Corollary

If $K$ is a (not necessarily commutative) field of characteristic $p > 0$, every function on $\mathscr{V}(K)$ is an ultrametric absolute value.

Every non-zero element $z = n.1$ ($n$ an integer $> 0$) belongs to the prime subfield $\mathbf{F}_p$ of $K$ and hence satisfies the relation $z^{p-1} = 1$, which implies $f(z) = 1$ and we may apply Proposition 3 (c).

Given a real number $c$ such that $0 < c < 1$, the formulae

therefore establish a one-to-one correspondence between ultrametric absolute values on $K$ and valuations on $K$ with real values. The improper valuation corresponds to the improper absolute value (General Topology, Chapter IX, § 3, no. 2). Let $v_1, v_2$ be two valuations on $K$ with real values and $f_1, f_2$ the corresponding absolute values; for $v_1$ and $v_2$ to be equivalent, it is necessary and sufficient that $f_1$ and $f_2$ be so: for to say that $v_1$ and $v_2$ are equivalent amounts to saying that the relations $v_1(x) \geq 0$ and $v_2(x) \geq 0$ are equivalent or again that the relations $f_1(x) \leq 1$ and $f_2(x) \leq 1$ are equivalent; it is therefore sufficient to apply Proposition 5 of General Topology, Chapter IX, § 3, no. 2. Moreover (loc. cit.) for the topologies defined on K by $f_1$ and $f_2$ to be identical, it is necessary and sufficient that $f_1$ and $f_2$ be equivalent.