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Let $K$ be a finite extension of $\mathbb{Q}_ p,$ and $\mathcal O$ the ring of integers of $K$. Let $\mathfrak p$ be the maximal ideal of $\mathcal O$.

Is it true that the quotient ring $\mathcal O/\mathfrak p^n$ is finite for all $n>0$?

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Alternatively, without using topology, this follows easily from the fact that $\mathcal{O}$ is a DVR. Namely, I’m sure you know that $k:=\mathcal{O}/\mathfrak{p}$ is finite (this is one of the axioms of a local field–depending on what your definition is). Let $\pi$ be a uniformizer for $\mathcal{O}$. Then, consider the map

$$\varphi:\mathcal{O}\to \mathfrak{p}^n/\mathfrak{p}^{n+1}:x\mapsto \pi^nx+\mathfrak{p}^{n+1}$$

This is clearly a surjective homomorphism of abelian groups. Moreover, it’s clear that $\ker\varphi=\mathfrak{p}$. Thus, we obtain an isomorphism of abelian groups $k=\mathcal{O}/\mathfrak{p}\cong \mathfrak{p}^n/\mathfrak{p}^{n+1}$. In fact, if you stare at this for a second, you see this an isomorphism of $k$-spaces. Since we have the filtration of $k$-subspaces

$$\{0\}\subseteq\mathfrak{p}^{n-1}/\mathfrak{p}^n\subseteq\cdots\subseteq\mathcal{O}/\mathfrak{p}^n$$

of $\mathcal{O}/\mathfrak{p}^n$. Since each subsequent quotient $(\mathfrak{p}^j/\mathfrak{p}^n)/(\mathfrak{p}^{j+1}/\mathfrak{p}^n)\cong \mathfrak{p}^j/\mathfrak{p}^{j+1}\cong k$ is one-dimensional this implies that $\dim_k \mathcal{O}/\mathfrak{p}^n=n$. Thus, $|\mathcal{O}/\mathfrak{p}^n|=|k|^n<\infty$.

Yes, because $\mathcal O$ is compact, and $\mathfrak p^n$ is open.

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