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  ,  whence   (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 $R$.