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Let $(K, |.|)$ be a non-archimedean field, with valuation ring $\mathring{K}$ and residue field $\tilde{K}$. I want to show that

$\mathring{K}$ is compact $\Leftrightarrow$ $K$ is complete, $\mathring{K}$ is a DVR and $\tilde{K}$ is finite.

I did one, sense ($\Leftarrow$) Consider $K$ complete, $\mathring{K}$ a DVR and $\tilde{K}=\mathring{K}/\mathcal{m}$ finite, with $\mathcal{m}$ maximum ideal of $\mathring{K}$

Since $\mathring{K}/\mathcal{m} \cong m^n/\mathcal{m^{n+1}} $ is finite (as $\mathring{K}$-modules) and $m^{n+1} \subset m^n \subset ... \subset \mathring{K} $, we get that $\mathring{K}/\mathcal{m^n}$ is finite. Now consider $a_1, ..., a_n$ to be a complete system of representatives mod $m^n$, then $\mathring{K} = \bigcup_i (a_i + m^n)$ and thus $\mathring{K}$ is totally bounded. since it is closed in $K$, it's complete and thus compact.

I have no idea for the other way, i know that in a metric space, completness + total boundedness $\Leftrightarrow$ compactness, but i can't see how to get finitenes of $\tilde{K}$ from total boundedness... nor the discreteness of the valuation.

LeLoupSolitaire
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1 Answers1

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If the valuation is not discrete, then its image is not a discrete set of the real numbers, say $a = |x|$ is a limit point. Note that we can assume $x=1$ and that we can assume that a sequence, whose absolute values approximize $1$, is a sequence in $\mathring K$ (Take any approximizing sequence and invert those elements not contained in the valuation ring).

Then the open cover $$m = \bigcup_{n \in \mathbb N} \{|y| \in K ~|~ |y| > \frac{1}{n}\}$$ does not admit a finite subcover.

For the finiteness of the residue field, note that if $A$ is complete system of representatives for $\mathring K/m$, we $a-b \in \mathring K^*$ for any distinct $a,b \in A$, thus $|a-b|=1$. Hence $A$ is a discrete subset of a compact space and thus finite.

MooS
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