What are the open normal subgroups of the inertia group of a local field? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:19:48Z http://mathoverflow.net/feeds/question/33710 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33710/what-are-the-open-normal-subgroups-of-the-inertia-group-of-a-local-field What are the open normal subgroups of the inertia group of a local field? Keenan Kidwell 2010-07-28T20:51:33Z 2010-07-29T12:48:35Z <p>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: </p> <p><strong>For every positive integer $e$ prime to $p$, there exists a unique open normal subgroup of $I_K$ of index $e$.</strong> </p> <p>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 $p$. </p> <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).</p> <p>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. </p> http://mathoverflow.net/questions/33710/what-are-the-open-normal-subgroups-of-the-inertia-group-of-a-local-field/33714#33714 Answer by Tony Scholl for What are the open normal subgroups of the inertia group of a local field? Tony Scholl 2010-07-28T21:12:12Z 2010-07-28T21:12:12Z <p>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. </p> http://mathoverflow.net/questions/33710/what-are-the-open-normal-subgroups-of-the-inertia-group-of-a-local-field/33782#33782 Answer by Keenan Kidwell for What are the open normal subgroups of the inertia group of a local field? Keenan Kidwell 2010-07-29T12:48:35Z 2010-07-29T12:48:35Z <p>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 <em>finite</em> (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 <em>some</em> 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. </p>