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.

finiteextension $K'/K$ generated by the coefficents of $\alpha$'s min. polynomial over $K_{un}$, $K'(\alpha)/K'$ is TTR... $\endgroup$ – KConrad Jul 28 '10 at 22:52