For a nonarchimedean field $K$, does the group of units of $\mathcal{O}_K$ being compact imply that $K$ is local?

Question

Let $(K,|\cdot|)$ be a nonarchimedean field (that is, a field equipped with a nontrivial nonarchimedean absolute value). Define

$$\mathcal{O}_K = \{x\in K:|x|\le 1\}, U_K = \{x\in K:|x|=1\}, \mathfrak{m}_K = \{x\in K:|x|<1\},$$

It is standard that the following three are equivalent:

(a) $K$ is locally compact;

(b) $\mathcal{O}_K$ is compact;

(c) $|\cdot|$ is discrete (that is to say, $|K^\times|$ is a discrete subgroup of $\mathbb{R}^+$), $(K,|\cdot|)$ is complete, and that $\mathcal{O}_K/\mathfrak{m}_K$ is finite.

A field satisfying one (hence all) of (a), (b), and (c) is exactly a nonarchimedean local field.

My question is, $K$ being a nonarchimedean local field implies that $U_K$ is compact because it is a closed subset of $\mathcal{O}_K$. But, if $U_K$ is compact, is it necessary that $K$ is a local field?

Answer 1

Take $v(a) > 0$. Then $1+a O_K$ is a closed subset of $O_K^\times$, so it must be compact.

So $O_K$ is compact.