Galois theory. Philippe H. Charmoy supervised by Prof Donna M. Testerman

Transcription

1 Galois theory Philippe H. Charmoy supervised by Prof Donna M. Testerman Autumn semester 2008

2 Contents 0 Preliminaries Soluble groups Field extensions Generalities Construction of finite field extensions Splitting field Algebraic closure Formal differentiation Finite fields Galois correspondence Introduction Normality and separability Normal extension Separable extensions Monomorphisms and automorphisms Normal closure Field extension degrees and group order Galois group and fixed fields The Galois correspondence The fundamental theorem of Galois theory Galois theory of finite fields Simple extensions Cyclotomic extensions Some examples A Galois correspondence of a splitting field of a polynomial An example of a symmetric group as Galois group The importance of separability Solutions by radicals Radical extensions Equations soluble by radicals

3 Symmetric polynomials The general polynomial equation and symmetric groups Equation of degree smaller than or equal to An insoluble quintic Inverse Galois problem Some results on realisability over the field of rational numbers.. 40 Finite abelian groups Symmetric groups Soluble groups An alternative proof for symmetric groups of prime order Another result on realisability over arbitrary fields

4 Preface I am extremely grateful to Prof Donna Testerman, first for accepting to supervise my work, and most importantly for her patience and expertise without which this project could never have been completed. Also, I would like to thank Prof Manuel Ojanguren for providing me with some additional material on which to work and some help when I had difficulties understanding it. Despite their help, some errors doubtless remain; they are of course mine. 3

5 Chapter 0 Preliminaries 0.1 Soluble groups Definition A group G is said to be soluble if there exists a finite sequence of subgroups of G such that {1} = G 0 G 1 G n = G, and G i+1 /G i is abelian for every i {0,..., n 1}. Theorem Let G be a group, H < G and N 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 ]. Theorem 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 G, if we delete repetitions. Therefore, G must be abelian. Since G cannot have any proper non-trivial subgroup it must have a prime order and thus be cyclic. The converse is obvious. Theorem If n 5, then the alternating group A n is simple. Proof. This proof is omitted here. See [10, pp ]. Corollary If n 5, then the symmetric group S n is not soluble. Proof. By contraposition, if S n were soluble, then A n would be as well (Theorem 0.1.2). But this cannot be true since A n is simple (Theorem 0.1.4) and does not have prime order (Theorem 0.1.3). 4

6 Lemma If G is a finite p-group of order p n, then it has a series of normal subgroups {1} = G 0 G 1 G n = G, such that Card G i = p i for every i {0,..., n}. Proof. Let us reason by induction on n. If n = 0, the result is clear. Let us now suppose the result holds for groups of order p n 1. It follows from the class equation that Z(G), the centre of G, is not trivial. As Card(Z(G)) = p m for some m n, the Classification of finitely generated abelian groups implies that there exists an element z Z(G) of order p. Furthermore, the subgroup H = z generated by z is a subgroup of Z(G) and is therefore normal in G. It follows that G/H is a group of order p n 1. By our induction assumption, there exists a series of normal subgroups of G/H {1} = H/H = G 1 /H G n /H = G/H, where for every i {1,..., n} we have Card(G i /H) = p i 1. It follows that for every i {1,..., n}, Card(G i ) = p i and, by the Correspondence theorem, that G i is normal in G. So it suffices to let G 0 = {1} to have the result. Theorem If G is a finite p-group, then it is soluble. Proof. The series of subgroups given by Lemma will a fortiori be of the form {1} = G 0 G 1 G n = G. Furthermore, as G i+1 /G i has prime order for every i {0,..., n 1}, the quotients are cyclic and hence abelian, as desired. 0.2 Field extensions In this section, we give some fundamental definitions about field extensions and present some basic results. Field extensions will provide the underpinnings from which to build Galois theory. Generalities Definition A field extension L of a field K is a non-zero homomorphism φ : K L. The extension is written L : K. Let us notice that such a homomorphism is always injective, for φ 1 (0) is a proper ideal of K and is thus equal to {0}. Therefore, φ embeds K into L and induces an isomorphism between K and φ(k). This definition is a generalisation of the more intuitive idea that L : K is a field extension if K is a subfield of L. However, these situations are identical up to isomorphism. In order to simplify notations we will henceforth assume, without loss of generality, that if L : K is a field extension, then K is a subfield of L. 5

7 Definition Let M K and L M be two field extensions. Then we write L M K and call this a tower of extensions or simply a tower. Definition The degree of an extension L : K, written [L : K], is the dimension of L as a K-vector space. If this dimension is finite, L : K is said to be a finite extension. Example The field extension C : R has degree 2, with basis {1, i}. Proposition (Tower law). If K M L is a tower of finite field extensions, then [L : K] = [L : M][M : K]. Definition Let f (t) = a 0 + a 1 t + + a n t n K[t] be a polynomial. An element k K is called a root of f (t) if the evaluation of f (t) at k, written f (k) = a 0 + a 1 k + + a n k n, is equal to 0. Definition Let L : K be a field extension. An element a L is called algebraic over K if it is a root of a non-zero polynomial f (t) K[t], and transcendental over K otherwise. If every element of L is algebraic over K, then the field extension L : K is itself called algebraic. Finally, let us define the notion of isomorphic field extensions. Definition Let L : K and L : K be two field extensions. They are called isomorphic field extensions if there exists a field isomorphism ψ : L L such that ψ k is also an isomorphism. Remark 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 be isomorphic, in general. Indeed, if we consider the extension Q[r 1, r 2, r 3 ] : Q[r 1 ], where r 1 = 3 2, and r 2, r 3 are the other two complex roots of the polynomial t 2 3 Q[t]. The automorphism of Q[r 1, r 2, r 3 ] defined by r 1 r 2, r 2 r 1 and r 3 r 3 is not an extension isomorphism. (It is not obvious that this defines an automorphism; this is proved in Section 2.2.) Notice though that if K = Q, then we must have ψ(1) = 1, and so Z and Q have to be fixed pointwise. Thus in this case, every isomorphism ψ : L M is also an field extensions isomorphism (This argument works whenever K is a prime field, i.e. F p with p prime or Q.) Construction of finite field extensions In this subsection we describe some procedures through which finite algebraic field extensions can be constructed. For a more thorough coverage, see [7, Chapter 1]. Definition 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 is called the minimal polynomial of a over K and noted min(a, K). 6

8 This definition makes sense, for K[t] is a principal ideal domain and the set of polynomials of which a is a root forms an ideal of K[t]. This ideal is thus generated by some polynomial, say m(t). To see that m(t) is irreducible; let us we write 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 it be monic and thus m(t) = min(a, K). Theorem Let L : K be a field extension and let us suppose a L is algebraic 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) 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,..., a n 1 } is a K-basis of K[a]. First, we check the independence. If λ 0 + λ 1 a + + λ n 1 a n 1 = 0, then λ i = 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], then using 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 by noticing 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 Two algebraic elements a 1 and a 2 over a field K are called conjugate if they have the same minimal polynomial. When a is transcendental, K[a] is much less interesting as the following theorem shows. 7

9 Theorem Let L : K be a field extension and suppose a L is transcendental over K. Then we have the following properties. 1. Evaluation at a induces a ring isomorphism 2. [K(a) : K] is infinite. K[t] K[a]. Proof. The first claim follows from the fact that f (t) f (a) is injective, as a is transcendental. Indeed, by definition of K[a], f (t) f (a) is a surjective homomorphism. The second comes a fortiori because the dimension of K[t] as a K-vector space is infinite and K(a) K[a], which is isomorphic to K[t]. Corollary Every finite field extension is algebraic. Example 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, where the arrows represent inclusions with corresponding extension degrees. E[a] F x=? y=? q E K[a] m r K The Tower law implies that n = qr so that (r, m) = 1, and also that both m and r divide [E[a] : K]. Hence mr divides [E[a] : K]]. Furthermore [E[a] : K] mr, for m(t) is a divisor of min(a, E), as m(a) = 0. Hence, x = r and y = m; in particular, a has degree r over E (see Theorem ). We close this subsection by noticing that given a field extension L : K and a subset 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 Let K be a field and {K i : i I} be a family of subfields of K. Then i I K i is also a subfield of K. Proof. This follows directly from the axioms. Definition Let L : K be a field extension and Y L. The subfield of L generated by K and Y, denoted K(Y) is the intersection of all the subfields of L containing Y. This notation does not create inconsistencies, for if K is a field, K(a) = K({a}). 8

10 Splitting field Definition Let K be a field and f (t) K[t]. We say that f (t) splits over K if there exist k, a 1,..., a n K, such that f (t) = k(t a 1 ) (t a n ). Let K be a field and f (t) K[t]. The field S is a splitting field of f (t) over K if 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 splitting fields exist in general. The following theorem settles this. Theorem Let K be a field and f (t) K[t]. Then there exists a splitting field S of f (t) over K. Furthermore, let φ : K K be a field isomorphism, f (t) K[t]. Let S be a splitting field for f (t) over K and S be a splitting field for φ( f (t)) over K. Then there exists an isomorphism ψ : S S such that ψ K = φ. In other words, ψ is such that K S φ 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 of f (t), which we will denote by n. For n = 1, we choose S = K. Let us now suppose that the result holds for polynomials of degree n 1. If the polynomial f (t) does not split over K, it factorises into irreducible monic polynomials and a constant ψ f (t) = km 1 (t) m k (t), of which at least one, say m 1 (t), has degree greater than one. Therefore, we can adjoin a root of m 1 (t), say a 1, to K to form K[a 1 ] and write f (t) = (t a 1 )g(t) K[a 1 ][t], where g(t) K[a 1 ][t] has degree n 1. By our induction assumption, there exists a splitting field S for g(t) over K[a 1 ]. And then S is also a splitting field for f (t) over K. Proving uniqueness up to isomorphism is somewhat more complicated and is omitted here. For complete proofs, see [3, pp ] or [10, pp ]. Example Let us now illustrate with two familiar cases how to construct splitting fields. 9

11 1. Let f (t) = (t 2 3)(t 3 + 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 same as Q[ 3, i]. 2. Proceeding in a similar fashion, we can show that the two irreducible polynomials t 2 3 and t 2 2t 2, in Q[t] both have the same splitting field over Q, 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 polynomials over K. Algebraic closure Definition A field K is said to be algebraically closed if every polynomial in 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 (Steinitz). Let K be a field. Then there exists an algebraic closure 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 ]. 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 proper subfield of C as it does not contain π. Formal differentiation Differentiation as it is defined in analysis requires a norm. However, it is convenient to define a similar notion for rings of polynomials in order to detect multiple roots. Definition Let f (t) K[t] be a polynomial, and let us write f (t) = a 0 + a 1 t + + a n t n. Then the formal derivative of f (t) is defined as D f (t) = f (t) = a 1 + 2a 2 t + + na n t n 1. It is easy to check that all the algebraic properties of differentiation known from analysis also apply to this formal definition. Definition Let f (t) K[t] and r K be a root of f (t). The multiplicity n of r is the greatest integer such that (x r) n divides f (t) in K[t]. 10

12 We see from this definition that a root might have multiplicity 0 in K (i.e. in fact, r might not be a root); but once r is a root with positive multiplicity in an extension L of K, then its multiplicity remains unchanged in any extension M of L. Proposition Let f (t) K[t], and r be a root of f (t). Then r is a multiple root 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) n g(t) and f (t) = n(t r) n 1 g(t) + (t r) n g (t), where g(r) 0. And so we see that f (r) = f (r) = 0 if and only if n > 1. Scholium 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, and f (n) (r) 0. Proof. It suffices to iterate the proof of Proposition This last result is false in general if char K 0. Indeed, in Z/2Z[t], the polynomial t = (t + 1) 2 is such that f (t) = f (t) = f (t) = 0. In particular, f (1) = 0, but the root 1 has multiplicity Finite fields Definition A field K with finite cardinality Card K = q is called a finite field and is written F q. Definition Let K be a field with characteristic p > 0. Then the mapping x x p is called the Fröbenius homomorphism. It when K is finite, the Fröbenius homomorphism is an automorphism, as is easily checked. However, the Fröbenius homomorphism need not be surjective for infinite fields; for example in F 2 (t) no polynomial is mapped to t. Theorem Let F q be a finite field. Then q = p r where p is a prime and r a positive integer. Moreover, F q is the splitting field of the polynomial t q t over F p and all finite fields with same order are isomorphic. Proof. Obviously every finite field has a prime characteristic. If we choose p > 0 then F p = Z/pZ is a finite field of order p and any finite field that has characteristic p is an extension of F p ; and hence has order q = p r for some r. If q = p r, then using the Fröbenius automorphism we can see that the roots of t q t seen as a polynomial over F p form a finite field, say S, which is the splitting field of t q t over F p. And using Proposition , we can see that t q t has no multiple root in S as D(t q t) = 1; thereby establishing that S has order q. Finally, as splitting fields are unique up to isomorphism (see Theorem ), so are finite fields. 11

13 Chapter 1 Galois correspondence 1.1 Introduction Let us discuss informally the idea of Galois theory. Given a field K and a polynomial f (t) K[t] we are interested in the group of permutations of the roots of f (t) (in its splitting field over K). This group will under certain circumstances correspond 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 intermediate fields of K and the splitting field of f (t) and the subgroups of the Galois group. In fact, we will treat this slightly more generally, as the polynomial in question will not be explicit and we will only mention the field extension. Such a bijection exists if and only if the extension is finite, normal and separable ; such extensions are also called Galois extensions. We will then be able to use this correspondence in Chapter 3 to determine necessary and sufficient conditions for a polynomial to be soluble by radicals; which was the original purpose of Galois. 1.2 Normality and separability Normal extension Definition A field extension L : K is called normal if every irreducible polynomial f (t) K[t] which has at least one root in L splits over L. Remark The motivation for the name normal extension is that in the Galois correspondence, the intermediate normal extensions correspond to normal subgroups of the Galois group. Example 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. 3 It is an algebraic extension of degree 3 with basis {1, 3 2, 4}. But it is not normal. 12

14 Indeed, in C, f (t) = t 3 2 has one root in Q[ 3 2] and two other complex conjugate roots, say a 1 and a 2, which are not in Q[ 3 2]. In fact, we have a 2 Q[a 1 ], but 3 2 Q[a 1 ] and Q[a 1 ] Q[ 3 2] (see Theorem ). And we notice that Q[a 1 ] : Q is therefore not a normal extension either. Definition and Example illustrate some similarities between the notions of splitting field and normal extensions. As we are about to prove, these concepts are in fact almost equivalent. Theorem A finite field extension L : K is normal if and only if L is the splitting 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 Theorem that there exist a 1,..., a s L, algebraic over K, such that L = K[a 1,..., a s ]. Let us define f (t) = min(a 1, K) min(a s, K) K[t]. Then as each min(a i, K) has a root in L, our normality assumption implies that each min(a i, K) splits in L and therefore, so does f (t). Moreover, as the roots of f (t) generate L, f (t) will not split over any proper subfield of L, thereby proving that L is the splitting field of f (t) over K. Conversely, let us suppose that L is the splitting field over K of a polynomial g(t) K[t]. As g(t) has finitely many roots, L : K is a finite extension. To prove that L : K also normal, let f (t) K[t] be an irreducible polynomial having a root in 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 r 1 and r 2 M be two roots of f (t). This situation is represented in the following diagram where the arrows point to field extensions. M L(r 1 ) L(r 2 ) L K(r 1 ) K(r 2 ) K It is sufficient to prove that if r 1 L, then L(r 1 ) = L(r 2 ) = L, for the roots we chose are arbitrary. Using the Tower law, let us observe that for i {1, 2}, we have [L(r i ) : L][L : K] = [L(r i ) : K] = [L(r i ) : K(r i )][K(r i ) : K]. 13

15 We know from Theorem that [K(r 1 ) : K] = [K(r 2 ) : K]. Furthermore, using the Tower law, we can check that [L(r 1 ) : L] = [L(r 2 ) : L]. So we conclude that if r 1 L, then [L(r 1 ) : L] = [L(r 2 ) : L] = 1, and so L(r 1 ) = L(r 2 ) = L and r 2 L. Thus f (t) splits over L; and L : K is a normal extension, as desired. Example 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 degree 2. 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 Example Let us illustrate the use of Theorem in four cases. 1. The extension Q[ 4 2] : Q is not normal. Indeed, t 4 2 Q[t] has two complex 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 t 3 3 = ( t 3 3 ) ( t t ) = ( t 3 3 ) t t Therefore, t 3 3 splits over Q[ 3 3, 3]. Furthermore, the splitting field, say S, of t 3 3 over Q must contain 3 3 and 3. Therefore Q[ 3 3, 3] is included in S, and the result follows from Theorem The extension F 2 [α] : F 2, where α 3 + α + 1 = 0, is normal. Indeed, f (t) = t 3 + t + 1 F 2 [t] is irreducible for it has no root and has degree 3. It is therefore the minimal polynomial of α over F 2. It follows from Theorem that F 2 [α] F 2 [t]/( f (t)), which is a field as well as an F 2 -vector space of dimension 3 and has therefore 8 elements. Hence F 2 [α] F 8, which is the splitting field for some polynomial over F 2 (see Section 0.3). 4. The extension Q[ 7 5, 5, 3] : Q is not normal. Indeed, the polynomial t 7 5 has two complex roots in its splitting field, and therefore does not split over Q[ 7 5, 5, 3]. Separable extensions Definition The adjective separable is used in four distinct definitions. 1. An irreducible polynomial f (t) K[t] is separable over K if it has no multiple roots in a splitting field over K. In other words, in the splitting field of f (t), we can write f (t) = k(t a 1 ) (t a n ),. 14

16 where k K and all the a i 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 are all separable. 3. Let L : K be an algebraic field extension. Then a L is separable if its minimal polynomial over K is separable. 4. An algebraic field extension L : K is separable if every element a L is separable. We will now provide a criterion allowing one to determine when irreducible polynomials are separable. Lemma A non-zero polynomial f (t) K[t] has multiple roots in a splitting field over K if and only if f (t) and f (t) have a common factor g(t) K[t] of degree 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 ). 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 in S [t] (see Proposition ), and hence in K[t]. Theorem Let K be a field. If char K = 0, every non-zero irreducible polynomial in K[t] is separable. On the other hand, if char K = p > 0, an irreducible polynomial in K[t] is inseparable if and only if it is an element of K[t p ]. Proof. Let f (t) K[t] be an irreducible polynomial and S be its splitting field over K. Thanks to Lemma 1.2.8, we know that f (t) is inseparable if and only if f (t) and f (t) have a common factor of degree 1. Let us write f (t) = a 0 + a 1 t + + a n t n and f (t) = a 1 + 2a 2 t + + na n t n 1. Then notice that as f (t) is irreducible in K[t] and deg f (t) > deg f (t), the polynomials 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 separable. If char K = p > 0, it means that for every k {1,..., n} either a k = 0 or p k; and so f (t) K[t p ]. 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 definition irreducible. 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 is illustrated in the next example. 15

17 Example Let us define K = F p (t) for some prime p, and consider the extension K[a] where a is a root of s p t K[s]. Then this extension is inseparable, as follows from Theorem Remark Galois did not realise the importance of separability as all the fields with which he worked had characteristic 0, where the notion of separability is identical to that of irreducibility. Lemma Let K M L be a tower such that L : K is a separable extension. Then both L : M and M : K are separable extensions. Proof. The fact that M : K is separable follows directly from the fact that elements of 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 separable over K, min(a, K) is separable over K. This implies min(a, K) cannot have multiple roots as a polynomial of L[t]. Therefore, neither can min(a, M), meaning it is separable over M. This implies L : M is a separable extension. Monomorphisms and automorphisms Definition Let L : K and L : K be two field extensions. A monomorphism 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 The only automorphism of R is the identity. To see that, let us recall that R is an extension of Q, and every automorphism of an extension of Q fixes Q pointwise. Furthermore, if φ Aut(R), it maps non-negative numbers to non-negative numbers. Indeed, if a 0, then a = b 2 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 ]), φ must be the identity. Theorem Let K M L be a tower such that L : K is a finite normal extension, and let θ : M L be a K-monomorphism. Then there exists a K-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)) over θ(m). Indeed, θ is a K-monomorphism, so f (t) = θ( f (t)). By Theorem , there exists an automorphism σ : L L, such that σ M = θ M. Furthermore, for every k K, we have σ(k) = θ(k) = k, because θ is a K-monomorphism; and so σ is a K-automorphism. Corollary Let us suppose L : K is a finite normal extension and r 1, r 2 L are roots of f (t) K[t], an irreducible polynomial. Then there exists a K- automorphism of L such that r 1 r 2. 16

18 Proof. There is an isomorphism K[r 1 ] (See ) K[r 2 ] that maps r 1 to r 2 and fixes K. Normal closure In this subsection we will see that algebraic extensions can always be extended to normal extensions; those in which we are particularly interested. Definition Let L : K be an algebraic field extension. A normal closure of L : K is an extension N : L such that N : K is normal and N is minimal in the sense of inclusion, i.e. if M N and M : K is normal, then M = N. Theorem Let L : K be a finite field extension. Then there exists a normal closure N of L : K which is a finite extension. Furthermore, N is unique up to K-isomorphism. Proof. Let {e 1,..., e n } be a K-basis of L. As L : K is algebraic, every e i has a minimal polynomial which we will write m i (t). Let us define f (t) = m 1 (t) m n (t) and let N be the splitting field of f (t) over K. Obviously, L N, and N : K is a finite extension. Moreover, Theorem ensures that N : K is a normal extension. Finally, N is minimal because if L P N with P normal over K, then {e 1,..., e n } 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. Then f (t) must split over M as well and over none of its subfield. So M is also a splitting field of f (t) over K and we conclude using uniqueness up to isomorphism of splitting fields (see Theorem ). More intuitively, normal closures are obtained by adding the missing roots of the minimal polynomials that were used to create the extension in the first place. Lemma Let K L N M be a tower such that L : K is finite and N is a normal closure of L : K, and let θ : L M be a K-monomorphism. Then θ(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-monomorphisms because roots cannot escape. Theorem Let L : K be a finite field extension. Then the following are equivalent: 1. L : K is normal; 2. there exists a normal extension N of L : K such that every K-monomorphism θ : L N is a K-automorphism of L; 17

19 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 closure of L : K. By Lemma , for every K-monomorphism θ, we have θ(l) L. Furthermore, as θ is injective and L is a finite dimensional K-vector space, dim K θ(l) = dim K (L); and so θ(l) = L. (3) = (2). A normal closure N of L : K exists (see Theorem ), and by (3) it has the desired properties. (2) = (1). Let f (t) K[t] be an irreducible polynomial and let r L be a root of f (t). By normality, f (t) splits over N; and for any other root r N, there exists a K-automorphism θ of N such that θ(r) = r (see Corollary ). But (2) implies that θ is an automorphism of L; and so r L as well. Therefore f (t) splits over L and L : K is normal. Theorem Let L : K be a finite separable extension of degree n. Then there 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 the only K-monomorphism is the embedding of K into N. Let us now suppose that [L : K] = n and that the result holds for field extensions of degree n 1. Let a L 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 and has k distinct roots r 1,..., r k, say. Let us write s = [L : K[a]] < n. By our induction assumption, there are exactly s distinct K[a]-monomorphisms ρ 1,..., ρ s : L N. By Corollary , there are k distinct K-automorphisms θ 1,..., θ k of N such that θ i (a) = r i. The maps θ i ρ j : L N therefore define sk = n (see Tower law ) distinct K-monomorphisms. To see that there is no other, suppose that φ : L N is a K-monomorphism, then φ(a) is a root of min(a, K), and so, φ(a) = r i for some i. Furthermore, it is easy to see that θi 1 φ is a K[a]-monomorphism, and so φ factorises into θ i θi 1 φ which has the form θ i ρ j. This concludes the proof. Corollary Let L : K be a finite separable normal extension, of degree n. Then there are precisely n distinct K-automorphisms of L. Theorem 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 (see Theorem ) and to every K-monomorphism L M corresponds one K- monomorphism L N (see Theorem ), and vice versa. We may and will therefore assume that M : L is normal, without loss of generality. 18

20 Let us now reason by induction on n as in the proof of Theorem and use the same notation. As we did not assume separability, the roots of min(a, K) in N need not be distinct. So there are at most s distinct K[a]-monomorphisms L N and 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 is a group under the composition of functions. It is therefore a subgroup of Aut(L), the group of automorphisms of L. Definition Let L : K be a field extension. The group of K-automorphisms is called the Galois group of the extension and is written Gal(L : K). Example The most simple example comes from the extension C : R. Indeed, in this case, it is easy to see that Gal(C : R) only has two elements: id and the complex conjugation z z. And so obviously, Gal(C : R) Z/2Z. Let us consider a field extension L : K, and let H < Gal(L : K). Then by definition, any automorphism φ H fixes K, but it may fix a bigger subset of L. If this is the case, it is easy to check that this subset is a subfield of L as it is closed under field operations. Using Lemma , we see that the intersection of all these 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 is clear that the set of all elements of Gal(L : K) that fix M pointwise constitute a subgroup of Gal(L : K). In order to deal with these considerations clearly, we introduce new definitions and notations. Definition Let L : K be a field extension. If H < Gal(L : K), then the fixed field of H is defined by and written fi(h) = {fixed field of φ}. φ H Symmetrically, if M is an intermediate field of the extension L : K, then the fixing 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 of Gal(L : K). 19

21 Proposition 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, then the mappings fi and gr are mutual inverses. In order to this, we will need some preliminary lemmas. Lemma (Dedekind). Let K and L be fields. Then every set of distinct monomorphisms K L is L-linearly independent. Proof. Let λ 1,..., λ n : K L be distinct monomorphisms. Let us assume by contraposition that there exist a 1,..., a n L not all 0 such that x K, a 1 λ 1 (x) + + a n λ n (x) = 0. (1.1) Of all the equations of this form, there must be one whose number of non-zero coefficients 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 that contains a null coefficient. As λ 1 λ n, there exists y K such that λ 1 (y) λ n (y). Evaluating Equation (1.1) at yx, we have x K, a 1 λ 1 (yx) + + a n λ n (yx) = a 1 λ 1 (y)λ 1 (x) + + a n λ n (y)λ n (x) = 0. (1.2) Equation (1.1) multiplied by λ 1 (y) minus Equation (1.2) then yields x K, a 2 (λ 1 (y) λ 2 (y))λ 2 (x) + + a 2 (λ 1 (y) λ n (y))λ n (x) = 0; which has fewer than n terms and has a non-zero coefficient since a 2 (λ 1 (y) λ n (y)) 0. This contradicts our assumption. Lemma Let K be a field. A system of m homogeneous equations in n > m unknowns of the form i {1,..., m}, a i,1 x a i,n x n = 0, where the a i, j are in K, has a non-trivial solution. Proof. The matrix corresponding to the system has a non-trivial kernel, by the nullity plus rank theorem (see for example [8, p. 59]). 20

22 Lemma Let G be a finite group with elements g 1,..., g n. Let x G. Then as i varies from 1 to n, the element xg i runs through the group G and each element of G occurs exactly once. Proof. The mapping g i xg i is bijective. Theorem Let K be a field, and G be a finite subgroup of Aut(K). Let K 0 = fi(g). Then [K : K 0 ] = Card(G). Proof. Let us introduce the notations n = Card G, m = [K : K 0 ] and G = {1 = g 1, g 2,..., g n }. First, let us suppose m < n. Let {x 1,..., x m } be a K 0 -basis of K. By Lemma 1.3.6, there exist y 1,..., y n K, not all zero such that So, for every where α 1,..., α m K 0, we have j {1,..., m}, g 1 (x j )y g n (x j )y n = 0. a = α 1 x α m x m K, m m g 1 (a)y g n (a)y n = g 1 α l x l y g n α l x l y n = l=1 l=1 m α l [g 1 (x l )y g n (x l )y n ] l=1 = 0. Hence the monomorphisms g 1,..., g n are linearly dependent, contradicting Lemma Therefore, m n Second, let us suppose that m > n. Then there exists a set {x 1,..., x n+1 } of n + 1 elements of K which are K 0 -linearly independent. By Lemma 1.3.6, there exist y 1,..., y n+1 K, not all zero such that j {1,..., n}, g j (x 1 )y g j (x n+1 )y n+1 = 0. Without loss of generality, we may and will suppose that the number of non-zero y i is minimal and that for some r {0,..., n + 1}, y 1,..., y r 0 and y r+1 = = y n+1 = 0. The previous equations may thus be written j {1,..., n}, g j (x 1 )y g j (x r )y r = 0. (1.3) 21

23 Let g G, we have j {1,..., n}, gg j (x 1 )g(y 1 ) + + gg j (x r )g(y r ) = 0 which is, by Lemma 1.3.7, equivalent to j {1,..., n}, g j (x 1 )g(y 1 ) + + g j (x r )g(y r ) = 0. (1.4) Equations (1.3) multiplied by g(y 1 ) minus Equations (1.4) yields j {1,..., n}, g j (x 2 )(y 2 g(y 1 ) g(y 2 )y 1 ) + + g j (x r )(y r g(y 1 ) g(y r )y 1 ) = 0. This is similar to Equations (1.3) but with fewer terms, which is a contradiction unless i {2,..., r}, y i g(y 1 ) y 1 g(y i ) = 0. If this happens, then g G, i {2,..., n}, y i y 1 1 = g(y i y 1 1 ) so that for every i {2,..., r}, y i y 1 1 K 0. Thus, there exist z 1,..., z r K 0 not all zero, and k K {0} such that for every i {2,..., r}, y i = kz i. It follows that when j = 1, Equations (1.3) become x 1 kz x r kz r = 0, which we may divide by k 0 thereby establishing that {x 1,..., x r } is K 0 -linearly dependent; a contradiction. Therefore m n; and the theorem is proved. Corollary Let L : K be a finite field extension, and let H be a finite subgroup 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 The most straightforward example is to choose L = C, and G = {id, z z}. Then obviously fi(g) = R, and we have [C : R] = Card G = 2. Theorem Let L : K be a finite separable normal extension of degree n. 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 using our new terminology of Galois group. Theorem Let L : K be a finite field extension. If L : K is normal and separable, then K is the fixed field of Gal(L : K). 22

24 Proof. Let us write G = Gal(L : K), let K 0 be the fixed field of G and let us set n = [L : K]. By Theorem , Card(G) = n; by Theorem 1.3.8, [L : K 0 ] = n. As K K 0, we have the result. Theorem Let L : K be a finite field extension such that K is the fixed field 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 , we conclude by contraposition that L : K is separable. Let us now prove normality. Let N be a normal extension of L : K and θ : L N be a K-monomorphism. As every element of G defines a K-monomorphism L N, there are n K-monomorphisms L N which are also K-automorphisms of L. By Theorem , θ is one of these monomorphisms. Hence θ is an automorphism of L. It follows from Theorem that L : K is normal. These results show that if the Galois correspondence is a bijection, then K must be the fixed field of Gal(L : K), and so the extension L : K has to be separable and normal. 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. In that pursuit, we will need the following lemma. Lemma Let L : K be a finite, normal and separable field extension, M be an 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, we have (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. 23

25 Theorem (Fundamental theorem of Galois theory). Let L : K be a finite, normal and separable field extension of degree n with Galois group G. Then using 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) is a normal subgroup of G. 5. If an intermediate field M is a normal extension of K, then the Galois group of M : K is isomorphic to the quotient group G/ gr(m). Proof. 1. This follows directly from Theorem Let M F. Then L : M is clearly normal, and is also separable thanks to Lemma Using Theorem we immediately have Symmetrically, let H G. Theorem yields fi gr(m) = M. (1.5) 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 our result. 3. Again, L : M is normal and separable, and so [L : M] = Card(gr(M)) by Theorem 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-monomorphism M L and a K-automorphism of M by Theorem Therefore θ(m) = M and by Lemma 1.4.1, gr(m) is stable by conjugation and thus a normal subgroup of G. Conversely, let H G and σ be a K-monomorphism M L. By Theorem , there exists a K-automorphism τ of L such that τ M = σ. As H is a normal subgroup, Theorem implies that gr(τ(m)) = gr(m). So the bijection (part 2) indicates that τ(m) = M. As σ is arbitrary, it follows from Theorem that M : K is normal. 24

26 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 (Theorem ), and it is surjective (Theorem ). Clearly, the kernel of Ψ is gr(m) and so the result follows from the first isomorphism theorem. Let us conclude this section by a remark regarding the first informal definition of the Galois group given at the beginning of the section. Remark Let K be a field and f (t) K[t] be an irreducible polynomial. Let us write S the splitting field of f (t) over K, G = Gal(S, K), and r 1,..., r n the roots of f (t) in its splitting field. Clearly, every g G permutes the roots r 1,..., r n. Conversely, as S is generated by the roots of f (t), i.e. S = K[r 1,..., r n ], every element s S may be written as k i1,...,i n r i1 1 r i n n, i 1,...,i n where all the coefficients are in K. As the elements of G are K-automorphisms, we have g G, g(s) = k i1,...,i n g(r 1 ) i1 g(r n ) i n. i 1,...,i n Therefore, an element of G is entirely determined by the permutation it induces on r 1,..., r n. It follows that there exists an injective homomorphism G S n. More precisely, G may be viewed as a subgroup of S n. Furthermore, this subgroup is transitive by Corollary A useful characterisation of Galois extensions Theorem Let L : K be a field extension. Then L : K is Galois if and only if 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[a 1,..., a n ] and we know from the proof of Theorem that L is the splitting field over K of f (t) = min(a 1, K) min(a n, K) K[t]. Furthermore, f (t) is separable, for every min(a i, 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[a 1,..., a k ], where all the a i are separable. Clearly, L : K is finite and normal. We know from Theorem that G = Gal(L : K) is finite. Let us write K 0 = fi(g). By Theorem , L : K 0 is Galois; and from the Fundamental theorem of Galois theory, [L : K 0 ] = Card G. Hence, to show that L : K is Galois it remains to prove that [L : K] = Card G, for K 0 K. 25

27 Let us reason by induction on n = [L : K]. For n = 1, the result is trivial. Let us suppose the result holds for extensions of degree < n. Let us pick u L, a root of f (t), let us write m(t) = min(u, K) and set s = deg m(t). Clearly, f (t) is a multiple of m(t), and m(t) has s distinct roots in L. Furthermore, [K[u] : K] = s. Let H = Aut K[u] (L) and let us define ψ : G/H {roots of m(t)} : σh σ(u). It is easy to verify that ψ is well-defined; and it is injective, for if σ, τ G map u to the same element, then τ 1 σ(u) = u, and so τ 1 σ 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 ). As we already know that L : K is normal, by Theorem , there exists an element g G such that g 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 Section 0.3). To simplify notations, the Fröbenius automorphism (see Definition 0.3.2) will be written φ. Let p be a prime number and q = p r for some integer r. It is easy to see that for any integer m 1, every root of the polynomial t q t F p [t] is a root of t qm t F p [t]. Thus, F q is isomorphic to a subfield of F q m. Proposition The field extension F q m : F q is finite, normal and separable. Proof. Clearly, F q m : F q is finite. It is also normal as a splitting field of the polynomial t qm t over F q (see Theorem 1.2.4). Finally, it is separable by Theorem Theorem The Galois group of the field extension F q m order m and generated by φ r, i.e. φ r (x) = x pr = x q. : F q is cyclic of Proof. Clearly φ r Gal(F q m : F q ) and it fixes exactly F q which may be viewed as the set of all the solutions to t q t = 0. For the same reason (φ r ) m = id Fq m. Using the fundamental theorem of Galois theory, we get [F q m : F q ] = m = Gal(F q m : F q ). 26

28 So it suffices to show that φ r has order m. To achieve this, let us choose k {1,..., m 1}. If (φ r ) k = id Fq m, we would have x qk x = 0 for every x F q m, which is impossible as only q k < q m elements can satisfy this. Corollary The field extension F q m : F q contains exactly one intermediate field isomorphic to F q d for every divisor d of m. Proof. This follows directly from the Galois correspondence, as there is a unique subgroup of Gal(F q m : F q ) Z/mZ of order m/d for every divisor d of m. This subgroup is generated by (φ r ) d. Finally the fixed field of this subgroup is the set of elements such that x qd x = 0 and is therefore isomorphic to F q d. Corollary For every finite cyclic group G, there exists a field extension whose Galois group is G. 1.6 Simple extensions Definition An algebraic field extension L : K is called simple, if there exists a L such that L = K[a]. The element a is called a primitive element of the extension. Theorem (Primitive element theorem). A field extension L : K is simple if and only if it has finitely many intermediate fields. Proof. This proof is omitted here. See [1, pp ]. Example If L : K is a finite extension of fields with characteristic 0, then there are finitely many intermediate fields. Indeed, let us set N to be the normal closure of L : K. Then N : K is a Galois extension. The Fundamental theorem of Galois theory therefore ensures that N : K only has finitely many intermediate field, 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 in general, as shown in Section Cyclotomic extensions Let us start with a preliminary result. Lemma Let K be a field and K be its multiplicative group. If G is a finite subgroup 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, g m = 1. It follows that t m 1 has at least n roots in 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, as desired. 27