# What are the open normal subgroups of the inertia group of a local field?

Let $K$ be a non-Archimedean local field, i.e., complete with respect to a non-trivial, non-archimedean discrete absolute value, with finite residue field $k$ of characteristic $p\neq 0$. Also let $K_s$ be a fixed separable closure of $K$, and $K_{un}$ (resp. $K_t$) the maximal unramified (resp. tamely ramified) extensions of $K$ inside $K_s$. Finally let $I_K=Gal(K_s/K_{un})$ be the inertia group of $K$ and $P_K=Gal(K_s/K_t)$. My basic question is whether or not the following statement is true:

For every positive integer $e$ prime to $p$, there exists a unique open normal subgroup of $I_K$ of index $e$.

I don't recall ever seeing this explicitly stated, but I think it's plausible for the following reason. An open normal subgroup of $I_K$ of index $e$ (with $e$ as above) is of the form $Gal(K_s/F)$ with $F/K_{un}$ Galois of degree $e$. Such an extension is necessarily totally tamely ramified (totally ramified because the residue field of $K_{un}$ is algebraically closed and tame because $e$ is prime to $p$). An example of such an extension is $K_{un}(\pi^{1/e})$, where $\pi$ is a uniformizer for $K$, which is Galois of degree $e$ since $K_{un}$ contains $\mu_e$ and $X^e-\pi$ is Eisenstein (over the integers of $K_{un}$). In fact, $K_t$ is the union of such extensions over integers prime to $p$.

If I knew (as in complete case) that every TTR extension of $K_{un}$ of degree $e$ had this form, it would imply that $K_{un}(\pi^{1/e})$ is necessarily the unique extension of $K_{un}$ of degree $e$ (since the unit group of $K_{un}$ is $e$-divisible by Hensel's lemma), which gives the statement I'm after (unless I've done something wrong).

My guess is that maybe the assertion relating TTR extensions and $e$-th roots of uniformizers really only requires a valuation ring where Hensel's lemma is valid (I guess these are called Henselian), but I've also never seen this asserted before, so I'm sort of skeptical.

• Yes, if $L$ is frac. field of hens. dvr (such as $K^{\rm{un}}$ above) then for finite sep. ext'n field $F/L$ (equipped with unique valuation extending one on $L$, uniqueness due to henselian hypothesis on $L$), the map $\widehat{L} \otimes_L F \rightarrow \widehat{F}$ is an isom, so $\widehat{L} \otimes_L F$ is a field. By using Krasner's Lemma, one then shows the functor $F \mapsto \widehat{L} \otimes_L F$ from finite etale $L$-algebras to finite etale $\widehat{L}$-algebras is equiv. of categories, so likewise for fields giving isom. of Galois gps. Thus, $I_K = G_ {\widehat{K^{\rm{un}}}}$. – BCnrd Jul 28 '10 at 21:11
• I should add that the invariance of Galois gp with respect to completion for henselian valued fields is a standard fact in the theory, even if you don't see it in some basic books. It is addressed in a self-contained manner in sources such as the BGR book on rigid-analytic geometry & Berkovich's IHES paper on etale cohomology for non-archimedean analytic spaces. But I recommend you figure out the proof for yourself in the discretely-valued case, following the sketch in the preceding comment. (That's what I did before I ever found a reference, and it was an instructive experience.) – BCnrd Jul 28 '10 at 21:15
• It's not as general, but for your specific question, you can reduce to your known case of TTR extns of local fields. Your TTR problem is unchanged if you replace $K$ by a finite unram. extn. of $K$: units in $K_{un}$ have all their $e$-th roots in $K_{un}$, so in your example $\pi$ can be any uniformizer of $K_{un}$, not just of $K$, without changing the example. Therefore if you pick a field generator $\alpha$ of your TTR extn. of $K_{un}$ and then replace $K$ by the finite extension $K'/K$ generated by the coefficents of $\alpha$'s min. polynomial over $K_{un}$, $K'(\alpha)/K'$ is TTR... – KConrad Jul 28 '10 at 22:52
• Thank you both for your comments. I appreciate it very much. @KConrad Your comment does answer my specific question. I'm a little bit ashamed I didn't think of it myself. Oh well. If you want to make it an answer I'd accept it, although I suppose there's not much else to add to it. – Keenan Kidwell Jul 28 '10 at 23:25
• I'm content to leave my answer where it is. – KConrad Jul 29 '10 at 1:54

As all the responses indicate, the answer to my question is "yes." The most direct route seems to be the one suggested by KConrad. Explicitly, if $F/K_{un}$ is Galois of degree $e$ (inside $K_s$), then the ring of integers of $F$ is a DVR, and if $\Pi$ is a uniformizer for $O_F$, then because $F/K_{un}$ is totally ramified, $F=K_{un}(\Pi)$ and the minimal polynomial $f$ for $\Pi$ over $K_{un}$ is Eisenstein (of degree $e$). Taking $K^\prime$ to be the finite (necessarily unramified) extension of $K$ obtained by adjoining the coefficients of $f$, $K^\prime(\Pi)/K^\prime$ is TTR of degree $e$. It is totally ramified because $f$ is still Eisenstein when viewed in $O_{K^\prime}$ (since $K^\prime$ is unramified over $K$). Thus by the result alluded to in my question, $K^\prime(\Pi)=K^\prime((\pi^\prime)^{1/e})$ for some uniformizer $\pi^\prime$ in $K^\prime$, and as a result, $F=K_{un}(\Pi)=K_{un}((\pi^\prime)^{1/e})$. The last extension is equal to $K_{un}(\pi^{1/e})$ since both $\pi$ and $\pi^\prime$ are uniformizers in $O_{K_{un}}$ and $O_{K_{un}}^\times$ is $e$-divisible.