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

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

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

$$|x+y|=\max \{|x|,|y|\}\mathit{\hspace{1em}}\mathrm{for}|x|\ne |y|$$

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

Theorem 3.

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

$$\{|1|,|1+1|,|1+1+1|,\mathrm{\dots }\}$$

of the “values” of the multiples of the unity is not bounded.

Proof.  If $|\cdot |$ is non-archimedean, then  $|n\cdot 1|=|1+\mathrm{\dots }+1|\leqq \max \{|1|\}=1$,  and the multiples are bounded.  Conversely, let .  Now one obtains, when  $|x|\leqq 1$:

or     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.