Proposition 2

Tag 04VB, permanent. Cite it as [Bourbaki, Tag 04VB].

Commutative Algebra, chapter VI, § 6. Absolute values

Let $f$ be a mapping of $K$ to $\mathbf{R}_+$ belonging to $\mathscr{V}(K)$ ; if there exists $C > 0$ such that $f(n.1) \leq C$ for every integer $n > 0$ , $f$ is an absolute value on $K$ .

By induction on $r > 0$ we deduce from $(\mathbf{U}_A)$ the relation

(1)

f(x_1 + x_2 + \cdots + x_{2^r}) \leq A^r \sup_{1 \leq i \leq 2^r} f(x_i)

for every family $(x_i)$ of $2^r$ elements of $K$ . We set $n = 2^r - 1$ ; for all $x \in K$ , we deduce from (1)

(f(1 + x))^n = f((1 + x)^n) = f\left( \sum_{i=0}^n \binom{n}{i} x^i \right) \leq A^r \sup \left( f\left( \binom{n}{i} \right) (f(x))^i \right)

\leq CA^r \sum_{i=0}^n \binom{n}{i} (f(x))^i = CA^r (1 + f(x))^n

for $f\left( \binom{n}{i} \right) \leq C \binom{n}{i}$ ; therefore

f(1 + x) \leq C^{1/n} A^{r/n} (1 + f(x)).

Letting $r$ tend to $+\infty$ , we obtain $f(1 + x) \leq 1 + f(x)$ for all $x \in K$ ; applying this inequality with $x$ replaced by $xy^{-1}$ (where $y \neq 0$ ) and taking account of $(\mathbf{VA}_{II})$ , we obtain relation $(\mathbf{VA}_{III})$ , which proves the proposition.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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