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 $K$