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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

  • 0
    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.2012-06-28
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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.2012-06-28

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