Proposition 2

Let $K$ be a not necessarily commutative field, $v$ a non-improper valuation on $K$, $A$ the ring of $v$ and $m$ the ideal of $v$. For $K$ with the topology $\mathscr{T}_v$ to be locally compact, it is necessary and sufficient that the following conditions be fulfilled:

(i) $K$ is complete; (ii) $v$ is discrete; (iii) the residue field $\kappa(A)$ is finite.

If so, $A$ is compact.

Suppose $K$ is locally compact. Then it is complete (General Topology, Chapter 111, $§ 3$, no. 3, Corollary 1 to Proposition 4); further there exists a compact neighbourhood of 0, which contains a neighbourhood $V'_a$, where $a$ belongs to the value group of $v$; in other words, there exists $a \neq 0$ in $K^*$ such that $A . a$ is compact and it follows that $A = (A . a)a^{-1}$ is compact. As every ideal $b \neq (0)$ of $A$ is open, $A/b$ is compact and discrete (General Topology, Chapter III, $§ 2$, no. 5, Proposition 14) and therefore finite and in particular $\kappa(A) = A/m$ is finite. Moreover, for $y \neq 0$ in $m$, the ring $A/Ay$ being finite, there is only a finite number of ideals of $A$ containing $Ay$ and the set $P$ of elements of the form $v(x)$ such that

is finite; as $v(K^*)$ is totally ordered, $P$ has a least element $y$. Then for all $x \in A$ such that $v(x) > 0$, either $v(x) > v(y) \geq y$, or $v(x) \leq v(y)$ and then $v(x) \geq y$ by definition, so that $y$ is the least of the elements $> 0$ of $v(K^*)$. As $P$ is finite, there is a greatest integer $m \geq 0$ such that $my \in P$, whence my \leq v(y) < (m+1)\gamma. We deduce that 0 \leq v(y) - my < y and by definition of y this implies $v(y) = my$. Therefore $v(K^*) = \mathbf{Z}.\gamma$ and the valuation $v$ is discrete.

Conversely, suppose conditions (i), (ii), (iii) hold. We may restrict our attention to the case where $v$ is normed; let $u$ be a uniformizer for $v$. Then $\kappa(A) = A/Au$ and hence $A/Au$ is finite. As $x \mapsto xu^n$ defines by taking quotients an isomorphism of the additive group $A/Au$ onto $Au^n/Au^{n+1}$, $A/Au^j$ is finite for all $j \geq 0$. As $A$ is closed in $K$, it is complete and hence isomorphic to the inverse limit of the $A/Au^j$ (General Topology, Chapter 111, § 7, no. 3, Proposition 2) and therefore compact. Since $A$ is open in $K$, it is therefore seen that $K$ is locally compact.