Galois Theory

Galois Theory

Galois Theory Dr P.M.H. Wilson1 Michaelmas Term 2000 1LATEXed by James Lingard — please send all comments and corrections to [email protected] These notes are based on a course of lectures given by Dr Wilson during Michaelmas Term 2000 for Part IIB of the Cambridge University Mathematics Tripos. In general the notes follow Dr Wilson’s lectures very closely, although there are certain changes. In particular, the organisation of Chapter 1 is somewhat different to how this part of the course was lectured, and I have also consistently avoided the use of a lower-case k to refer to a field — in these notes fields are always denoted by upper-case roman letters. These notes have not been checked by Dr Wilson and should not be regarded as official notes for the course. In particular, the responsibility for any errors is mine — please email me at [email protected] with any comments or corrections. James Lingard October 2001 Contents 1 Revision from Groups, Rings and Fields 2 1.1 Field extensions . 2 1.2 Classification of simple algebraic extensions . 3 1.3 Tests for irreducibility . 3 1.4 The degree of an extension . 4 1.5 Splitting fields . 5 2 Separability 7 2.1 Separable polynomials and formal differentiation . 7 2.2 Separable extensions . 8 2.3 The Primitive Element Theorem . 10 2.4 Trace and norm . 11 3 Algebraic Closures 12 3.1 Definitions . 12 3.2 Existence and uniqueness of algebraic closures . 12 4 Normal Extensions and Galois Extensions 16 4.1 Normal extensions . 16 4.2 Normal closures . 17 4.3 Fixed fields and Galois extensions . 19 4.4 The Galois correspondence . 20 4.5 Galois groups of polynomials . 21 5 Galois Theory of Finite Fields 24 5.1 Finite fields . 24 5.2 Galois groups of finite extensions of finite fields . 24 6 Cyclotomic Extensions 27 7 Kummer Theory and Solving by Radicals 30 7.1 Introduction . 30 7.2 Cubics . 32 7.3 Quartics . 33 7.4 Insolubility of the general quintic by radicals . 34 1 1 Revision from Groups, Rings and Fields 1.1 Field extensions Suppose K and L are fields. Recall that a non-zero ring homomorphism θ : K ! L is necessarily injective (since ker θ ¢ K and so ker θ = f0g) and satisfies θ(a=b) = θ(a)/θ(b). Therefore θ is a homomorphism of fields. Definition A field extension of K is given by a field L and a non-zero homomorphism θ : K,! L. Such a θ will also be called an embedding of K into L. Remark In fact, we often identify K with its image θ(K) ⊆ L, since θ : K ! θ(K) is an isomor- phism, and denote the extension by L=K or K,! L. Lemma 1.1 T If fKigi2I is any collection of subfields of a field L, then i2I Ki is also a subfield of L. Proof Easy exercise from the axioms. Definition Given a field extension L=K and an arbitrary subset S ⊆ L, the subfield of L generated by K and S is \ K(S) = fsubfields M ⊆ L j M ¶ K; M ¶ Sg: The lemma above implies that it is a subfield — it is the smallest subfield containing K and S. Notation If S = f®1; : : : ; ®ng we write K(®1; : : : ; ®n) for K(S). Definition A field extension L=K is finitely generated if for some n there exist ®1; : : : ; ®n 2 L such that L = K(®1; : : : ; ®n). If L = K(®) for some ® 2 L, the extension is simple. Definition Given a field extension L=K, an element ® 2 L is algebraic over K if there exists a non-zero polynomial f 2 K[X] such that f(®) = 0 in L. Otherwise, ® is transcendental over K. If ® is algebraic, the monic polynomial n n¡1 f = X + an¡1X + ¢ ¢ ¢ + a1X + a0 of smallest degree such that f(®) = 0 is called the minimal polynomial of f. Clearly such an f is unique and irreducible. 2 Definition A field extension L=K is algebraic if every ® 2 L is algebraic over K. It is pure transcen- dental if every ® 2 L n K is transcendental over K. 1.2 Classification of simple algebraic extensions Given a field K and an irreducible polynomial f 2 K[X], recall that the quotient ring K[X]=(f) is a field. Therefore we have a simple algebraic field extension K,! K(®) = K[X]=(f), ® denoting the image of X under the quotient map. Also, for any simple algebraic field extension K,! K(®) let f be the minimal polynomial of ® over K. We then have a commutative diagram / K D K[X] DD DD DD D! ² K(®) inducing an isomorphism of fields K[X]=(f) =» K(®). Thus up to field isomorphisms, any simple algebraic extension of K is of the form K,! K[X]=(f) for some irreducible f 2 K[X]. Therefore, classifying simple algebraic extensions of K (up to isomorphism) is equivalent to classifying irreducible monic polynomials in K[X]. 1.3 Tests for irreducibility Let R be a UFD and K its field of fractions, e.g. R = Z, K = Q. Lemma 1.2 (Gauss’ Lemma) A polynomial f 2 R[X] is irreducible in R[X] iff it is irreducible in K[X]. Theorem 1.3 (Eisenstein’s Criterion) Suppose n n¡1 f = anX + an¡1X + ¢ ¢ ¢ + a1X + a0 2 R[X] 2 and there exists an irreducible p 2 R such that p - an, p j ai for i = n¡1;:::; 0 and p - a0. Then f is irreducible in R[X] and hence irreducible in K[X]. Proofs See ‘Groups, Rings and Fields’. 3 1.4 The degree of an extension Definition If L=K is a field extension, then L has the structure of a vector space over K. The dimension of the vector space is called the degree of the extension, written [L : K]. We say that L is finite over K if [L : K] is finite. Theorem 1.4 Given a field extension L=K and an element ® 2 L, ® is algebraic over K iff K(®)=K is finite. When ® is algebraic, [K(®): K] is the degree of the minimal polynomial of ®. Proof (() If [K(®): K] = n, then 1; ®; : : : ; ®n are linearly dependent over K, so there exists a polynomial f 2 K[X] with f(®) = 0, as claimed. ()) If ® is algebraic over K with minimal polynomial f, then n n¡1 f(®) = ® + an¡1® + ¢ ¢ ¢ + a1® + a0 = 0 (¤) in L. Suppose g 2 K[X] with g(®) 6= 0. Since f is irreducible we have hcf(f; g) = 1. Euclid’s algorithm implies that there exist x; y 2 K[X] such that xf + yg = 1 and so y(®)g(®) = 1 in L (since f(®) = 0). So g(®)¡1 2 h1; ®; ®2;:::i, the subspace of L generated by powers of ®. Now K(®) consists of all elements of the form h(®)=g(®) for h; g 2 K[X] polynomials, g(®) 6= 0, and so K(®) is spanned as a K-vector space by 1; ®; ®2;::: and hence from relation (¤) by 1; ®; : : : ; ®n¡1. Minimality of n implies that the spanning set 1; ®; : : : ; ®n¡1 is a basis and hence [K(®): K] = n. Proposition 1.5 (Tower Law) Given a tower of field extensions K,! L,! M, [M : K] = [M : L][L : K]: Proof Let (ui)i2I , be a basis for M over L and let (vj)j2J , be a basis for be a basis for L over K. We shall show that (uivj)i2I;j2J is a basis for M over K, from which the result follows. First we show that the uivj span M over K. Now any vector x 2 M may be written as a linear combination of the ui, that is X x = ¹iui i2I for some ¹i 2 L. But since the vj span L over K we can write each ¹i as a linear combination of the vj, that is X ¹i = ¸ijvj j2J 4 for some ¸ij 2 K. But then X x = ¸ijuivj i2I j2J as required. Now we shall show that the uivj are linearly independent over K. Suppose that we have X ¸ijuivj = 0 i2I j2J for some ¸ij 2 L. But then 0 1 X X @ ¸ijvjA ui = 0 i2I j2J and then since the ui are linearly independent over L we must have X ¸ijvj = 0 j2J for each j 2 J. But then since the vj are linearly independent over K we must have that ¸ij = 0 for each i 2 I; j 2 J, as required. Corollary 1.6 If L=K is finitely generated, L = K(®1; : : : ; ®n), with each ®i algebraic over K, then L=K is a finite extension. Proof Each ®i is algebraic over K(®1; : : : ; ®i¡1) and so by (1.4) we have that for each i, [K(®1; : : : ; ®i): K(®1; : : : ; ®i¡1)] is finite. Induction and the Tower Law give the required result. 1.5 Splitting fields Recall that if L=K is a field extension and f 2 K[X] we say that f splits (completely) over L if it may be written as a product of linear factors f = k(X ¡ ®1) ¢ ¢ ¢ (X ¡ ®n); where k 2 K and ®i 2 L. L is called a splitting field for f if f fails to split over any proper subfield of L, that is, if L = K(®1; : : : ; ®n). Remark Splitting fields always exist. For if g is any irreducible factor of f, then K[X]=(g) = K(®) is an extension of K for which g(®) = 0, where ® denotes the image of X.

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