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# 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 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 5). Consequently, the inverse $c^{{-1}}$ belongs to $R$, and we conclude that $R$ is a valuation domain. The presentations $a=\frac{a}{1}$ and $c=\frac{1}{c^{{-1}}}$ make evident that $K$ is the field of fractions of $R$.

## Mathematics Subject Classification

13F30*no label found*13A18

*no label found*12J20

*no label found*11R99

*no label found*

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