004.

[Djo83] Dragomir Z. Djokovic. Classification of trivectors of an eight-dimensional real vector space.Linear and Multilinear Algebra, 13(1):3–39, 1983.

[J93] Klaus Janich. Funktionentheorie. Springer-Lehrbuch. [Springer Textbook]. Springer-Verlag,Berlin, third edition, 1993. Eine Einfuhrung. [An introduction].

[Tig01] Jean-Pierre Tignol. Galois’ Theory of Algebraic Equations. World Scientific Publishing, 2001.