Exercise 9, § 10

Let K be a field of characteristic p, $\Omega$ an algebraically closed extension of K, x an element of $\Omega$ such that $x \notin K$ and M a maximal element of the set of subextensions of $\Omega$ not containing x (V, p. 148, Ex. 9), so that $\Omega$ is an algebraic extension of M (\emph{loc. cit.}). a) Show that if M is not a perfect field, $\Omega$ is a p-radical extension of M, union of the fields $M(x^{p^{-n}})$ for $n \geq 0$ (observe that for every element $y \in \Omega$ p-radical over M, the field $M(y)$ contains $M(x)$; deduce that we necessarily have $[M(x) : M] = p$, because every element of $\Omega$ separable over M is necessarily in M). b) Show that if M is perfect, $M(x)$ is a Galois extension of M, of prime degree q, and for every subextension N of $\Omega$ which is Galois of finite degree over M, the degree $[N : M]$ is a power of q. (Take for q a prime number dividing $[M(x) : M]$; note that every extension of M distinct from M and contained in $\Omega$ contains $M(x)$, and use Ex. 8, a).)