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

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

2012-06-25T18:40:01Z

First of all, I find it hard to formulate a good title for this question. Sorry that it is so vague.

Let's move on te the question itself. Lately I have been studying the article "Good reduction of abelian varieties" by Serre and Tate.

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.

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 $\bar{v}$.

Let $L$ be the fixed field of the inertia group $I$, and $O_{L}$ the ring of $\bar{v}$-integers in $L$.

As far as I can see, no other assumptions are made.

Why is the ring $O_{L}$ henselian?

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.

Does anyone know a reference for this question? Or a direct answer? (Thanks in advance.)

Answer by Will Sawin for Good reduction of abelian varieties [S-T] -- Why is this ring henselian?

2012-06-25T18:55:36Z

$O_L$ is not complete. The completion is usually uncountable, but if $K$ is countable then $K_s$ is countable.

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 $L$.

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

2012-06-25T19:11:10Z

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.