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I got this question while I am studying Galois correspondence.

Let $K/F$ be an infinite extension and $G = \mathrm{Aut}(K/F)$. Let $H$ be a subgroup of $G$ with finite index and $K^H$ be the fixed field of $H$. Is it true that $[K^H:F]= (G:H)$?

For finite extension, I verified this is true. Is it true in infinite case also?

Thanks

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The Galois correspondence extends to infinite Galois extensions, but to get a correspondence you need to restrict attention to closed subgroups (see for example Theorem 1.3.11 in Szamuely's Galois Groups and Fundamental Groups). I don't know whether every subgroup of finite index is necessarily closed. –  Qiaochu Yuan Jun 28 '12 at 3:57
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A subgroup of finite index is closed iff it is open. Every Galois group is a profinite group, and every profinite group is a Galois group of some algebraic extension of fields. There are some profinite groups that contain finite-index subgroups that are not closed. For example, the countable product $\prod_{n \geq 1} {\mathbf Z}/2{\mathbf Z}$ has uncountably many index-two subgroups but only countably many index-two open subgroups, so it has many index-two subgroups that are not closed. –  KCd Jun 28 '12 at 4:04

1 Answer 1

KCd's comment answers the question, but let's be a little more explicit. Let $F = \mathbb{Q}(x_1, x_2, ...)$ and let $K = \mathbb{Q}(\sqrt{x_1}, \sqrt{x_2}, ...)$. The Galois group $G = \text{Aut}(K/F)$ is $\prod_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ with the product topology. The subgroup $\bigoplus_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ is countable, so is a proper subgroup of $G$. Conditional on the axiom of choice (but actually we only need the ultrafilter lemma here) this subgroup is contained in a maximal subgroup $H$. The quotient $G/H$ is a simple group in which every element has order $2$, so must be $\mathbb{Z}/2\mathbb{Z}$.

So $H$ has index $2$. But since it contains the topological generators of $G$ we conclude that $K^H = F$.

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