2011
DOI: 10.1007/s13369-011-0049-5
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Prüfer Conditions in Commutative Rings

Abstract: This article explores several extensions of the Prüfer domain notion to rings with zero divisors. These extensions include semihereditary rings, rings with weak global dimension less than or equal to 1, arithmetical rings, Gaussian rings, locally Prüfer rings, strongly Prüfer rings, and Prüfer rings. The renewed interest in these properties, due to their connection to Kaplansky's Conjecture, has resulted in a large body of results shedding new light on the area. We survey the work done in this direction in the… Show more

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Cited by 18 publications
(12 citation statements)
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References 45 publications
(75 reference statements)
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“…where w. dim(R) denotes the weak global dimension of R. Recall that all these properties are identical to the notion of Prüfer domain if R has no zero-divisors, and that the above implications are irreversible, in general, as shown by examples provided in [1,4,5,6,8,9,10,12,13,22,23,24]. Very recently, these conditions (among other Prüfer conditions) were thoroughly investigated in various contexts of duplications [12].…”
Section: Introductionmentioning
confidence: 97%
“…where w. dim(R) denotes the weak global dimension of R. Recall that all these properties are identical to the notion of Prüfer domain if R has no zero-divisors, and that the above implications are irreversible, in general, as shown by examples provided in [1,4,5,6,8,9,10,12,13,22,23,24]. Very recently, these conditions (among other Prüfer conditions) were thoroughly investigated in various contexts of duplications [12].…”
Section: Introductionmentioning
confidence: 97%
“…Let k be a field and let t and u be indeterminates over k. [10,Example 4.4], R is a local ring with maximal ideal m with m 2 = 0, and so every P-flat ideal of R is flat but R is not an arithmetical ring.…”
Section: Complementmentioning
confidence: 99%
“…However, in this form, Gauss's lemma is not true for general commutative rings R. Indeed, if R is an integral domain, it's only true if R is a Prüfer domain (which in the Noetherian case means that R is a Dedekind domain). For this and more on Gauss's lemma in rings with zero-divisors, see the survey article [GS11].…”
Section: Introductionmentioning
confidence: 99%