# 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 = { x^{2} + 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_{Q}E, then σ^{2} = 1_{E}. Therefore, the group Aut_{Q}E is actually a vector space over Z_{2}

(c) Aut_{Q}E is infinite and not denumerable.

(d) If B is a basis of Aut_{Q}E over Z_{2}, then B is infinite and not denumerable.

(e) Aut_{Q}E 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_{Q}E is denumerable.

(h) [E: Q] ≤ N_{0} , whence [E: Q] ≤ |Aut_{Q}EI.

**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(v_{1},, ... , 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(u_{1}, ... , 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_{K}E 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_{K}F 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_{K}F; 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 g_{1}, ... , g_{i} ϵ T such that F is a splitting field of {g_{1}, ... , g_{t}} over K (otherwise F would be infinite dimensional over K). Furthermore Aut_{K}F is a finite group by Lemma 2.8. If K_{o} is the fixed field of Aut_{K}F, then F is a Galois extension of K_{o} with[F: K_{o}] = IAut_{K}FI 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_{K}FI. 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 g_{1}, 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 g_{1}; then (K(u): K] = deg g_{i} = s by Theorem1.6 and the number of distinct roots of g_{1} is s since g_{1} 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_{K}F to the set of all roots of g_{1} in F, given by σH|→σu.Therefore, [Aut_{K}F : 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 {g_{1}, ... , g_{t}) over K(u) and over K(v) (Exercise 2), τ extends to an automorphism σ ϵ Aut_{K}F with σ(u) = v (Theorem 3.8). Therefore, every root of g_{1} is the image of some coset of Hand [Aut_{K}F : 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.

## Step by Step Answer:

**Related Book For**

## Algebra Graduate Texts In Mathematics 73

**ISBN:** 9780387905181

8th Edition

**Authors:** Thomas W. Hungerford

**Question Details**

**5**- Fields and Galois Theory

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