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We study the question up to which power an irreducible integer-valued polynomial that is not absolutely irreducible can factor uniquely. For example, for integer-valued polynomials over principal ideal domains with square-free denominator, already the third power has to factor non-uniquely or the element is absolutely irreducible. Recently, it has been shown that for any $N\in\mathbb{N}$, there exists a discrete valuation domain $D$ and a polynomial $F\in\operatorname{Int}(D)$ such that the minimal $k$ for which $F^k$ factors non-uniquely is greater than $N$. In this paper, we show that, over principal ideal domains with infinitely many maximal ideals of finite index, the minimal power for which an irreducible but not absolutely irreducible element has to factor non-uniquely depends on the $p$-adic valuations of the denominator and cannot be bounded by a constant.

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 $R$.

Regarding non-unique factorization of integer-valued polynomials over a discrete valuation domain $(R,M)$ with finite residue field, it is known that there exist absolutely irreducible elements, that is, irreducible elements all of whose powers factor uniquely, and non-absolutely irreducible elements. We completely and constructively characterize the absolutely irreducible elements among split integer-valued polynomials. They correspond bijectively to finite sets, which we call \emph{balanced}, characterized by a combinatorial property regarding the distribution of their elements among residue classes of powers of $M$. For each such balanced set as the set of roots of a split polynomial, there exists a unique vector of multiplicities and a unique constant so that the corresponding product of monic linear factors times the constant is an absolutely irreducible integer-valued polynomial. This also yields sufficient criteria for integer-valued polynomials over Dedekind domains to be absolutely irreducible., Comment: This version comes with the shortened introduction as it appears in the journal version. [v2] contains an extended introduction, the remaining sections are identical to [v3]

An irreducible element of a commutative ring is absolutely irreducible if no power of it has more than one (essentially different) factorization into irreducibles. In the case of the ring $\text{Int}(D)=\{f\in K[x]\mid f(D)\subseteq D\}$, of integer-valued polynomials on a principal ideal domain $D$ with quotient field $K$, we give an easy to verify graph-theoretic sufficient condition for an element to be absolutely irreducible and show a partial converse: the condition is necessary and sufficient for polynomials with square-free denominator., Comment: To appear in Comm. Algebra

Let $R$ be a commutative ring with identity. An element $r \in R$ is said to be absolutely irreducible in $R$ if for all natural numbers $n>1$, $r^n$ has essentially only one factorization namely $r^n = r \cdots r$. If $r \in R$ is irreducible in $R$ but for some $n>1$, $r^n$ has other factorizations distinct from $r^n = r \cdots r$, then $r$ is called non-absolutely irreducible. In this paper, we construct non-absolutely irreducible elements in the ring $\text{Int}(\mathbb{Z}) = \{f\in \mathbb{Q}[x] \mid f(\mathbb{Z}) \subseteq \mathbb{Z}\}$ of integer-valued polynomials. We also give generalizations of these constructions.

Let $D$ be a Dedekind domain with infinitely many maximal ideals, all of finite index, and $K$ its quotient field. Let $\operatorname{Int}(D) = \{f\in K[x] \mid f(D) \subseteq D\}$ be the ring of integer-valued polynomials on $D$. Given any finite multiset $\{k_1, \ldots, k_n\}$ of integers greater than $1$, we construct a polynomial in $\operatorname{Int}(D)$ which has exactly $n$ essentially different factorizations into irreducibles in $\operatorname{Int}(D)$, the lengths of these factorizations being $k_1$, \ldots, $k_n$. We also show that there is no transfer homomorphism from the multiplicative monoid of $\operatorname{Int}(D)$ to a block monoid.