Proposition 5

Let $K$ be a not necessarily commutative field, $v$ a valuation on $K$ and $G$ the group $v(K^*)$ with the discrete topology.

(a) The completion ring $\hat{K}$ of $K$ (with $\mathscr{T}_v$) is a topological field.

(b) The mapping $v : K^* \to G$ can be extended uniquely to a continuous mapping $8 : \hat{K}^* \to G$. The mapping $0$ (extended by $\hat{v}(0) = +\infty$) is a valuation on $\hat{K}$ and $\hat{v}(\hat{K}^*) = v(K^*)$.

(c) The topology on $\hat{K}$ is the topology defined by the valuation $0$.

(d) For all $a \in G$ let $V_\alpha, V'_\alpha$ be the subgroups of $K$ defined by the conditions $v(x) > a, v(x) \geq a$. Then the closures $\overline{V}_\alpha, \overline{V}'_\alpha$ of $V_\alpha, V'_\alpha$ in $\hat{K}$ are defined by the conditions $\hat{v}(x) > a, \hat{v}(x) \geq a$ respectively.

(e) The ring of $0$ is the completion $\hat{A}$ of the ring $A$ of $v$; the ideal of $0$ is the completion $\hat{m}$ of the ideal $m$ of $v$.

(f) $\hat{A} = A + m$; the residue field of $8$ is canonically identified with that of $v$.

To prove (a) it suffices (General Topology, Chapter III, § 6, no. 8, Proposition 7) to show the following: let $\mathfrak{F}$ be a Cauchy filter (with respect to the additive uniform structure) on $K^*$ for which $0$ is not a cluster point; then the image of $\mathfrak{F}$ under the bijection $x \mapsto x^{-1}$ is a Cauchy filter (with respect to the additive uniform structure). For since $0$ is not a cluster point of $\mathfrak{F}$, there exists $M \in \mathfrak{F}$ and $\beta \in G$ such that $\beta$ is an upper bound of $v(M)$. Let $a \in G$. If $M'$ is an element of $\mathfrak{F}$ contained in $M$ and such that $v(x - y) > \sup(\alpha + 2\beta, \beta)$ for $x \in M'$ and $y \in M'$, then $v(x^{-1} - y^{-1}) > a$ for $x \in M'$ and $y \in M'$ (no. 1, Lemma 1). Whence (a).

By Proposition 1 of no. 1, $v|K^*$ is a continuous homomorphism from $K^*$ to $G$ and hence can be extended uniquely to a continuous homomorphism $0$ from $K^*$ to $G$. the Relation

holds in $K^*$ and hence also holds in $\hat{K}^*$ by continuity. Thus 0 (extended by $\hat{v}(0) = +\infty$) is a valuation on $\mathbf{K}$ and (b) is proved.

We now show (d). Let $a \in G$ and $x \in \mathscr{S}, -\{0\}$. For $y$ in $V$, sufficiently close to $x$, $\hat{v}(x) = \hat{v}(y) = v(y)$ and hence $\hat{v}(x) > a$. Conversely, let $x \in \hat{K}^*$ be such that $\hat{v}(x) > a$; for $y$ in $K^*$ sufficiently close to $x$, $v(y) = \hat{v}(y) = \hat{v}(x)$ and therefore $y \in V$, whence $x \in \mathscr{S}$. Thus $\mathscr{S}$ is the set of $x \in \hat{K}$ such that $\hat{v}(x) > a$. The argument is analogous for $V'_a$. This proves (d).

Taking account of Proposition 7 of General Topology, Chapter III, § 3, no. 4, assertion (c) is a consequence of (d). Assertion (e) is a special case of (d). Finally let $x \in \hat{A}$; there exists $y \in A$ such that $\hat{v}(x - y) > 0$; then $z = x - y \in \hat{m}$ and hence $x = y + z \in A + \hat{m}$; thus $\hat{A} = A + \hat{m}$, which shows (f).