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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.

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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

2 Answers 2

Are you asking whether the structure of the tame inertia group is the same in the henselian case as in the complete case? The answer is "yes", and in fact similar results hold for local henselian rings of higher dimension (with the appropriate definition of tame ramification). See for example SGA1, Expose XIII, Appendice I "Variations sur le lemme d'Abhyankar", Cor. 5.3, which in dimension 1 reduces to your case.

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Thank you for the reference. –  Keenan Kidwell Jul 29 '10 at 12:49
up vote 2 down vote accepted

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.

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Tamely ramified extensions of local fields are classified in Hasse's Number Theory, Chapter 16. See also Albert, Annals, 1940 (jstor.org/stable/info/1968740). –  Chandan Singh Dalawat Nov 30 '13 at 10:25

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