# 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 $ab$ 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 KrullValuationDomain 2013-03-22 14:55:01 2013-03-22 14:55:01 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 13F30 msc 13A18 msc 12J20 msc 11R99 ValuationDeterminedByValuationDomain