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"](http://wstein.org/papers/bib/Serre-Tate-Good_Reduction_of_Abelian_Varieties.pdf) 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 $\bar{v}$.
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> 
> Let $L$ be the fixed field of the inertia group $I$, and $O\_{L}$ the ring of $\bar{v}$-integers in $L$.
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> 


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



> 
> Why is the ring $O\_{L}$ henselian?
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> 


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.)



Good reduction of abelian varieties [S-T] -- Why is this ring henselian?
$O\_L$ 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 $L$.


