invertible ideal is finitely generated
Theorem. Let be a commutative ring containing regular elements. Every invertible (http://planetmath.org/FractionalIdealOfCommutativeRing) fractional ideal of is finitely generated and regular (http://planetmath.org/RegularIdeal), i.e. regular elements.
Proof. Let be the total ring of fractions of and the unity of . We first show that the inverse ideal of has the unique quotient presentation (http://planetmath.org/QuotientOfIdeals) where . If is an inverse ideal of , it means that . Therefore we have
Since , there exist some elements of and the elements of such that . Then an arbitrary element of satisfies
because every belongs to the ring . Accordingly, . Since the converse inclusion is apparent, we have seen that is a finite of the invertible ideal .
Since the elements belong to the total ring of fractions of , we can choose such a regular element of that each of the products belongs to . Then
and thus the fractional ideal contains a regular element of , which obviously is regular in , too.
- 1 R. Gilmer: Multiplicative ideal theory. Queens University Press. Kingston, Ontario (1968).
|Title||invertible ideal is finitely generated|
|Date of creation||2015-05-06 14:44:03|
|Last modified on||2015-05-06 14:44:03|
|Last modified by||pahio (2872)|