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Let $E/F$ be infinite Galois extension.

Associate to each intermediate field $L$, the subgroup $\mbox{Gal}(E/L)$ of $\mbox{Gal}(E/F)$.

It is mentioned in almost all the resources, which introduce this theory, that the above association may not be surjective, and discuss 2-3 common examples.

My question is very basic, but I did not find comments in the direction of question in any reference.

Question: The association$L\mapsto \mbox{Gal}(E/L)$ is (sometimes not surjective) or (never surjective)?

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    $\begingroup$ I think that the question you have asked might be more complicated than it appears at first glance. I attempted to answer it by looking at whether there exists an infinite Galois extension $E/F$ such that every subgroup of $Gal(E/F)$ is closed in the Krull topology. By the infinite Galois correspondence, $L \mapsto Gal(E/L)$ would then be a bijection from the collection of intermediate field extensions $F \leq L \leq E$ to the collection of subgroups of $Gal(E/F)$. But I did not come across any examples of such a Galois group. Perhaps someone more knowledgeable than I can clarify this point. $\endgroup$
    – Menander I
    Commented Apr 1, 2022 at 9:18
  • $\begingroup$ @Menander: Thanks for comments; I am very new in infinite Galois theory. But, when looking several notes or books on this topic, including standard algebra texts, I did not find any comment considering the natural question I posed. I think, this question will be raised with probability 0.9 in classroom where the difference between finite and infinite Galois correspondence is stated. $\endgroup$
    – Maths Rahul
    Commented Apr 1, 2022 at 9:22

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It is never surjective. To prove this, recall that $\mbox{Gal}(E/F)$ always has the natural structure of a profinite group (which is infinite if $[E:F]$ is infinite), and any subgroup of the form $\mbox{Gal}(E/L)$ is a closed subgroup. So in particular, $\mbox{Gal}(E/F)$ is an infinite compact Hausdorff space with no isolated points (since if it had isolated points it would be discrete since it is homogeneous and thus finite by compactness). Thus $\mbox{Gal}(E/F)$ must be uncountable, and by the same argument any subgroup of the form $\mbox{Gal}(E/L)$ must be either finite or uncountable. So, if you take a countably infinite subset of $\mbox{Gal}(E/F)$, then the subgroup it generates is countably infinite so cannot have the form $\mbox{Gal}(E/L)$.

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