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?
