Exercise 10, § 10

Let K be a perfect field such that for every integer $n > 1$ there exists a Galois extension N of K such that $\mathrm{Gal}(N/K)$ is isomorphic to the alternating group $\mathfrak{A}_n$; for example, for every field k of characteristic 0 the field $K = k(X_e)_{e \in \mathbf{N}}$ of rational fractions in an infinity of indeterminates has this property (V, p. 59, Example 5). a) Let A be a finite set of integers $> 1$ and let $\Omega$ be an algebraic closure of K. An element $x \in \Omega$ is said to be A-constructible if there exists an increasing sequence $K = E_0 \subset E_1 \subset \ldots \subset E_r \subset \Omega$ of extensions of K such that each of the degrees $[E_j : E_{j-1}]$ belongs to A for $1 \leq j \leq r$ and $x \in E_r$. Show that there exists a maximal element L in the set of subextensions $E \subset \Omega$ such that each $x \in E$ is A-constructible. For each $x \notin L$ show that the degree $[L(x) : L]$ belongs to $N - A$. b) Let a be the greatest element of A and let $n$ be an integer such that $n \geq \max(a + 1, 5)$. Show that there exists an extension M of L such that $[M : L] = n$. (Observe firstly that there exists an irreducible polynomial $f \in K[K]$ of degree $n$ such that if N is the splitting field of f, $\mathrm{Gal}(N/K)$ is isomorphic to $\mathfrak{A}_n$; let $y \in \Omega$ be a root of f in $\Omega$ and let $E = K(y)$. Show that we necessarily have $E \cap L = K$ and therefore $L(E)$ is the desired field. Argue by contradiction, noting that the elements of $E \cap L$ are A-constructible and that the relation $E \cap L \neq K$ would imply the existence of a normal subgroup $\Gamma$ of $\mathfrak{A}_n$ such that $1 < (\mathfrak{A}_n : \Gamma) < n$, which would contradict the simplicity of $\mathfrak{A}_n$ (I, p. 143, Ex. 16).)