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valuation determined by valuation domain

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\, Every valuation domain determines a Krull valuation of the field of fractions.

{\em 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 $K$.