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 where . If is an inverse ideal of , it means that . Therefore we have
This implies that , and because is a cancellation ideal, it must mean that , i.e. is the unique inverse of the ideal .
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 generator system 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).