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 Aut_QE, then σ² = 1_E. Therefore, the group Aut_QE is actually a vector space over Z₂ 

(c) Aut_QE is infinite and not denumerable. 

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

(e) Aut_QE 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 Aut_QE is denumerable.

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

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 {u_i 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 u_i. The preceding remarks show that each ∫_i 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,, ... , v_n) with each v_i a root of some ∫_i ϵT by the definition of a splitting field and Theorem 1 .3(vii). Thus u ϵ E = K(u₁, ... , u_r) where the u_i 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 ∫= ∫₁∫₂· · · ∫_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 Aut_KE such that τ(u) ≠ u. Since F is a splitting field of T over E (Exercise 2), τ extends to an automorphism σ ϵ Aut_KF 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 Aut_KF; 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, ... , g_i ϵ T such that F is a splitting field of {g₁, ... , g_t} over K (otherwise F would be infinite dimensional over K). Furthermore Aut_KF is a finite group by Lemma 2.8. If K_o is the fixed field of Aut_KF, then F is a Galois extension of K_o with[F: K_o] = IAut_KFI by Artin's Theorem 2.15 and the Fundamental Theorem. Thus in order to show that F is Galois over K (that is, K = K_o) it suffices to show that [F: K] = IAut_KFI. We proceed by induction on n = [F: K], with the case n = 1 being trivial. If n > I, then one of the g_i, say g₁, has degree s > 1 (otherwise all the roots of the g_i lie in K and F = K). Let u ϵ F be a root of g₁; then (K(u): K] = deg g_i = 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 ∫ = g_i) shows that there is an injective map from the set of all left cosets of H = Aut_K(u)F in Aut_KF to the set of all roots of g₁ in F, given by σH|→σu.Therefore, [Aut_KF : H] ≤ s. Now if v ϵ F is any other root of g_i, there is an isomorphism τ : K(u) ≅ K(v) with τ(u) = v and τ I K = I_K by Corollary 1.9. Since Fis a splitting field of {g₁, ... , g_t) over K(u) and over K(v) (Exercise 2), τ extends to an automorphism σ ϵ Aut_KF with σ(u) = v (Theorem 3.8). Therefore, every root of g₁ is the image of some coset of Hand [Aut_KF : H] = s. Furthermore, F is a splitting field over K(u) of the set of all irreducible factors h_i (in K(u)[x]) of the polynomials g_i (Exercise 4). Each h_i is clearly separable since it divides some g_i• Since [F: K(u)] = n/ s K(u)FI = IHI. Therefore, and the proof is complete. 

Transcribed Image Text:

X = {√√pp = -1 or p is a prime integer). | P