Complete DVR with finite residue field is compact? 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).
 A: 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.
A: 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.
