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.
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 $n \in \mathbb{Z}^+$, $\# \mathcal{O}_v/\mathcal{M}_v^n = (\# \mathbb{F})^n$.