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This answer states that $\mathbb{C}/\mathbb{Q}$ is a Galois extension, and from this wikipedia page I deduced that there is a correspondence between intermediate field extensions of $\mathbb{C}/\mathbb{Q}$ and the closed subgroups of $\text{Gal}(\mathbb{C}/\mathbb{Q})$ under the Krull topology. But it also mention that the Galois group becomes a profinite group.

On the other hand, this answer mention that any element of $\text{Aut}(\mathbb{C}$) always fixes $\mathbb{Q}$. And from what I can understand in this discussion, $\text{Aut}(\mathbb{C})$ is not profinite making $\text{Gal}(\mathbb{C}/\mathbb{Q})$ also not profinite.

My question is: does the fundamental theorem mentioned in the wikipedia page still holds for $\mathbb{C}/\mathbb{Q}$ even if $\text{Gal}(\mathbb{C}/\mathbb{Q})$ is not profinite? And if so, is the definition of the topology on the Galois group still the same?

AyamKiwi
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    Usually $\Bbb C\mid \Bbb Q$ is not considered to be a Galois extension. It is not algebraic. But some results do hold nevertheless (not all results, though). For the infinite version of the fundamental theorem, I don't see how it works, because the Krull topology turns out to be equivalent to the profinite topology. – Dietrich Burde Dec 22 '24 at 14:20

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