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Galois theory
Philippe H. Charmoysupervised by Prof Donna M. Testerman
Autumn semester 2008
Contents
0 Preliminaries 40.1 Soluble groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.2 Field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Construction of finite field extensions . . . . . . . . . . . . . . . 6Splitting field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Algebraic closure . . . . . . . . . . . . . . . . . . . . . . . . . . 10Formal differentiation . . . . . . . . . . . . . . . . . . . . . . . . 10
0.3 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1 Galois correspondence 121.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2 Normality and separability . . . . . . . . . . . . . . . . . . . . . 12
Normal extension . . . . . . . . . . . . . . . . . . . . . . . . . . 12Separable extensions . . . . . . . . . . . . . . . . . . . . . . . . 14Monomorphisms and automorphisms . . . . . . . . . . . . . . . . 16Normal closure . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.3 Field extension degrees and group order . . . . . . . . . . . . . . 19Galois group and fixed fields . . . . . . . . . . . . . . . . . . . . 19
1.4 The Galois correspondence . . . . . . . . . . . . . . . . . . . . . 23The fundamental theorem of Galois theory . . . . . . . . . . . . . 23
1.5 Galois theory of finite fields . . . . . . . . . . . . . . . . . . . . 261.6 Simple extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 271.7 Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . 27
2 Some examples 312.1 A Galois correspondence of a splitting field of a polynomial . . . 312.2 An example of a symmetric group as Galois group . . . . . . . . 322.3 The importance of separability . . . . . . . . . . . . . . . . . . . 33
3 Solutions by radicals 353.1 Radical extensions . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 Equations soluble by radicals . . . . . . . . . . . . . . . . . . . . 36
1
Symmetric polynomials . . . . . . . . . . . . . . . . . . . . . . . 36The general polynomial equation and symmetric groups . . . . . . 36Equation of degree smaller than or equal to 4 . . . . . . . . . . . 38An insoluble quintic . . . . . . . . . . . . . . . . . . . . . . . . . 38
4 Inverse Galois problem 404.1 Some results on realisability over the field of rational numbers . . 40
Finite abelian groups . . . . . . . . . . . . . . . . . . . . . . . . 40Symmetric groups . . . . . . . . . . . . . . . . . . . . . . . . . . 42Soluble groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 46An alternative proof for symmetric groups of prime order . . . . . 47
4.2 Another result on realisability over arbitrary fields . . . . . . . . . 48
2
Preface
I am extremely grateful to Prof Donna Testerman, first for accepting to supervisemy work, and most importantly for her patience and expertise without which thisproject could never have been completed.
Also, I would like to thank Prof Manuel Ojanguren for providing me withsome additional material on which to work and some help when I had difficultiesunderstanding it.
Despite their help, some errors doubtless remain; they are of course mine.
3
Chapter 0
Preliminaries
0.1 Soluble groups
Definition 0.1.1. A group G is said to be soluble if there exists a finite sequenceof subgroups of G such that
{1} = G0 C G1 C · · · C Gn = G,
and Gi+1/Gi is abelian for every i ∈ {0, . . . , n − 1}.
Theorem 0.1.2. Let G be a group, H < G and N C G. Then:
1. if G is soluble, so is H;
2. if G is soluble, so is G/N;
3. If N and G/N are soluble, so is G.
Proof. This proof is omitted here. See [10, pp. 126–127]. �
Theorem 0.1.3. A soluble group is simple if and only if it is cyclic of prime order.
Proof. If G is simple and soluble, the only possible sequence will be 1 C G, if wedelete repetitions. Therefore, G must be abelian. Since G cannot have any propernon-trivial subgroup it must have a prime order and thus be cyclic.
The converse is obvious. �
Theorem 0.1.4. If n ≥ 5, then the alternating group An is simple.
Proof. This proof is omitted here. See [10, pp. 129–131]. �
Corollary 0.1.5. If n ≥ 5, then the symmetric group S n is not soluble.
Proof. By contraposition, if S n were soluble, then An would be as well (Theorem0.1.2). But this cannot be true since An is simple (Theorem 0.1.4) and does nothave prime order (Theorem 0.1.3). �
4
Lemma 0.1.6. If G is a finite p-group of order pn, then it has a series of normalsubgroups
{1} = G0 ⊂ G1 ⊂ · · · ⊂ Gn = G,
such that Card Gi = pi for every i ∈ {0, . . . , n}.
Proof. Let us reason by induction on n. If n = 0, the result is clear. Let us nowsuppose the result holds for groups of order pn−1. It follows from the class equationthat Z(G), the centre of G, is not trivial. As Card(Z(G)) = pm for some m ≤ n, the‘Classification of finitely generated abelian groups’ implies that there exists anelement z ∈ Z(G) of order p. Furthermore, the subgroup H = 〈z〉 generated by z isa subgroup of Z(G) and is therefore normal in G. It follows that G/H is a group oforder pn−1.
By our induction assumption, there exists a series of normal subgroups of G/H
{1} = H/H = G1/H ⊂ · · · ⊂ Gn/H = G/H,
where for every i ∈ {1, . . . , n} we have Card(Gi/H) = pi−1. It follows that for everyi ∈ {1, . . . , n}, Card(Gi) = pi and, by the ‘Correspondence theorem’, that Gi isnormal in G. So it suffices to let G0 = {1} to have the result. �
Theorem 0.1.7. If G is a finite p-group, then it is soluble.
Proof. The series of subgroups given by Lemma 0.1.6 will a fortiori be of the form
{1} = G0 C G1 C · · · C Gn = G.
Furthermore, as Gi+1/Gi has prime order for every i ∈ {0, . . . , n − 1}, the quotientsare cyclic and hence abelian, as desired. �
0.2 Field extensions
In this section, we give some fundamental definitions about field extensions andpresent some basic results. Field extensions will provide the underpinnings fromwhich to build Galois theory.
Generalities
Definition 0.2.1. A field extension L of a field K is a non-zero homomorphismφ : K ↪→ L. The extension is written L : K.
Let us notice that such a homomorphism is always injective, for φ−1(0) is aproper ideal of K and is thus equal to {0}. Therefore, φ embeds K into L andinduces an isomorphism between K and φ(K).
This definition is a generalisation of the more intuitive idea that L : K is afield extension if K is a subfield of L. However, these situations are identical upto isomorphism. In order to simplify notations we will henceforth assume, withoutloss of generality, that if L : K is a field extension, then K is a subfield of L.
5
Definition 0.2.2. Let M ↪→ K and L ↪→ M be two field extensions. Then wewrite L ↪→ M ↪→ K and call this a tower of extensions or simply a tower.
Definition 0.2.3. The degree of an extension L : K, written [L : K], is thedimension of L as a K-vector space. If this dimension is finite, L : K is said tobe a finite extension.
Example 0.2.4. The field extension C : R has degree 2, with basis {1, i}. ^
Proposition 0.2.5 (Tower law). If K ↪→ M ↪→ L is a tower of finite field exten-sions, then
[L : K] = [L : M][M : K].
Definition 0.2.6. Let f (t) = a0 + a1t + · · · + antn ∈ K[t] be a polynomial. Anelement k ∈ K is called a root of f (t) if the evaluation of f (t) at k, writtenf (k) = a0 + a1k + · · · + ankn, is equal to 0.
Definition 0.2.7. Let L : K be a field extension. An element a ∈ L is calledalgebraic over K if it is a root of a non-zero polynomial f (t) ∈ K[t], and tran-scendental over K otherwise. If every element of L is algebraic over K, thenthe field extension L : K is itself called algebraic.
Finally, let us define the notion of isomorphic field extensions.
Definition 0.2.8. Let L : K and L′ : K′ be two field extensions. They are calledisomorphic field extensions if there exists a field isomorphism ψ : L → L′ suchthat ψ|k is also an isomorphism.
Remark 0.2.9. Let L : K and M : K be two field extensions. The existence of �
an isomorphism ψ : L → M is not a sufficient condition for the extensions to beisomorphic, in general.
Indeed, if we consider the extension Q[r1, r2, r3] : Q[r1], where r1 =3√2,
and r2, r3 are the other two complex roots of the polynomial t2 − 3 ∈ Q[t]. Theautomorphism of Q[r1, r2, r3] defined by r1 7→ r2, r2 7→ r1 and r3 7→ r3 is not anextension isomorphism. (It is not obvious that this defines an automorphism; thisis proved in Section 2.2.)
Notice though that if K = Q, then we must have ψ(1) = 1, and so Z and Q haveto be fixed pointwise. Thus in this case, every isomorphism ψ : L → M is also anfield extensions isomorphism (This argument works whenever K is a prime field,i.e. Fp with p prime or Q.)
Construction of finite field extensions
In this subsection we describe some procedures through which finite algebraic fieldextensions can be constructed. For a more thorough coverage, see [7, Chapter 1].
Definition 0.2.10. Let L : K be a field extension and a ∈ L be algebraic over K.The unique monic and irreducible polynomial f (t) ∈ K[t] of which a is a root iscalled the minimal polynomial of a over K and noted min(a,K).
6
This definition makes sense, for K[t] is a principal ideal domain and the setof polynomials of which a is a root forms an ideal of K[t]. This ideal is thusgenerated by some polynomial, say m(t). To see that m(t) is irreducible; let us wewrite m(t) = f (t)g(t) with f (a) = 0. This implies that f (t) ∈ (m(t)) and so g(t) = 1.Therefore m(t) must be irreducible. Further, m(t) is unique if one imposes that itbe monic and thus m(t) = min(a,K).
Theorem 0.2.11. Let L : K be a field extension and let us suppose a ∈ L isalgebraic over K. Then we have the following properties.
1. Evaluation at a induces a ring isomorphism
K[t]/(min(a,K)) � K[a],
where K[a] is the set of all polynomial of K[t] evaluated at a.
2. K[a] is a field.
3. [K[a] : K] = deg(min(a,K)).
Proof.
1. This follows directly from the first isomorphism theorem, because f (t) 7→f (a) is a surjective homomorphism with kernel (min(a,K)).
2. As min(a,K) is irreducible, (min(a,K)) is a prime ideal. Recalling that K[t]is a principal ideal domain, we get that (min(a,K)) is also a maximal ideal.Hence the result.
3. To simplify notation, we set n = deg(min(a,K)). Let us show that B =
{1, a, . . . , an−1} is a K-basis of K[a]. First, we check the independence. If
λ0 + λ1a + · · · + λn−1an−1 = 0,
then λi = 0 for all i ∈ {0, . . . , n − 1}, for min(a,K) has degree n.
Let us now see that B generates K[a]. Let us suppose f (a) ∈ K[a], thenusing Euclidean division f (t) = q(t) min(a,K) + r(t), where q(t), r(t) ∈ K[t],and deg(r(t)) < n. Evaluating at a, we get f (a) = r(a). We conclude bynoticing that r(a) is necessarily in the span of B. �
It follows directly from this proposition that for every algebraic element a,K[a] = K(a), the fraction field of K[a].
Definition 0.2.12. Two algebraic elements a1 and a2 over a field K are calledconjugate if they have the same minimal polynomial.
When a is transcendental, K[a] is much less interesting as the following theo-rem shows.
7
Theorem 0.2.13. Let L : K be a field extension and suppose a ∈ L is transcen-dental over K. Then we have the following properties.
1. Evaluation at a induces a ring isomorphism
K[t] � K[a].
2. [K(a) : K] is infinite.
Proof. The first claim follows from the fact that f (t) 7→ f (a) is injective, as a istranscendental. Indeed, by definition of K[a], f (t) 7→ f (a) is a surjective homomor-phism. The second comes a fortiori because the dimension of K[t] as a K-vectorspace is infinite and K(a) ⊃ K[a], which is isomorphic to K[t]. �
Corollary 0.2.14. Every finite field extension is algebraic.
Example 0.2.15. Let E and F be finite extensions of a field K, such that [E : K] =
m, [F : K] = n, and (m, n) = 1. Let us suppose that a ∈ F has degree r over K;then a has degree r over E as well (i.e. min(a, E) has degree r as well).
Indeed, if we denote min(a,K) by m(t), we have the following diagram, wherethe arrows represent inclusions with corresponding extension degrees.
E[a] F
E
x=?=={{{{{{{{
K[a]
y=?ccFFFFFFFF
q=={{{{{{{{
K
maaCCCCCCCC
r;;wwwwwwwww
The ‘Tower law’ implies that n = qr so that (r,m) = 1, and also that both m and rdivide [E[a] : K]. Hence mr divides [E[a] : K]]. Furthermore [E[a] : K] ≤ mr, form(t) is a divisor of min(a, E), as m(a) = 0. Hence, x = r and y = m; in particular, ahas degree r over E (see Theorem 0.2.11). ^
We close this subsection by noticing that given a field extension L : K and asubset Y of L, it is always possible to create a minimal (in the sense of inclusion)intermediate field M containing Y . To do this, we need the following lemma.
Lemma 0.2.16. Let K be a field and {Ki : i ∈ I} be a family of subfields of K.Then ∩i∈IKi is also a subfield of K.
Proof. This follows directly from the axioms. �
Definition 0.2.17. Let L : K be a field extension and Y ⊂ L. The subfield of Lgenerated by K and Y, denoted K(Y) is the intersection of all the subfields of Lcontaining Y.
This notation does not create inconsistencies, for if K is a field, K(a) = K({a}).
8
Splitting field
Definition 0.2.18. Let K be a field and f (t) ∈ K[t]. We say that f (t) splits overK if there exist k, a1, . . . , an ∈ K, such that
f (t) = k(t − a1) · · · (t − an).
Let K be a field and f (t) ∈ K[t]. The field S is a splitting field of f (t) over Kif S : K is a field extension and f (t) splits over S but over none of its subfields.
The question that arises naturally at this point is to know whether splittingfields exist in general. The following theorem settles this.
Theorem 0.2.19. Let K be a field and f (t) ∈ K[t]. Then there exists a splittingfield S of f (t) over K.
Furthermore, let φ : K → K′ be a field isomorphism, f (t) ∈ K[t]. Let S be asplitting field for f (t) over K and S ′ be a splitting field for φ( f (t)) over K′. Thenthere exists an isomorphism ψ : S → S ′ such that ψ|K = φ. In other words, ψ issuch that
K
φ
��
// S
ψ
��
K′ // S ′
commutes.In particular, all splitting fields are isomorphic.
Sketch of proof. To prove the existence, let us reason by induction on the degree off (t), which we will denote by n. For n = 1, we choose S = K. Let us now supposethat the result holds for polynomials of degree n − 1. If the polynomial f (t) doesnot split over K, it factorises into irreducible monic polynomials and a constant
f (t) = km1(t) · · ·mk(t),
of which at least one, say m1(t), has degree greater than one. Therefore, we canadjoin a root of m1(t), say a1, to K to form K[a1] and write
f (t) = (t − a1)g(t) ∈ K[a1][t],
where g(t) ∈ K[a1][t] has degree n − 1. By our induction assumption, there existsa splitting field S for g(t) over K[a1]. And then S is also a splitting field for f (t)over K.
Proving uniqueness up to isomorphism is somewhat more complicated and isomitted here. For complete proofs, see [3, pp. 112–113] or [10, pp. 88–89]. �
Example 0.2.20. Let us now illustrate with two familiar cases how to constructsplitting fields.
9
1. Let f (t) = (t2 − 3)(t3 + 1) ∈ Q[t]. We see that in, C[t], f (t) factorises as
f (t) = (t +√
3)(t −√
3)(t + 1)(t + (1 + i√
3)/2)(t − (1 + i√
3)/2).
And so, the splitting field of f (t) is Q[√
3, (−1 + i√
3)/2], which is the sameas Q[
√3, i].
2. Proceeding in a similar fashion, we can show that the two irreducible poly-nomials t2 − 3 and t2 − 2t − 2, in Q[t] both have the same splitting field overQ, namely Q[
√3]; thereby proving there is in general no injective mapping
from the set of irreducible polynomials in K[t] to splitting fields of polyno-mials over K. ^
Algebraic closure
Definition 0.2.21. A field K is said to be algebraically closed if every polynomialin K[t] splits over K.
If L : K is an algebraic field extension such that L is algebraically closed,we say that L is an algebraic closure of K.
Theorem 0.2.22 (Steinitz). Let K be a field. Then there exists an algebraicclosure L : K. Furthermore, the algebraic closure is unique up to isomorphism.
Proof. This proof not central to Galois theory and is thus omitted here. See [5,Chapter 5, pp. 231–233]. �
Obviously, the algebraic closure of a field K has the same characteristic as K.However, fields with the same characteristic may have different algebraic closures.Indeed, the algebraic closure of R is C while that of Q, denoted by Q is a propersubfield of C as it does not contain π.
Formal differentiation
Differentiation as it is defined in analysis requires a norm. However, it is convenientto define a similar notion for rings of polynomials in order to detect multiple roots.
Definition 0.2.23. Let f (t) ∈ K[t] be a polynomial, and let us write
f (t) = a0 + a1t + · · · + antn.
Then the formal derivative of f (t) is defined as
D f (t) = f ′(t) = a1 + 2a2t + · · · + nantn−1.
It is easy to check that all the algebraic properties of differentiation known fromanalysis also apply to this formal definition.
Definition 0.2.24. Let f (t) ∈ K[t] and r ∈ K be a root of f (t). The multiplicity nof r is the greatest integer such that (x − r)n divides f (t) in K[t].
10
We see from this definition that a root might have multiplicity 0 in K (i.e. infact, r might not be a root); but once r is a root with positive multiplicity in anextension L of K, then its multiplicity remains unchanged in any extension M of L.
Proposition 0.2.25. Let f (t) ∈ K[t], and r be a root of f (t). Then r is a multipleroot if and only if f (r) = f ′(r) = 0.
Proof. Let n ≥ 1 be the multiplicity of r. We can write
f (t) = (t − r)ng(t) and f ′(t) = n(t − r)n−1g(t) + (t − r)ng′(t),
where g(r) , 0. And so we see that f (r) = f ′(r) = 0 if and only if n > 1. �
Scholium 0.2.26. Let f (t) ∈ K[t], where char K = 0, and let r be a root of f (t).Then r has multiplicity n if and only if f (r) = f ′(r) = · · · = f (n−1)(r) = 0, andf (n)(r) , 0.
Proof. It suffices to iterate the proof of Proposition 0.2.25. �
This last result is false in general if char K , 0. Indeed, in Z/2Z[t], the poly- �
nomial t2 + 1 = (t + 1)2 is such that f ′(t) = f ′′(t) = f ′′′(t) = 0. In particular,f ′′′(1) = 0, but the root 1 has multiplicity 2.
0.3 Finite fields
Definition 0.3.1. A field K with finite cardinality Card K = q is called a finite fieldand is written Fq.
Definition 0.3.2. Let K be a field with characteristic p > 0. Then the mappingx 7→ xp is called the Frobenius homomorphism.
It when K is finite, the Frobenius homomorphism is an automorphism, as iseasily checked. However, the Frobenius homomorphism need not be surjective forinfinite fields; for example in F2(t) no polynomial is mapped to t.
Theorem 0.3.3. Let Fq be a finite field. Then q = pr where p is a prime and r apositive integer. Moreover, Fq is the splitting field of the polynomial tq − t overFp and all finite fields with same order are isomorphic.
Proof. Obviously every finite field has a prime characteristic. If we choose p > 0then Fp = Z/pZ is a finite field of order p and any finite field that has characteristicp is an extension of Fp; and hence has order q = pr for some r.
If q = pr, then using the Frobenius automorphism we can see that the roots oftq − t seen as a polynomial over Fp form a finite field, say S , which is the splittingfield of tq − t over Fp. And using Proposition 0.2.25, we can see that tq − t has nomultiple root in S as D(tq − t) = −1; thereby establishing that S has order q.
Finally, as splitting fields are unique up to isomorphism (see Theorem 0.2.19),so are finite fields. �
11
Chapter 1
Galois correspondence
1.1 Introduction
Let us discuss informally the idea of Galois theory. Given a field K and a poly-nomial f (t) ∈ K[t] we are interested in the group of permutations of the roots off (t) (in its splitting field over K). This group will under certain circumstancescorrespond to the formal definition of the Galois group (see Definition 1.3.1).
The purpose of the theory will then be to construct a bijection between inter-mediate fields of K and the splitting field of f (t) and the subgroups of the Galoisgroup. In fact, we will treat this slightly more generally, as the polynomial inquestion will not be explicit and we will only mention the field extension. Such abijection exists if and only if the extension is finite, ‘normal’ and ‘separable’; suchextensions are also called Galois extensions.
We will then be able to use this correspondence in Chapter 3 to determine nec-essary and sufficient conditions for a polynomial to be soluble by radicals; whichwas the original purpose of Galois.
1.2 Normality and separability
Normal extension
Definition 1.2.1. A field extension L : K is called normal if every irreduciblepolynomial f (t) ∈ K[t] which has at least one root in L splits over L.
Remark 1.2.2. The motivation for the name ‘normal extension’ is that in the Ga-lois correspondence, the intermediate normal extensions correspond to normal sub-groups of the Galois group.
Example 1.2.3. Of course, C : R is a normal extension, for it is in fact much more:the algebraic closure of R.
However, not all algebraic extensions are normal. Let us consider Q[ 3√2] : Q.It is an algebraic extension of degree 3 with basis {1, 3√2, 3√4}. But it is not normal.
12
Indeed, in C, f (t) = t3 − 2 has one root in Q[ 3√2] and two other complex conjugateroots, say a1 and a2, which are not in Q[ 3√2]. In fact, we have a2 ∈ Q[a1], but 3√2 <Q[a1] and Q[a1] � Q[ 3√2] (see Theorem 0.2.11). And we notice that Q[a1] : Q istherefore not a normal extension either. ^
Definition 1.2.1 and Example 1.2.3 illustrate some similarities between the no-tions of splitting field and normal extensions. As we are about to prove, theseconcepts are in fact almost equivalent.
Theorem 1.2.4. A finite field extension L : K is normal if and only if L is thesplitting field over K for some polynomial f (t) ∈ K[t].
Proof. Let us first suppose that L : K is normal and finite. It follows from Theorem0.2.11 that there exist a1, . . . , as ∈ L, algebraic over K, such that L = K[a1, . . . , as].Let us define
f (t) = min(a1,K) · · ·min(as,K) ∈ K[t].
Then as each min(ai,K) has a root in L, our normality assumption implies thateach min(ai,K) splits in L and therefore, so does f (t). Moreover, as the roots off (t) generate L, f (t) will not split over any proper subfield of L, thereby provingthat L is the splitting field of f (t) over K.
Conversely, let us suppose that L is the splitting field over K of a polynomialg(t) ∈ K[t]. As g(t) has finitely many roots, L : K is a finite extension. To provethat L : K also normal, let f (t) ∈ K[t] be an irreducible polynomial having a rootin L; we want to check that f (t) splits over L. If deg f (t) = 1, the result is obvious,so let us suppose that deg f (t) ≥ 2. Let M : L be an extension such that f (t)g(t)splits over M (so that both polynomials split over M), and let r1 and r2 ∈ M betwo roots of f (t). This situation is represented in the following diagram where thearrows point to field extensions.
M
L(r1)
<<yyyyyyyyL(r2)
bbEEEEEEEE
L
bbEEEEEEEEE
<<yyyyyyyyy
K(r1)
OO
K(r2)
OO
K
bbEEEEEEEE
OO
<<yyyyyyyy
It is sufficient to prove that if r1 ∈ L, then L(r1) = L(r2) = L, for the roots wechose are arbitrary. Using the ‘Tower law’, let us observe that for i ∈ {1, 2}, wehave
[L(ri) : L][L : K] = [L(ri) : K] = [L(ri) : K(ri)][K(ri) : K].
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We know from Theorem 0.2.11 that [K(r1) : K] = [K(r2) : K]. Furthermore, usingthe ‘Tower law’, we can check that [L(r1) : L] = [L(r2) : L].
So we conclude that if r1 ∈ L, then [L(r1) : L] = [L(r2) : L] = 1, and soL(r1) = L(r2) = L and r2 ∈ L. Thus f (t) splits over L; and L : K is a normalextension, as desired. �
Example 1.2.5. Any extension L : K of degree 2 is normal. Indeed, in this case,there must exist an element a ∈ L whose minimal polynomial, say m(t) has degree2. So m(t) factorises as (t − a) f (t) over L where f (t) must have degree 1. So m(t)splits over L and the result follows from Theorem 1.2.4. ^
Example 1.2.6. Let us illustrate the use of Theorem 1.2.4 in four cases.
1. The extension Q[ 4√2] : Q is not normal. Indeed, t4 − 2 ∈ Q[t] has twocomplex roots in its splitting field, and therefore does not split over Q[ 4√2].
2. The extension Q[ 3√3,√−3] : Q is normal. Indeed, we have
t3 − 3 =(t −
3√3) (
t2 +3√3t +
3√9)
=(t −
3√3) t − − 3√3 +
√−3 3√3
2
t − − 3√3 −√−3 3√3
2
.Therefore, t3 − 3 splits over Q[ 3√3,
√−3]. Furthermore, the splitting field,
say S , of t3 − 3 over Q must contain 3√3 and√−3. Therefore Q[ 3√3,
√−3]
is included in S , and the result follows from Theorem 1.2.4.
3. The extension F2[α] : F2, where α3 + α + 1 = 0, is normal. Indeed, f (t) =
t3 + t + 1 ∈ F2[t] is irreducible for it has no root and has degree 3. It istherefore the minimal polynomial of α over F2. It follows from Theorem0.2.11 that
F2[α] � F2[t]/( f (t)),
which is a field as well as an F2-vector space of dimension 3 and has there-fore 8 elements. Hence F2[α] � F8, which is the splitting field for somepolynomial over F2 (see Section 0.3).
4. The extension Q[ 7√5,√
5,√
3] : Q is not normal. Indeed, the polynomialt7 − 5 has two complex roots in its splitting field, and therefore does not splitover Q[ 7√5,
√5,√
3]. ^
Separable extensions
Definition 1.2.7. The adjective separable is used in four distinct definitions.
1. An irreducible polynomial f (t) ∈ K[t] is separable over K if it has nomultiple roots in a splitting field over K. In other words, in the splittingfield of f (t), we can write
f (t) = k(t − a1) · · · (t − an),
14
where k ∈ K and all the ai are distinct.
If f (t) ∈ K[t] is not separable, it is said to be inseparable.
2. A polynomial f (t) ∈ K[t] is separable over K if its irreducible factors areall separable.
3. Let L : K be an algebraic field extension. Then a ∈ L is separable if itsminimal polynomial over K is separable.
4. An algebraic field extension L : K is separable if every element a ∈ L isseparable.
We will now provide a criterion allowing one to determine when irreduciblepolynomials are separable.
Lemma 1.2.8. A non-zero polynomial f (t) ∈ K[t] has multiple roots in a splittingfield over K if and only if f (t) and f ′(t) have a common factor g(t) ∈ K[t] ofdegree ≥ 1.
Proof. Let us first suppose that f (t) has a multiple root r in a splitting field S .Then t − r is a common factor of f (t) and f ′(t) in S [t] (see Proposition 0.2.25).And therefore min(r,K) is a common factor of f (t) and f ′(t) in K[t].
Conversely, if f (t) has no multiple root in S [t], then it and f ′(t) are coprime inS [t] (see Proposition 0.2.25), and hence in K[t]. �
Theorem 1.2.9. Let K be a field. If char K = 0, every non-zero irreduciblepolynomial in K[t] is separable. On the other hand, if char K = p > 0, anirreducible polynomial in K[t] is inseparable if and only if it is an element ofK[tp].
Proof. Let f (t) ∈ K[t] be an irreducible polynomial and S be its splitting field overK. Thanks to Lemma 1.2.8, we know that f (t) is inseparable if and only if f (t) andf ′(t) have a common factor of degree ≥ 1.
Let us write
f (t) = a0 + a1t + · · · + antn and f ′(t) = a1 + 2a2t + · · · + nantn−1.
Then notice that as f (t) is irreducible in K[t] and deg f (t) > deg f ′(t), the polyno-mials f (t) and f ′(t) have a common factor if and only if f ′(t) = 0.
If char K = 0, it means f (t) is constant, which is absurd; and so f (t) is separa-ble. If char K = p > 0, it means that for every k ∈ {1, . . . , n} either ak = 0 or p | k;and so f (t) ∈ K[tp]. �
If we consider an algebraic field extension L : K where char L = char K = 0,we see that every element is separable as its minimal polynomial is by definitionirreducible. So the notion of separability is not very insightful in characteristic 0.As we will see in Section 1.5 finite extensions of finite fields are also separable.
Inseparable extensions exist for infinite fields of characteristic p > 0 as isillustrated in the next example.
15
Example 1.2.10. Let us define K = Fp(t) for some prime p, and consider theextension K[a] where a is a root of sp−t ∈ K[s]. Then this extension is inseparable,as follows from Theorem 1.2.9. ^
Remark 1.2.11. Galois did not realise the importance of separability as all thefields with which he worked had characteristic 0, where the notion of separabilityis identical to that of irreducibility.
Lemma 1.2.12. Let K ↪→ M ↪→ L be a tower such that L : K is a separableextension. Then both L : M and M : K are separable extensions.
Proof. The fact that M : K is separable follows directly from the fact that elementsof M can be seen as elements of L, which are all separable.
Now let a ∈ L. Clearly, min(a,M) divides min(a,K) in M[t]. As a is separableover K, min(a,K) is separable over K. This implies min(a,K) cannot have multipleroots as a polynomial of L[t]. Therefore, neither can min(a,M), meaning it isseparable over M. This implies L : M is a separable extension. �
Monomorphisms and automorphisms
Definition 1.2.13. Let L : K and L′ : K be two field extensions. A monomor-phism L ↪→ L′ that fixes K pointwise is called a K-monomorphism. If L = L′
and the K-monomorphism is surjective, we call it a K-automorphism.
Example 1.2.14. The only automorphism of R is the identity. To see that, let usrecall that R is an extension of Q, and every automorphism of an extension of Qfixes Q pointwise.
Furthermore, if φ ∈ Aut(R), it maps non-negative numbers to non-negativenumbers. Indeed, if a ≥ 0, then a = b2 for some b ∈ R, and thus φ(a) = φ(b)2 ≥ 0.It follows that φ preserves the order of R.
Finally, as every real number x may be viewed as the set of rational numbers≤ x (see for example [4, pp. 51–52]), φ must be the identity. ^
Theorem 1.2.15. Let K ↪→ M ↪→ L be a tower such that L : K is a finite normalextension, and let θ : M → L be a K-monomorphism. Then there exists aK-automorphism σ of L such that σ|M = θ.
Proof. By Theorem 1.2.4, L is a splitting field for some polynomial f (t) ∈ K[t],over K. Therefore, it is also a splitting field for f (t) over M and for θ( f (t)) overθ(M). Indeed, θ is a K-monomorphism, so f (t) = θ( f (t)). By Theorem 0.2.19,there exists an automorphism σ : L → L, such that σ|M = θ|M. Furthermore, forevery k ∈ K, we have σ(k) = θ(k) = k, because θ is a K-monomorphism; and so σis a K-automorphism. �
Corollary 1.2.16. Let us suppose L : K is a finite normal extension and r1, r2 ∈
L are roots of f (t) ∈ K[t], an irreducible polynomial. Then there exists a K-automorphism of L such that r1 7→ r2.
16
Proof. There is an isomorphism K[r1]�→ K[r2] that maps r1 to r2 and fixes K.
(See 1.2.15.) �
Normal closure
In this subsection we will see that algebraic extensions can always be extended tonormal extensions; those in which we are particularly interested.
Definition 1.2.17. Let L : K be an algebraic field extension. A normal closureof L : K is an extension N : L such that N : K is normal and N is minimal in thesense of inclusion, i.e. if M ⊂ N and M : K is normal, then M = N.
Theorem 1.2.18. Let L : K be a finite field extension. Then there exists anormal closure N of L : K which is a finite extension. Furthermore, N is uniqueup to K-isomorphism.
Proof. Let {e1, . . . , en} be a K-basis of L. As L : K is algebraic, every ei has aminimal polynomial which we will write mi(t). Let us define f (t) = m1(t) · · ·mn(t)and let N be the splitting field of f (t) over K. Obviously, L ⊂ N, and N : Kis a finite extension. Moreover, Theorem 1.2.4 ensures that N : K is a normalextension. Finally, N is minimal because if L ⊂ P ⊂ N with P normal over K, then{e1, . . . , en} ⊂ P and therefore f (t) must split over P, which implies that P = N.
To prove uniqueness up to isomorphism, let M be another normal closure. Thenf (t) must split over M as well and over none of its subfield. So M is also a split-ting field of f (t) over K and we conclude using uniqueness up to isomorphism ofsplitting fields (see Theorem 0.2.19). �
More intuitively, normal closures are obtained by adding the missing roots ofthe minimal polynomials that were used to create the extension in the first place.
Lemma 1.2.19. Let K ⊂ L ⊂ N ⊂ M be a tower such that L : K is finite and Nis a normal closure of L : K, and let θ : L → M be a K-monomorphism. Thenθ(L) ⊂ N.
Proof. Let us choose a ∈ L, then, θ(a) is also a root of min(a,K) = θ(min(a,K)).And θ(a) ∈ N as N is a normal closure. It follows that θ(L) ⊂ N. �
Informally, this lemma states that normal extensions are stable by K-mono-morphisms because roots cannot ‘escape’.
Theorem 1.2.20. Let L : K be a finite field extension. Then the following areequivalent:
1. L : K is normal;
2. there exists a normal extension N of L : K such that every K-mono-morphism θ : L→ N is a K-automorphism of L;
17
3. for every normal extension M of L : K, every K-monomorphism θ : L →N is a K-automorphism of L.
Proof. (1) =⇒ (3). If L : K is normal, then L is obviously a normal closureof L : K. By Lemma 1.2.19, for every K-monomorphism θ, we have θ(L) ⊂L. Furthermore, as θ is injective and L is a finite dimensional K-vector space,dimK θ(L) = dimK(L); and so θ(L) = L.
(3) =⇒ (2). A normal closure N of L : K exists (see Theorem 1.2.18), and by(3) it has the desired properties.
(2) =⇒ (1). Let f (t) ∈ K[t] be an irreducible polynomial and let r ∈ L be aroot of f (t). By normality, f (t) splits over N; and for any other root r′ ∈ N, thereexists a K-automorphism θ of N such that θ(r) = r′ (see Corollary 1.2.16). But (2)implies that θ is an automorphism of L; and so r′ ∈ L as well. Therefore f (t) splitsover L and L : K is normal. �
Theorem 1.2.21. Let L : K be a finite separable extension of degree n. Thenthere are exactly n distinct K-monomorphisms of L into a normal closure N.
Proof. Let us use induction on n. For n = 1, L = K and the result is clear since theonly K-monomorphism is the embedding of K into N.
Let us now suppose that [L : K] = n and that the result holds for field exten-sions of degree n − 1. Let a ∈ L r K and let us write
k = deg min(a,K) = [K[a] : K] > 1.
As min(a,K) is separable and N is a normal extension, min(a,K) splits over N andhas k distinct roots r1, . . . , rk, say.
Let us write s = [L : K[a]] < n. By our induction assumption, there are exactlys distinct K[a]-monomorphisms ρ1, . . . , ρs : L → N. By Corollary 1.2.16, thereare k distinct K-automorphisms θ1, . . . , θk of N such that θi(a) = ri.
The maps θi ◦ ρ j : L → N therefore define sk = n (see ‘Tower law’) distinctK-monomorphisms. To see that there is no other, suppose that φ : L → N is aK-monomorphism, then φ(a) is a root of min(a,K), and so, φ(a) = ri for somei. Furthermore, it is easy to see that θ−1
i ◦ φ is a K[a]-monomorphism, and so φfactorises into θi ◦ θ
−1i ◦ φ which has the form θi ◦ ρ j. This concludes the proof. �
Corollary 1.2.22. Let L : K be a finite separable normal extension, of degreen. Then there are precisely n distinct K-automorphisms of L.
Theorem 1.2.23. Let K ↪→ L ↪→ M be a tower such that M : K is finite and[L : K] = n. Then there are at most n distinct K-monomorphisms L→ M.
Proof. Let N be a normal closure of M : K. The extension N : K is finite (seeTheorem 1.2.18) and to every K-monomorphism L → M corresponds one K-monomorphism L → N (see Theorem 1.2.15), and vice versa. We may and willtherefore assume that M : L is normal, without loss of generality.
18
Let us now reason by induction on n as in the proof of Theorem 1.2.21 and usethe same notation. As we did not assume separability, the roots of min(a,K) in Nneed not be distinct. So there are at most s distinct K[a]-monomorphisms L → Nand at most k distinct K-automorphisms of N. Hence the result. �
1.3 Field extension degrees and group order
Galois group and fixed fields
If L : K be a field extension. It is easy to check that the set of K-automorphisms isa group under the composition of functions. It is therefore a subgroup of Aut(L),the group of automorphisms of L.
Definition 1.3.1. Let L : K be a field extension. The group of K-automorphismsis called the Galois group of the extension and is written Gal(L : K).
Example 1.3.2. The most simple example comes from the extensionC : R. Indeed,in this case, it is easy to see that Gal(C : R) only has two elements: id and thecomplex conjugation z 7→ z. And so obviously, Gal(C : R) � Z/2Z. ^
Let us consider a field extension L : K, and let H < Gal(L : K). Then bydefinition, any automorphism φ ∈ H fixes K, but it may fix a bigger subset of L. Ifthis is the case, it is easy to check that this subset is a subfield of L as it is closedunder field operations. Using Lemma 0.2.16, we see that the intersection of allthese subsets over φ ∈ H is itself a subfield of L containing K.
Conversely, if we choose an intermediate field M in the extension L : K, it isclear that the set of all elements of Gal(L : K) that fix M pointwise constitute asubgroup of Gal(L : K).
In order to deal with these considerations clearly, we introduce new definitionsand notations.
Definition 1.3.3. Let L : K be a field extension. If H < Gal(L : K), then the fixedfield of H is defined by and written
fi(H) =⋂φ∈H
{fixed field of φ}.
Symmetrically, if M is an intermediate field of the extension L : K, then thefixing Galois subgroup of M is defined by and written
gr(M) = {φ ∈ Gal(L : K) : φ fixes M}.
In sum, given a field extension L : K, we have defined two mappings
fi : G → F and gr : F → G,
where F is the set of intermediate fields of L : K and G the set of subgroups ofGal(L : K).
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Proposition 1.3.4. Let L : K be a field extension, M be an intermediate field,and H be a subgroup of Gal(L : K). Then we have
H ⊂ gr ◦fi(H) and M ⊂ fi ◦ gr(M).
Proof. It follows directly from the definitions. �
Our next goal will be to show that if L : K is finite, separable and normal, thenthe mappings fi and gr are mutual inverses. In order to this, we will need somepreliminary lemmas.
Lemma 1.3.5 (Dedekind). Let K and L be fields. Then every set of distinctmonomorphisms K → L is L-linearly independent.
Proof. Let λ1, . . . , λn : K → L be distinct monomorphisms. Let us assume bycontraposition that there exist a1, . . . , an ∈ L not all 0 such that
∀x ∈ K, a1λ1(x) + · · · + anλn(x) = 0. (1.1)
Of all the equations of this form, there must be one whose number of non-zerocoefficients is minimal. Hence, we may and will assume, without loss of generality,that no coefficient of Equation (1.1) is zero and that there is no such equation thatcontains a null coefficient.
As λ1 , λn, there exists y ∈ K such that λ1(y) , λn(y). Evaluating Equation(1.1) at yx, we have
∀x ∈ K, a1λ1(yx) + · · · + anλn(yx)
= a1λ1(y)λ1(x) + · · · + anλn(y)λn(x)
= 0.
(1.2)
Equation (1.1) multiplied by λ1(y) minus Equation (1.2) then yields
∀x ∈ K, a2(λ1(y) − λ2(y))λ2(x) + · · · + a2(λ1(y) − λn(y))λn(x) = 0;
which has fewer than n terms and has a non-zero coefficient since a2(λ1(y) −λn(y)) , 0. This contradicts our assumption. �
Lemma 1.3.6. Let K be a field. A system of m homogeneous equations in n > munknowns of the form
∀i ∈ {1, . . . ,m}, ai,1x1 + · · · + ai,nxn = 0,
where the ai, j are in K, has a non-trivial solution.
Proof. The matrix corresponding to the system has a non-trivial kernel, by thenullity plus rank theorem (see for example [8, p. 59]). �
20
Lemma 1.3.7. Let G be a finite group with elements g1, . . . , gn. Let x ∈ G. Thenas i varies from 1 to n, the element xgi runs through the group G and eachelement of G occurs exactly once.
Proof. The mapping gi 7→ xgi is bijective. �
Theorem 1.3.8. Let K be a field, and G be a finite subgroup of Aut(K). LetK0 = fi(G). Then
[K : K0] = Card(G).
Proof. Let us introduce the notations
n = Card G, m = [K : K0] and G = {1 = g1, g2, . . . , gn}.
First, let us suppose m < n. Let {x1, . . . , xm} be a K0-basis of K. By Lemma 1.3.6,there exist y1, . . . , yn ∈ K, not all zero such that
∀ j ∈ {1, . . . ,m}, g1(x j)y1 + · · · + gn(x j)yn = 0.
So, for everya = α1x1 + · · · + αmxm ∈ K,
where α1, . . . , αm ∈ K0, we have
g1(a)y1 + · · · + gn(a)yn = g1
m∑`=1
α`x`
y1 + · · · + gn
m∑`=1
α`x`
yn
=
m∑`=1
α`[g1(x`)y1 + · · · + gn(x`)yn]
= 0.
Hence the monomorphisms g1, . . . , gn are linearly dependent, contradictingLemma 1.3.5. Therefore, m ≥ n
Second, let us suppose that m > n. Then there exists a set {x1, . . . , xn+1} ofn + 1 elements of K which are K0-linearly independent. By Lemma 1.3.6, thereexist y1, . . . , yn+1 ∈ K, not all zero such that
∀ j ∈ {1, . . . , n}, g j(x1)y1 + · · · + g j(xn+1)yn+1 = 0.
Without loss of generality, we may and will suppose that the number of non-zeroyi is minimal and that for some r ∈ {0, . . . , n + 1},
y1, . . . , yr , 0 and yr+1 = · · · = yn+1 = 0.
The previous equations may thus be written
∀ j ∈ {1, . . . , n}, g j(x1)y1 + · · · + g j(xr)yr = 0. (1.3)
21
Let g ∈ G, we have
∀ j ∈ {1, . . . , n}, gg j(x1)g(y1) + · · · + gg j(xr)g(yr) = 0
which is, by Lemma 1.3.7, equivalent to
∀ j ∈ {1, . . . , n}, g j(x1)g(y1) + · · · + g j(xr)g(yr) = 0. (1.4)
Equations (1.3) multiplied by g(y1) minus Equations (1.4) yields
∀ j ∈ {1, . . . , n}, g j(x2)(y2g(y1) − g(y2)y1) + · · · + g j(xr)(yrg(y1) − g(yr)y1) = 0.
This is similar to Equations (1.3) but with fewer terms, which is a contradictionunless
∀i ∈ {2, . . . , r}, yig(y1) − y1g(yi) = 0.
If this happens, then
∀g ∈ G, ∀i ∈ {2, . . . , n}, yiy−11 = g(yiy−1
1 )
so that for every i ∈ {2, . . . , r}, yiy−11 ∈ K0. Thus, there exist z1, . . . , zr ∈ K0 not all
zero, and k ∈ K r {0} such that for every i ∈ {2, . . . , r}, yi = kzi. It follows thatwhen j = 1, Equations (1.3) become
x1kz1 + · · · + xrkzr = 0,
which we may divide by k , 0 thereby establishing that {x1, . . . , xr} is K0-linearlydependent; a contradiction. Therefore m ≤ n; and the theorem is proved. �
Corollary 1.3.9. Let L : K be a finite field extension, and let H be a finitesubgroup of Gal(L : K). Then
[fi(H) : K] = [L : K]/Card(H).
Proof. By the theorem, Card(H) = [L : fi(H)]. And so the result follows from the‘Tower law’. �
Example 1.3.10. The most straightforward example is to choose L = C, and G =
{id, z 7→ z}. Then obviously fi(G) = R, and we have [C : R] = Card G = 2. ^
Theorem 1.3.11. Let L : K be a finite separable normal extension of degreen. Then there are precisely n distinct K-automorphisms of L. In other words,Card(Gal(L : K)) = n.
Proof. This is just a restatement of Corollary 1.2.22 using our new terminology ofGalois group. �
Theorem 1.3.12. Let L : K be a finite field extension. If L : K is normal andseparable, then K is the fixed field of Gal(L : K).
22
Proof. Let us write G = Gal(L : K), let K0 be the fixed field of G and let us setn = [L : K]. By Theorem 1.3.11, Card(G) = n; by Theorem 1.3.8, [L : K0] = n.As K ⊂ K0, we have the result. �
Theorem 1.3.13. Let L : K be a finite field extension such that K is the fixedfield of Gal(L : K). Then L : K is normal and separable.
Proof. Let us write G = Gal(L : K). By Theorem 1.3.8, [L : K] = Card G = n,say. If L : K were not separable, there would be strictly fewer than n distinct K-automorphisms of L; hence, as in the proof of Theorem 1.2.23, we conclude bycontraposition that L : K is separable.
Let us now prove normality. Let N be a normal extension of L : K andθ : L → N be a K-monomorphism. As every element of G defines a K-mo-nomorphism L → N, there are n K-monomorphisms L → N which are alsoK-automorphisms of L. By Theorem 1.2.21, θ is one of these monomorphisms.Hence θ is an automorphism of L. It follows from Theorem 1.2.20 that L : K isnormal. �
These results show that if the Galois correspondence is a bijection, then K mustbe the fixed field of Gal(L : K), and so the extension L : K has to be separable andnormal. We will see in the next section that those conditions also suffice.
1.4 The Galois correspondence
The fundamental theorem of Galois theory
This section is devoted to proving the fundamental theorem of Galois theory. Inthat pursuit, we will need the following lemma.
Lemma 1.4.1. Let L : K be a finite, normal and separable field extension, M bean intermediate field, and τ ∈ Gal(L : K). Then
gr(τ(M)) = τ gr(M)τ−1.
Proof. Let us choose y ∈ τ(M) and φ ∈ gr(M). As y = τ(x) for some x ∈ M, wehave
(τφτ−1)(y) = τφ(x) = τ(x) = y,
from which we conclude that
τ gr(M)τ−1 ⊂ gr(τ(M)).
Interchanging gr(M) and gr(τ(M)), we have τ−1 gr(τ(M))τ ⊂ gr(M); and hence
τ gr(M)τ−1 ⊃ gr(τ(M)),
which completes the proof. �
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Theorem 1.4.2 (Fundamental theorem of Galois theory). Let L : K be a finite,normal and separable field extension of degree n with Galois group G. Thenusing the same notation as above, we have the following properties.
1. The Galois group G has order n.
2. The mappings fi and gr are mutual inverses.
3. If M is an intermediate field of L : K, then
[L : M] = Card(gr(M)) and [M : K] = Card(G)/Card(gr(M)).
4. An intermediate field M is a normal extension of K if and only if gr(M) isa normal subgroup of G.
5. If an intermediate field M is a normal extension of K, then the Galoisgroup of M : K is isomorphic to the quotient group G/ gr(M).
Proof.
1. This follows directly from Theorem 1.3.8.
2. Let M ∈ F . Then L : M is clearly normal, and is also separable thanks toLemma 1.2.12. Using Theorem 1.3.12 we immediately have
fi ◦ gr(M) = M. (1.5)
Symmetrically, let H ∈ G. Theorem 1.3.8 yields
Card(H) = [L : fi(H)] and Card(gr ◦fi(H)) = [L : fi ◦ gr ◦fi(H))].
But Equation (1.5) implies fi ◦ gr ◦fi(H) = fi(H) and so we get
Card(H) = Card(gr ◦fi(H)).
And so as H ⊂ gr ◦fi(H) (see Proposition 1.3.4) and H is finite, we have ourresult.
3. Again, L : M is normal and separable, and so [L : M] = Card(gr(M)) byTheorem 1.3.8. The other equation is derived similarly if we also use the‘Tower law’.
4. Suppose M : K is normal and let θ ∈ G. Then θ|M is a K-monomorphismM → L and a K-automorphism of M by Theorem 1.2.20. Therefore θ(M) =
M and by Lemma 1.4.1, gr(M) is stable by conjugation and thus a normalsubgroup of G.
Conversely, let H C G and σ be a K-monomorphism M → L. By Theorem1.2.15, there exists a K-automorphism τ of L such that τ|M = σ. As H isa normal subgroup, Theorem 1.2.15 implies that gr(τ(M)) = gr(M). So thebijection (part 2) indicates that τ(M) = M. As σ is arbitrary, it follows fromTheorem 1.2.20 that M : K is normal.
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5. Let us define Ψ : Gal(L : K)→ Gal(M : K) by
Ψ(τ) = τ|M.
This is a group homomorphism as τ|M is a K-automorphism of M (Theorem1.2.20), and it is surjective (Theorem 1.2.15). Clearly, the kernel of Ψ isgr(M) and so the result follows from the first isomorphism theorem. �
Let us conclude this section by a remark regarding the first informal definitionof the Galois group given at the beginning of the section.
Remark 1.4.3. Let K be a field and f (t) ∈ K[t] be an irreducible polynomial. Letus write S the splitting field of f (t) over K, G = Gal(S ,K), and r1, . . . , rn the rootsof f (t) in its splitting field. Clearly, every g ∈ G permutes the roots r1, . . . , rn.Conversely, as S is generated by the roots of f (t), i.e. S = K[r1, . . . , rn], everyelement s ∈ S may be written as∑
i1,...,in
ki1,...,inri11 · · · r
inn ,
where all the coefficients are in K. As the elements of G are K-automorphisms, wehave
∀g ∈ G, g(s) =∑
i1,...,in
ki1,...,ing(r1)i1 · · · g(rn)in .
Therefore, an element of G is entirely determined by the permutation it induces onr1, . . . , rn. It follows that there exists an injective homomorphism G → S n. Moreprecisely, G may be viewed as a subgroup of S n. Furthermore, this subgroup istransitive by Corollary 1.2.16.
A useful characterisation of Galois extensions
Theorem 1.4.4. Let L : K be a field extension. Then L : K is Galois if and onlyif it is the splitting field over K of a separable polynomial f (t) ∈ K[t].
Proof. Let us first suppose that L : K is Galois. We may thus write L =
K[a1, . . . , an] and we know from the proof of Theorem 1.2.4 that L is the splittingfield over K of
f (t) = min(a1,K) · · ·min(an,K) ∈ K[t].
Furthermore, f (t) is separable, for every min(ai,K) is, as L : K is Galois.Conversely, let us suppose that L is the splitting field over K of some separable
polynomial f (t) ∈ K[t]. Then again, we may write L = K[a1, . . . , ak], where all theai are separable. Clearly, L : K is finite and normal. We know from Theorem 1.2.23that G = Gal(L : K) is finite. Let us write K0 = fi(G). By Theorem 1.3.13, L : K0 isGalois; and from the ‘Fundamental theorem of Galois theory’, [L : K0] = Card G.Hence, to show that L : K is Galois it remains to prove that [L : K] = Card G, forK0 ⊃ K.
25
Let us reason by induction on n = [L : K]. For n = 1, the result is trivial. Let ussuppose the result holds for extensions of degree < n. Let us pick u ∈ L, a root off (t), let us write m(t) = min(u,K) and set s = deg m(t). Clearly, f (t) is a multipleof m(t), and m(t) has s distinct roots in L. Furthermore, [K[u] : K] = s.
Let H = AutK[u](L) and let us define
ψ : G/H → {roots of m(t)} : σH 7→ σ(u).
It is easy to verify that ψ is well-defined; and it is injective, for if σ, τ ∈ G map u tothe same element, then τ−1σ(u) = u, and so τ−1σ ∈ H as it will fix K[u] pointwise;equivalently σH = τH.
Now if v ∈ L is another root of m(t), there is an isomorphism θ : K[u] → K[v]such that θ(u) = v and θ|K = id (see Theorem 0.2.11). As we already know thatL : K is normal, by Theorem 1.2.15, there exists an element g ∈ G such thatg|K[u] = θ. Therefore, every root in of m(t) is the image of some element gH by ψ,and thus ψ is surjective. It follows that [G : H] = s.
Since [L : K(u)] = n/s < n, our induction assumption implies that [L : K(u)] =
Card H. Therefore,
[L : K] = [L : K(u)][K(u) : K] = Card H[G : H] = Card G,
and the proof is complete. �
1.5 Galois theory of finite fields
In this section, we apply the results of Galois theory to finite fields (see Section0.3). To simplify notations, the Frobenius automorphism (see Definition 0.3.2)will be written φ.
Let p be a prime number and q = pr for some integer r. It is easy to see thatfor any integer m ≥ 1, every root of the polynomial tq − t ∈ Fp[t] is a root oftqm− t ∈ Fp[t]. Thus, Fq is isomorphic to a subfield of Fqm .
Proposition 1.5.1. The field extension Fqm : Fq is finite, normal and separable.
Proof. Clearly, Fqm : Fq is finite. It is also normal as a splitting field of the poly-nomial tqm
− t over Fq (see Theorem 1.2.4). Finally, it is separable by Theorem1.2.9. �
Theorem 1.5.2. The Galois group of the field extension Fqm : Fq is cyclic oforder m and generated by φr, i.e. φr(x) = xpr
= xq.
Proof. Clearly φr ∈ Gal(Fqm : Fq) and it fixes exactly Fq which may be viewed asthe set of all the solutions to tq − t = 0. For the same reason (φr)m = idFqm .
Using the fundamental theorem of Galois theory, we get
[Fqm : Fq] = m = Gal(Fqm : Fq).
26
So it suffices to show that 〈φr〉 has order m. To achieve this, let us choose k ∈{1, . . . ,m−1}. If (φr)k = idFqm , we would have xqk
− x = 0 for every x ∈ Fqm , whichis impossible as only qk < qm elements can satisfy this. �
Corollary 1.5.3. The field extension Fqm : Fq contains exactly one intermediatefield isomorphic to Fqd for every divisor d of m.
Proof. This follows directly from the Galois correspondence, as there is a uniquesubgroup of Gal(Fqm : Fq) � Z/mZ of order m/d for every divisor d of m. Thissubgroup is generated by (φr)d.
Finally the fixed field of this subgroup is the set of elements such that xqd−x = 0
and is therefore isomorphic to Fqd . �
Corollary 1.5.4. For every finite cyclic group G, there exists a field extensionwhose Galois group is G.
1.6 Simple extensions
Definition 1.6.1. An algebraic field extension L : K is called simple, if thereexists a ∈ L such that L = K[a]. The element a is called a primitive element ofthe extension.
Theorem 1.6.2 (Primitive element theorem). A field extension L : K is simpleif and only if it has finitely many intermediate fields.
Proof. This proof is omitted here. See [1, pp. 64–65]. �
Example 1.6.3. If L : K is a finite extension of fields with characteristic 0, thenthere are finitely many intermediate fields. Indeed, let us set N to be the normalclosure of L : K. Then N : K is a Galois extension. The ‘Fundamental theoremof Galois theory’ therefore ensures that N : K only has finitely many intermediatefield, and a fortiori, the same is true for L : K. ^
If the assumption of 0 characteristic is not made, the result ceases to be true ingeneral, as shown in Section 2.3.
1.7 Cyclotomic extensions
Let us start with a preliminary result.
Lemma 1.7.1. Let K be a field and K∗ be its multiplicative group. If G is a finitesubgroup of K∗, then it is cyclic.
Proof. Let us set n = Card G. Let m be the least common multiple of {Card〈g〉 :g ∈ G}. Clearly, for every g ∈ G, gm = 1. It follows that tm − 1 has at least n rootsin K and that m ≥ n. But ‘Lagrange’s theorem’ tells us that m | n, and so m = n.
As G is finite and abelian, there is an element of order m, and so G � Z/mZ, asdesired . �
27
Definition 1.7.2. Let m be a natural number and K be a field. The mth cyclo-tomic extension of K is the splitting field S of tm − 1 over K.
Since mtm−1 and tm−1 have no common root, the roots of tm−1 are all distinct(see Proposition 0.2.25); in particular, S : K is finite, normal and separable (seeTheorem 1.4.4). It follows that the roots of tm − 1 in S form a multiplicative groupof order m which we will denote by µm(K).
Therefore, µm(K) is a finite subgroup of K∗, the multiplicative group of theinvertible elements of K; and so µm(K) is cyclic too (see Lemma 1.7.1).
Hence, there exists an element ξ ∈ L such that µm(K) = 〈ξ〉 and it follows fromTheorem 0.2.11 that L = K[ξ].
Definition 1.7.3. An element ξ′ ∈ µm(K) is called a primitive mth root of unity ifµm(K) = 〈ξ′〉.
Each primitive mth root of unity ξ determines an isomorphism
Z/mZ�→ µm(K) : i 7→ ξi.
As ξi is a generator if and only if (m, i) = 1, the primitive roots correspond to theinvertible elements (Z/mZ)∗ of Z/mZ, which we will denote by U(m).
Given a primitive mth root of unity ξ, we can define an injective mapping
θ : Gal(L : K)→ U(m) : g 7→ i,
where g(ξ) = ξi.In fact, θ is a group homomorphism. Indeed, if g, h ∈ Gal(L : K), θ(g) = i and
θ(h) = j, then θ(gh) = i j, for gh(ξ) = ξi j. Via this homomorphism, Gal(L : K) maybe considered as a subgroup of U(m).
Definition 1.7.4. Let K be a field and m be as above. The mth cyclotomicpolynomial over K is
Φm(t) =∏
i∈U(m)
(t − ξi) ∈ K[t],
where ξ is a primitive mth root of unity.
This means the roots of Φ(t) are exactly the primitive mth roots of unity. It doesnot follow from the definition that the coefficients of Φm(t) always are in the fieldK, but this is a consequence of the next proposition.
Proposition 1.7.5. Let K be a field and m be as above. Then
1. we have the equalitytm − 1 =
∏d|m
Φd(t);
2. the the coefficients of Φm(t) are in the prime subfield of K (i.e. the subfieldisomorphic to Fp if char K = p > 0 and the subfield isomorphic to Q ifchar K = 0).
28
Proof.
1. Every mth root of unity is a primitive root of unity for some d | m. Con-versely, every primitive root of unity for some d | m is an mth root of unity.Hence the result.
2. Let us denote by P the prime subfield of K, and let us reason by induction onm. For m = 1, we have Φ1(t) = t − 1 ∈ P[t]. Let us now suppose the resultholds for every kth cyclotomic polynomial, where k < m.
Thus, we may definef (t) =
∏d|md<m
Φd(t) ∈ P[t].
On the one hand, we know from (1) that
tm − 1 = f (t)Φm(t) ∈ P[t].
On the other hand, the euclidean division in P[t] yields
tm − 1 = f (t)q(t) + r(t),
where q(t), r(t) ∈ P[t]. By uniqueness of quotient and remainder, it followsthat Φm(t) = q(t) ∈ P[t], and r(t) = 0. �
Example 1.7.6. If K = Q, then we have
Φ1(t) = t − 1, Φ2(t) = t + 1 and Φ4(t) = t2 + 1. ^
Theorem 1.7.7. Let K be a field and Φm(t) be the mth cyclotomic polynomialover K and L the mth cyclotomic extension. Then, the homomorphism
θ : Gal(L : K)→ U(m) : g 7→ i
defined above is an isomorphism (which means the action of Gal(L : K) onµm(K) is transitive) if and only if Φm(t) is irreducible over K.
Proof. If Φm(t) is irreducible θ is surjective by Corollary 1.2.16.Otherwise, as Φm(t) is monic, it is the product of monic irreducible polynomi-
als that are minimal polynomials for some elements. As all the roots of Φm(t) aredistinct, there exist two roots r1 and r2 that do not have the same minimal polyno-mial. And so, there cannot be an automorphism mapping r1 to r2. �
Lemma 1.7.8. Let ξn be a primitive nth root of unity in C, p - n be a prime andlet us suppose that m(t) = min(ξn,Q) ∈ Z[t]. Suppose ξ is another root primitiveroot of unity such that m(ξ) = 0. Then m(ξp) = 0.
29
Proof. Let us denote by O the ring of algebraic integers ofQ[ξ], and let us considera factorisation in O[t]
tn − 1 = m(t)g(t),
for some polynomial g(t) ∈ Z[t]. Let P be a prime ideal of O containing (p). Letus note that tn − 1 does not have any repeated roots modulo P as ntn−1 and tn − 1are coprime modulo P for p - n (see Proposition 0.2.25). Therefore,
f (ξ) ≡ 0 (mod P) =⇒ g(ξ) . 0 (mod P).
Applying the homomorphism y 7→ yp to both sides, we get g(ξp) . 0 (mod P);where we used that x ≡ xp (mod P). In particular, g(ξp) , 0 in C. However, ξp isa root of tn − 1. It follows that f (ξp) = 0, which completes the proof. �
Theorem 1.7.9. Let n ≥ 2, and Φn(t) be the nth cyclotomic polynomial over Q.Then Φn(t) is irreducible.
Proof. Let us first recall that
Φn(t) =∏
i∈U(m)
(t − ξi) =∏
i∈Z/nZ(i,n)=1
(t − ξi),
where ξ is an nth primitive root. And so the roots of Φn are exactly the primitiventh roots of unity. Furthermore, the splitting field of Φn(t) over Q is Q[ξ].
We know that the coefficients of Φn(t) are in Q. Moreover, by definition ofΦn(t), its coefficients are algebraic integers, and are thus in Z.
Let us define m(t) = min(ξ,Q). Again, as ξ is an algebraic integer, m(t) ∈ Z[t].Every other primitive root can be reached by raising ξ by prime powers pk - n afinite number of times. Therefore, it follows from Lemma 1.7.8 that
∀i ∈ Z/nZ such that (i, n) = 1, m(ξi) = 0,
and so m(t) = Φn(t). In particular, Φn(t) is irreducible. �
Corollary 1.7.10 (Galois group of a cyclotomic extension over Q). Let ξ be anmth primitive root of unity over Q. Then Gal(Q[ξ] : Q) � U(m).
Proof. This follows directly from Theorems 1.7.7 and 1.7.9. �
Proposition 1.7.11. Let L be the mth cyclotomic extension of Fq, where q = pr
and p - m. Then G = Gal(L : Fq) is isomorphic to the cyclic subgroup of U(m)generated by q.
Proof. The Galois group G is generated by the Frobenius automorphism x 7→ xq
(see Theorem 1.5.2). So using the same notation as above, we have
G � θ(G) = 〈q〉,
as desired. �
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Chapter 2
Some examples
In this chapter, we will use the shorter notation G for the Galois group, unlessstated otherwise.
2.1 A Galois correspondence of a splitting field of a poly-nomial
Let f (t) = t4 − 5t2 + 6 ∈ Q[t] and let us denote by S it splitting field over Q. First,we see that in C[t],
f (t) = (t −√
2)(t +√
2)(t −√
3)(t +√
3).
And so the splitting field in question is S = Q[√
2,√
3]. The extension S : Q hasdegree 4, with Q-basis {
1,√
2,√
3,√
6},
since the minimal polynomial of√
3 over Q[√
2] remains t2 − 3. Furthermore,S : Q is Galois, which implies that Card G = 4.
The elements of G may be found among the permutations of {−√
2,√
2,−√
3,√
3},and using Corollary 1.2.16. They are listed in the following table.
Name Image of√
2 Image of√
3id = (1)
√2
√3
σ = (1 2)(3 4) −√
2 −√
3τ = (1 2) −
√2
√3
στ = (3 4)√
2 −√
3
Let us notice that it is sufficient to determine the image of −√
3 and√
3, sinceany Q-automorphism fixes Q pointwise, and
√6 =
√2√
3. Therefore, for φ ∈ G.we have
∀s ∈ S , φ(s) = φ(q1 + q2√
2 + q3√
3 + q4√
6)
= q1 + q2φ(√
2) + q3φ(√
3) + q4φ(√
2)φ(√
3),
31
where the qi’s are selected appropriately.Let us now verify that all the element listed in the table do define Q-automo-
rphisms. To see that τ and στ are, we use Corollary 1.2.16. Indeed, Q[√
2,√
3] :Q[√
3] is a Galois extension, and t2 − 2 is irreducible over Q[√
3]. Therefore,there exists a Q[
√3]-automorphism of Q[
√2,√
3] that maps√
2 to −√
2; thisautomorphism is obviously also a Q-automorphism of Q[
√2,√
3] and is τ. Usingthe same argument, we see that στ is a Q-automorphism of Q[
√2,√
3]. Finally,σ = σττ, and is therefore a Q-automorphism as well.
Clearly,
fi({id}) = Q[√
2,√
3], fi(〈σ〉) = Q[√
6],
fi(〈τ〉) = Q[√
3], fi(〈στ〉) = Q[√
2],
fi(〈σ, τ〉) = Q;
and those are the only subgroups of G. As all subgroups have order 2, they havetwo cosets and are thus normal. It follows that all the intermediate fields definenormal extensions.
2.2 An example of a symmetric group as Galois group
Let E = Q[ 3√2, e2iπ/3], and let us consider the field extension E : Q. As we haveseen in Example 1.2.3, 3√2 has minimal polynomial m(t) = t3 − 2 over Q. Let usnotice that in C[t], we can write
m(t) =(t −
3√2) t − − 3√2 + i
√3 3√2
2
t − − 3√2 − i√
3 3√22
=
(t −
3√2)
(t − α)(t − α),
where α has been introduced to simplify notations.Furthermore, n(t) = t2 + t + 1 is the minimal polynomial of e2iπ/3 over Q[ 3√2].
Indeed, n(e2iπ/3) = 0, and n(t) is irreducible over Q[ 3√2] for it has degree 3 and noreal roots. (It suffices to start with t3 − 1 and to suppress the real root by dividingby t − 1.) Let us notice that in C[t], we can write
n(t) =
t − −1 + i√
32
t − −1 − i√
32
.It follows that E : Q is an extension of degree 6 (see ‘Tower law’), and that{
1,3√2,
3√4, i√
3, i3√2√
3, i3√4√
3},
is a Q-basis of E.Clearly E : Q is finite and separable. To use the Galois correspondence, it
remains to show that it is normal, which follows from the fact that E is the splittingfield of m(t) over Q, as is easily seen.
32
Reasoning as above, the elements of the Galois group may be found among thepermutations of the roots of m(t): { 3√
2, α, α},
and it is sufficient to know the image of 3√2 and i√
3 to determine an element of G.Recalling Remark 1.4.3, we see that the G = Gal(E : Q) is a subgroup of S 3 andhas order 6. It is therefore the whole group S 3.
To determine the elements of G explicitly, it is useful to see that
i√
3 =2α3√2
+ 1 = −2α3√2− 1.
The elements of G are described explicitly in the following table.
Name Image of 3√2 Image of i√
3id = (1) 3√2 i
√3
τ1 = (1 2) α −i√
3τ2 = (1 3) α −i
√3
σ = (2 3) 3√2 −i√
3τ1σ = (1 3 2) α i
√3
τ2σ = (1 2 3) α i√
3
The corresponding fixed fields are
fi({(1)}) = E, fi(〈(1 2)〉) = Q[α],
fi(〈(1 3)〉) = Q[α], fi(〈(2 3)〉) = Q[3√2],
fi(〈(1 2 3)〉) = Q[i√
3].
We see that the only intermediate field which is normal is fi(〈(1 2 3)〉) (see Ex-amples 1.2.3 and 1.2.5); and it corresponds to 〈(1 2 3)〉, the only normal subgroupof S 3, as expected.
2.3 The importance of separability
Let K be a field of characteristic p > 0, and let L = K(v,w), where v and w areindeterminates. Let a be a root of tp − v and b be a root of tp − w, where bothpolynomials are in L[t].
Proposition 2.3.1. The degree [L[a, b] : L] = p2.
Proof. Let us first notice that both tp − v and tp − w are irreducible over L byEisenstein’s criterion.
33
There is no k ≥ 0 such that bk ∈ L[a]. If it were the case, we would be able tofind `0, . . . , `n ∈ L such that
wk = (bk)p
= (`0 + `1a + · · · + `nan)p
= `p0 + `
p1 ap + · · · + `
pn anp
= `p0 + `
p1 v + · · · + `
pn vn,
which would induce an algebraic relation between v and w. This implies that nodivisor (t−b)k = tk + · · ·+ (−b)k of tp−w can be in L[a][t] and therefore that tp−wis irreducible over L[a]. The result follows. �
Proposition 2.3.2. The extension L[a, b] : L has infinitely many intermediatefields.
Proof. Let us choose ` ∈ L. Then L[a + `b] , L[a, b]. Indeed, b < L[a + `b] forif it were, there would be an algebraic relation between v and w (see the proof ofProposition 2.3.1). For the same reason, if ` , `′ ∈ L, then L[a + `b] , L[a + `′b].As Card L = ∞, we have the desired result. �
Remark 2.3.3. It also follows from the ‘Primitive element theorem’ that the ex-tension L[a, b] does not have a primitive element.
This is a counterexample to the Galois correspondence when the extensionswith which we are dealing are not separable.
34
Chapter 3
Solutions by radicals
In this chapter, we examine the use of the correspondence theorem to answer theoriginal question that Galois sought to solve. Namely, we will determine when apolynomial equation admits a solution by radicals, i.e. informally a solution thatcan be expressed in terms of basic algebraic operations and root extractions.
3.1 Radical extensions
Definition 3.1.1. A field extension L : K is called radical if we can writeL = K[a1, . . . , am] and there exist integers n1, . . . , nm such that for everyi ∈ {1, . . . ,m}, ani
i ∈ K[a1, . . . , ai−1]. The numbers a1, . . . , am form a radicalsequence.
Informally, this definition implies that for every i ∈ {1, . . . ,m}, we will be ableto write ai = ni
√α, where α ∈ K[a1, . . . , ai−1].
Definition 3.1.2. Let K be a field, f (t) ∈ K[t] and let S be the splitting field off (t) over K. We say that f (t) is soluble by radicals if S is contained in a radicalextension M : K.
Definition 3.1.3. Let K be a field and f (t) ∈ K[t] with splitting field S over K.The Galois group of f (t) over K is the Galois group of the extension S : K. Wewill denote it by Gal( f (t),K).
Theorem 3.1.4. Let K be a field of characteristic 0 and f (t) ∈ K[t]. Then f (t)is soluble by radicals if and only if the Galois group of f (t) over K is a solublegroup.
Proof. This proof is omitted here. See [10, Chapter 14–15, pp.141–145 and 159–160]. �
35
3.2 Equations soluble by radicals
In this section we will limit ourselves to polynomials over Q. The discussion canbe generalised to any field of characteristic 0. However, the field considered shouldnot be ‘too large’ for the results to be interesting. Indeed, if we choose to studypolynomials over R, all polynomials can be solved by radicals as C : R is a radicalextension and C is algebraically closed.
It is important to emphasise that in our framework, we will be able to use the‘Fundamental theorem of Galois theory’ as the extensions with which we will workwill be normal (as the splitting field of a polynomial) and separable (as an extensionof a field with characteristic 0).
Let us notice that it is enough to study irreducible polynomials. Indeed, it isobvious that f (t) ∈ K[t] is soluble by radicals if an only if all its irreducible factorsare. The remaining difficulty is therefore to determine whether a polynomial isirreducible.
Symmetric polynomials
Definition 3.2.1. Let K be a field and f (t1, . . . , tn) ∈ K[t1, . . . , tn] be a polyno-mial. We say that f (t1, . . . , tn) is symmetric if for every permutation σ ∈ S n
f (tσ(1), . . . , tσ(n)) = f (t1, . . . , tn).
Definition 3.2.2. The rth elementary symmetric polynomial
sr(t1, . . . , tn),
in the indeterminates t1, . . . , tn is the sum of all possible products, taken r at atime, of the elements t1, . . . , tn. Explicitly,
sr(t1, . . . , tn) =∑ ∏
1≤i1,...,ir≤n
tik .
Theorem 3.2.3 (Fundamental theorem on symmetric polynomials). Let K bea field. A symmetric polynomial f (t1, . . . , tn) ∈ K[t1, . . . , tn] can be expressedas a polynomial of smaller or equal degree in the corresponding elementarysymmetric polynomials.
Proof. This proof is omitted here. See [2, §11, pp. 9–12]. �
The general polynomial equation and symmetric groups
In this subsection we prove that some polynomials cannot be solved by radicals.To do this we will introduce a polynomial called the ‘general polynomial’, whichhas coefficients in a transcendental extension of Q.
36
Definition 3.2.4. Let L : K be a field extension and x1, . . . , xn ∈ L. Furthermore,let us define a homomorphism
ψ : K[t1, . . . , tn]→ L : f (t1, . . . , tn) 7→ f (x1, . . . , xn).
The elements x1, . . . , xn are said to be algebraically independent over K if ψ isinjective; in other words, if there exists no non-zero polynomial in K[t1, . . . , tn]whose evaluation at (x1, . . . , xn) is zero.
Let us emphasise that this definition implies, in particular that all the elementsxi are transcendental over Q.
Lemma 3.2.5. For every integer n, there exist n algebraically independent com-plex numbers over Q.
Proof. Let us start by noticing that both Q, and its algebraic closure Q are count-able. So we can pick x1 ∈ C r Q. Similarly, Q[x1] is countable and so we can pickx2 ∈ C r Q[x1].
Iterating this process, we get a sequence with the desired properties. �
Definition 3.2.6. Let x1, . . . , xn ∈ C be algebraically independent algebraic ele-ments over Q, and let us write N = Q[x1, . . . , xn]. The polynomial
g(t) = (t − x1) · · · (t − xn) ∈ N[t],
is called the general polynomial of degree n (without precise reference to thexi). And the general polynomial equation of degree n is defined as
g(t) = a0 + a1t + · · · + an−1tn−1 + tn = 0,
where the ai are defined appropriately and g(t) ∈ K[t] with K = Q[a0, . . . , an].
Lemma 3.2.7. Let K be a field of characteristic 0 and f (t) ∈ K[t]. Let us definen = deg f (t) and denote by S the splitting field of f (t) over K. Then [S : K]divides n!. In particular, [S : K] ≤ n!.
Proof. Let us reason by induction on n. If n = 1, the result is trivial. Let us supposethat the result holds for polynomials of degree n − 1.
If f (t) is irreducible, and r ∈ S is a root of f (t) in S , then, [K[r] : K] = n andby the induction assumption, [S : K[r]] | (n − 1)!, for S is the splitting field off (t)/(t − r) over K[r]; and the result follows from the ‘Tower law’.
If f (t) has an irreducible factor g(t) of degree k < n, let us denote by T itssplitting field over K. Then, by the induction assumption, [T : K] | k! and [S : T ] |(n− k)!, for S is the splitting field of f (t)/g(t) over T . Consequently, by the ‘Towerlaw’, [S : K] | k!(n − k)! and thus [S : K] | n!. �
37
Theorem 3.2.8. Let x1, . . . , xn ∈ C be algebraically independent elements overQ, let N = Q[x1, . . . , xn], and let us write the general polynomial equation ofdegree n as
g(t) = a0 + a1t + · · · + an−1tn−1 + tn ∈ K[t],
with the same notation as above. Then we have Gal(g(t),K) = S n.
Proof. Let σ ∈ S n. As the xi have no algebraic relation, σ defines a unique Q-automorphism φ of N given by
φ(xi) = xσ(i), i ∈ {1, . . . , n},
since every element of N can be seen as a polynomial in Q[t1, . . . , tn] evaluated atx1, . . . , xn (or xσ(1), . . . , xσ(n)). Let us define L = fi(S n), where S n is seen as a groupof Q-automorphisms of N. By Theorem 1.3.8, [N : L] = n!.
Firstly, it is easy to check that for every i ∈ {1, . . . , n − 1}, ai is a symmetricpolynomial of x1, . . . , xn, and so ai ∈ L. This implies that K ⊂ L and thus [N :K] ≥ n!. Secondly, as N is the splitting field of g(t) over K we have [N : K] ≤ n!by Lemma 3.2.7.
Hence [N : K] = n!, K = L, and
Gal(N : L) = Gal(N : K) = Gal(g(t),K) = S n. �
Equation of degree smaller than or equal to 4
The group theory presented in Section 0.1 enables us to show that every polynomialequation of degree ≤ 4 is soluble by radicals.
Theorem 3.2.9. Let K be a field of characteristic 0 and f (t) ∈ K[t] be an irre-ductible polynomial of degree ≤ 4. Then f (t) is soluble by the radicals.
Proof. Clearly, f (t) has four distinct roots in its splitting field as it is irreducible andchar K = 0. As every element of the Galois group of f (t) over K maps roots of f (t)to roots of f (t) it must be a subgroup of S 4, which is soluble. Therefore the Galoisgroup of f (t) over K is soluble (see Theorem 0.1.2) and the result follows. �
An insoluble quintic
In this subsection we exhibit a polynomial of degree 5 which cannot be solved byradicals.
Lemma 3.2.10. Let p be a prime number and f (t) ∈ Q[t] be an irreduciblepolynomial of degree p. Let’s suppose f (t) has precisely two non-real roots inC. Then the Galois group of f (t) over Q is the symmetric group S p.
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Proof. By the ‘Fundamental theorem of algebra’, the splitting field of f (t) over Qis a subfield of C. Let us write G = Gal( f (t),Q) which is a subgroup of S p, anddenote by S the splitting field of f (t) over Q.
Recalling the construction of splitting fields and the ‘Tower law’, we have
[S : Q] = [S : Q[t]/( f (t))][Q[t]/( f (t)) : Q] = [S : Q[t]/( f (t))] · p.
And so by the ‘Fundamental theorem of Galois theory’ and Sylow’s theorems, Ghas an element of order p, i.e. a p-cycle. Furthermore, z 7→ z is also an element ofG, of order 2. These elements generate S p. �
Proposition 3.2.11. The polynomial f (t) = t5 − 6t + 3 ∈ Q[t] is not soluble byradicals.
Proof. By Eisentein’s criterion, f (t) is irreducible over Q. Furthermore, usingbasic analysis, it is possible to prove that f (t) has exactly three reals root, andthanks to Lemma 3.2.10, we conclude that Gal(S f (t) : Q) � S 5 which is not soluble(see Corollary 0.1.5). �
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Chapter 4
Inverse Galois problem
The ‘inverse Galois problem’ poses the following question. Given a finite groupG and a field K, is there a Galois extension L : K whose Galois group is G? (Themost important case is when K = Q.) Only a partial answer to this question isknown and we will present some basic results in this chapter.
Most of the results presented here are taken from [6, pp. 55–57], [9] and [11,Section 8.10].
Let us start with some terminology.
Definition 4.0.12. Let K be a field, and G be a finite group. We say that G isrealisable, or realised, over K if there exists a Galois extension L : K whoseGalois group is G.
4.1 Some results on realisability over the field of rationalnumbers
Finite abelian groups
Theorem 4.1.1. Every cyclic group is realised as a Galois group over Q.
Proof. Let G = Z/nZ. By Dirichlet’s theorem, there exists an infinite number ofprimes p such that p ≡ 1 (mod n). Let us choose one. Let ξ be a primitive pth rootof unity so that Q[ξ] is the pth cyclotomic extension of Q.
Recalling that U(p) = Fp r {0} has cardinality p − 1, Corollary 1.7.10 impliesthat Gal(Q[ξ]/Q) � U(p), which is Z/(p−1)Z (see Lemma 1.7.1). Since n | (p−1)there exists a subgroup H of G which has order (p − 1)/n.
Finally, as Q[ξ] : Q is a Galois extension, the ‘Fundamental theorem of Galoistheory’ and the ‘Tower law’ ensure that the Galois extension fi(H) : Q has degreen. Using the ‘Fundamental theorem of Galois theory’ once more, Gal(fi(H) : Q) isa subgroup of G of order n and is thus isomorphic to Z/nZ. �
Definition 4.1.2. Let L and M be two subfields of a field T . The compositumof L and M, written LM is the intersection of all the subfields of T containing
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L ∪ M. The compositum of any finite number of subfields of a field is definedby induction.
Let us note that if L and M are finite extensions of a field K, then they mayboth be viewed as subfields of K, the algebraic closure of K and therefore, LM isalways well defined. In particular, this is the case if L and M are Galois extensionsof K.
Theorem 4.1.3 (Compositum of two Galois extensions). Let L : K and M : Kbe two Galois extensions of a field K such that L,M are included in an algebraicclosure of K. Then LM is Galois over K and Gal(LM : K) is isomorphic to thesubgroup H of
G = Gal(L : K) × Gal(M : K)
defined byH = {(σ, ψ) ∈ G : σ|L∩M = ψ|L∩M}.
Proof. First, let us prove that LM : K is Galois. As both L : K and M : Kare Galois extensions, by Theorem 1.4.4, there exist separable polynomials `(t),respectively m(t) in K[t] whose splittings field over K are L, respectively M. ThenLM is a splitting field for `(t)m(t), which is a separable polynomial. Thus the resultfollows from Theorem 1.4.4.
It is easy to check that H is a group. Let us define
θ : Gal(LM : K)→ Gal(L : K) × Gal(M : K) : σ 7→ (σ|L, σ|M).
This mapping is a group homomorphism. Its kernel is formed by the elementsthat fix both L and M pointwise. Thus if φ is in ker θ its fixed field contains both Land M and is therefore LM by minimality. Hence θ is injective.
Clearly, the image of θ is included in H. Furthermore, M = K[a1, . . . , a j], forM : K is finite. It is then easy to check that LM = L[a1, . . . , a j], whose elementshave the form `1m1 + · · ·+ `nmn, where `i ∈ L and mi ∈ M, for every i ∈ {1, . . . , n}.Let (σ, ψ) ∈ H, we may define φ ∈ Gal(LM : K) by
φ(`1m1 + · · · + `nmn) = σ(`1)ψ(m1) + · · · + σ(`n)ψ(mn),
which is well-defined as σ|L∩M = ψ|L∩M. Indeed, it is easy to check that φ is aK-automorphism. Finally θ(φ) = (σ, ψ); and so we conclude that the image of θ isexactly H. Therefore, θ is an isomorphism. �
Corollary 4.1.4. Let L and M be two Galois extensions of a field K such thatL ∩ M = K. Then
Gal(LM : K) � Gal(L : K) × Gal(M : K).
We now have the necessary tools to prove the following theorem.
Theorem 4.1.5. Every finite abelian group G is realised over Q.
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Proof. By the ‘Fundamental theorem of finitely generated abelian groups’ thereexist integers d1, . . . , ds such that
G = Z/d1Z × · · · × Z/dsZ.
By Dirichlet’s theorem, there exist distinct primes p1, . . . , ps such that pi ≡ 1(mod di). For i ∈ {1, . . . , s} we set Ei to be the pith cyclotomic extension of Q.Therefore, if i , j, then Ei ∩ E j = Q. It follows from Theorem 4.1.1 and Corollary4.1.4 that
Gal(E1 · · · Es) = Z/d1Z × · · · × Z/dsZ.
Hence the result. �
Symmetric groups
The method presented here is based on [11, Section 8.10]. It gives an algorithm toconstruct explicitly a polynomial in Q[t] whose Galois group over Q is the sym-metric group. Other methods using number theory exist; they are not presentedhere.
Proposition 4.1.6. Let K be a field and a be an algebraic element over K. Then
AutK(K[a]) � AutK(t)(K(t)[a]).
Proof. Let us introduce the notation
G = AutK(K[a]) and Γ = AutK(t)(K(t)[a]).
We define ψ : Γ→ G by
ψ(γ)(k0 + k1a + · · · + knan) = γ(k0 + k1a + · · · + knan),
where ki ∈ K for every i ∈ {0, . . . , n}. Clearly, ψ is a homomorphism, and we provethat it is bijective.
To see that ψ is injective, let us suppose that γ, γ′ ∈ Γ have the same image inG; in particular γ(a) = γ′(a). Then, for every k0(t)+k1(t)a+ · · ·+kn(t)an ∈ K(t)[a],we have
γ(k0(t) + k1(t)a + · · · + kn(t)an) = k0(t) + k1(t)γ(a) + · · · + kn(t)γ(a)n
= k0(t) + k1(t)γ′(a) + · · · + kn(t)γ′(a)n
= γ′(k0(t) + k1(t)a + · · · + kn(t)an),
from which it follows that γ = γ′.To see that ψ is surjective, let us choose g ∈ G, and define
γ(k0(t) + k1(t)a + · · · + kn(t)an) = k0(t) + k1(t)g(a) + · · · + kn(t)g(a)n
for every k0(t) + k1(t)a + · · ·+ kn(t)an ∈ K(t)[a]. It is easy to check that γ ∈ Γ. Andclearly, ψ(γ) = g. This completes the proof. �
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Definition 4.1.7. Let K be a field, f (t) ∈ K[t], and a1, . . . , an be the roots of f (t)in its splitting field over K. Let u1, . . . , un be indeterminates and
θ = u1a1 + · · · + unan.
Let us set u = (u1, . . . , un), and for every σ ∈ S n, the symmetric group of ordern, let us define a left group action ∗ of S n on {a1uρ(1) + · · · anuρ(n) : ρ ∈ S n} by
σ ∗ (a1u1 + · · · + anun) = a1uσ(1) + · · · + anuσ(n).
Then the polynomial
F(t,u) =∏σ∈S n
(t − σ ∗ θ) ∈ K(u)[t]
is called the symmetrised polynomial1 of f (t) over K
The fact that all the coefficients of F(t,u) are in K(u) follows from the fact thatit is a symmetric polynomial of the roots of f (t) (see for example [11, pp. 99–101]).
Theorem 4.1.8. Let K be a field, f (t) ∈ K[t] be a separable polynomial, and letus write the symmetrised polynomial of f (t) over K as a product of its irreduciblefactors over K(u)
F(t,u) = F1(t,u) · · · Fr(t,u).
Then, using the same notation as above, the subgroup of S n generated by thepermutations σ that leave a fixed factor, say F1(t,u), invariant is isomorphic toGal( f (t),K).
Proof. Let us write F1(t,u) as
F1(t,u) = (t − σ1,1 ∗ θ) · · · (t − σ1,k ∗ θ);
and let us introduce the shorter notation
G = Gal(F(t,u),K(u)) � Gal(S ,K(u)).
Let us denote by S the splitting field of F(t,u) over K(u) and by Σ the splittingfield of f (t) over K. As f (t) is separable, so is F1(t,u), and therefore the extensionS : K(u) and Σ : K are Galois by Theorem 1.4.4.
As the numbering of the irreducible factors is arbitrary, we may and will sup-pose, without loss of generality, that t − θ is a factor of F1(t,u) in S [t]. Thereforewe may assume that σ1,1 = id and so
F1(t,u) = (t − θ)(t − σ1,2 ∗ θ) · · · (t − σ1,k ∗ θ).
The strategy of the rest of the proof is as follows:
1This is not standard terminology, but its introduction here is convenient.
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• show that a permutation σ ∈ S n satisfies
(t − σ ∗ θ)(t − σ ∗ σ1,2 ∗ θ) · · · (t − σ ∗ σ1,k ∗ θ)
(i.e. leaves F1(t,u) invariant) if and only if (t − σ ∗ θ) is a linear factor ofF1(t,u) in S [t];
• show that that there exists a bijection between the elements g ∈ G and the setof permutations σ for which (t − σ ∗ θ) is a linear factor of F1(t,u) in S [t].
Firstly, let σ be a permutation that leaves F1(t,u) invariant. It must map t − θto another linear factor of F1(t,u). Conversely, if σ maps t − θ to another linearfactor of F1(t,u), it will map F1(t,u) to another irreducible factor of F(t,u) withwhich it has a common linear factor. As F1(t,u) is irreducible and separable, theonly possible image is F1(t,u), and therefore σ leaves F1(t,u) invariant.
Secondly, it is clear that a given g ∈ G maps t − θ to another linear factorof F1(t,u), as this polynomial is irreducible over K(u). Furthermore, there existsσ−1
g ∈ S n such that g(ai) = aσg(i) for every i ∈ {1, . . . , n}. So σg ∗ θ = g(θ) and σg
maps t − θ to another linear factor of F1(t,u).Conversely, if σ ∈ S n maps t − θ to another linear factor of F1(t,u) by transi-
tivity of the action of the Galois group (see Corollary 1.2.16) there exists gσ ∈ Gsuch that gσ(ai) = aσ−1(i). And so gσ(θ) = σ ∗ θ.
In summary, we have defined a bijection
G → {σ ∈ S n : σ leaves F1(t,u) invariant} by g 7→ σg.
This bijection is an isomorphism since
∀g ∈ G,∀h ∈ G, gh(θ) = σgσh ∗ θ.
It follows that
{σ ∈ S n : σ leaves F1(t,u) invariant} � G � Gal(S ,K(u)),
which was to be proved. �
Theorem 4.1.9. Let R be a unique factorisation domain, P be a prime ideal of R,and R = R/P. Let us write K, respectively K the fraction fields of R, respectivelyR. Finally, let f (t) = a0 + a1t + · · · + antn ∈ R[t] and f be the image of f by theprojection R→ R. Let us suppose that both f (t) are f (t) are separable.
Then Gal( f (t), K) is isomorphic to a subgroup of Gal( f (t),K).
Proof. As above, let us write the symmetrised polynomial of f (t) over K(t) as aproduct of its irreducible factors over K(u)
F(t,u) = F1(t,u) · · · Fr(t,u).
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By Gauss’s lemma, this factorisation takes place within R[u][t], and so, the projec-tion R→ R/P maps this factorisation to
F(t,u) = F1(t,u) · · · Fr(t,u) ∈ R(u)[t],
where the Fi(t,u) need not be irreducible.Let us write F1(t,u) as a product of its irreducible factors
F1(t,u) = G1,1(t,u) · · · G1,k(t,u).
Arguing as in the proof of Theorem 4.1.8, we know that the Galois group off (t) over K is isomorphic to the group of permutations σ that map linear factorsof G1,i(t,u) to linear factors of G1,i(t,u) for every i ∈ {1, . . . , k}. Thus, a fortiori,those permutations map linear factors of F1(t,u) to linear factors of F1(t,u); andtherefore, they also map linear factors of F1(t,u) to linear factors of F1(t,u).
In other words, every permutation σ corresponding to an element of the Galoisgroup of f (t) over K leaves F1(t,u) invariant, and thus corresponds to an elementof the Galois group of f (t) over K too (see Theorem 4.1.8).
It follows that every permutation σ corresponding to an element of Gal( f (t), K)also corresponds to an element of Gal( f (t),K). We conclude that we can embedGal( f (t), K) into Gal( f (t), K) as pictured in the diagram
Gal( f (t), K) � //
∃ embedding��
G < S n
ı
��
Gal( f (t),K) � // G < S n,
where ı denotes the inclusion, and G and G are appropriate subgroups of S n. Thiscompletes the proof. �
These results can be used when computing a Galois group. The most importantcase is when we consider a polynomial f (t) ∈ Z[t] and consider its Galois groupG over Q. Reducing modulo (p) for some prime p, we will be able to determinecyclic subgroups contained in G; namely, the Galois group of f (t) which is alwayscyclic as it is the Galois group of a Galois extension of finite fields (see Theorem1.5.2).
Let us now prove a lemma from group theory.
Lemma 4.1.10. Let G be a transitive subgroup of the symmetric group S n. Letus suppose that G contains a transposition and an (n − 1)-cycle. Then G = S n.
Proof. As the numbering of {1, . . . , n} is arbitrary, we may and will suppose thatthe (n − 1)-cycle is σ = (1 2 · · · n − 1) and will write the transposition (i j), wherei, j ∈ {1, . . . , n}. By transitivity, there exists g ∈ G such that g(i) = n. Therefore,gτg−1 = (k n) where k = g( j).
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It follows that products of the form
(1 2 · · · n − 1)m(k n)(1 2 · · · n − 1)−m, m ∈ {0, . . . , n − 2}
are in G. Those are all the transpositions (1 n), (2 n), . . . , (n − 1 n), which generateS n. �
The previous lemma tells us that in order to have a symmetric group of degreen as a Galois group over Q we need to choose a polynomial f (t) ∈ Q[t] and makesure using Theorems 4.1.8 and 4.1.9 that its Galois group is transitive and has atransposition as well as an (n − 1)-cycle.
We can now prove the following theorem.
Theorem 4.1.11. For every integer n, the group S n is realised as a Galois groupover Q.
Proof. For n ≤ 2, the result is clear; for n = 3 we may refer to Section 2.2. Letus suppose n > 3. Firstly, we choose a polynomial f1(t) ∈ Z[t] of degree n which,modulo 2, yields an irreducible polynomial of degree n. Secondly, we choose apolynomial f2(t) ∈ Z[t] which factorises, modulo 3, as an irreducible factor ofdegree n − 1 and a linear factor. Thirdly, we choose a polynomial f3(t) ∈ Z[t] ofdegree n which factorises, modulo 5, as a quadratic factor and one or two factorsof odd degree.
All of this is possible, for there exist polynomials of any degree which areirreducible modulo any prime number.
Finally, we set f (t) ∈ Z[t] as
f (t) = 15 f1(t) + 10 f2(t) + 6 f3(t).
We notice that f (t) is irreducible as, modulo 2, it is irreducible and of degree n.Therefore, its Galois group G over Q is transitive (see Corollary 1.2.16 and Re-mark 1.4.3). Furthermore, by Theorem 4.1.9, G contains an (n − 1)-cycle, and atransposition multiplied by cycles of odd power, say g (see Section 1.5). Raising gto an appropriate odd power yields a pure transposition. And therefore, the resultfollows from Lemma 4.1.10. �
Soluble groups
Theorem 4.1.12 (Shafarevich’s theorem). Every finite soluble group is realis-able over Q.
Proof. This long proof is omitted here. It is the object of [9]. �
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An alternative proof for symmetric groups of prime order
For symmetric group of prime order, we may use Lemma 3.2.10 to construct amore elementary algorithm to build polynomials whose Galois groups over Q aresymmetric. The method presented here is taken from [6, pp. 55–57].
Theorem 4.1.13. Let p be a prime. Then there exists a polynomial f (t) ∈ Q[t]whose Galois group over Q is S p.
Proof. For p = 2, we can choose f (t) = t2 + 1.Let us now suppose that p ≥ 3. By Lemma 3.2.10, it suffices to construct an
irreducible polynomial of degree p with exactly 2 complex roots. If p = 3, we canchoose f (t) = t3 − 2.
Let us now suppose that p ≥ 5. Let a1 > · · · > ap−2 be p− 2 even integers, andlet us define
g(t) = (t2 + b)p−2∏j=1
(t − a j) ∈ Q[t],
for some positive even integer b and
∀k ∈ {1, . . . , p − 3}, tk =ak + ak+1
2,
which are clearly all integers. It is easy to see that we have the following fork ∈ {1, . . . , p − 3}
t2k + b ≥ 2,
∀ j ∈ {1, . . . , p − 2}, |tk − a j| ≥ 1, and
∃ j ∈ {1, . . . , p − 2}, |tk − a j| > 1.
And therefore, for every k ∈ {1, . . . , p − 3}, |g(tk)| > 2.As the a j and ±i
√b are the only roots of g(t),
∀ j ∈ {1, . . . , p − 3}, x ∈]a j+1, a j[ =⇒ g(x) , 0.
More precisely, g(x) > 0 if j is even, and < 0 if j is odd. It follows that fork ∈ {1, . . . , p − 3}, g(tk) − 2 > 0 if k is even, and < 0 if j is odd. Furthermore, forevery real x sufficiently large, g(x) − 2 > 0, and for every real y ≤ ap−2, we haveg(y) − 2 < 0.
Let us definef (t) = g(t) − 2 ∈ Q[t].
Then f (t) has at least p − 2 real roots r1, . . . , rp−2 such that r1 > t1; for everyj ∈ {1, . . . , p − 4} we have ti > ri+1 > ti+1, and rp−2 < tp−3.
Let us denote by rp−1 and rp the two remaining roots of f (t) in C. Then thecoefficients of degree p − 1 of f (t) and g(t) are equal, which can be written as
p∑j=1
r j =
p−2∑j=1
a j,
47
for√−b is imaginary. This quantity we will denote by s. Similarly, coefficients of
degree p − 2 of f (t) and g(t) are equal, which yields∑1≤ j<k≤p
r jrk = b +∑
1≤ j<k≤p−2
a jak,
which we will denote by b + m. It follows that
p∑j=1
r2j = s2 − 2(b + m).
Therefore, if we choose b large enough to have s2−2(b+m) < 0, which will alwaysbe possible, we see that it will not be possible for all the roots to be real, implyingthat rp−1 and rp will both be complex conjugates.
Finally, if we write
f (t) = tp +
p∑j=1
c jtp− j,
where the c j are even integers defined appropriately, it is easy to verify that
∀i ∈ {1, . . . , p}, 2 | ci and 4 - cp.
It then follows from Eisenstein’s criterion that f (t) is irreducible over Q, and hasexactly two complex roots, as desired. �
4.2 Another result on realisability over arbitrary fields
Theorem 4.2.1. Every finite group G is realised as the Galois group of somefield extension L : K, where K is a finite extension of Q.
Proof. Let n = Card(G). We know from ‘Cayley’s theorem’ that G is isomorphicto some subgroup of S n. Furthermore, thanks to Theorem 4.1.11, there exists aGalois field extension L : Q whose Galois group is isomorphic to S n. Thus thereexists G′ < Gal(L : Q) such G′ � G. By the ‘Fundamental theorem of Galoistheory’
Gal(L : fi(G′)) = G′ � G.
So choosing K = fi(G′), we have the desired result. �
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[1] Emil Artin, Galois Theory, 1998: Dover Books on Mathematics.
[2] Harlod M. Edwards, Galois Theory, 1984: Springer GTM.
[3] Jean-Pierre Escofier, Galois Theory, 1997: Springer GTM.
[4] Roger Godement, Analyse mathematique, 2001: Springer.
[5] Serge Lang, Algebra, 1993: Addison Wesley.
[6] M. Pavaman Murthy, K. G. Ramanathan, C. S. Seshadri, U. Shukla, R. Srid-haran, Galois Theory, 1965: Tata Institute of Fundamental Research, Bom-bay.
[7] Mikhail M. Postnikov, Foundations of Galois theory, 2004: Dover Books onMathematics.
[8] Steven Roman, Advanced Linear Algebra, 2000: Springer GTM.
[9] Alexander Schmidt and Kay Wingberg, Safarevich’s Theorem onSolvable Groups as Galois Groups, http://www.math.uiuc.edu/Algebraic-Number-Theory/0136/.
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