invertibility of regularly generated ideal
Lemma. β Let be a commutative ring containing regular elements. βIf , and are three ideals of such that β,β β andβ β are invertible (http://planetmath.org/FractionalIdealOfCommutativeRing), then also their sum idealβ β is .
Proof. βWe may assume that has a unity, therefore the product of an ideal and its inverse (http://planetmath.org/FractionalIdealOfCommutativeRing) is always .β Now, the idealsβ ,β β andβ β have the , and , respectively, so that
Becauseβ β andβ ,β we obtain
Theorem.β Let be a commutative ring containing regular
elements.β If every ideal of generated by two regular elements is , then in also every ideal generated by a finite set of regular elements is .
Proof. βWe use induction on , the number of the regular elements of the generating set.β We thus assume that every ideal of generated by regular elementsβ (β is .β Let β be any set of regular elements of .β Denote
The sums β, ββ andβ β are, by the assumptions, .β Then the ideal
is, by the lemma, , and the induction proof is complete.
References
- 1 R. Gilmer: Multiplicative ideal theory. βQueens University Press. Kingston, Ontario (1968).
Title | invertibility of regularly generated ideal |
---|---|
Canonical name | InvertibilityOfRegularlyGeneratedIdeal |
Date of creation | 2015-05-06 15:27:47 |
Last modified on | 2015-05-06 15:27:47 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 17 |
Author | pahio (2872) |
Entry type | Theorem |
Classification | msc 13A15 |
Classification | msc 11R04 |
Related topic | IdealMultiplicationLaws |
Related topic | PruferRing |
Related topic | InvertibleIdealIsFinitelyGenerated |