1986
DOI: 10.2140/pjm.1986.124.333
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Some results on Prüfer rings

Abstract: For a domain D, D is a Prufer domain if and only if D/P is aPrufer domain for every prime ideal P of D. The same result does not hold for rings with zero divisors. In this paper it is shown that for a Prufer ring R with prime ideal P, R/P is a Prufer ring if P is not properly contained in an ideal consisting entirely of zero divisors. An example is provided to show that, in general, this is the best possible result. According to M. Boisen and P. Sheldon, a pre-Prufer ring is defined to be a ring for which ever… Show more

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Cited by 20 publications
(6 citation statements)
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“…An appeal to Theorem 3.9 gives us that I itself is invertible, hence R is a Prüfer ring. This inclusion was also proved by different methods in [9] and in [40]. An example of a Prüfer ring that is not locally Prüfer was given in Example 3.16.…”
Section: R R R Locally Prüfer R R R Prüfermentioning
confidence: 89%
See 1 more Smart Citation
“…An appeal to Theorem 3.9 gives us that I itself is invertible, hence R is a Prüfer ring. This inclusion was also proved by different methods in [9] and in [40]. An example of a Prüfer ring that is not locally Prüfer was given in Example 3.16.…”
Section: R R R Locally Prüfer R R R Prüfermentioning
confidence: 89%
“…Therefore, by Krull's characterization, a domain R is Prüfer if and only if it is locally Prüfer. The locally Prüfer condition first appeared in Lucas's 1986 article [40], which also provides an example of a Prüfer ring that is not locally Prüfer. The example below is due to Boynton [9]: Example 3.16 [9] A Prüfer ring which is not locally Prüfer.…”
Section: Is Von Neumann Regular and R P Is A Valuation Domain For Evementioning
confidence: 99%
“…To see this, apply the above proposition to get a quasi-Prüfer extension R ⊆ S if each R/P ⊆ S/Q is Prüfer. Actually, this situation already occurs for Prüfer rings and their factor domains, as Lucas's paper [29] shows. More precisely, [29, Proposition 2.7] and the third paragraph of [29, p. 336] shows that if R is a ring with Tot(R) absolutely flat, then R is a quasi-Prüfer ring if R/P is a Prüfer domain for each P ∈ Spec(R).…”
Section: Properties Of Quasi-prüfer Extensionsmentioning
confidence: 92%
“…A total quotient ring (which is a (quasi-)Prüfer ring) need not to be quasi-Prüferian. To see this we consider the following example given by Lucas [31,Example 2.11]. There is a total quotient ring R which is not locally Prüfer because there is a prime ideal P of R such that R P is not Prüfer but is integrally closed (actually, an UFD).…”
Section: Quasi-prüferian Ringsmentioning
confidence: 99%
“…We recall the main notions, given by M. Manis in [lo] : if T is a commutative ring with identity, a valuation v on T is a map from T onto G u I-}, where G is a totally ordered abelian group, such -that, for all x,y E T: Many authors used Manis rings to get results on PrUfer rings, which extend the classical notion of PrUfer dcmains (see Griffin [ 6 ] , Hinkle and Huckaba [ 7 ] , Alajbegovic [I], Lucas [9], Anderson and Pascual [ 3 ] ) .…”
Section: Introductionmentioning
confidence: 99%