invertibility of regularly generated ideal
Lemma. Let be a commutative ring containing regular elements. If , and are three ideals of such that , and are invertible, then also their sum ideal is invertible.
Proof. We may assume that has a unity, therefore the product of an ideal and its inverse is always . Now, the ideals , and have the inverses , and , respectively, so that
Because and , we obtain
Theorem. Let be a commutative ring containing regular
elements. If every ideal of generated by two regular elements is invertible, then in also every ideal generated by a finite set of regular elements is invertible.
Proof. We use induction on , the number of the regular elements of the generating set. We thus assume that every ideal of generated by regular elements ( is invertible. Let be any set of regular elements of . Denote
The sums , and are, by the assumptions, invertible. Then the ideal
is, by the lemma, invertible, and the induction proof is complete.
- 1 R. Gilmer: Multiplicative ideal theory. Queens University Press. Kingston, Ontario (1968).