Open normal subgroups of Galois

Let $K$ be a number field, $v$ a finite place of $K$. Let $K'$ be the maximal extension of $K$ unramified at $v$.

For instance, for $K = \mathbf{Q}$ and $v = p$ a prime, $K' = \mathbf{Q}(\{\zeta_{n}, p\nmid n\})$, $\zeta_n$ primitive $n$-th roots of unity.

Let $\overline{K}$ an algebraic closure of $K$. Is $\text{Gal}(\overline{K}/K')$ an open normal subgroup of $\text{Gal}(\overline{K}/K)$?

For profinite groups $G$, open subgroups $H$ always have finite index, because the map $G \to G/H$ is continuous, $G/H$ has the discrete topology, and $G$ is compact. In your question, this index $[K' :K]$ is infinite, so the answer is no.
• Is it clear that $[K' : K]$ is always infinite? It certainly "often" is, but is this provably true for all $K$ and $v$? – David Loeffler Mar 16 '18 at 10:44
• Good question, I had only given it cursory thought. Can't one just add $n$-th roots of unity for $v(n) = 0$, as in the $\mathbb{Q}$-case? – Johann Haas Mar 16 '18 at 11:48
• Then you get an extension ramified at all primes of $K$ dividing $n$ (and there may be more than one). – David Loeffler Mar 16 '18 at 15:43
• I am missing something obvious, I'm afraid. Isn't this extension still unramified at $v$, which is what we care about? – Johann Haas Mar 16 '18 at 15:46
• Oops, I had misread the question! I was thinking of the maximal extension ramified only at $v$ in which case the question of whether or not $[K' : K]$ is finite is much more subtle. – David Loeffler Mar 16 '18 at 16:33