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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.)

flag	asked Jun 25 at 18:40 Johan Commelin 460●1●9

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Even if $K=\mathbf{Q}_p$, the field $L$ is not complete. When learning local fields, it is a good exercise to prove that an algebraic extension of $\mathbf{Q}_p$ is complete if and only if it is finite. – François Brunault Jun 26 at 15:50

Ok. I knew the if part, but not the only if. Thanks. – Johan Commelin Jun 28 at 17:28

Will Sawin · Accepted Answer · 2012-06-26 06:44:39Z UTC

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$O_L$ 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 $L$.

edited Jun 26 at 6:44

answered Jun 25 at 18:55

Will Sawin
18.5k●1●24●48

	Thanks. Apperently the straightforward approach was the way I should have taken. I had not expected that. But then, I don't have very much experience with henselian rings (-; – Johan Commelin Jun 25 at 19:31
	@Will: you mean "if $K$ is countable then $K_s$ is countable"! – Laurent Moret-Bailly Jun 26 at 6:28
	@Will: Actually I have onee question about your last sentence of the proof. I agree that the inertia group preserves residues of the roots. But how does it follow that therefore the roots are fixed? – Johan Commelin Jun 26 at 13:45
	Since the root under consideration is a simple root modulo $\bar{v}$ by assumption. Its residue doesn't change, and it he has to remain a root of the polynomial (since the polynomial is defined over $L$) so the root itself must be fixed. – Will Sawin Jun 26 at 17:43
	Ah. Because the polynomial is monic, the degree of the reduction does not go down. So there is a "kind of" 1-1 correspondence between the roots (if we take multiplicities into account). And thus roots reducing to simple roots are fixed by I. Got it. – Johan Commelin Jun 28 at 17:30

Xarles · Answer 2 · 2012-06-25 19:11:10Z UTC

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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.

answered Jun 25 at 19:11

Xarles
193●1●6

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.) – Johan Commelin Jun 25 at 19:33

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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 $L$. – Will Sawin Jun 25 at 19:55

Will Sawin explained exactly the proof I had in mind. Thanks! – Xarles Jun 25 at 20:32

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

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