Proposition 12

— The function mod is continuous on K, and $\mathrm{mod}(ab) = \mathrm{mod}(a)\mathrm{mod}(b)$ for all $a, b$ in K.

The last assertion is obvious. Prop. 4 of No. 4 shows that the function mod is continuous at every point of $K^*$. It remains only to show its continuity at 0. This is obvious for discrete K; we shall therefore assume K nondiscrete. Let $\alpha$ be a Haar measure on $K^+$ and let C be a compact subset of K such that $\alpha(C) > 0$; for $a \in K^*$, we have $\alpha(aC) = \mathrm{mod}(a)\alpha(C)$.

(1) Corps (A, I, §9, No. 1), also translated as "division ring" (GT, III, §6, No. 7).

Since K is not discrete, $\alpha(\{0\}) = 0$ (No. 2, Prop. 2); therefore, for every $\varepsilon > 0$ there exists an open neighborhood U of 0 such that $\alpha(U) \leq \varepsilon$. Since the product in K is continuous, $aC \subset U$ for $a$ sufficiently near 0, and then $\mathrm{mod}(a) \leq \varepsilon / \alpha(C)$.