Compacity of the unit group of discrete valuation ring.

Question

I was working on global class field theory, and I blocked on a small problem... Let $K/\mathbb{Q}$ number field, and $\mathfrak{p}$ a finite place of $K$. If we denote by $K_\mathfrak{p}$ the completion of K with respect to $\mathfrak{p}$, and $U_\mathfrak{p}$ the unit group of the valuation ring of $K_\mathfrak{p}$, why can we say that $U_\mathfrak{p}$ is compact? Which topology do we use to prove it?

Answer 1

The field $K_\mathfrak{p}$ has the metric topology induced by the $\mathfrak{p}$-adic absolute value; any subset of $K_\mathfrak{p}$ then carries the subspace topology (this applies in particular to the valuation ring and its unit group). Regarding the compactness of $U_\mathfrak{p}$ in this topology, as Mindlack says, one usually first proves compactness of the valuation ring itself; since $U_\mathfrak{p}$ is easily seen to be a closed subset of the valuation ring, it too is compact. The compactness of the valuation ring requires more work, but the key ingredient for the proof is the finiteness of the residue field.