GALOIS THEORY I (Supplement to Chapter 4)

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1 GALOIS THEORY I (Supplement to Chapter 4) 1 Automorphisms of Fields Lemma 1 Let F be a eld. The set of automorphisms of F; Aut (F ) ; forms a group (under composition of functions). De nition 2 Let F be a eld and let E be an extension of F: We say that an automorphism 2 Aut (E) xes F if j F = Id F (that is (a) = a for each a 2 F ): Lemma 3 Automorphisms of Q 1. Aut (Q) = fid Q g ; and 2. If E is an extension eld of Q and 2 Aut (E) ; then xes Q: Proof. Homework. Theorem 4 Let F be a eld and let E be an extension of F: 1. Let p 2 F [x] be an irreducible polynomial in F. If ; 2 E are two roots of p; then there exists an automorphism 2 Aut (E) such that xes F and () = : 2. For any polynomial p 2 F [x] ; if 2 E is a root of p; then, for any 2 Aut (E) ; if xes F; we have that () is a root of p (so permutes the roots of p) Proof. Part (1) follows from the work we did to prove the uniqueness of splitting elds. For part (2), suppose p = b 0 + b 1 x + + b n x n where each b i 2 F and let 2 E be a root of p: Then p ( ()) = b 0 + b 1 () + + b n ( ()) n = (b 0 ) + (b 1 ) () + + (b n ) ( ()) n since j F = Id F = (b 0 + b b n n ) = (p ()) = (0) = 0: 1

2 Example 5 Determine Aut Q p 2 : Example 6 Is Q p 2 a splitting eld for p (x) = min p 2; Q = x 2 2? Example 7 Determine Aut Q 3p 2 where 3p 2 is the real cube root of 2. Example 8 Is Q 3p 2 a splitting eld for p (x) = min 3p 2; Q = x 3 2? Remark 9 From the above theorem we have that if p 2 F [x] and E is an extension eld of F in which p splits, then 1. if p is irreducible over F; then for any two roots of p; there is an automorphism of E which xes F and sends one of the two roots to the other, and 2. any automorphism of E which xes F will permute the roots of p: (proposition 4.1) Question: Is every permutation of roots allowed? irreducible polynomial. NO, not even if you choose an Example 10 Consider p (x) = x 4 + x 3 + x 2 + x Q [x] : Then p (x) is an irreducible polynomial in Q [x] (you ll be verifying this in homework). Let E be the splitting eld for p over Q: Since p (x) (x 1) = x 5 1; the roots of p are all fth roots of unity. In fact, they are the four primitive fth roots of unity. If is one of the primitive fth roots of unity, then the other primitive fth roots of unity are 2 ; 3 ; and 4 : Let 1 = ; 2 = 2 ; 3 = 3 and 4 = 4 ; the four roots of p: We want to show that not every permutation of the roots of p will produce an automorphism of E: Consider the permutation ( ) ; so 1! 2! 3! 4! 1 : Then ( 2 ) = 3 = 3 and ( 2 ) = 2 = ( ) = ( 1 1 ) = ( 1 ) ( 1 ) = 2 2 = 2 2 = 4 : This is a contradiction since 3 6= 4 : Example 11 Continuing with the above example. What permutations of the roots are allowed? First, note that E = Q () = r 0 + r 1 + r r 3 3 : r 0; r 1 ; r 2 ; r 3 2 Q : (Homework.) So, if 2 E; then = r 0 + r 1 + r r 3 3 for some r 0; r 1 ; r 2 ; r 3 2 Q: So, for any 2 Aut (E) ; we have () = r 0 + r 1 + r r 3 3 = r 0 + r 1 () + r 2 ( ()) 2 + r 3 ( ()) 3 : 2

3 This means that, for 2 Aut (E) ; the value of () completely determines : We know that, for any root of p (x) ; there is an automorphism which takes to that root. Thus there are four automorphisms on E : 1.! 2.! 2 3.! 3 4.! 4 What permutation of the roots does each of the above automorphisms correspond to? What well-known group is Aut (Q ()) isomorphic to. Lemma 12 (lemma 4.2) Let E = k (z 1 ; z 2 ; : : : ; z n ) : If is an automorphism of E which xes k and xes each z i ; then is the identity mapping on E: Proof. Since E = (k (z 1 ; z 2 ; : : : ; z n extension (i.e. for n = 1): 1 )) (z n ) ; it su ces to prove this is true for a simple Let E = k (z) and let 2 Aut (E) which xes k and xes z: Let u 2 E: Then u = f(z) g(z) = (f (z)) (g (z)) 1 for some f; g 2 k [x] : Note that (f (z)) = f ( (z)) and (g (z)) = g ( (z)) (see proof of theorem 4). So, since (z) = z; we have (f (z)) = f (z) and (g (z)) = g (z) : Thus, (u) = (f (z)) ( (g (z))) 1 = f (z) (g (z)) 1 = u: Therefore is the identity mapping. 1.1 Problems 1. Let F be a eld. De ne : F [x]! F [x] by (p (x)) = p (x + 1) ; i.e. nx a i (x + 1) i : i=0! nx a i x i = i=0 (a) Prove that is an automorphism on F [x] : (b) Let p (x) = x 4 + x 3 + x 2 + x + 1: Determine (p (x)) : 3

4 (c) Prove p (x) is irreducible in Q [x] : 2. In the setting of example 10, explain how you know each of the following: (a) E = Q () ; and (b) Q () = r 0 + r 1 + r r 3 3 : r 0; r 1 ; r 2 ; r 3 2 Q 3. Prove lemma 3. 2 Translating into Group Theory The Galois Group De nition 13 Let E be an extension eld of the eld F: The Galois group of E over F, denoted Gal (E=F ) ; is the set of all automorphisms of E which x F: If p (x) 2 F [x] and E is a splitting eld for p (x) over F; then the Galois group for p (x) over F is de ned to be Gal (E=F ) : Lemma 14 Gal(E=F ) < Aut (E) : Proof. Homework. Theorem 15 (Theorem 4.3) Let p 2 F [x] with deg (p) = n and let E be a splitting eld for p over F: Then Gal(E=F ) is isomorphic to a subgroup of S n : Proof. Let X = fz 1 ; z 2 ; : : : ; z n g E be the set of zeros for p. So, E = F (z 1 ; z 2 ; : : : ; z n ) : By 4(2) we have that any element of Gal (E=F ) will permute the elements of X: That is, if 2 Gal (E=F ) ; then j F 2 S X : De ne : Gal (E=F )! S X by () = j F : It can be shown that is a group homomorphism. (Homework) Therefore (Gal (E=F )) < S X : Suppose 2 ker : Then () = Id X : That is (z i ) = z i for each z i 2 X: By lemma 12, this means that is the identity mapping on E: So, ker = fid E g : This means (Gal (E=F )) = (Gal (E=F )) = ker = Gal (E=F ) : Thus, Gal (E=F ) is isomorphic to (Gal (E=F )) < S X : Since jxj = n; we have S X = Sn and the theorem is proven. Example 16 Determine Gal (C=R) : (Compare jgal (C=R)j and [C : R]) Example 17 Determine Gal Q p 2 =Q ; Gal Q a primitive 5th root of unity. See examples 5, 7 and 11 from the rst section. 3p 2 =Q and Gal (Q () =Q) where is 4

5 2.1 Problems 1. Prove lemma Fixed Fields De nition 18 Let S be a set of automorphisms of a eld E: So, S Aut (E) : The xed eld of S is E S = fx 2 E : (x) = x for all 2 Sg : Lemma 19 E S is a sub eld of E: Proof. Homework. Lemma 20 If S T; then E T E S : Proof. Homework. Theorem 21 (Lemma 4.31) If S Aut (E) and jsj = n; then E : E S n: Proof. Let S Aut (E) with jsj = n: Then S = f 1 ; 2 ; : : : ; n g for some 1 ; 2 ; : : : ; n 2 Aut (E) : Suppose E : E S = r < n: Then, there is a basis f 1 ; 2 ; : : : ; r g E for E over E S : Consider the following homogeneous system of equations in E: 1 ( 1 ) x n ( 1 ) x n = 0 (1) 1 ( 2 ) x n ( 2 ) x n = 0 : : : = : : : 1 ( r ) x n ( r ) x n = 0 This is a homogenous system with r equations and n unknowns. Since r < n; there is a nontrivial solution to this system of equations: Choose a nontrivial solution (c 1 ; : : : ; c k ; 0; : : : ; 0) have the smallest number, k; of nonzero components (by reindexing the i s, we may assume that the nonzero terms come rst). Substituting our solution into the system of equations above gives: 1 ( 1 ) c k ( 1 ) c k = 0 (2) 1 ( 2 ) c k ( 2 ) c k = 0 = 1 ( r ) c k ( r ) c k = 0 5

6 Now, let 2 E: Then = b b b r r for some b 1 ; b 2 ; : : : ; b r 2 E S : Let 2 S: (So, = i for some i:) Then () = (b 1 ) ( 1 ) + (b 2 ) ( 2 ) + + (b r ) ( r ) (3) = b 1 ( 1 ) + b 2 ( 2 ) + + b r ( r ) since xes elements of E S Returning to the system of equations (equation 2), we will multiply the rst equation by b 1 ; the second by b 2 ; and so on, and then add to get b 1 1 ( 1 ) c b 1 n ( 1 ) c k +b 2 1 ( 2 ) c b 2 n ( 2 ) c k + +b r 1 ( r ) c b r n ( r ) c k = 0 Collecting like terms gives 1 () c k () c k = 0: (4) Since 2 E was arbitrary, we have that equation 4 holds for each 2 E: Note that, by the way in which (c 1 ; c 2 ; : : : ; c n ) was chosen, there is no shorter relation of this form (equation 4) which holds. Clearly, we can now see that k 6= 1; since otherwise we have 1 () = 0 for all 2 E which is impossible. Since k > 1; we have 1 6= k ; thus 1 () 6= k () for some 2 E: For arbitrary 2 E; we have 2 E: So, since equation 4 holds for every element of E; it holds for : That is 1 () c k () c k = 0: So, 1 () 1 () c k () k () c k = 0: (5) Multiply equation 4 by k () and subtract equation 5 from it to get 1 () ( k () 1 ()) c () ( k () 2 ()) c k () ( k () k ()) c k = 0: In the above equation, the coe cient of c 1 is not zero since 1 () 6= k () ; but the 6

7 coe cient of c k is zero. De ne d i = ( k () i ()) c i for each i. Then, the above equation becomes 1 () d k 1 () d k 1 = 0 since d k = 0: This is a contradiction because by the way in which (c 1 ; c 2 ; : : : ; c n ) was chosen, there is no shorter relation of the form of equation 4 which holds and this is a shorter equation of that form. Therefore, E : E S n: Corollary 22 If S Aut (E) is an in nite set, then E : E S = 1: Proof. Homework. Hint: Prove E : E S n for all n 2 N: Theorem 23 (Proposition 4.32) If G < Aut (E) ; then E : E G = jgj : Proof. If G is an in nite group, then, from the corollary above, E : E G = 1: So, assume jgj = n < 1: Let G = f 1 ; 2 ; : : : ; n g : From the above theorem, we know E : E G n: Suppose E : E G > n: Then there exists a set f! 1 ;! 2 ; : : : ;! n+1 g of linearly independent vectors in E over E G : Consider the homogeneous system of n equations in n + 1 unknowns: 1 (! 1 ) x (! n+1 ) x n+1 = 0 = n (! 1 ) x n (! n+1 ) x n+1 = 0: Since n + 1 > n; this system has a nontrivial solution over E: Choose a solution ( 1 ; 2 ; : : : ; r ; 0; : : : ; 0) having the smallest number r of nonzero components (by reindexing the! i ; we may assume that the nonzero terms come rst). If r = 1; we have 1 (! 1 ) 1 = 0: Note:! 1 6= 0; so 1 (! 1 ) 6= 0: Thus, 1 = 0 which is a contradiction. Thus, r 6= 1: Multiplying by 1 r ; if necessary, we may assume r = 1: Also note that, since G is a group, one of the i is the identity mapping. Plugging our solution ( 1 ; 2 ; : : : ; 1; 0; : : : ; 0) into the row above corresponding to the identity mapping, we have! 1 1 +! ! r 1 r 1 +! r = 0: 7

8 Since f! 1 ;! 2 ; : : : ;! n+1 g is linearly independent over E G ; this means that one of the i in the equation above must not be in E G : WLOG, suppose 1 62 E G : Since 1 62 E G ; there exists some k 2 G such that k ( 1 ) 6= 1 : Substituting our solution into the original system of equations, we have 1 (! 1 ) (! r 1 ) r (! r ) = 0 = n (! 1 ) n (! r 1 ) r 1 + n (! r ) = 0 Now, we apply k to this system. For each j; the jth row will be k j (! 1 ) k ( 1 ) + + k j (! r 1 ) k r 1 + k j (! r ) = 0: (6) Since G is a group, k 1 ; k 2 ; : : : ; k n is just a permutation of 1 ; 2 ; : : : ; n : Setting k j = i and reordering the rows so that this row will be in the ith position, the ith row is i (! 1 ) k ( 1 ) + + i (! r 1 ) k r 1 + i (! r ) = 0: Subtracting this from the ith row of the system as described in equation 6, we obtain a new system whose ith row is i (! 1 ) [ 1 k ( 1 )] + + i (! r 1 ) r 1 k r 1 = 0: Since 1 k ( 1 ) 6= 0; we have found a nontrivial solution 1 k ( 1 ) ; : : : ; r 1 k r 1 ; 0; : : : ; 0 for the original system of equations having few than r nonzero components which is a contradiction. Therefore, E : E G = jgj : Corollary 24 If E is an extension eld of F, then jgal (E=F )j = E : E Gal(E=F ) : Remark 25 When will we have E G = F: Is it true that E G = F when Gal (E=F )? Not in general. See the example below. Example 26 By the corollary above, if E G = F where G = Gal (E=F ) ; we would have jgal (E=F )j = [E : F ] : This is not always true. Consider F = Q and E = Q 3p 2 : We saw in example 7 that Gal Q 3p 2 =Q has only one element. So, jgal (E=F )j = 1: However, [E : F ] = Q 3p 2 : Q is easily seen to be 3. 8

9 3.1 Problems 1. Prove lemma Prove lemma Prove corollary Normal Extensions De nition 27 Let E be an extension of the eld F: We say that E is a normal extension of F if [E : F ] is nite and F is the xed eld of Gal (E=F ). That is E Gal(E=F ) = F. We write F C E: Example 28 In example 26, we showed that Q 3p 2 is not a normal extension of Q. Example 29 Show Q C Q p 2 : Theorem 30 Let F C E: Let p (x) be an irreducible polynomial in F [x] : Further suppose that E contains at least one root of p (x) : Then E contains all of the roots of p (x) and those roots are all distinct (so p (x) splits into distinct linear factors in E [x]): Proof. Let 2 E with p () = 0: For each i 2 Gal (E=F ) ; let i = i () : Then, since i permutes the roots of p (x) ; we have that i is a root of p (x) : Let 1 = ; 2 ; : : : ; k be the distinct images of under the members i of Gal (E=F ) : Then p (x) = (x 1 ) (x 2 ) (x k ) r (x) for some r (x) 2 E [x] : Let s (x) = (x 1 ) (x 2 ) (x k ) : So, p (x) = s (x) r (x) : Since each element i 2 Gal (E=F ) permutes the roots of p; we have that i (s (x)) = s (x) (where i denotes the induced homomorphism on E [x]). Thus, each i xes the coe cients of s (x) ; so s (x) 2 F [x] : Note that this means, that for any i 2 Gal (E=F ) ; we have p (x) = i (p (x)) = i (s (x)) i (r (x)) = s (x) i (r (x)) : Subtracting this from p (x) = s (x) r (x) gives 0 = s (x) r (x) i (r (x)) : F [x] : Since s (x) 6= 0; this means r (x) = i (r (x)) for each i 2 Gal (E=F ) and, thus, r (x) 2 9

10 So, p (x) = s (x) r (x) for some s (x) ; r (x) 2 F [x] : Remember, p (x) is irreducible in F [x] ; so this means that r (x) is a constant (since s (x) is not a constant). Therefore, p (x) = c (x 1 ) (x 2 ) (x k ) in E [x] and, all of the roots of p (x) are distinct (by de nition of the i s). De nition 31 Let F be a eld and let p (x) 2 F [x] : We say that p (x) is a separable polynomial if each of its irreducible factors in F [x] has no repeated roots in a splitting eld for p (x) : An element of an extension eld E is said to be separable if either is transcendental over F or if min (; F ) is a separable polynomial. Lemma 32 (Lemma 4.4) If F is a eld of characteristic 0, every polynomial in F [x] is separable. Proof. We need to show that if p (x) is irreducible in F [x] ; then p (x) has no repeated roots in a splitting eld. Hint: look at p 0 (x) : Homework. Theorem 33 F C E if and only if E is a splitting eld of a separable poynomial in F [x]. Proof. Let F be a sub eld of E: ( =) ) : Suppose F C E: Then [E : F ] is nite. Let h = [E : F ] : Then there is a basis f 1 ; 2 ; : : : ; h g for E over F: Each i is algebraic over F since [E : F ] is nite. So, for each i ; there is an irreducible polynomial p i (x) 2 F [x] with p ( i ) = 0: By theorem 30, the roots of p i (x) are all distinct, since F C E: hy Let f (x) = p i (x) : Then f (x) is a separable polynomial and E is a splitting eld for f (x) : i=1 ( (= ) : Let E be the splitting eld for f (x) = polynomial in F [x] with no repeated roots in E: hy p i (x) where each p i (x) is an irreducible Then E = F (a 1 ; : : : ; a n ) where a 1 ; : : : ; a n are the distinct roots of f (x) in E: Thus, [E : F ] < 1: Now, we induct on [E : F ] : If [E : F ] = 1; then E = F and, clearly F C E: Now, for the induction hypothesis, we assume that whenever E is a splitting eld for a separable polynomial g (x) 2 K [x] and [E : K] n 1; then K C E: 10 i=1

11 Since F (a 1 ; : : : ; a n ) = E 6= F; we have that at least one of the a i s is not in F: WLOG assume a 1 62 F: Also, we may assume a 1 is a root of p 1 (x) : Let K = F (a 1 ) : Then K 6= F; so [K : F ] > 1: Thus, [E : K] < n: Let t = deg (p 1 (x)) : Then by assumption p 1 (x) has t distinct roots in E: These roots are listed amongst a 1 ; : : : ; a n : After suitable renumbering, we may assume a 1 ; : : : ; a t are the distinct roots of p 1 (x) in E: Now, to apply the induction hypothesis, note that f (x) 2 F [x] K [x] : Thus, f (x) is a separable polynomial in K [x] : So, by the induction hypothesis, K C E: Therefore, E Gal(E=K) = K: Let 2 E Gal(E=F ) : Since Gal (E=K) Gal (E=F ) ; we have E Gal(E=F ) E Gal(E=K) : Thus, 2 E Gal(E=K) = K: Since K = F (a 1 ) ; we have = b 0 + b 1 a 1 + b 2 (a 1 ) b t 1 (a 1 ) t 1 for some b 0 ; b 1 ; : : : ; b t 1 2 F: Since a 1 ; : : : ; a t are the roots of an irreducible polynomial in F [x] ; there is, for each j (1 j t) ; an automorphism j 2 Gal (E=F ) such that j (a 1 ) = a j : Then, for each j; = j () since 2 E Gal(E=F ) = b 0 + b 1 a j + b 2 (a j ) b t 1 (a j ) t 1 : De ne g (x) = (b 0 ) + b 1 x + + b t 1 x t 1 : Since each b i 2 F K and 2 K; we have g (x) 2 K [x] : Note that, for each j; (a j ) = (b 0 ) + b 1 a j + + b t 1 (a j ) t 1 = b 0 b 0 + b 1 a j + b 2 (a j ) b t 1 (a j ) t 1 + b 1 a j + + b t 1 (a j ) t 1 = 0 Therefore, has t distinct roots. This is impossible unless g (x) = 0 (since otherwise, deg (g (x)) t 1): Thus, g (x) = 0: This means that all of the coe cients of g (x) must be 0. In particular, b 0 = 0: Thus, = b 0 2 F: Thus, E Gal(E=F ) F: So, E Gal(E=F ) = F: Therefore, F C E: Theorem 34 If F C E; then there are only a nite number of intermediate elds between 11

12 E and F: Proof. Assume F C E: Then, by the previous theorem, E is the splitting eld of some separable polynomial f (x) 2 F [x] : Let a 1 ; a 2 ; : : : ; a n be the roots of f (x) : Then E = F (a 1 ; a 2 ; : : : ; a n ) : Let K be an intermediate eld, so F < K < E: Then K C E: (You ll be proving this in homework!) Therefore, E Gal(E=K) = K: Note also that Gal (E=K) Gal (E=F ) : Thus, each intermediate eld, K; corresponds to a subgroup Gal (E=K) of Gal (E=F ) : We have jgal (E=F )j = [E : F ] ; by theorem 23 and we have [E : F ] < 1 since F C E: Thus, Gal (E=F ) is a nite group. Thus, Gal (E=F ) has a nite number of subgroups and thus there are nite number of intermediate elds K between F and E: Theorem 35 Let E be an extension of F with [E : F ] < 1: Then E is a simple extension of F if and only if there are only a nite number of intermediate elds between F and E: Proof. Let E be an extension of F with [E : F ] < 1: ( =) ) : Let E = F () and let f (x) = min (; F ) : Let B be an intermediate eld between F and E and let g (x) = min (; B) : Then g (x) = b 0 + b 1 x + + b m 1 x m 1 + x m for some b 0 ; b 1 ; : : : ; b m 1 2 B: At this point, we have [E : B] = deg (g (x)) = m: Let B 0 = F (b 0 ; b 1 ; : : : ; b m 1 ) : Then F B 0 B E: Also, since g (x) is irreducible over B; we have that g (x) is irreducible over B 0 : Also, notice that E = F () B 0 () E () = E: Thus, E = B 0 () : So, [E : B 0 ] = deg (g (x)) = m: Therefore, [E : B] = m = [E : B 0 ] : So, each intermediate eld produces a unique polynomial g (x) : However, g (x) is a monic divisor of f (x) and there are only a nite number of monic divisors of f (x) : Thus, there are only a nite number of intermediate elds between F and E: ( (= ) : Assume that there are only a nite number of intermediate elds between F and E: If jf j is nite, then jej is nite, since [E : F ] < 1: Since E is a nite elds, E = E f0g is a cyclic group (under multiplication). So, E = () for some 2 E and thus, E = F () : So, E is a simple extension of F: Now, suppose F is an in nite eld. Then jej = 1: Claim: For each choice of ; 2 E; there exists 2 E such that F (; ) = F () : 12

13 Let ; 2 E: Let T = f = + a : a 2 F g : Consider all elds of the form F () where 2 T: There are only a nite number of elds of this form, F () ; since F F () E; but there are an in nite number of a s to choose from to create the elements of T: Thus, there must be a 1 ; a 2 2 F such that F ( + a 1 ) = F ( + a 2 ) but a 1 6= a 2 : Let B = F ( + a 1 ) = F ( + a 2 ) : Note B F (; ) since + a 1 2 F (; ) : Then ( + a 1 ) ( + a 2 ) 2 B: So, (a 1 a 2 ) 2 B: We have a 1 a 2 2 F B: Thus, 2 B: So, since + a 1 2 B; we have 2 B: Therefore, F (; ) B: So, F (; ) = B and we have proven the claim. Now, consider all simple extension of F in E: Choose 2 E such that [F () : F ] is maximal. If F () 6= E; then there exists " 2 E F (). By the claim above, F (; ") = F () for some 2 E and then [F () : F ] > [F () : F ] which is a contradiction to the maximality of : Thus, F () = E and we have that E is a simple extension. Corollary 36 If F C E; then E is a simple extension of F: Corollary 37 If E = F ( 1 ; : : : ; n ) is a nite extension such that each i 2 E is separable over F; then there exists 2 E such that E = F () : ny Proof. For each i ; let p i (x) = min ( i ; F ) : Let f (x) = p i (x) (so f (x) is separable) and let B be a splitting eld for f (x) over F: Then, F C B: Thus, there are only a nite number of intermediate elds between F and B: Note that any splitting eld of f (x) must contain 1 ; : : : ; n ; so E B: Therefore, there are a nite number of intermediate elds between F and E: Therefore, E is a simple extension of F: Remark 38 Now we are going to see why this type of extension is called normal. Theorem 39 Suppose F < K < E is a chain of eld extensions such that F C E: Then i=1 F C K () Gal (E=K) C Gal (E=F ) : Proof. Suppose F < K < E is a chain of eld extensions such that F C E: ( =) ) : Suppose F C K: We know Gal (E=K) Gal (E=F ) ; so we simply need to prove normality. 13

14 Let 2 Gal (E=K) and 2 Gal (E=F ) : We need to prove 1 2 Gal (E=K) : Since 1 is easily seen to be an automorphism of E; this means that we need to show 1 xes K: That is, 1 (k) = k for all k 2 K: Claim: (K) K: Since F C K; we have that K is a splitting eld for some separable polynomial f (x) 2 F [x] : Let 1 ; : : : ; m be the roots of f (x) in K: Then K = F ( 1 ; : : : ; m ) : We know that must permute the roots of f (x) : Therefore, (K) = (F ( 1 ; : : : ; m )) = F ( ( 1 ) ; : : : ; ( m )) = F ( 1 ; : : : ; m ) = K and we have proven the claim. Now, let k 2 K: Then (k) 2 K: Thus, since 2 Gal (E=K) xes K; we have ( (k)) = (k) : Therefore, 1 (k) = 1 ( (k)) = k: Thus, Gal (E=K) C Gal (E=F ) : ( (= ) : Suppose Gal (E=K) C Gal (E=F ) : Since F C E; there are only a nite number of intermediate elds between F and E: So, since K E; there are only a nite number of intermediate elds between K and F: Thus, K is a simple extension of F: So, K = F () for some 2 K: Let f (x) = min (; F ) : Then, since one root of f (x) ; ; is in E and F C E; we have that all roots of f (x) are distinct and in E. So, f (x) is separable. Let 2 Gal (E=F ) : Claim: (K) = K: Since Gal (E=K) C Gal (E=F ) ; we have that 1 2 Gal (E=K) for all 2 Gal (E=K) : Thus, for each 2 Gal (E=K) ; we have 1 (k) = k for all k 2 K: Therefore, for each 2 Gal (E=K) ; we have ( (k)) = (k) for all k 2 K: Therefore, (k) 2 E Gal(E=K) : We have E Gal(E=K) = K since K C E: Therefore, (k) 2 K: Thus, (K) K: However, (K) and K both have the same nite dimension as a vector space over F (since is an automorphism). Thus, (K) = K: Returning to f (x) : For each i ; a root of f (x) ; there is an element i 2 Gal (E=F ) such that i () = i : Since i (K) = K and since 2 K; this means all roots of f (x) are in K: Therefore, K is a splitting eld for f (x) over F and thus, F C K: 4.1 Problems 1. Suppose F < K < E with F C E: Prove K C E: 2. Let E be an extension eld of F: Prove that E is a splitting eld of a separable polynomial in F [x] if and only if E is a splitting eld of a polynomial in F [x] with no 14

15 repeated roots. Conclude: F C E if and only if E is the splitting eld of a polynomial in F [x] with no repeated roots. 3. Suppose [E : F ] = 2: Prove F C E: 4. Prove lemma Suppose and are algebraic over Q: Prove Q (; ) = Q ( + r) for some r 2 Q: 5 The Fundamental Theorem of Galois Theory We are now ready to see the fundamental theorem of Galois theory. However, in truth, we have already seen most of this theorem in the previous sections. Be careful when applying this theorem. It can only be applied if F C E: If you use the fundamental theorem, make sure you rst argue how you know F C E: Theorem 40 Let F C E and let G = Gal (E=F ) : Then, 1. There is a one-to-one correspondence between intermediate elds between F and E and subgroups of Gal (E=F ) : The correspondences are K (an intermediate eld)! Gal (E=K) E H H (a subgroup of Gal (E=F )): (Note: Under this correspondence E corresponds to (1) G and F corresponds to G:) E (1) j K j F j H j G 2. (inclusion reversing) If K 1 ; K 2 are intermediate elds corresponding to subgroups H 1 ; H 2 of Gal (E=F ) (so K i = E H i ); then K 1 K 2 if and only if H 2 H 1 : 3. If the intermediate eld K corresponds to the subgroup H of G; then [E : K] = [H : (1)] = jhj and [K : F ] = [G : H] : 4. K C E. 15

16 5. F C K if and only if H C G (where H = Gal (E=K) is the subgroup corresponding to the eld K): In this case, Gal (K=F ) = G=H = Gal (E=F ) =Gal (E=K) : Proof. Let F C E and let G = Gal (E=F ) : 1. Let S be the set of intermediate elds between F and E and let T be the set of subgroups of G = Gal (E=F ) : De ne : S! T by (K) = Gal (E=K) : De ne : T! S by (H) = E H : Let K 2 S: Then (K) = (Gal (E=K)) = E Gal(E=K) : Since F K E and F C E; we have K C E: (yes, I m using a later part of the theorem part 4 here, but it s a part that you already proved in homework.). Therefore, E Gal(E=K) = K: So, (K) = K: Thus, = Id S : Let H 2 T: Then (H) = E H = Gal E=E H : Let J = Gal E=E H : We have F E H E; so, since F C E; we have E H C E: Thus, E J = E Gal(E=EH ) = E H and E : E H is nite. Thus, since jhj = E : E H ; H is a nite group. Similarly, E J C E; so jjj = E : E J is nite. We have E J = E H where H and J are nite subgroups of G = Gal (E=F ) : This gives jhj = E : E H = E : E J = jjj : It is easy to see that, by de nition, H Gal E=E H = J: so, since H and J are nite groups of equal size with H J; we have H = J: Therefore, (H) = J = H: So, = Id T : 2. ( =) ) : Need to show: K 1 K 2 =) Gal (E=K 2 ) Gal (E=K 1 ) Exercise. ((=) : Need to show H 1 H 2 =) E H 2 E H 1 : Exercise. 3. Since K corresponds to the subgroup H; we have K = E H : So, [E : K] = E : E H = jhj : Also, since F C E; we have F = E G where G = Gal (E=F ) : Thus, [E : F ] = E : E G = jgj : So, since [E : F ] = [E : K] [K : F ] ; we have jgj = jhj [K : F ] : Thus, [K : F ] = [G : H] : 16

17 4. You ve already shown K C E for homework. To see that Gal (E=K) = H where H is the subgroup corresponding to K; note that K = E H and then look at the proof of part (1). 5. F C K if and only if H C G (where H is the subgroup corresponding to the eld K) is just a restatement of theorem 39. Now, we want to show Gal (K=F ) = G=H where G = Gal (E=F ) and H = Gal (E=K) is the subgroup corresponding to K: So, we want to show Gal (K=F ) = Gal (E=F ) =Gal (E=K) : To prove this, use the fundamental homomorphism theorem on : Gal (E=F )! Gal (K=F ) de ned by () = j K for each 2 Gal (E=F ) : Homework. Example 41 Determine all sub elds of Q () where is a primitive 5th root of unity. How do you know you ve found them all? (Refer to example 11) 5.1 Problems 1. Suppose F C E and Gal (E=F ) is Abelian. Further, suppose F < K < E: Prove F C K: 2. Prove part 2 of the fundamental theorem. 3. Finish proving part 5 of the fundamental theorem. 17