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invertibility of regularly generated ideal
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(Theorem)
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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 1 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 ready.
- 1
- R. GILMER: Multiplicative ideal theory. Queens University Press. Kingston, Ontario (1968).
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"invertibility of regularly generated ideal" is owned by pahio.
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Cross-references: sums, generating set, induction, finite set, ideal generated by, generated by, product, unity, sum ideal, ideals, regular elements, commutative ring
There are 3 references to this entry.
This is version 12 of invertibility of regularly generated ideal, born on 2005-04-30, modified 2008-01-14.
Object id is 6984, canonical name is InvertibilityOfRegularlyGeneratedIdeal.
Accessed 1337 times total.
Classification:
| AMS MSC: | 11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers) | | | 13A15 (Commutative rings and algebras :: General commutative ring theory :: Ideals; multiplicative ideal theory) |
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Pending Errata and Addenda
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