Characterizing absolutely irreducible integer-valued polynomials over discrete valuation domains

TL;DR

This paper characterizes absolutely irreducible integer-valued polynomials over discrete valuation domains, providing explicit criteria and bounds for their factorization properties and connecting these to linear algebraic structures.

Contribution

It introduces a method to determine absolute irreducibility via a fixed divisor kernel and establishes bounds for when powers of polynomials become uniquely factorable.

Findings

Explicit characterization of absolute irreducibility using the fixed divisor kernel.

Construction of non-trivial factorizations for powers of polynomials beyond a certain bound.

Identification of bounds depending on valuation and residue field size, with proof of their optimality.

Abstract

Rings of integer-valued polynomials are known to be atomic, non-factorial rings furnishing examples for both irreducible elements for which all powers factor uniquely (\emph{absolutely irreducibles}) and irreducible elements where some power has a factorization different from the trivial one. In this paper, we study irreducible polynomials $F \in \operatorname{Int}(R)$ where $R$ is a discrete valuation domain with finite residue field and show that it is possible to explicitly determine a number $S\in \mathbb{N}$ that reduces the absolute irreducibility of $F$ to the unique factorization of $F^S$. To this end, we establish a connection between the factors of powers of $F$ and the kernel of a certain linear map that we associate to $F$. This connection yields a characterization of absolute irreducibility in terms of this so-called \emph{fixed divisor kernel}. Given a non-trivial element $\boldsymbol{v}$ of this kernel, we explicitly construct non-trivial factorizations of $F^k$, provided that $k\ge L$, where $L$ depends on $F$ as well as the choice of $\boldsymbol{v}$. We further show that this bound cannot be improved in general. Additionally, we provide other (larger) lower bounds for $k$, one of which only depends on the valuation of the denominator of $F$ and the size of the residue class field of $R$.