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Let $k$ be a number field, $v$ a discrete (non-archimedean) valuation on $k$. Let $k_v$ be the completion of $k$ with respect to $v$. Also let $\mathcal{O}_v$ be the valuation ring of $k_v$ and $\mathcal{M}_v$ its unique maximal ideal.

My question is this:

Is the factor ring $\mathcal{O}_v/\mathcal{M}_v^n$ finite for each positive integer $n$?

Many thanks.

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up vote 4 down vote accepted

Yes. When $n = 1$, $\mathcal{O}_v/\mathcal{M}_v = \mathbb{F}$ is a finite field. (This can be seen by considering $k$ as a finite extension of $\mathbb{Q}$. The valuation $v$ on $k$ restricts to a $p$-adic valuation on $\mathbb{Q}$. Since $\mathcal{O}_v$ is finitely generated as a $\mathbb{Z}_p$-module, $\mathbb{F}$ is finite-dimensional over $\mathbb{F}_p$, i.e., finite.)

For $n \geq 1$, let $\pi_v$ be a uniformizing element. Then the key observation is that multiplication by $\pi_v^n$ gives an isomorphism $\mathcal{O}_v/\mathcal{M}_v \rightarrow \mathcal{M}_v^n/\mathcal{M}_v^{n+1}$.

From these two observations and a simple inductive argument it follows that for all $n \in \mathbb{Z}^+$, $\# \mathcal{O}_v/\mathcal{M}_v^n = (\# \mathbb{F})^n$.

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I see! I guessed that there is an isomorphism $\mathcal{O}_v/\mathcal{M}_v \cong \mathcal{M}_v^n/{\mathcal{M}_v^{n+1}}$ but couldn't figure out why $\mathcal{O}_v/\mathcal{M}_v$ is finite. Many thanks (again). – Mark Rodriguez Aug 10 '13 at 17:32
@Mark: Let me add that when I say that $\mathcal{O}_v$ is finitely generated as a $\mathbb{Z}_p$-module, that's meant to be a reminder of a known fact, not something which is obvious. The finite generation here is a local analogue of the argument which shows that the ring of integers of a number field is finitely generated over $\mathbb{Z}$. Really the exact same argument shows: if $L/K$ is a finite degree separable field extension, $R$ is a domain with fraction field $K$ and $S$ is the integral closure of $R$ in $L$, then $S$ is finitely generated as an $R$-module. – Pete L. Clark Aug 10 '13 at 17:40
For a proof, see e.g. $\S$ 18.1 of – Pete L. Clark Aug 10 '13 at 17:41
Isn't $\mathbb{Z}_p$ a PID, so by Theorem 18.1, $\S 18.1$ of your book, $\mathcal{O}_v$ would be a free $\mathbb{Z}_p$-module of rank $[k_v:\mathbb{Q}_p]$, provided that $k_v/\mathbb{Q}_p$ is a finite, separable field extension? Would the latter be true since char$(\mathbb{Q}_p)=0$ and $k/\mathbb{Q}$ is finite? Or, by the local analogue of the argument of theorem 18.1, do you mean a different method applicable only for completions? – Mark Rodriguez Aug 10 '13 at 18:18
@Mark: Yes. Yes. No. – Pete L. Clark Aug 10 '13 at 18:39

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