Proposition 3

*Let $v$ and $v'$ be two non-improper valuations on the same field $K$. For $v$ and $v'$ to define the same topology on $K$, it is necessary and sufficient that they be dependent.

Suppose that the topologies $\mathscr{T}_v$ and $\mathscr{T}_{v'}$, defined by $v$ and $v'$, are identical. Since $\mathscr{T}_v$ is Hausdorff, the diagonal of $K^2$ is closed and hence $v$ and $v'$ are dependent (Corollary 2 to Theorem 1).

Conversely, suppose that $v$ and $v'$ are dependent. Then their rings $A$ and $A'$ are contained in the same ring $A''$ distinct from $K$ and $A''$ is the ring of a valuation $v''$ ($§ 4$, no. 1, Proposition 1). It suffices to show that the topology $\mathscr{T}_{v''}$ is identical with $\mathscr{T}_v$. Let $\Gamma$ and $\Gamma''$ be the order groups of $v$ and $v''$. There exists an increasing homomorphism $\lambda$ of $\Gamma$ onto $\Gamma''$ such that $v'' = A \circ v$ ($§ 4$, no. 3). If $a'' \in I?''$, let $a \in \lambda^{-1}(a'')$; the condition $v(x) \geq \alpha$ implies $v''(x) \geq a''$. Let $\beta \in \Gamma$ and $\beta'' = \lambda(\beta)$; the condition $v(x) \leq \beta$ implies $v''(x) \leq \beta''$ and hence the condition $v''(x) > \beta''$ implies $v(x) > \beta$. As $v$ and $v''$ are not improper, the inequalities in question define fundamental systems of neighbourhoods of 0 for $\mathscr{T}_v$ and $\mathscr{T}_{v''}$. Hence $\mathscr{T}_v = \mathscr{T}_{v''}$, which completes the proof.

Remarks (1) Proposition 3 shows that the relation '"$v$ and $v'$ are dependent"' is an equivalence relation. (2) Taking account of the relations between valuations of height 1 and ultrametric absolute values (§ 6, no. 2), Proposition 3 also follows, in the case of valuations of height 1, from the characterization of equivalent absolute values (General Topology, Chapter IX, § 3, no. 2, Proposition 5).