ultrametric triangle inequality

Proof. The value $y = 1$ in the ultrametric triangle inequality gives the (*) as result. Secondly, let's assume the condition (*). Let $x$ and $y$ be non-zero elements of the field $K$ (if $xy =0$ then 3 is at once verified), and let e.g. $|x| \leqq |y|$ . Then we get $\displaystyle|\frac{x}{y}| = |x|\cdot|y|^{-1}\leqq 1$ , and thus according to (*), $$|x\!+\!y|\cdot|y|^{-1} \;=\; \left|\frac{x\!+\!y}{y}\right| \;=\; \left|\frac{x}{y}+1\right|\leqq 1.$$ So we see that $|x\!+\!y|\leqq |y| = \max\{|x|,\,|y|\}$ .

Proof. Let e.g. $|x| > |y|$ . Surely $|x\!+\!y| \leqq |x|$ , but also $|x| = |(x\!+\!y)\!-\!y| \leqq \max\{|x\!+\!y|,\,|y|\}$ ; this maximum is $|x\!+\!y|$ since otherwise one would have $|x| \leqq |y|$ . Thus the result is: $|x\!+\!y| = |x|$ .

Note. The metric defined by a non-archimedean valuation of the field $K$ is the ultrametric of $K$ . Theorem 2 implies, that every triangle of $K$ with vertices $A$ , $B$ , $C$ ($\in K$ ) is isosceles: if $|B\!-\!C| \neq |C\!-\!A|$ , then $|A\!-\!B| = \max\{|B\!-\!C|,\,|C\!-\!A|\}$ .

Proof. If $|\cdot|$ is non-archimedean, then $|n\cdot 1| = |1\!+\ldots+\!1| \leqq\max\{|1|\} = 1$ , and the multiples are bounded. Conversely, let $|n\cdot1| < M \,\, \forall n\in\mathbb{Z}_+$ . Now one obtains, when $|x|\leqq 1$ : $$|x\!+\!1|^n \;\leqq\; \sum_{j = 0}^n \left|{n\choose j}\right|\cdot|x|^j \;<\; (n+1)M,$$ or $|x\!+\!1| < \sqrt[n]{(n\!+\!1)M}$ for all $n$ . As $n$ tends to infinity, this $n^\mathrm{th}$ root has the limit 1. Therefore one gets the limit inequality $|x\!+\!1| \leqq 1$ , i.e. the valuation is non-archimedean.

Bibliography

1

EMIL ARTIN: Theory of Algebraic Numbers. Lecture notes. Mathematisches Institut, Göttingen (1959).