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How do I go about proving this? Do I have to show total boundedness (I don't know how to use the finiteness of the residue field, and this seems like something that it might pertain to).

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    I'm only guessing but since it is complete, can you mimic the proof that the ring of p-adic integers is compact?2012-10-18

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Since this was asked in the comments: Compactness refers to the topology on $R$, which is induced by the absolute value, which is again induced by the valuation.

Here is a hint for the solution: Think of power series with coefficients in the residue field and apply Tychonov's Theorem.

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Let $A$ be a DVR. Let $P$ be its maximal ideal.

Lemma 1 $P^n/P^{n+1}$ is, as an $A$-module, isomorphic to $A/P$ for every integer $n > 0$.

Proof: Let $\pi$ be a generator of $P$. Let $\phi\colon P^n \rightarrow P^n/P^{n+1}$ be the canonical $A$-homomorphism. Let $g\colon A \rightarrow P^n$ be the $A$-homomorphism defined by $g(x) = \pi^n x$. Let $f\colon A \rightarrow P^n/P^{n+1}$ be $\phi\circ g$. Clearly $f$ is surjective. Suppose $f(x) = 0$. Since $\pi^n x \in P^{n+1}$, there exists $y \in A$ such that $\pi^n x = \pi^{n+1} y$. Hence $x = \pi y$. Hence $Ker(f) = P$. Hence $P^n/P^{n+1}$ is isomorphic to $A/P$. QED

Lemma 2 Suppose $A/P$ is finite. Then $A/P^n$ is finite for every integer $n > 0$.

Proof: This follows immediately from Lemma 1 and the follwoing series.

$A \supset P \supset P^2 \supset \cdots \supset P^{n-1} \supset P^n$. QED

Lemma 3 Suppose $A/P$ is finite. Then $A$ is totally bounded in the $P$-adic topology.

Proof: Let $n > 0$ be an integer. By Lemma 2, $A/P^n$ is finite. Let $a_1, \dots, a_m$ be a complete system of representatives modulo $P^n$. Then $A = \bigcup_i (a_i + P^n)$. Hence $A$ is totally bounded. QED

Proposition Suppose $A$ is complete and $A/P$ is finite. Then $A$ is compact.

Proof: This follows immediately from Lemma 3.