Krull valuation domain
Proof. We first see that since . Let then be any two elements of . The non-archimedean triangle inequality shows that , i.e. that the difference belongs to . Using the multiplication rule (http://planetmath.org/OrderedGroup) 4 of inequalities we obtain
which shows that also the product is element of . Thus, is a subring of the field , and so an integral domain. Let now be an arbitrary element of not belonging to . This implies that , whence (see the inverse rule (http://planetmath.org/OrderedGroup) 5). Consequently, the inverse belongs to , and we conclude that is a valuation domain. The and make evident that is the field of fractions of .
|Title||Krull valuation domain|
|Date of creation||2013-03-22 14:55:01|
|Last modified on||2013-03-22 14:55:01|
|Last modified by||pahio (2872)|