ultrametric triangle inequalityUltrametricTriangleInequality2004-12-16 17:40:222008-12-23 10:56:41TheoremKrull valuationnon-archimedean triangle inequality% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{thmplain}{Theorem}\begin{thmplain}
\, Let $K$ be a field and $G$ an ordered group equipped with zero.\, Suppose that the function\, $|\cdot|: K\to G$\, satisfies the postulates 1 and 2 of Krull valuation.\, Then the {\em non-archimedean} or {\em ultrametric triangle inequality}
3. \quad\quad\quad\quad $|x+y|\leqq\max\{|x|,\,|y|\}$
in the field is \PMlinkescapetext{equivalent} with the condition
(*) $\quad\quad\quad |x|\leqq 1 \,\,\,\, \Rightarrow \,\,\,\, |x+1|\leqq 1.$
\end{thmplain}
{\em 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\, $|x+y|\leqq |y| = \max\{|x|,\,|y|\}$.\\
\begin{thmplain}
\, The Krull valuation (and any \PMlinkname{non-archimedean valuation}{Valuation})\, $|\cdot|$\, of the field $K$ satisfies the sharpening
$$|x+y| = \max\{|x|,\,|y|\}\quad\mathrm{for}\,\,\,|x| \neq |y|$$
of the ultrametric triangle inequality.
\end{thmplain}
{\em 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:\, $|x+y| = |x|$.\\
\textbf{Note.}\, The metric defined by a non-archimedean valuation of the field $K$ is the {\em 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\, $|A-B| = \max\{|B-C|,\,|C-A|\}$.\\
\begin{thmplain}
\, The \PMlinkname{valuation}{Valuation}\, $|\cdot|: K\to \mathbb{R}$\, of the field $K$ is archimedean if and only if the set
$$\{|1|,\,|1+1|,\,|1+1+1|,\,\ldots\}$$
of the ``values'' of the multiples of the unity is not bounded.
\end{thmplain}
{\em 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.
\begin{thebibliography}{9}
\bibitem{Artin} {\sc Emil Artin}: {\em Theory of Algebraic Numbers}.\, Lecture notes.\, Mathematisches Institut, G\"ottingen (1959).
\end{thebibliography}