# valuation determined by valuation domain

Proof.  Let $R$ be a valuation domain, $K$ its field of fractions and $E$ the group of units of $R$. Then $E$ is a normal subgroup  of the multiplicative group  $K^{*}=K\!\smallsetminus\!\{0\}$.  So we can form the factor group$K^{*}/E$, consisting of all cosets $aE$ where  $a\in K^{*}$,  and attach to it the additional “coset” $0E$ getting thus a multiplicative group  $K/E$  equipped with zero.  If  $\mathfrak{m}=R\!\smallsetminus\!E$  is the maximal ideal   of $R$ (any valuation domain has a unique maximal ideal — cf. valuation domain is local), then we denote  $\mathfrak{m}^{*}=\mathfrak{m}\!\smallsetminus\!\{0\}$  and  $S=\mathfrak{m}^{*}/E=\{aE:\,\,a\in\mathfrak{m}^{*}\}$.  Then the subsemigroup $S$ of $K/E$ makes $K/E$ an ordered group equipped with zero.  It is not hard to check that the mapping

 $x\mapsto|x|:=xE$

from $K$ to  $K/E$  is a Krull valuation of the field $K$.

 Title valuation determined by valuation domain Canonical name ValuationDeterminedByValuationDomain Date of creation 2013-03-22 14:54:58 Last modified on 2013-03-22 14:54:58 Owner pahio (2872) Last modified by pahio (2872) Numerical id 10 Author pahio (2872) Entry type Theorem Classification msc 13F30 Classification msc 13A18 Classification msc 12J20 Classification msc 11R99 Related topic ValuationDomainIsLocal Related topic KrullValuationDomain Related topic PlaceOfField