3
$\begingroup$

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.)

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$
    – François Brunault
    Commented Jun 26, 2012 at 15:50
  • $\begingroup$ Ok. I knew the if part, but not the only if. Thanks. $\endgroup$
    – jmc
    Commented Jun 28, 2012 at 17:28

2 Answers 2

Reset to default
10
$\begingroup$

$O_L$ 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 $L$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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 (-; $\endgroup$
    – jmc
    Commented Jun 25, 2012 at 19:31
  • $\begingroup$ @Will: you mean "if $K$ is countable then $K_s$ is countable"! $\endgroup$
    – Laurent Moret-Bailly
    Commented Jun 26, 2012 at 6:28
  • $\begingroup$ @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? $\endgroup$
    – jmc
    Commented Jun 26, 2012 at 13:45
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Jun 26, 2012 at 17:43
  • $\begingroup$ 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. $\endgroup$
    – jmc
    Commented Jun 28, 2012 at 17:30
7
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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.) $\endgroup$
    – jmc
    Commented Jun 25, 2012 at 19:33
  • 1
    $\begingroup$ 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 $L$. $\endgroup$
    – Will Sawin
    Commented Jun 25, 2012 at 19:55
  • $\begingroup$ Will Sawin explained exactly the proof I had in mind. Thanks! $\endgroup$
    – Xarles
    Commented Jun 25, 2012 at 20:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .