PlanetMath: ultrametric triangle inequality

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ultrametric triangle inequality

(Theorem)

Theorem 1 Let

be a field and

an ordered group equipped with zero. Suppose that the function $\vert\cdot\vert: K\to G$ satisfies the postulates 1 and 2 of Krull valuation. Then the non-archimedean or ultrametric triangle inequality

3. $\vert x+y\vert\leqq\max\{\vert x\vert,\,\vert y\vert\}$

in the field is equivalent with the condition

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

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

$\displaystyle \vert x+y\vert\cdot\vert y\vert^{-1} = \left\vert\frac{x+y}{y}\right\vert = \left\vert\frac{x}{y}+1\right\vert\leqq 1.$

So we see that $\vert x+y\vert\leqq \vert y\vert = \max\{\vert x\vert,\,\vert y\vert\}$ .

Theorem 2 The Krull valuation (and any non-archimedean valuation) $\vert\cdot\vert$ of the field

satisfies the sharpening

$\displaystyle \vert x+y\vert = \max\{\vert x\vert,\,\vert y\vert\}\quad\mathrm{for}\,\,\,\vert x\vert \neq \vert y\vert$

of the ultrametric triangle inequality.

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

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

Theorem 3 The valuation $\vert\cdot\vert: K\to \mathbb{R}$ of the field

is archimedean if and only if the set

$\displaystyle \{\vert 1\vert,\,\vert 1+1\vert,\,\vert 1+1+1\vert,\,\ldots\}$

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

Proof. If $\vert\cdot\vert$ is non-archimedean, then $\vert n\cdot 1\vert = \vert 1+\ldots+1\vert \leqq\max\{\vert 1\vert\} = 1$ , and the multiples are bounded. Conversely, let $\vert n\cdot1\vert < M \,\, \forall n\in\mathbb{Z}_+$ . Now one obtains, when $\vert x\vert\leqq 1$ :

$\displaystyle \vert x+1\vert^n \leqq \sum_{j = 0}^n \left\vert{n\choose j}\right\vert\cdot\vert x\vert^j < (n+1)M,$

or $\vert x+1\vert < \sqrt[n]{(n+1)M}$ for all

. As

tends to infinity, this $n^\mathrm{th}$ root has the limit 1. Therefore one gets the limit inequality $\vert x+1\vert \leqq 1$ , i.e. the valuation is non-archimedean.

Bibliography

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

"ultrametric triangle inequality" is owned by pahio.

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non-archimedean triangle inequality

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Cross-references: inequality, limit, root, bounded, unity, multiples, archimedean, isosceles, vertices, triangle, implies, ultrametric, valuation, metric, non-archimedean, Krull valuation, postulates, function, ordered group equipped with zero, field
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This is version 20 of ultrametric triangle inequality, born on 2004-12-16, modified 2008-06-22.
Object id is 6587, canonical name is UltrametricTriangleInequality.
Accessed 4294 times total.

Classification:

AMS MSC:	11R99 (Number theory :: Algebraic number theory: global fields :: Miscellaneous)
	12J20 (Field theory and polynomials :: Topological fields :: General valuation theory)
	13A18 (Commutative rings and algebras :: General commutative ring theory :: Valuations and their generalizations)
	13F30 (Commutative rings and algebras :: Arithmetic rings and other special rings :: Valuation rings)

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