Timeline for Existence of maximal totally ramified extensions of an arbitrary CDVF
Current License: CC BY-SA 2.5
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| Jul 31, 2019 at 16:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 27, 2012 at 15:18 | answer | added | IPies | timeline score: -1 | |
| Mar 12, 2010 at 4:45 | comment | added | Chandan Singh Dalawat | 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 $\neq0$. | |
| Mar 11, 2010 at 14:11 | comment | added | BCnrd | 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 $K^{\rm{un}}/K$. | |
| Mar 11, 2010 at 10:07 | comment | added | Pete L. Clark | 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.... | |
| Mar 11, 2010 at 10:00 | comment | added | Qing Liu | 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. | |
| Mar 11, 2010 at 9:54 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Mar 11, 2010 at 9:30 | history | asked | Pete L. Clark | CC BY-SA 2.5 |