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 (http://planetmath.org/FractionalIdealOfCommutativeRing), then also their sum ideal is .
Proof. We may assume that has a unity, therefore the product of an ideal and its inverse (http://planetmath.org/FractionalIdealOfCommutativeRing) is always . Now, the ideals , and have the , and , respectively, so that
Because and , we obtain
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 . Let be any set of regular elements of . Denote
The sums , and are, by the assumptions, . Then the ideal
is, by the lemma, , and the induction proof is complete.
- 1 R. Gilmer: Multiplicative ideal theory. Queens University Press. Kingston, Ontario (1968).
|Title||invertibility of regularly generated ideal|
|Date of creation||2015-05-06 15:27:47|
|Last modified on||2015-05-06 15:27:47|
|Last modified by||pahio (2872)|