# Let F be an algebraic closure of the field Q of rational numbers and let E F be a

## Question:

Let F be an algebraic closure of the field Q of rational numbers and let E ⊂ F be a splitting field over Q of the set S = { x2 + a | a ϵ Q} so that E is algebraic and Galois over Q (Theorem 3.11).

(a) E = Q(X) where (b) If σ ϵ AutQE, then σ2 = 1E. Therefore, the group AutQE is actually a vector space over Z2

(c) AutQE is infinite and not denumerable.

(d) If B is a basis of AutQE over Z2, then B is infinite and not denumerable.

(e) AutQE has an infinite nondenumerable number of subgroups of index 2.

(f) The set of extension fields of Q contained in E of dimension 2 over Q is denumerable.

(g) The set of closed subgroups of index 2 in AutQE is denumerable.

(h) [E: Q] ≤ N0 , whence [E: Q] ≤ |AutQEI.

data from theorem 3.11

If F is an extension field of K, then the following statements are equivalent.

(i) F is algebraic and Galois over K;

(ii) F is separable over K and F is a splitting field over K of a set S of polynomials in K[x]; (iii)

F is a splitting field over K of a set T of separable polynomials in K[x].

REMARKS.

If F is finite dimensional over K, then statements (ii) and (iii) can be slightly sharpened. In particular (iii) may be replaced by: F is a splitting field over K of a polynomial ∫ ϵ K[x] whose irreducible factors are separable (Exercise 13).

PROOF OF 3.11.

(i) => (ii) and (iii). If u ϵ F has irreducible polynomial ∫, then the first part of the proof of Lemma 2.13 (with E = F) carries over verbatim and shows that ∫ splits in F[x] into a product of distinct linear factors. Hence u is separable over K. Let {ui I i ϵI} be a basis of F over K and for each i ϵ I let ∫i ϵ K[x] be the irreducible polynomial of ui. The preceding remarks show that each ∫is separable and splits in F[x]. Therefore F is a splitting field over K of S ={∫i| i ϵ I). (ii)==> (iii) Let ∫ ϵ S and let g ϵ K[x] be a monic irreducible factor of ∫. Since ∫ splits in F[x], g must be the irreducible polynomial of some u ϵ F. Since F is separable over K, g is necessarily separable. It follows that F is a splitting field over K of the set T of separable polynomials consisting of all monic irreducible factors (in K[x]) of polynomials in S (see Exercise 4). (iii) ==> (i) F is algebraic over K since any splitting field over K is an algebraic extension. If u ϵ F - K, then u ϵ K(v1,, ... , vn) with each va root of some ∫i ϵT by the definition of a splitting field and Theorem 1 .3(vii). Thus u ϵ E = K(u1, ... , ur) where the ui are all the roots of ∫i, ... , ∫n in F. Hence [E: K] is finite by Theorem 1.12. Since each ∫i splits in F, E is a splitting field over K of the finite set {∫i, ... , ∫n} or equivalently, of ∫= ∫12· · · ∫n. Assume for now that the theorem is true in the finite dimensional case. Then E is Galois over K and hence there exists τ ϵ AutKE such that τ(u) ≠ u. Since F is a splitting field of T over E (Exercise 2), τ extends to an automorphism σ ϵ AutKF such that σ(u) = τ(u) ≠ u by Theorem 3.8. Therefore, u (which was an arbitrary element of F - K) is not in the fixed field of AutKF; that is, F is Galois over K. The argument in the preceding paragraph shows that we need only prove the theorem when [F : K] is finite. In this case there exist a finite number of polynomials g1, ... , gi ϵ T such that F is a splitting field of {g1, ... , gt} over K (otherwise F would be infinite dimensional over K). Furthermore AutKF is a finite group by Lemma 2.8. If Ko is the fixed field of AutKF, then F is a Galois extension of Ko with[F: Ko] = IAutKFI by Artin's Theorem 2.15 and the Fundamental Theorem. Thus in order to show that F is Galois over K (that is, K = Ko) it suffices to show that [F: K] = IAutKFI. We proceed by induction on n = [F: K], with the case n = 1 being trivial. If n > I, then one of the gi, say g1, has degree s > 1 (otherwise all the roots of the gi lie in K and F = K). Let u ϵ F be a root of g1; then (K(u): K] = deg gi = s by Theorem1.6 and the number of distinct roots of g1 is s since g1 is separable. The second paragraph of the proof of Lemma 2.8 (with L = K, M = K(u) and ∫ = gi) shows that there is an injective map from the set of all left cosets of H = AutK(u)F in AutKF to the set of all roots of g1 in F, given by σH|→σu.Therefore, [AutKF : H] ≤ s. Now if v ϵ F is any other root of gi, there is an isomorphism τ : K(u) ≅ K(v) with τ(u) = v and τ I K = IK by Corollary 1.9. Since Fis a splitting field of {g1, ... , gt) over K(u) and over K(v) (Exercise 2), τ extends to an automorphism σ ϵ AutKF with σ(u) = v (Theorem 3.8). Therefore, every root of g1 is the image of some coset of Hand [AutKF : H] = s. Furthermore, F is a splitting field over K(u) of the set of all irreducible factors h(in K(u)[x]) of the polynomials gi (Exercise 4). Each hi is clearly separable since it divides some gi• Since [F: K(u)] = n/ s K(u)FI = IHI. Therefore, and the proof is complete.