On Prüfer rings

An integral domain D with identity is a Prüfer domain if each nonzero finitely generated ideal of D is invertible. Many equivalent forms of this condition are known; for example, see [2, Exercise 12, p. 93] or [4, Chapter 4]. The concept of a Prüfer domain has been generalized to a commutative ring R with identity (R is a Prüfer ring if each nonzero finitely generated regular ideal of R is invertible), and the analogues of conditions equivalent in the domain case have been investigated; see [3], [8], [5]. One question in this vein that has resisted solution is : If each ideal ofR generated by a finite set of regular elements is invertible, is R a Prüfer ring! In this note, we give an example of a ring R0 that shows the answer to the preceding question is negative. Such a ring R0 must, of necessity, fail to have what Marot in [11] refers to as property (P): The commutative ring S has property (P) if each regular ideal of S is generated by a set of regular elements. Our ring R0 will have the property that each ideal of R0 generated by a set of regular elements is principal. Hence our ring R0 will show that Marot's assertion in [11] that such a ring must have property (P) is false. We proceed to define R0. Let D = K[X, Y] be a polynomial ring in two indeterminates over a field K, and let {MA}AeA be the set of maximal ideals of D not containing Y We let JV be the weak direct sum of the family {D/Mx}XeA of D-modules, and we define R0 = D + N to be the idealization of D and N [12, p. 2]. The set of zero divisors on AT is (JA MA. We observe that if p is a prime element of D not an associate of % then because D is a Hubert ring [7, p. 18], Y is not in some maximal ideal of D containing p, and hence p e (JA Mx. It follows that the set of regular elements on N is {(xY\aeK — {0}, m ^ 0}, and since elements of AT are nilpotent in R0, {ocY + n\aeK{0}, m ^ 0, neN}