Non-Archimedean absolute value

Remarks

Clause 3 implies the triangle inequality, so a non-Archimedean absolute value is an absolute value, and $(K, \abs{\cdot})$ is a normed field. Example: for a prime $p$, the p-adic absolute value on $\Q$, given by $\abs{x}_p = p^{-v_p(x)}$ and $\abs{0}_p = 0$, is non-Archimedean. Clause 2 is $p^{-v_p(xy)} = p^{-v_p(x)} p^{-v_p(y)}$, and clause 3 follows from $v_p(x + y) \ge \min(v_p(x), v_p(y))$. Non-example: the usual absolute value on $\Q$ satisfies clauses 1 and 2 but fails clause 3, since $\abs{1 + 1} = 2 > 1 = \max(\abs{1}, \abs{1})$. An absolute value on $K$ is non-Archimedean if and only if $\abs{n \cdot 1} \le 1$ for all integers $n$. If $\abs{x} \ne \abs{y}$, then $\abs{x + y} = \max(\abs{x}, \abs{y})$. For $0 < c < 1$, the rule $\abs{x} = c^{v(x)}$ turns a valuation $v$ with values in $\R$ into a non-Archimedean absolute value, and every non-Archimedean absolute value arises this way.