All profinite groups are Galois groups (Thm 3.3.2 in J. Wilson's Profinite groups)

Question

I am reading about infinite Galois theory in Wilson's Profinite groups, and I have a problem in understanding the proof of Lemma 3.3.1 and Theorem 3.3.2 (here you can see them). In particular, I don't understand why, in the proof of the theorem, we can say that $K$ is an algebraic field extension of $k$ in order to apply the previous Lemma.

Secondly, where exactly is the hypothesis that $K/k$ is algebraic used in the proof of Lemma 3.3.1? I think that it is necessary in order to show that the index of the stabilizer of any element $x$ of $K$ under the action of $G$ is finite, because $G$ acts as a group of $k$-automorphisms of $K$, so it can map $x$ only in another zero of its minimum polynomial over $k$. (So also the indices of their normal cores are finite, and so also is the index of their finite intersection).

So having an algebraic field extension seems necessary to me to prove the finiteness condition which leads to the application of Artin's theorem in the proof of the Lemma.

Am I right or is there another proof which works also with an arbitrary field extension, so that we can omit to verify, in the proof of the Theorem, that $K$ over $k$ is algebraic?

Thanks in advance for sharing your knowledge.

Answer 1

Since nobody answered this old question, I give myself the answer, using an insight from a useful discussion with my teacher.

The point is that the statement of the Lemma is wrong, in the sense that it simply assumes more than it is required, and too much to be applied in the subsequent theorem; in fact, the extension does not have to be algebraic. It can simply be an arbitrary field extension.

To show that the index of the intersection of the normal cores is finite it is not necessary to do the above reasoning with the minimal polynomial (which obviously isn't defined if the extension is not algebraic!); instead, it suffices to show that this intersection is an open subgroup of $G$, which is profinite by hypothesis, hence compact. And by the properties of the group topology the index of an open subgroup of a compact topological group must be finite.