# ultrametric triangle inequality

###### Theorem 1.

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 non-archimedean or ultrametric triangle inequality

3.     $|x\!+\!y|\;\leqq\;\max\{|x|,\,|y|\}$

in the field is with the condition

(*) $\quad\quad\quad|x|\leqq 1\,\,\,\,\Rightarrow\,\,\,\,|x\!+\!1|\leqq 1.$

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  $|x\!+\!y|\leqq|y|=\max\{|x|,\,|y|\}$.

###### Theorem 2.

The Krull valuation (and any non-archimedean valuation (http://planetmath.org/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.

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  $|A\!-\!B|=\max\{|B\!-\!C|,\,|C\!-\!A|\}$.

###### Theorem 3.

The valuation (http://planetmath.org/Valuation)  $|\cdot|:K\to\mathbb{R}$  of the field $K$ is archimedean   if and only if the set

 $\{|1|,\,|1\!+\!1|,\,|1\!+\!1\!+\!1|,\,\ldots\}$

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\cdot 1|.  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.

## References

• 1 Emil Artin: .  Lecture notes.  Mathematisches Institut, Göttingen (1959).
 Title ultrametric triangle inequality Canonical name UltrametricTriangleInequality Date of creation 2013-03-22 14:54:15 Last modified on 2013-03-22 14:54:15 Owner pahio (2872) Last modified by pahio (2872) Numerical id 25 Author pahio (2872) Entry type Theorem Classification msc 13F30 Classification msc 13A18 Classification msc 12J20 Classification msc 11R99 Related topic MaximalNumber Related topic PAdicCanonicalForm Related topic UltrametricSpace Related topic MinimalAndMaximalNumber Related topic ExponentValuation2 Defines non-archimedean triangle inequality