# Open subgroup of $Gal(k_s| k)$

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.

• Yes, that’s part of the definition of what it means to be open. – Lubin Feb 20 at 22:05
• 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 – reuns Feb 20 at 22:22
• @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. – Simonsays Feb 21 at 6:26