Section 14.2. Let K=F be a Galois extension. The main goal of this lecture is to prove the Funda- mental Theorem of Galois Theory, which makes precise the relation between the subgroups H ⊂ Gal(K=F ) and the subfields F ⊂ E ⊂ K. Let us do some preparatory work first. Lemma 1. A finite-dimensional vector space over an infinite field can not be presented as a union of a finite number of its proper subspaces. s Proof. Let V = [i=1Vs, where Vs are proper subspaces of the vector space V . For every Vi let us choose a non-zero linear function li on V , which turns into zero on Vi. Now, let us Qs consider a polynomial p = i=1 li. Then for any v 2 V we must have p(v) = 0 which would imply that p is constantly zero, and we arrive at a contradiction. Theorem 2. Let K=F be a field extension of degree n, and G ⊂ Aut(K=F ) be a subgroup of Aut(K=F ). Denote by KG the fixed field of G. Then KG = F if and only if jGj = n. Moreover, if KG = F then for any fields P and Q such that F ⊂ P ⊂ Q ⊂ K, any homomorphism ': P ! K extends to a homomorphism : Q ! K in precisely jQ : P j ways. Proof. By the definition of a fixed field, we have G ⊂ Aut(K=KG). Therefore, G jGj 6 jK : K j 6 jK : F j = n: Then jGj = n implies KG = F . Conversely, let KG = F . We shall first show that G = Aut(K=F ), and then prove the equality j Aut(K=F )j = n. For any element α 2 K let fα1; : : : ; αmg ⊂ K be its G-orbit. Then m Y G f(x) = (x − αi) 2 K [x] = F [x](∗) i=1 is the minimal polynomial of the element α 2 K. Note that for any ' 2 Aut(K=F ), the ele- ment '(α) is also a root of f(x). Then, there exists an element gα 2 G such that '(α) = gα(α). Now, if K is a finite field, we can choose α to be the generator of the cyclic group K×. Then ' = gα 2 G. If K is an infinite field, then for any g 2 G let us set Kg = fα 2 K j '(α) = g(α)g ⊂ K: Clearly, Kg is a vector subspace (in fact, even a subfield) of K, and it follows from the above discussion that [ K = Kg: g2G By previous lemma we have K = Kg for some element g 2 G, which in turn implies G = Aut(K=F ). Now, let us pause for a second and prove the second statement of the theorem. Since any finite field extension can be obtained as a chain of simple extensions, it is enough to prove the statement in the case when Q = P (α) is a simple field extension. Let p(x) be the minimal polynomial for α over P . Then p(x) divides f(x) in the ring P [x], where f(x) is the minimal polynomial of α over F . Now, let p'(x) 2 K[x] be the polynomial obtained from p(x) by applying the homomorphism ' to its coefficients. Then p'(x) divides f(x) in the ring K[x] = '(P )[x] and therefore decomposes into linear factors in K[x]. The latter implies that the homomorphism ': P ! K extends to a homomorphism : Q ! K in precisely jQ : P j ways. Applying the second statement of the theorem to the case P = F and Q = K, we obtain j Aut(K=F )j = n. 1 2 Remark 3. Note that in the course of proving the previous theorem we showed that if K is a field, G ⊂ Aut(K) is a finite subgroup, and KG is the fixed subfield of G, then (1) K is algebraic over KG; (2) G = Aut(K=KG). Corollary 4. If K is a field, G ⊂ Aut(K) is a finite subgroup, and F = KG is the fixed subfield of G, then K is a finite extension of F . Proof. Assume that jK : F j = 1. Since K is algebraic over F , there exists an infinite chain F = F0 ⊂ F1 ⊂ F2 ⊂ ::: of simple extensions Fi+1=Fi, where Fi ⊂ K for all i. Note that for any N 2 Z+ there exists a number i such that jFi : F j > N. Since G = Aut(K=F ) and j Aut(K=F )j > j Aut(Fi=F )j for any i, we have j Aut(G)j = 1 which contradicts our assumption. Corollary 5. If K is a field, G ⊂ Aut(K) is a finite subgroup, and KG is the fixed subfield of G, then K=KG is a Galois extension and G is its Galois group. Proof. Since K=KG is a finite extension, we can apply Theorem 2 and conclude that G G G G jGj = jK : K j. Since jGj 6 j Aut(K=K ))j 6 jK : K j, we have G = Aut(K=K ) G G and j Aut(K=K )j = jK=K j in the statement of the Corollary follows. Corollary 6. Let K be a field and G1;G2 ⊂ Aut(K) be a pair of distinct subgroups. Then the fixed fields KG1 and KG2 are also distinct. G G G G Proof. If K 1 = K 2 , then K 1 is fixed by G2 and G2 ⊂ G1 = Aut(K=K 1 ). Similarly, we get G1 ⊂ G2. Then we have G1 = G2 which contradicts our assumption. Last time we proved that the splitting field K of a separable polynomial f(x) 2 F [x] is a Galois extension of F . Let us now prove the converse, which will characterize Galois extensions. Theorem 7. The extension K=F is Galois if and only if K is the splitting field of some separable polynomial f(x) 2 F [x]. Furthermore, if K=F is Galois then every irreducible polynomial p(x) 2 F [x] which has a root in K is separable and has all its roots in K. Proof. As we just mentioned, we only need to prove the \only if" statement. Let K=F be a Galois extension, G = Gal(K=F ), and α 2 K be a root of an irreducible polynomial p(x) 2 F [x]. Then we have p(x) = f(x) 2 KG[x] = F [x], where the polynomial f(x) is defined by (∗), which implies the second statement of the Theorem. Now, let β1; : : : ; βn be a basis of K over F . For each i = 1; : : : ; n, let pi(x) 2 F [x] be the minimal polynomial of βi. Define g(x) 2 F [x] to be a polynomial obtained by removing any multiple factors from the product p1(x) : : : pn(x). Then g(x) is separable and K is its splitting field. Recall that an extension K=F is normal if every irreducible polynomial p(x) 2 F [x] which has a root in K splits into linear factors over K. Then previous results yield the following equivalent descriptions of Galois extensions. Corollary 8. Let K=F be a field extension with j Aut(K=F )j < 1. Then the following 4 statements are equivalent: (1) K=F is a Galois extension, that is j Aut(K=F )j = jK : F j; (2) K is a splitting field of a separable polynomial in F [x]; (3) F is the fixed field of Aut(K=F ); 3 (4) K is a finite, normal, and separable extension of F . Now we are ready to prove the Fundamental Theorem of Galois Theory. Let K=F be a Galois extension, and G = Gal(K=F ). To any subfield F ⊂ E ⊂ K we associate a subgroup GE = fg 2 G j g(α) = α 8α 2 Eg ⊂ G of elements which act trivially on E. On the other hand, to any subgroup H ⊂ G we associate its fixed subfield KH = fα 2 K j h(α) = α 8h 2 Hg ⊂ K: Theorem 9 (Fundamental Theorem of Galois Theory). Let K=F be a Galois extension, H and G = Gal(K=F ). Then mappings E 7! GE and H 7! K are mutually inverse, and therefore establish a bijction between the subfields of K which contain F and the subgroups of G. Moreover, under this bijection subfields E ⊂ K, which are Galois over F correspond to normal subgroups of G. Proof. First, let us note that by Theorem 2 we have H jK : K j = jHj and jGEj = jK : Ej: G G Now, on one hand we have E ⊂ K E , on the other jK : K E j = jGEj = jK : Ej which GE H implies E = K . Similarly, H ⊂ GKH , and jGKH j = jK : K j = jHj implies H = GKH . This proves the first statement. Now, by Theorem 2 every automorphism ' 2 Aut(E=F ) extends to an automorphism 2 Aut(K=F ) in exactly jK : Ej = jGEj ways, namely one can compose ' with any element σ 2 GE to obtain σ ◦ ' 2 G. In general, j Aut(E=F )j 6 jE : F j and the condi- tion that E=F is Galois is equivalent to the statement that the subgroup which preserves (but not necessarily fixes) E induces exactly jE : F j distinct automorphisms on E. Now, jE : F j = jK : F j=jK : Ej = jG : GEj, and therefore if E=F is Galois the subgroup which preserves E would have jG : GEj · jGEj = jGj elements. Thus we conclude that the whole Galois group G = Gal(K=F ) has to preserve E in order for E=F to be Galois.