Good reduction of abelian varieties [S-T] -- Why is this ring henselian? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:55:37Z http://mathoverflow.net/feeds/question/100620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100620/good-reduction-of-abelian-varieties-s-t-why-is-this-ring-henselian Good reduction of abelian varieties [S-T] -- Why is this ring henselian? Johan Commelin 2012-06-25T18:40:01Z 2012-06-26T06:44:39Z <p>First of all, I find it hard to formulate a good title for this question. Sorry that it is so vague.</p> <p>Let's move on te the question itself. Lately I have been studying the article <a href="http://wstein.org/papers/bib/Serre-Tate-Good_Reduction_of_Abelian_Varieties.pdf" rel="nofollow">"Good reduction of abelian varieties"</a> by Serre and Tate.</p> <p>At a certain point (in the proof of Lemma 2) they claim that a ring is henselian, and I don't see why. I will introduce the notation, so that I can specify my question.</p> <blockquote> <p>Let $K$ be a field, $v$ a discrete valuation of $K$, $K_{s}$ a seperable closure of $K$ and $\bar{v}$ an extension of $v$ to $K_{s}$. Let $I$ and $D$ denote the inertia group and the decomposition group of $\bar{v}$.</p> <p>Let $L$ be the fixed field of the inertia group $I$, and $O_{L}$ the ring of $\bar{v}$-integers in $L$.</p> </blockquote> <p>As far as I can see, no other assumptions are made.</p> <blockquote> <p>Why is the ring $O_{L}$ henselian?</p> </blockquote> <p>If I am not mistaken $L$ is the maximal unramified extension of $K$. I have searched Serre's "Local fields" for reasons why $O_{L}$ might be complete (hence henselian) but I could not find them.</p> <p>Does anyone know a reference for this question? Or a direct answer? (Thanks in advance.)</p> http://mathoverflow.net/questions/100620/good-reduction-of-abelian-varieties-s-t-why-is-this-ring-henselian/100621#100621 Answer by Will Sawin for Good reduction of abelian varieties [S-T] -- Why is this ring henselian? Will Sawin 2012-06-25T18:55:36Z 2012-06-26T06:44:39Z <p>$O_L$ is not complete. The completion is usually uncountable, but if $K$ is countable then $K_s$ is countable.</p> <p>I think the easiest way is just to prove it. Let $f$ be an irreducible polynomial with a simple root mod $\bar{v}$. Then the derivative of $f$ is nonzero mod $\bar{v}$, so it's nonzero, so $f$ is separable, so its roots are in $K_S$. Every root that doesn't disappear mod $\bar{v}$ is a $\bar{v}$-integer. Look at the Galois action on those roots. The inertia group preserves each root's residue mod $\bar{v}$, so it fixes that root, so that root lies in $L$.</p> http://mathoverflow.net/questions/100620/good-reduction-of-abelian-varieties-s-t-why-is-this-ring-henselian/100625#100625 Answer by Xarles for Good reduction of abelian varieties [S-T] -- Why is this ring henselian? Xarles 2012-06-25T19:11:10Z 2012-06-25T19:11:10Z <p>In fact, $O_L$ is the strict henselianization of $O_K$ (with residue field a fixed algebraic closure of the residue field of $K$). It is not complete, but it is Henselian, and it is "the minimum" of all the Henselian rings containing $O_K$ and with residue field algebraically closed. </p>