# Chapter 4. Fields and Galois Theory

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1 Chapter 4 Fields and Galois Theory 63

2 64 CHAPTER 4. FIELDS AND GALOIS THEORY 4.1 Field Extensions K[u] and K(u) Def. A field F is an extension field of a field K if F K. Obviously, F K = 1 F = 1 K. Def. When F K, let [F : K] := dim K F denote the dim of F over K. Def. Let K be a field. K[x] := the polynomial ring of K, { } f K(x) := f, g K[x], g 0 g = the rational function field of K = the quotient field of K[x]. If F K and u F, denote K[u] := the subring of F generated by K and u = {f(u) f K[x]}, K(u) := the subfield of F generated by K and u Def. Suppose F K and u F. = {f(u)/g(u) f, g K[x], g(u) 0}. u is called algebraic over K if g(u) = 0 for some nonzero polyn g K[x]; otherwise, u is called transcendental over K. Every u F induces a ring homom φ u : K[x] F, φ u (f) := f(u). Since K[x] is a PID, Ker φ u = {f K[x] f(u) = 0} = (p u ) for a monic polyn p u K[x]. Thm 4.1. Suppose F K. 1. if u F is algebraic over K, then

3 4.1. FIELD EXTENSIONS 65 (a) p u (x) is irreducible in K[x], called the irreducible polynomial of u over K, denoted by irr(u, K) = p u. For f K[x], f(u) = 0 iff p u f, (b) The ring K[u] = K(u) K[x]/(p u ) is a field, and [K[u] : K] = [K(u) : K] = deg p u =: deg K (u) (the degree of u over K.) Ex. Indeed, K[u] = K(u) has a K-basis {1, u, u 2,, u n 1 } for n = deg p u. 2. if u F is transcendental over K, then (a) p u = 0, (b) K[u] K[x] and K(u) K(x), both have infinite dim over K. 1. R Q, u = R is algebraic over Q. Then u 3 = 3 2 = (u 3) 3 = 2 = u 3 + 9u 8 = 3 3(u 2 + 1) = (u 3 + 9u 8) 2 = 27(u 2 + 1) 2 = u 6 9u 4 16u u 2 144u + 37 = 0. Then p(x) = x 6 9x 4 16x x 2 144x + 37 is the irred polyn of u = in Q. Any f Q[x] satisfies f(u) = 0 iff p f. Q[u] = Q(u), [Q(u) : Q] = 6, and {1, u,, u 5 } is a basis of Q(u) in Q. 2. R Q, π R is transcendental over Q. Then Q[π] Q[x] and Q(π) Q(x), both have infinite dim Field Extensions Def. F is a finite extension of K if [F : K] <, F is an infinite extension of K if [F : K] is infinite. Thm 4.2. (proved) If F E K, then [F : K] = [F : E][E : K]. Moreover, [F : K] is finite iff [F : E] and [E : K] are finite.

4 66 CHAPTER 4. FIELDS AND GALOIS THEORY It implies the following theorem: Thm 4.3. F K, u F. The foll are equiv: 1. u is algebraic over K, 2. K(u) is a finite extension of K, 3. every v K(u) is algebraic over K, and deg K (v) deg K (u). Def. F K and X F. Let K[X] (resp. K(X)) denote the subring (resp. the subfield) of F generated by K X. Def. F K is a simple extension of K if F = K(u) for some u F ; finitely generated extension of K if F = K(u 1,, u n ) for some u 1,, u n F. Ex. Every fin ext is a fin gen ext. The converse is false. e.g. K(x) is a fin gen ext of K but not a fin ext of K. Def. F K is an algebraic extension if every element of F is algebraic over K. Thm 4.4. F K is a finite extension iff F = K[u 1,, u n ] where each u i is algebraic over K. In particular, finite extensions are algebraic extensions. Thm 4.5. F E K. Then F is alg ext of K iff F is alg ext of E and E is alg ext of K. Ex. Q( 2) is algebraic extension over Q, and Q( 2, 5 3) is an algebraic extension over Q( 2). Then Q( 2, 5 3) is an algebraic extension over Q. For example, both and are algebraic numbers over Q. Thm 4.6. F K. The set of all elements of F that are algebraic over K forms an intermediate field K between F and K (F K K), called the algebraic closure of K in F. Moreover, every element of F K is transcendental over K. Remark. 1. Given a field K and an irreducible monic polynomial p(x) K[x], we can always construct an algebraic extension F K such that the irred polyn of certain u F in K is p(x):

5 4.1. FIELD EXTENSIONS 67 (a) The quotient ring F := K[x]/(p(x)) is a field since p(x) is irreducible. (b) Let ι : K K[x] be the canonical inclusion, and π : K[x] K[x]/(p(x)) the canonical projection. Then πι(k) K and F πι(k). (c) The element u := π(x) F has irred polyn p(x) in πι(k) K. 2. Any field K can be extended to an algebraic closure field K that contains the roots of all irreducible polynomials of K[x], using Zorn s Lemma and the above remark. Any two algebraic closures of K are K-isomorphic (Hungerford, Thm V.3.6).

6 68 CHAPTER 4. FIELDS AND GALOIS THEORY 4.2 Galois Theory We will focus on Galois theory for finite extensions (i.e., fin gen alg exts) K-automorphism Thm 4.7. F K and u, v F. Then φ u,v : K(u) K(v) def by φ u,v K = id K and φ u,v (u) = v is a field isomorphism iff one the followings holds: 1. Both u and v are algebraic and irr(u, K) = irr(v, K). u and v are said to be conjugate over K. 2. Both u and v are transcendental over K. Def. Let E K and F K. A map σ : E F is a K-isomorphism if σ is both a field isomorphism and a K-mod isomorphism. If E = F, then σ is a K-automorphism. All K-automorphisms of F form a group G(F/K) = Aut K F, called the Galois group of F over K. Remark. 1. σ : E F is a K-isomorphism iff σ is a field isomorphism that acts as identity map on K. { Q, if char F = 0, 2. Let B := be the base field of F. Then a chain of Z p, if char F = p, fields F F 1 F 2 B induces a chain of automorphism groups {1} = G(F/F ) G(F/F 1 ) G(F/F 2 ) G(F/B) = Aut (F ). Thm 4.8. Let F K be an algebraic extension, and σ G(F/K). Then irr(u, K) = irr(σ(u), K) for every u F. Proof. A special case of Thm 4.7. Remark. This important theorem can be used to determined all elements of G(F/K) when F = K(u 1,, u n ), in particular when [F : K] <.

7 4.2. GALOIS THEORY 69 Ex. Consider K = Q and F = Q( 2, 3). Clearly 3 / Q( 2) and so [Q( 2, 3) : Q] = 4. Then Q( 2, 3) = 1Q + 2Q + 3Q + 6Q. Then G(F/K) consists of 4 elements: (classified by their actions on generators 2 and 3) G(F/K) 1 σ τ στ = τσ image of image of For example, σ(a + b 2 + c 3 + d 6) = a + b 2 c 3 d 6. Remark. G(F/K) stabilizes the algebraic closure of K in F. Ex. Let F = Q( 2, 3, x) Q = K, where x is transcendental over Q. The algebraic closure of K in F is K = Q( 2, 3). Suppose σ G(F/K). Then σ(q( 2, 3)) = Q( 2, 3). So σ sends 2 to ± 2, sends 3 to ± 3, and sends x to ax + b cx + d for a, b, c, d Q( 2, 3) and ad bc 0 (since from u = σ(x) we should get a rational function expression of x). The group (Z 2 Z 2 ) G(F/K) P GL(2, K) Splitting Field Def. A polyn f F [x] is split over F (or to split in F [x]) if f can be written as a product of degree one polyns in F [x]. Ex. F = Q( 2, 3 3). Then x 2 2 is split over F, but x 2 3 is not split over F. Def. Suppose K F K. Then 1. Let {f i i I} be a set of polyns in K[x]. F is the splitting field over K of {f i i I} if F is generated over K by the roots of all f i. 2. F is a splitting field over K if F is the splitting field of some set of polynomials in K[x]. Ex. Let α 1,, α n be the roots of f K[x] in K. The splitting field over K of f is F := K(α 1,, α n ). Note that [F : K] deg f(x). Ex. K is a splitting field over K of K[x]. Every f K[x] is split over K.

8 70 CHAPTER 4. FIELDS AND GALOIS THEORY Ex. E := Q( 2, 3) splitting over Q of {x 2 2, x 2 3}. The polynomials x 2 2, x 2 3, x 4 5x 2 + 6, are split over E. Ex. What is the splitting field F over Q of x 3 2? What are the elements of G(F/Q)? Thm 4.9. Any two splitting fields of S K[x] over K are K-isomorphic. (c.f. Any two algebraic closures of K are K-isomorphic.) Thm F K, [F : K] <. Let σ : K K 1 be a field isomorphism, and K 1 an algebraic closure of K 1. The number of extensions of σ to a field isomorphism τ of F onto a subfield of K 1 is finite, and is completely determined by F and K (so the number is not relative to K 1, K 1, and σ.) Proof. It suffices to prove for simple extension F = K(u) and then apply induction. Suppose irr(u, K) = p(x) = c 0 + c 1 x + + c n x n K[x]. Any extension of σ to τ : F F 1 with F 1 K 1 is uniquely determined by τ(u), which is a root of irr(τ(u), K 1 ) = p σ (x) = σ(c 0 ) + σ(c 1 )x + + σ(c n )x n K 1 [x]. Therefore, the number of extensions of σ to an isomorphism of F onto a subfield of K 1 equals to the number of distinct roots of p σ (x), or the number of distinct roots of p(x), which is completely determined by F and K. Def. Let K F K with [F : K] <. The number of K-isomorphisms of F onto a subfield of K is the index of F over K, denoted by {F : K}. Remark. 1. {F : K} is called the separable degree [F : K] s of F over K in [Hungerford, Def. V.6.10]. 2. G(F/K) {F : K}. In fact, G(F/K) divides {F : K}. 3. {F : K} [F : K]. In fact, {F : K} divides [F : K]. Thm If F L K and [F : K] <, then {F : K} = {F : L}{L : K}

9 4.2. GALOIS THEORY 71 Proof. There are {L : K} many K-isomorphisms of L onto a subfield of K. By Theorem 4.10, each such K-isomorphism has {F : L} many extensions to a K-isomorphisms of F onto a subfield of K. Remark. 1. Compare: If F L K, then [F : K] = [F : L][L : K]. 2. By the Theorem, if F = K(α 1,, α r ) is a finite extension of K, let F i := K(α 1,, α i ) so that F i = F i 1 (α i ). Then where {F : K} = {F : F r 1 }{F r 1 : F r 2 } {F 1 : K} = {F r 1 (α r ) : F r 1 } {K(α 1 ) : K} {F i 1 (α i ) : F i 1 } = the number of distinct roots in irr(α i, F i 1 ). 3. When α is algebraic over K, the index {K(α) : K} equals the number of distinct roots of irr(α, K). So {K(α) : K} [K(α) : K] and thus {F : K} [F : K] in general. Ex. F = Q( 2, 3 3) Q = K. Compute the order G(F/K), the index {F : K}, and the degree [F : K]. Thm F K with [F : K] <. F is a splitting field over K iff every K-isomorphism of F onto a subfield of K is a K-automorphism of F (i.e., in G(F/K)), iff G(F/K) = {F : K}. Cor F K splitting. Then every irred polyn in K[x] having a zero in F splits in F [x] Separable Extension Def. 1. A polyn f K[x] is separable if in some splitting field of f over K every root of f is a simple root. 2. F K. u F is called separable over K if irr(u, K) is separable. 3. F K is called a separable extension of K if every element of F is separable over K.

10 72 CHAPTER 4. FIELDS AND GALOIS THEORY Thm If p(x) K[x] is an irred polyn, then every root of p(x) has the same multiplicity. Proof. Suppose α and β are two roots of p(x) with multiplicities m α and m β respectively, so that (x α) mα and (x β) m β are factors of p(x). The K-isomorphism φ α,β : K(α) K(β) that sends α to β can be extended (by Zorn s Lemma) to a K-automorphism φ : K K, and be further extended to a ring automorphism φ : K[x] K[x]. One has φ(p(x)) = p(x) since φ fixes every element of K. Then φ((x α) mα ) = (x β) m β and so m α = m β. Remark. p(x) K[x] monic irreducible. Then p(x) = [ i (x α i)] m, where m is the multiplicity of a root of p(x), and α i are distinct. 1. The multiplicity m divides deg p(x). 2. {K(α i ) : K} = 1 m deg p(x) divides [K(α i) : K] = deg p(x). 3. If char K = 0, then m 1 and {F : K} = [F : K] for any F K. So any extension is a separable extension. 4. If char K = p, then [K(α i) : K] {K(α i ) : K} = m = pr for some r and 4. can be proved by derivative technique. Thm A finite extension F K is a separable extension of K iff {F : K} = [F : K]. F K fin ext. Then G(F/K) {F : K} [F : K]: F is splitting over K iff G(F/K) = {F : K}, F is separable over K iff {F : K} = [F : K]. Thm F L K, [F : K] <. Then F is separable over K iff F is separable over L and L is separable over K. Ex. Let K = Q and F = Q( 2, 3). Then {Q( 2, 3) : Q} = 4 = [Q( 2, 3) : Q]. So Q( 2, 3) is a separable extension over Q. Thm 4.17 (Primitive Element Theorem). A finite separable extension F K is always a simple extension, i.e. F = K(u) for some u F. Ex. Q( 2, 3) = Q( 2 + 3) is a simple extension over Q.

11 4.2. GALOIS THEORY 73 Proof of Thm If K <, then F and K are finite fields. Then F = u for some u F (to be shown later). Hence F = K(u). Suppose K is an infinite field. By induction, we may assume that F = K(v, w) for some v, w F. Let p = irr(v, K) and q = irr(w, K). Let v 1 = v, v 2,, v m and w 1 = w, w 2,, w n be the roots of p and q in F. Then v i s are distinct since F is separable. Similarly, w j s are distinct. As K is infinite, there is a K such that a v i v w w j for all i and all j 1. Let u := v + aw and E := K(u). Then F E K. We show that w E. Obviously, irr(w, E) irr(w, K) = q. Define f := p(u ax) E[x]. Then f(w) = p(v) = 0 and so irr(w, E) f. However, f(w j ) 0 for j 1. Therefore, gcd(q, f) = x w and irr(w, E) = x w. So w E. Then v = u aw E so that F = E = K(u) is a simple extension. So inseparable extensions exist in some characteristic p infinite fields. Def. F K. u F is purely inseparable over K if irr(u, K) = (x u) m (m must be a power of p). F is a purely inseparable extension of K if every element of F is purely inseparable over K. Prop If F L K, then F is purely insep over K iff F is purely insep over L and L is purely insep over K. Thm Let F K, [F : K] <, and char K = p. Then α F is purely insep iff α pr K for some r N. Ex. Let K := Z p (y) (p prime, y transcendental over Z p ). The polyn x p y is inseparable irreducible over K. If u K is a root of x p y, then u is purely insep over K. Remark. F K. The set of all elements of F purely insep over K forms a field T, the purely inseparable closure of K in F. Then F T K, F is separable over T, and T is purely inseparable over K. Similarly, there exists L, the separable closure of K in F, such that F L K, F is purely inseparable over L, and L is separable over K Galois Theory Def. F K. For a subgroup H G(F/K), the set F H := {u F σ(u) = u for every σ H} is an intermediate field between F and K, called the fixed field of H in F.

12 74 CHAPTER 4. FIELDS AND GALOIS THEORY Lem Let F K. Then 1. a field L F = L F G(F/L) ; 2. a subgroup H G(F/K) = H G(F/F H ). (exercise) Def. A finite extension F K is a finite normal extension of K if G(F/K) = {F : K} = [F : K], i.e., F is a separable splitting field of K. Let F K be a finite separable extension. Then F = K(u) for some u F. The splitting field L over K of irr(u, K) satisfies that L F K, where L K is a normal extension. Lem K F L K. If F is a finite normal extension of K, then F is a finite normal extension of L. The group G(F/L) G(F/K). Moreover, two K-automorphisms σ, τ G(F/K) induce the same K-isomorphism of L onto a subfield of K iff σ and τ are in the same left coset of G(F/L) in G(F/K). Proof. If F is the splitting field of a set of polynomials of K[x] over K, then F is the splitting field of the same set of polynomials of L[x] over L. So F is splitting over L. Moreover, F is separable over K implies that F is separable over L. Thus F is a finite normal extension over L. Two automorphisms σ, τ G(F/K) satisfy that σ L = τ L iff (σ 1 τ) L = 1 L, iff σ 1 τ G(F/L), iff τ σ G(F/L), that is, σ and τ are in the same left coset of G(F/L). Thm 4.22 (Fundamental Theorem of Galois Theory). Let F be a finite normal extension of K (i.e. G(F/K) = {F : K} = [F : K]). Let L denote an intermediate field (F L K). Then L G(F/L) is a bijection of the set of all intermediate fields between F and K onto the set of all subgroups of G(F/K). Moreover, 1. L = F G(F/L) for every intermediate field L with F L K. 2. H = G(F/F H ) for every subgroup H G(F/K). 3. L is a normal extension of K if and only if G(F/L) is a normal subgroup of G(F/K). In such situation, G(L/K) G(F/K)/G(F/L)

13 4.2. GALOIS THEORY The subgroup diagram of G(F/K) is the inverted diagram of the intermediate field diagram of F over K. Galois theory implies that: To understand the field extensions in F K, it suffices to understand the group structure of G(F/K). Proof. (Sketch) 1. Every automorphism in G(F/L) leaves L fixed. So L F G(F/L). Note that F is normal over L. Given α F L, there is another root β F L of the polynomial irr(α, L). By Thm 4.7, There is an automorphism in G(F/L) that sends α to β. So every α F L is not fixed by G(F/L). Hence F G(F/L) L. So L = F G(F/L). 2. Let H G(F/K). Every element of H leaves F H fixed. So H G(F/F H ). It remains to prove that H G(F/F H ) (= [F : F H ]) so that H = G(F/F H ). Since F is a finite normal (=separable+splitting) extension over F H, we can write F = F H (α) for some α F F H. Suppose H := {σ 1,, σ H }. Denote H f(x) := (x σ i (α)) F [x]. i=1 Every σ k H G(F/K) induces a ring automorphism of F [x], with H σ k (f(x)) = (x σ k σ i (α)) = (x σ i (α)) = f(x). i=1 So the coefficients of f(x) are in F H and f(x) F H [x]. The group H contains identity automorphism. So there is some σ i (α) = α. So f(α) = 0. Then irr(α, F H ) f(x). So G(F/F H ) = [F : F H ] = [F H (α) : F H ] = deg(α, F H ) deg f(x) = H. Therefore H = G(F/F H ). 3. L is a normal extension over K iff L is splitting (and separable) over K; iff σ(α) L for any α L; (notice that L = F G(F/L) ) iff τσ(α) = σ(α) for every τ G(F/L); iff σ 1 τσ(α) = α; H i=1

14 76 CHAPTER 4. FIELDS AND GALOIS THEORY iff σ 1 τσ G(F/L) for every τ G(F/L) and σ G(F/K); iff G(F/L) is a normal subgroup of G(F/K). Suppose L is a normal extension over K. We show that G(L/K) G(F/K)/G(F/L). Since L is splitting over K, if σ G(F/K) then σ L G(L/K). Define φ : G(F/K) G(L/K) by φ(σ) := σ L. Then φ is a group homomorphism. On one hand, every γ G(L/K) can be extended to an element γ G(F/K), with φ(γ) = γ L = γ. So φ is onto. On the other hand, Ker (φ) = G(F/L). Therefore, G(L/K) G(F/K)/G(F/L). 4. The statements 1. and 2. build up the bijection between the set of intermediate fields of F over K and the set of subgroups of G(F/K) in desired order. The following Lagrange s Theorem on Natural Irrationalities discloses further relations on Galois correspondence. Thm If L and M are intermediate fields between F and K such that L is a finite normal extension of K, then the field (L, M) is finite normal extension of M and G((L, M)/M) G(L/L M). (show by graph) Proof. The idea is to show that: if L is the splitting field over L M of an irred polyn f (L M)[x], then (L, M) is the splitting field over M of f. This makes the correspondence. Ex. (HW) Let K := Q and F := Q( 2, 3). Then G(F/K) consists of 4 elements {ι, σ, τ, στ} Z 2 Z 2 : ι(a + b 2 + c 3 + d 6) := a + b 2 + c 3 + d 6 σ(a + b 2 + c 3 + d 6) := a + b 2 c 3 d 6 τ(a + b 2 + c 3 + d 6) := a b 2 + c 3 d 6 στ(a + b 2 + c 3 + d 6) := a b 2 c 3 + d 6 The intermediate field diagram of F over K and the subgroup diagram of G(F/K) are inverted to each other. Ex. Let F = Q( 3 2, i 3) be the splitting field of (x 3 2) over K = Q. 1. Describe the six elements of G(F/K) by describing their actions on 3 2 and i 3. (done)

15 4.2. GALOIS THEORY To what group we have seen before is G(F/K) isomorphic? (done) 3. Give the diagrams for the subfields of F and for the subgroups of G(F/K).

16 78 CHAPTER 4. FIELDS AND GALOIS THEORY 4.3 Illustration of Galois Theory Some Examples Def. The Galois group of a polynomial f K[x] over a field K, denoted by G(f/K), is the group G(F/K) where F is a splitting field over K of f. When K = Q, the preceding examples show 1. the Galois group of (x 2 2)(x 2 3) Q[x] is Z 2 Z 2, and 2. the Galois group of x 3 2 Q[x] is S 3 = D 3. Here S n denotes the group of permutations of n letters, and D n denotes the symmetric group of a regular n-gon. Ex. The Galois group of x 3 1 Q[x] is Z 2, which is totally different from the Galois group of x 3 2 Q[x]. Ex. Find the Galois groups of the following polynomials in Q[x]: 1. x (Z 2 Z 2 ) 2. x 4 1. (Z 2 ) 3. x 4 2. (D 4. See p.275 of [Hungerford, V.4]) The Galois group G of an irreducible separable polynomial f(x) K[x] of degree n = 2, 3, 4 has been classified [see Hungerford, V.4]. 1. n = 2, then G must be S 2 Z n = 3, then G could be S 3 or A 3 Z n = 4, then G could be S 4, A 4, D 4, Z 4, or Z 2 Z 2. Let us discuss the Galois group of irreducible f(x) Q[x] of degree 3. Def. char K 2; f K[x] a polyn with distinct roots u 1,, u n. F = K(u 1,, u n ) the splitting field over K of f. Denote = i<j(u i u j ) F ; define the discriminant of f as D = 2.

17 4.3. ILLUSTRATION OF GALOIS THEORY 79 Prop Let K, f, F and be as in preceding definition. 1. For each σ G(F/K) S n, σ is an even [resp. odd] permutation iff σ( ) = [resp. σ( ) = ]. 2. The discriminant 2 K. Cor F K( ) K. Consider G = G(F/K) S n. In the Galois correspondence, the subfield K( ) corresponds to the subgroup G A n. In particular, G consists of even permutations iff K. Cor Given a degree 3 irred separable polyn f(x) = x 3 + bx 2 + cx + d K[x], let g(x) = f(x b/3) = x 3 + px + q. Then the discriminant of f(x) is 2 = 4p 3 27q 2 K. 1. If 4p 3 27q 2 is a square in K, then G(f/K) = A 3 Z 3 ; 2. If 4p 3 27q 2 is not a square in K, then G(f/K) = S 3. Ex. Consider the foll irred polyns in Q[x]: 1. f(x) = x 3 2. Then 2 = is not a square in Q. So G(f/Q) = S 3 (as we have proved). 2. f(x) = x 3 + 3x 2 x 1. Then g(x) = f(x 3/3) = x 3 4x + 2 is irreducible. The discriminant of f(x) is 4( 4) 3 27(2) 2 = 148, which is not a square in Q. Thus G(f/Q) = S f(x) = x 3 3x + 1 is irreducible. The discriminant is 4( 3) 3 27(1) 2 = 81, which is a square in Q. So G(f/Q) = A 3 Z 3. In general, it is difficult to compute the Galois group of an irreducible polynomial of degree n 5. There is a special result: Thm p is prime, f(x) Q[x] an irred polyn of deg p with exactly two nonreal roots in C, then G(f/Q) = S p Finite Groups as Galois Groups Thm Let G be the Galois group of an irreducible separable polynomial f(x) K[x] of degree n. Then G S n and n G n!.

18 80 CHAPTER 4. FIELDS AND GALOIS THEORY Next we show that every finite group is the Galois group of a finite normal extension. Let y 1,, y n be indeterminates. The field F := Q(y 1,, y n ) consists of all rational functions of y 1,, y n. Every permutation σ S n induces a map σ G(F/Q) by σ ( ) f(y1,, y n ) := f(y σ(1),, y σ(n) ) g(y 1,, y n ) g(y σ(1),, y σ(n) ). Denote S n := {σ σ S n } G(F/Q). The subfield of F fixed by S n is K = Q(s 1,, s n ), where s 1,, s n are the following symmetric functions of y 1,, y n over Q: s 1 := y 1 + y y n, s 2 := y 1 y 2 + y 1 y y n 1 y n, s n := y 1 y 2 y n Now F = K(y 1,, y n ), and f(x) := (x y 1 )(x y 2 ) (x y n ) = x n s 1 x n 1 + s 2 x n ( 1) n s n K[x] So F = Q(y 1,, y n ) is the splitting field of f(x) over K = Q(s 1,, s n ), where y 1,, y n are the roots of f(x). Every element of G(F/K) permutes the n roots of f(x). This shows that: 1. G(F/K) = S n S n and G(F/K) = n!. 2. (Every finite group is the Galois group of a finite normal extension) By Cayley s Theorem, every finite group H is isomorphic to a subgroup of certain S n. By Galois theory, there is an intermediate field F 0 such that Q F F 0 K, and H G(F/F 0 ). 3. It is an open problem that which finite group is the Galois group of a finite normal extension over a given field (e.g. Q).

19 4.4. CYCLOTOMIC EXTENSIONS Cyclotomic Extensions Def. The splitting field F of x n 1 over K is the cyclotomic extension of K of order n. Def. An element u K is a primitive n-th root of unity if u n = 1 and u k 1 for any positive integer k < n. The cyclotomic extension of order n is related to the Euler function ϕ(n), where ϕ(n) is the number of integers i such that 1 i n and gcd(i, n) = 1. If n = p m 1 1 p m k k = k i=1 pm i i is the prime factorization of n (where p i are distinct primes), then ϕ(n) = k [p m i 1 i i=1 (p i 1)] = n k (1 1/p i ) When d n, ϕ(d) equals to the number of integers i such that 1 i n and gcd(i, n) = n/d. Therefore, d n ϕ(d) = n Cyclotomic extensions over Q Let Q C. Consider the splitting field of x n 1 over Q. There exists a primitive n-th root of unity ζ C. All the other primitive n-th roots of unity are ζ i where 1 i n and gcd(i, n) = 1. So there are ϕ(n) elements conjugate to ζ over Q. Denote g n (x) = (x ζ i ). 1 i n gcd(i,n)=1 Then g n (x) = irr(ζ, Q) Q[x] has degree ϕ(n). g n (x) is called the n-th cyclotomic polynomial over Q. Thm Let F = Q(ζ) be a cyclotomic extension of order n of the field Q, and g n (x) the n-th cyclotomic polynomial over Q. Then 1. g n (x) Z[x] and g n (x) is irreducible in Z[x] and Q[x]. Moreover, x n 1 = d n i=1 g d (x). 2. [F : Q] = ϕ(n), where ϕ is the Euler function.

20 82 CHAPTER 4. FIELDS AND GALOIS THEORY 3. The Galois group G(F/Q) of x n 1 is isomorphic to the multiplicative group of units in the ring Z n. Ex. If p is a prime, then the Galois group of x p 1 Q[x] is isomorphic to the cyclic group Z p 1. Ex. Consider the cyclotomic extension F n of degree n over Q: 1. n = 9 = 3 2. Then ϕ(9) = 3 2 = 6. The multiplicative group in Z 9 is A = {1, 2, 4, 5, 7, 8}. Notice that 2 generates A. So G(F 9 /Q) A Z n = 12 = Then ϕ(12) = 2 2 = 4. The multiplicative group in Z 12 is A = {1, 5, 7, 11} Z 2 Z 2. So G(F 12 /Q) A Z 2 Z Likewise, G(F 8 /Q) Z 2 Z 2 and G(F 14 /Q) Z Cyclotomic Extensions over K If char K = p 0 and n = mp t with gcd(p, m) = 1, then x n 1 = (x m 1) pt, so that a cyclotomic extension of order n coincides with one of order m. Now we consider the cyclotomic extensions where char K = 0 or char K does not divide n. Let ζ denote a primitive n-th root of unity over K. Then all primitive n-th root of unity over K are ζ i for 1 i n and gcd(i, n) = 1. However, some ζ i may not be conjugate to ζ over K anymore. We have irr(ζ, K) g n (x) = (x ζ i ), 1 i n gcd(i,n)=1 where g n (x) is called the n-th cyclotomic polynomial over K. Moreover, the following theorem says that deg(ζ, K) deg g n (x) = ϕ(n). Thm Let K be a field such that char K does not divide n, and F a cyclotomic extension of K of order n. 1. F = K(ζ), where ζ F is a primitive n-th root of unity. 2. G(F/K) is isomorphic to a subgroup of order d of the multiplicative group of units of Z n. In particular, [F : K] = G(F/K) = d divides ϕ(n). 3. x n 1 = d n g d(x). Moreover, deg g n (x) = ϕ(n), and the coefficients of g n (x) are integers (in Z or Z p, depending on char K).

21 4.4. CYCLOTOMIC EXTENSIONS 83 Ex. If ζ is a primitive 5th root of unity in C, then 1. Q(ζ) is a cyclotomic extension of Q of order 5, with G(Q(ζ)/Q) Z R(ζ) is a cyclotomic extension of R of order 5, with G(R(ζ)/R) Z 2. ζ satisfies that ζ + 1/ζ = 2Re(ζ). So irr(ζ, R) = x 2 2Re(ζ) x + 1.

22 84 CHAPTER 4. FIELDS AND GALOIS THEORY 4.5 Galois Theory on Finite Fields Structure of Finite Fields Examples of finite fields include Z p for primes p. We will see that every finite field F has a prime characteristic p, and F Z p (α) where α Z p is a primitive root of x pn 1 1 in Z p [x]. The characteristic of a field F is either 0 or a prime p. 1. If char F = 0, then F is an extension of Q. 2. If char F = p for a prime p, then F is an extension of Z p. A finite field F is simply a finite extension of Z p. Thm Let E be a finite extension of F with [E : F ] = m, where F is a finite field of q elements. Then E has q m elements. In particular, the finite field E contains exactly p n elements for p = char E and n = [E : Z p ]. Thm For every prime power p n, there is a unique (up to isomorphism) finite field GF(p n ) which contains exactly p n elements. If Z p GF(p n ) Z p, the elements of GF(p n ) are precisely the roots of x pn x Z p [x] in Z p. The multiplicative group F, of nonzero elements of a finite field F is cyclic. A finite extension E of a finite field F is a simple extension of F. Because if E = p n (i.e. E is a finite extension of F := Z p ), let α Z p be a primitive (p n 1)-th root of unity, then E = Z p (α) = F (α). For α Z p, the degree deg(α, Z p ) = n iff Z p (α) = GF(p n ), iff α is a primitive (p n 1)-root of unity in Z p. Ex. That are ϕ(p n 1) many primitive (p n 1)-roots of unity in GF(p n ). So the number of degree n irreducible polynomials in Z p [x] is equal to ϕ(pn 1) n. Moreover, x(x pn 1 1) = x pn x Z p [x] is the product of all degree m irreducible polynomials for m n. If GF(p n ) GF(p m ) Z p, then m n. So it is easy to draw the intermediate field diagram of GF(p n ). Moreover, every GF(p n ) is a normal extension over Z p and GF(p m ) for m n.

23 4.5. GALOIS THEORY ON FINITE FIELDS Galois Groups of Finite Fields Thm If F is a field of characteristic p and r is a positive integer, then σ r : F F given by σ r (u) = u pr is a Z p -monomorphism of fields. It is clear that σ r σ s = σ r+s. Cor G(GF(p n )/Z p ) = {1, σ 1, σ 2,, σ n 1 } Z n. 2. When m n, G(GF(p n )/GF(p m )) = {1, σ m, σ 2m,, σ ( n m 1)m } Z n/m. 3. G(Z p /Z p ) {, σ 2, σ 1, 1, σ 1, σ 2, } Z. Ex. x 3 3x + 1 Z 5 [x] is irreducible. The Galois group of x 3 3x + 1 over Z 5 is {1, σ 1, σ 2 } Z 3. Ex. Describe the Galois correspondence between the intermediate fields of GF(p 12 ) over Z p and the subgroups of G(GF(p 12 )/Z p ). If α Z p has deg(α, Z p ) = [Z p (α) : Z p ] = n, then all the Z p -conjugates of α in Z p are {1(α), σ 1 (α), σ 2 (α),, σ n 1 (α)} = {α, α p1, α p2,, α pn 1 }. The irreducible polynomial of α over Z p is irr(α, Z p ) = n 1 k=0 (x α pk) Similarly, when m n, the irreducible polynomial of α over GF(p m ) is irr(α, GF(p m )) = n m 1 (x α pkm) k=0

24 86 CHAPTER 4. FIELDS AND GALOIS THEORY 4.6 Radical Extensions Solvable Groups Def. A finite group G is solvable if there exists a subgroup sequence {1} = G 0 G 1 G 2 G n = G (4.1) such that G i G i+1 and G i+1 /G i is an abelian group for i = 0,, n 1. Remark. If H is a finite abelian group, then there exists a subgroup sequence {1} = H 0 H 1 H t such that H i+1 /H i is a cyclic group of prime order. So after making a refinement on the equence (4.1), we may assume that G i G i+1 and G i+1 /G i is an (abelian) cyclic group of prime order. We give another definition of solvable groups using derived subgroups. The commutator subgroup G of G is the subgroup of G generated by the set {aba 1 b 1 a, b G}. Lem G G, and G is the minimal normal subgroup of G such that G/G is an abelian group. Let G (0) = G, G (1) = G,, G (i+1) = (G (i) ), Then G = G (0) G (1) G (2) and G (i) G (i+1) for i = 0, 1, 2,. The group G (i) is called the i-th derived subgroup of G. Lem A finite group G is solvable iff G (n) = {1} for some n. Ex. 1. D 4 is solvable. Every finite group of order p n for a prime p is solvable. 2. A 4 is solvable. 3. A 5 is insolvable. Indeed, A 5 is the smallest insolvable group. Thm If G is a finite solvable group, then every subgroup and every quotient group of G are solvable. Equivalently, if G contains an insolvable subgroup or quotient group, then G is also insolvable. A remarkable result by W. Feit and J. Thompson claims that every finite group of odd order is solvable.

26 88 CHAPTER 4. FIELDS AND GALOIS THEORY is constructible. In fact, every constructible number can be obtained by a sequence of field extensions Q = F 0 F 1 F 2, [F i+1 : F i ] = 2. So α is constructible iff α is algebraic over Q and deg(α, Q) = 2 m. Therefore, using straightedge and compass, 1. trisecting a generic angle is impossible; 2. doubling the volume of a cube is impossible; 3. (Gauss) a regular n-gon is constructible iff cos 2π n is constructible, iff a primitive n-root of unity, say ζ, is constructible, iff irr(ζ, Q) = ϕ(n) = 2 m, iff n is a product of a power of 2 and some distinct odd primes of the form p = 2 t + 1. However, if t has an odd factor s > 1, then 2 t/s + 1 divides 2 t + 1. So t = 2 k and thus p = 2 2k + 1 (called a Fermat prime). Overall, a regular n-gon is constructible iff n = 2 l p 1 p 2 p q, where p i are distinct Fermat primes. The following regular n-gons are constructible using straightedge and compass: n = 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20,, 120,, 256, However, the regular 9-gon is inconstructible.

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