Timeline for Good reduction of abelian varieties [S-T] -- Why is this ring henselian?

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Jun 25, 2012 at 20:32	comment	added	Xarles		Will Sawin explained exactly the proof I had in mind. Thanks!
Jun 25, 2012 at 19:55	comment	added	Will Sawin		The straightforward approach should work. The idea is that if $\alpha \in O_L$ is a root of an irreducible polynomial $f$, then $\bar{v}$ restricted to $K(\alpha)$ is unramified, so $\alpha$ modulo $\bar{v}$ is a simple root of $f$ modulo $\bar{v}$ and lies in the algebraic closure of the residue field of $O_k$ mod $v$, so any strictly henselian ring containing $O_K$ has an algebraically closed residue field, so $\alpha$ mod $\bar{v}$ is a simple root of $f$, and is Henselian, so that extends to a root, so it contains $\alpha$. It contains every element of $O_L$, so it contains $L$.
Jun 25, 2012 at 19:33	comment	added	jmc		Thanks for your answer. I read about henselinizations and strict henselinizations. One question remains: How do you prove your claim? Probably by the same straightforward approach as above, right? (I did not try that much yet, on proving your claim, I should add.)
Jun 25, 2012 at 19:11	history	answered	Xarles	CC BY-SA 3.0