$x^{-1} - y^{-1} = x^{-1}(y - x)y^{-1}$ and hence

If $v(x - y) > v(y)$, Proposition 1 of § 3, no. 1 implies that $v(x) = v(y)$, since $x = y + (x - y)$. Moreover, if $v(x - y) > a + 2v(y)$, then

Let $x \in K^*$ and $a = v(x)$; then $x \notin V_a$ which shows that $\mathscr{T}_v$ is Hausdorff. For all $x_0 \in K$ and $a \in G$, there exists $\beta \in G$ such that $x_0 V_\beta \subset V_\alpha$ and $V_\beta x_0 \subset V$, (it is sufficient to take $\beta \geqslant a - v(x_0)$). On the other hand, if $a \geqslant 0$, then $V_\alpha V_\alpha \subset V_v$. The axioms (AV_I) and (AV_{II}) of General Topology, Chapter III, § 6, no. 3 being thus satisfied, $\mathscr{T}_v$ is compatible with the ring structure on $K$. Let $x_0 \in K^*$; if $x \in K^*$ satisfies $v(x - x_0) > \sup(\alpha + 2v(x_0), v(x_0))$, then $v(x^{-1} - x_0^{-1}) > a$ (Lemma 1), which shows that $x \mapsto x^{-1}$ is continuous and that $\mathscr{T}_v$ is therefore compatible with the field structure on $K$. Finally, the single condition $v(x - x_0) > v(x_0)$ implies $v(x) = v(x_0)$ ($§ 3$, no. 1, Proposition 1) and hence the mapping $v : K^* \to G$ is continuous if $G$ is given the discrete topology.

Let $\alpha \in G$ and $V'_\alpha$ be the set of $x \in K$ such that $v(x) \geq \alpha$. If $\beta < a$, then $V_\beta \supset V'_\alpha \supset V_v$. If $v$ is not improper, it is therefore seen that the $V'_\alpha$ form a fundamental system of neighbourhoods of 0 for $\mathscr{T}_v$.

The $V_v$ and the $V'_\alpha$ are open additive subgroups and therefore closed in $K$ and therefore the topological field $K$ is totally disconnected. As every non-zero ideal of the ring of $v$ contains a $V_v$, it is open and closed in $K$. The quotient topology on the residue field of $v$ is therefore discrete.

Let $A$ be the ring of $u$. If $v$ is discrete, Proposition 8 of $§ 3$, no. 6 shows that the topology induced by $\mathscr{T}_v$ on $A$ is the $m(A)$-adic topology. This is not so in general (Exercise 4).

(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.

This mapping is a continuous algebraic isomorphism. It is sufficient to show that it is bicontinuous. Let $\alpha \in G$. We need to show that there exists a neighbourhood $V$ of 0 in $E$ such that the relation $ax_0 \in V$ implies $v(a) > a$. Let $a, \in K^*$ be such that $v(a_0) = a$. As $E$ is Hausdorff, there exists a neighbourhood $W$ of 0 in $E$ such that $a_0x_0 \notin W$. As $v$ is not improper, there exist a neighbourhood $W'$ of 0 in $E$ and an element $\beta$ of $G$ such that the relations $y \in W'$, $v(a) \geq \beta$ imply $ay \in W$. Let $a, \in K^*$ be such that $v(a_1) = -\beta$. The relations $ax_0 \in a_1^{-1}W'$ and $v(a) \leq a$ imply $a_1ax_0 \in W$ and $v(a_0a^{-1}a_1^{-1}) = \alpha + \beta - v(a) \geq \beta$ and hence $a_0x_0 = a_0a^{-1}a_1^{-1}(a_1ax_0) \in W$, which is absurd; in other words, the relation $ax, \in a_1^{-1}W$ implies $v(a) > a$.

Taking account of Propositions 1 and 3, the proof is the same as that of Topological Vector Spaces, Chapter I, § 2, Corollary 2 to Theorem 1.

Taking account of Proposition 3 and its corollary, the proof is the same as that of Topological Vector Spaces, Chapter I, § 2, Theorem 2.

$F$ is complete.

(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).