Let D be a domain with quotient field K and A a D-algebra. A polynomial with coefficients in K that maps every element of A to an element of A is called integer-valued on A. For …
S Frisch - Monatshefte für Mathematik, 2017 - Springer
There are two kinds of polynomial functions on matrix algebras over commutative rings: those induced by polynomials with coefficients in the algebra itself and those induced by …
Let D be a commutative domain with field of fractions K and let A be a torsion-free D-algebra such that A ∩ K= DA∩ K= D. The ring of integer-valued polynomials on A with coefficients in …
Let D be a domain with fraction field K, and let M n (D) be the ring of n× n matrices with entries in D. The ring of integer-valued polynomials on the matrix ring M n (D), denoted Int K …
NJ Werner - Rings, polynomials, and modules, 2017 - Springer
Given a commutative integral domain D with fraction field K, the ring of integer-valued polynomials on D is Int (D)={f∈ K [x]∣ f (D)⊆ D}. In recent years, attention has turned to …
S Evrard, K Johnson - Journal of Algebra, 2015 - Elsevier
Let M n (Z) denote the ring of n× n matrices with integer entries and Int Q (M n (Z))⊆ Q [x] the algebra of polynomials that preserve M n (Z), ie polynomials for which f (M)∈ M n (Z) if M∈ …
G Peruginelli - Monatshefte für Mathematik, 2014 - Springer
Let DD be an integrally closed domain with quotient field KK and n na positive integer. We give a characterization of the polynomials in KXKX which are integer-valued over the set of …
G Peruginelli, NJ Werner - … algebra: Recent advances in commutative rings …, 2014 - Springer
Let D be an integrally closed domain with quotient field K. Let A be a torsion-free D-algebra that is finitely generated as a D-module. For every a in A we consider its minimal polynomial …
G Peruginelli - Journal of Number Theory, 2014 - Elsevier
Let K be a number field of degree n with ring of integers O K. By means of a criterion of Gilmer for polynomially dense subsets of the ring of integers of a number field, we show that …