TITLE: Are finite index subgroups of inertia closed?
QUESTION [7 upvotes]: Let $K$ be a finite extension of the $p$-adic numbers. $G_K$ be its absolute Galois group and $I_K$ the inertia subgroup. Are finite index subgroups of $I_K$ closed in its profinite topology?
By a result of  Nikolov/Segal it suffices to show, that inertia is topologically finitely generated.

What has been proved in the direction of this or the analogous question on wild inertia?

Do the results of Jannsen/Koch/Wingberg on the finite generator rank of the local Galois group help? I couldn't find anything on this and considering its importance, it surely would be mentioned somewhere, if it were always known. So maybe someone has some conditional results?

REPLY [7 votes]: Here is another way to see that the wild inertia group $P_K=\mathrm{Gal}(\bar K|T)$ (where $T$ is the maximal tamely ramified extension of $K$ (a finite extension of $\mathbf{Q}_p$) in $\bar K$, an algebraic closure of $K$, is not finitely generated (as a profinite group).  One doesn't need class field theory, only Kummer theory and the fact that a pro-$p$ group $G$ is finitely generated if and only if the $\mathbf{F}_p$-space $G/D$ is finite, where $D$ is the closed subgroup generated by all commutators and all $p$-th powers in $D$.
In the case at hand, $P_K/D=\mathrm{Gal}(T(\root p\of{T^\times})|T)$, and it can be easily seen that the $\mathbf{F}_p$-space $T^\times/T^{\times p}$ is infinite.