Krull valuation domain

Theorem.

Any Krull valuation $|\cdot|$ of a field $K$ determines a unique valuation domain $R=\{a\in K:\,\,|x|\leqq 1\}$ , whose field of fraction is $K$ .

Proof. We first see that $1\in R$ since $|1|=1$ . Let then $a,\,b$ be any two elements of $R$ . The non-archimedean triangle inequality shows that $|a-b|\leqq\max\{|a|,\,|b|\}\leqq 1$ , i.e. that the difference $a-b$ belongs to $R$ . Using the multiplication rule (http://planetmath.org/OrderedGroup) 4 of inequalities we obtain

|ab|=|a|\cdot|b|\leqq 1\cdot 1=1

which shows that also the product $a b$ is element of $R$ . Thus, $R$ is a subring of the field $K$ , and so an integral domain. Let now $c$ be an arbitrary element of $K$ not belonging to $R$ . This implies that $1<|c|$ , whence $|c^{-1}|=|c|^{-1}<1$ (see the inverse rule (http://planetmath.org/OrderedGroup) 5). Consequently, the inverse $c^{-1}$ belongs to $R$ , and we conclude that $R$ is a valuation domain. The $a=\frac{a}{1}$ and $c=\frac{1}{c^{-1}}$ make evident that $K$ is the field of fractions of $R$ .

Title	Krull valuation domain
Canonical name	KrullValuationDomain
Date of creation	2013-03-22 14:55:01
Last modified on	2013-03-22 14:55:01
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	8
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13F30
Classification	msc 13A18
Classification	msc 12J20
Classification	msc 11R99
Related topic	ValuationDeterminedByValuationDomain