The \(n\)-generator property in rings of integer-valued polynomials determined by finite sets
Let \(D\) denote an integral domain and \(Q\) its field of fractions. If \(E\) is a non-empty finite subset of \(D\). Recall that \(\text{Int}(E,E)= \{ f \in Q[X]~| ~f(E) \subseteq D \} \) is called the ring of integer-valued polynomials on \(D\) with respect to \(E\). Let \(n \geq 2\). Then the following are shown to be equivalent: (1) \(D\) has the \(n\)-generator property; (2) \(\text{Int}(E,D)\) has the \(n\)-generator property and (3) \(\text{Int}(E,D)\) has the strong \((n+1)\)-generator property. It is also shown that there is a Prüfer domain with the \(n\)-generator property but not the strong \(n\)-generator property and that if \(D\) is a Bézout domain but not a field, then \(\text{Int}(E,D)\) is a Bézout domain iff \(| E| =1.\) An element \(d \in D\) is said to be a strong \(n\)-generator in \(D\) if, for every finitely generated ideal \(I\) of \(D\) containing \(d\), there are elements \(d_{2}, \dots , d_{n} \in I\) such that \(I=(d, d_{2}, \dots d_{n})\) as an ideal of \(D\). For \(D\) an integral domain, \(E\) a finite non-empty subset of \(D\) and \(n \geq 2,\) it is shown that any nonzero polynomial \(f \in \text{Int}(E,D)\) is a strong \(n\)-generator in \(\text{Int}(E,D)\) if and only if \(f(a)\) is a strong \(n\)-generator in \(D\) for every \(a \in E\).
scientific article; zbMATH DE number 5013983