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).
Complete DVR with finite residue field is compact?
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1I'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
2 Answers
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.
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.