1992
DOI: 10.1112/jlms/s2-45.3.417
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Commutative Semigroup Rings with Two-Generated Ideals

Abstract: Let R be a commutative ring with identity. If an ideal I of R can be generated by two elements then we say that / is two-generated, and if every ideal of R is twogenerated we say that R has the two-generator property. It is well known that Dedekind domains have the two-generator property and if R has the two-generator property then R has Krull dimension at most 1. In [1] Bass began the study of onedimensional rings which have the two-generator property, and showed that if such a ring R is reduced and has finit… Show more

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Cited by 8 publications
(12 citation statements)
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“…(For this see [20,21] and the references listed there.) Similar studies of when a group ring A G satisfies d A G ≤ n for small values of n were given in [2,3,[15][16][17].…”
Section: Introductionmentioning
confidence: 90%
“…(For this see [20,21] and the references listed there.) Similar studies of when a group ring A G satisfies d A G ≤ n for small values of n were given in [2,3,[15][16][17].…”
Section: Introductionmentioning
confidence: 90%
“…APPLICATION 2. In [8] the commutative one-dimensional monoid rings R[G] which have the two-generator property were determined except in the case that G is a finite abelian group and R is a one-dimensional ring which has some maximal ideals which are also minimal. Since in this case it follows easily that R = R\XR 2 where maximal ideals of R[ have height 1 and R 2 is Artinian, and a product of two rings has the two-generator property if and only if each factor does, it remains to determine the commutative Artinian group rings which have the two-generator property.…”
Section: Proof Let M = (Xy) Then By [2 Theorems 9 and 12] We Can mentioning
confidence: 99%
“…Let (R, M) be a local Artinian ring and let G be a finite abelian group of order n. Then R[G] has the twogenerator property if and only if R has the two-generator property and (i) n is a unit in R if R is not a principal ideal ring, (ii) if R is a principal ideal ring and the characteristic of R/ M is a prime integer p which divides n, then the/?-Sylow subgroup G p of G is = Z/ p l Z and if M 2 ^ 0 then / = 1, unless p -2 where the additional case M = 0 and G 2 = Z/ 2 l Z 0 Z/ 2Z occurs. The proof that if M is a unit of R and R has the two-generator property then R [G] has the two-generator property proceeds by writing R as a homomorphic image of a one-dimensional ring R with the two-generator property as above, and using the characterization in [8] of one-dimensional commutative group rings with the two-generator property.…”
Section: Proof Let M = (Xy) Then By [2 Theorems 9 and 12] We Can mentioning
confidence: 99%
“…If an ideal I of R can be generated by « elements, then we say that / is n-generated; and, if every ideal of R is «-generated, we say that R has the n-generator property. Determining when a group or monoid ring has the «-generator property has been studied in [1], [3], [4], [6], [7], [8], [9], [11] and [12]. The case n = 1 can be found in [4] or Chapter 19 of [2].…”
mentioning
confidence: 99%
“…The case n = 1 can be found in [4] or Chapter 19 of [2]. Monoid (and group) rings with the two-generator property were determined in [8] and [9]. In this note we consider the problem of determining when a one-dimensional monoid ring R [S] has the «-generator property where R is an artinian ring and 5 is a commutative cancellative monoid.…”
mentioning
confidence: 99%