Let $K$ be a complete, discretely valued field with (let's say) perfect residue field $k$. We have a unique maximal unramified extension $K^{unr}$ of $K$ and a unique maximal tamely ramified extension $K^{tame}$ of $K$ and hence short exact sequences

$1 \rightarrow Gal(K^{sep}/K^{unr}) \rightarrow Gal(K^{sep}/K) \rightarrow Gal(K^{unr}/K) \rightarrow 1$


$1 \rightarrow Gal(K^{tame}/K^{unr}) \rightarrow Gal(K^{tame}/K) \rightarrow Gal(K^{unr}/K) \rightarrow 1$.

In the second case, the normal subgroup is abelian and I know exactly what the action of the quotient on it is: the tame cyclotomic character. Therefore if it splits, I know its structure as an explicit semidirect product.

In the most famous case, $k$ is finite, so $Gal(K^{unr}/K) = Gal(k^{sep}/k) \cong \widehat{\mathbb{Z}}$ is a projective profinite group, so both sequences certainly split. This means that I (and lots of other people) do know the structure of the tame Galois group explicitly: it is $\prod_{\ell \neq p} \mathbb{Z}_{\ell}(1) \rtimes \widehat{\mathbb{Z}}$. Similarly the first sequence splits so there is a totally ramified extension $L/K$ such that $K^{sep}/L$ is unramified. Moreover, this is a very useful fact: it follows for instance that any abelian variety over $K$ with potentially good reduction acquires good reduction over a totally ramified base extension.

What is known in general? We have $Gal(K^{unr}/K) = Gal(k^{sep}/k)$, but if $k$ is almost anything else reasonable -- e.g. a local or global field, or the function field of a variety -- then its absolute Galois group certainly will not be projective. What is known about the splitting of these two short exact sequences in general, and especially about the class $\eta \in H^2(Gal(K^{unr}/K),Gal(K^{tame}/K^{unr}))$ defined by the second sequence? Is there information on how the analogues of the above results do / do not work out if these sequences do not split?

  • $\begingroup$ Pete, as for the good reduction of abelian variety, you mean an abelian variety with good reduction over a tamely ramified extension ? Then it is always true that the a.v. has good reduction over a totally ramified extension. You just take any totally ramified extension $L/K$ of the right degree. Then the abelian variety has good reduction over an étale extension of $L$, so it already has good reduction over $L$ (because the Néron model commutes with étale base change). This holds also for semi-abelian reduction, and $K$ is any disc. val. field. No idea about the splitting. $\endgroup$ – Qing Liu Mar 11 '10 at 10:00
  • $\begingroup$ Qing: no I'm not assuming that the reduction becomes good over a tamely ramified extension. The point is that this is using the splitting of the sequence. I have no reason to think it is true in general.... $\endgroup$ – Pete L. Clark Mar 11 '10 at 10:07
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    $\begingroup$ The 2nd sequence always splits (similar to Q. Liu's comment). Take a "non-Galois Kummer extension" $K'/K$ generated by a compatible system of $e$th roots of a fixed uniformizer $\pi$ as $e$ varies through all integers $\ge 1$ not divisible by residue char. This is totally tame, and any tame finite $L/K$ is generated over an unram. subextension by $e$th root of $u \pi$ where $u$ is unit of that unram. subextension. So $LK'/K'$ is generated by an $e$th root of a unit in an "unramified" extension of $K'$. That is, $K^{\rm{tame}}$ is compositum of linearly disjoint $K'/K$ and $K^{\rm{un}}/K$. $\endgroup$ – BCnrd Mar 11 '10 at 14:11
  • $\begingroup$ You could also consider a third exact sequence $1 \rightarrow Gal(K^{sep}/K^{tame}) \rightarrow Gal(K^{sep}/K) \rightarrow Gal(K^{tame}/K) \rightarrow 1$, and ask for the existence of a maximal totally and wildly ramified extension. Of course this is intersting only when the residual characteristic is $\neq0$. $\endgroup$ – Chandan Singh Dalawat Mar 12 '10 at 4:45

The second sequence always split. This is proved in a paper by Kuhlmann, Pank, and Roquette, manuscripta mathematica 55 (1986), 39-77.


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