The n-generator property in rings of integer-valued polynomials determined by finite sets

Abstract

Let D be an integral domain and E a non-empty finite subset of D. For n 2, we show that D has the n-generator property if and only if Int(E, D) has the n-generator property if and only if Int(E, D) has the strong (n + 1)-generator property. Thus, iterating the Int(E, D) construction cannot produce Prüfer domains whose finitely generated ideals require an ever larger number of generators. We also show that, for n 2, a non-Throughout this paper, let D be an integral domain with field of fractions Q. If I is an ideal of D, we say that I is n-generated if it can be generated by n elements; we say that I is strongly n-generated if it is n-generated and the first of the n generators can be chosen at random from the non-zero elements of I . We say that D has the n-generator property if each finitely generated ideal of D is n-generated; we say that D has the strong n-generator property if each finitely generated ideal is strongly n-generated. (One might also refer to this as the (n − 1 2 )-generator property, at least in the case where D has zero Jacobson radical [7].) Proposition 1. If every n-generated ideal of D is strongly n-generated, then every finitely generated ideal of D is strongly n-generated.

Atomicity and the fixed divisor in certain pullback constructions

“…For example, [8] proves that Int(E, D) has the strong 2-generator property if and only if D is a Bézout domain. A similar result for Int(E, D) can be found in [4] for a larger number of generators. Also, [15] uses the representation above to show that Int(E, D) is a Prüfer domain if and only if D is a Prüfer domain.…”

“…The results of [3] and [4] provide some investigation toward Problem 1.1. The results of [4] show that for n ≥ 2, there exists a Prüfer domain D with the n-generator property but not the (n − 1)-generator property.…”

“…The results of [3] and [4] provide some investigation toward Problem 1.1. The results of [4] show that for n ≥ 2, there exists a Prüfer domain D with the n-generator property but not the (n − 1)-generator property. In addition, Int({0}, D) = D + xK[x] shares the same property as D. It follows from [15] and [3] that the answer to Problem 1.2 is affirmative.…”

Pullbacks of Prüfer rings

“…It is worth noting that Prüfer conditions (1)- (4) are preserved under localization while condition (5) is not.…”

“…, e r } be any finite subset of D. [19,Corollary 7]). Also, for n 2, the ring Int(E, D) has the n-generator property for finitely generated ideals if and only if D has the same property [4,Corollary 4].…”