Polynomial-mappings and \(M\)-equivalence
Let \(R\) be an integral domain with quotient field \(K\), and for \(A\subset R\) let \(\text{Int}(A,R)\) be the set of all polynomials \(f\in K[X]\) with \(f(A)\subset R\). \textit{R. Gilmer} and \textit{W. Smith} [Arch. Math. 73, 355--365 (1999; Zbl 0955.13009)] defined two subsets \(S,T\) of \(R\) to be polynomially equivalent if \(\text{Int}(S,R)= \text{Int}(T,R)\), and asked, whether the following holds: if \(R\) is a Dedekind domain, \(E\subset R\) is infinite, \(f\in \text{Int}(E,R)\) and the sets \(E\) and \(f(E)\) are polynomially equivalent, then \(f\) is linear. The author gives a positive answer in the following two cases: a) \(K\) is either an algebraic number field or a finitely generated extension of a finite field, and \(R\) is finitely generated, and b) \(R\) has only finitely many invertible elements, and all its residue fields are finite. Moreover he introduces the degree of exceptionality (\(\text{dex}(R)\)) of a ring \(R\) defined as the least positive integer \(d\) with the property that for every \(R\)-polynomial of degree exceeding \(d\) the following holds: if \(E\subset R\) has the property that for all distinct \(a,b\in E\) the ratio \((f(a)-f(b))/(a-b)\) is invertible, then \(E\) is finite. Various sufficient conditions for \(\text{dex}(R)=1\) are established, and it is shown, in particular, that this holds if \(A\) is a finitely generated domain over a finite field.
scientific article; zbMATH DE number 5078153