invertibility of regularly generated ideal
Lemma. β Let R be a commutative ring containing regular elements. βIf π, π and π are three ideals of R such that βπ+π ,β π +πβ andβ π+πβ are invertible
(http://planetmath.org/FractionalIdealOfCommutativeRing), then also their sum idealβ π+π+π β is .
Proof. βWe may assume that R has a unity, therefore the product of an ideal and its inverse
(http://planetmath.org/FractionalIdealOfCommutativeRing) is always R.β Now, the idealsβ π+π ,β π +πβ andβ π+πβ have the π£1, π£2 and π£3, respectively, so that
(π+π )π£1=(π +π)π£2=(π+π)π£3=R. |
Becauseβ ππ£2βRβ andβ π π£1βR,β we obtain
(π+π+π )(ππ£2π£3+π π£1π£2) | β=(π+π)ππ£2π£3+π (ππ£2)π£3+π(π π£1)π£2+(π+π )π π£1π£2 | ||
β=ππ£2+π π£2=(π +π)π£2 | |||
β=R. |
Theorem.β Let R be a commutative ring containing regular
elements.β If every ideal of R generated by two regular elements is , then in R also every ideal generated by a finite set of regular elements is .
Proof. βWe use induction on n, the number of the regular elements of the generating set.β We thus assume that every ideal of R generated by n regular elementsβ (nβ§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 |