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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?

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    $\begingroup$ Dear Konrad, Let $K_n$ be the degree $n$ unramified extension of $K$. Then $I_K = I_{K_n}$. Class field theory shows that $I_{K_n}^{ab} = \mathcal O_{K_n}^{\times}$, whose pro-$p$-part, modulo its torsion subgroup (the $p$th power roots of unity contained in $K_n$), is isomorphic to $\mathbb Z_p^{nd}$, if $d$ denotes the degree of $K$ over $\mathbb Q_p$. Thus $I_K^{\mathrm{wild}}$ admits a surjection onto $\mathbb Z_p^{nd}$ for all $n$, and so is not topologically finitely generated. Regards, Matthew $\endgroup$
    – Emerton
    Commented Apr 24, 2013 at 10:30
  • $\begingroup$ @Emerton: Thank you for your nice comment. Maybe in the light of this mathoverflow.net/questions/82177/… question and its answers, one can't hope for much more concerning the closedness of finite index normal subgroups. I will think some more about it. Thanks again! $\endgroup$
    – Konrad
    Commented Apr 24, 2013 at 12:05
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    $\begingroup$ @Konrad: Refining Emerton's comment, since $I_K$ is the absolute Galois group of a complete discretely-valued field with alg. closed residue field, Serre's geometric local class field theory applies to $I_K$. Thus, the commutative profinite $I_K^{\rm{ab}}$ admits $p$-torsion Hausdorff quotients finite-dimensional over $\mathbf{F}_p$ with arbitrarily large dimension, so it admits a continuous homomorphism onto the countably infinite direct product $P = \prod_{n=1}^{\infty} \mathbf{F}_p$. But $P$ has finite-index subgroups that are not closed, so the answer to your question is negative. $\endgroup$
    – user29283
    Commented Apr 24, 2013 at 14:48
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    $\begingroup$ Oops, when I wrote that the answer to your question is "negative" I overlooked the form in which you posed it. I should have said that the answer to your question is "affirmative": there are finite-index subgroups of $I_K$ that are not closed (for reasons indicated in my preceding comment, since a countably infinite direct product of $\mathbf{F}_p$'s has only countably many closed subgroups of finite index but has uncountable dimension over $\mathbf{F}_p$ and hence has uncountably many finite-index subgroups, so "most" of them are not closed). $\endgroup$
    – user29283
    Commented Apr 24, 2013 at 16:15
  • $\begingroup$ @xuhan, I will accept Chandan's answer, since yours was written as a comment. I will surely have a look at the reference. Thank you for it. $\endgroup$
    – Konrad
    Commented Apr 24, 2013 at 19:32

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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.

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    $\begingroup$ Note that proving that a profinite group is not topologically finitely generated does not imply that there are non-closed finite index subgroups. (Example: $\prod_p(\mathbf{Z}/p\mathbf{Z})^p$ where $p$ ranges over primes is not t.f.g. but finite index subgroups are closed.) But it's indeed true for a pro-$p$-group. $\endgroup$
    – YCor
    Commented Aug 10, 2013 at 5:21
  • $\begingroup$ @YvesCornulier, thanks. Fortunately the profinite group in question (the wild inertia group) is a pro-$p$-group. $\endgroup$
    – Chandan Singh Dalawat
    Commented Aug 10, 2013 at 6:05

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