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Let $k$ be a field with fixed separable closure $k_s$. Consider the profinite group $Gal(k_s|k)=: G$ with the corresponding topology. Furthermore, let $U \subset G$ be an open subgroup. Every subgroup of G can be naturally identified with a group of the form $Gal(k_s|L)$, w./ $L|k$ some subextension of $k$. So far so good, just the following thing buggs me:

My question: Is the extenstion $L|k$ in this particular case ($U$ open subgroup) necessarily finite? Thanks in advance.

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    $\begingroup$ Yes, that’s part of the definition of what it means to be open. $\endgroup$ – Lubin Feb 20 at 22:05
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    $\begingroup$ You may want to look first at $k = \mathbb{Q}$ and replace $k_s$ by $F=\bigcup_n \mathbb{Q}(\zeta_{p^n})$ then $Gal(F/k) = \mathbb{Z}_p^\times= \varprojlim (\mathbb{Z}_p/p^n)^\times = \varprojlim Gal(\mathbb{Q}(\zeta_{p^n})/\mathbb{Q})$ where $\mathbb{Q}(\zeta_{p^n})/k$ is a tower of Galois extensions limiting to $F$ and $L/k$ is finite iff $L$ is contained in one of those $\mathbb{Q}(\zeta_{p^n})$. So $H = (\mathbb{Z}_p^\times)^2$ isn't an open subgroup and $F^H/k$ isn't a finite extension $\endgroup$ – reuns Feb 20 at 22:22
  • $\begingroup$ @Lubin I know that openness is equivalent to being a closed subgroup of finite index. Does finite index imply that $L$ is finite? I don't see it. $\endgroup$ – Simonsays Feb 21 at 6:26

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