Related papers: Factorization in the self-idealization of a PID

Let $F$ be a field, $p$ a prime number, $X$ an indeterminate over $F$, $D_n =F[X^{\frac{1}{p^n}}, X^{-\frac{1}{p^n}}]$ for each integer $n \geq 0$ and $D = \bigcup\limits_{n\in\mathbb{N}_0}D_n.$ Then $D$ is a one-dimensional B{\'e}zout…

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…

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…

For an integral domain $R$ and a commutative cancellative monoid $M$, the ring consisting of all polynomial expressions with coefficients in $R$ and exponents in $M$ is called the monoid ring of $M$ over $R$. An integral domain is called…

An integral domain is atomic if every nonzero nonunit factors into irreducibles. Let $R$ be an integral domain. We say that $R$ is a bounded factorization domain if it is atomic and for every nonzero nonunit $x \in R$, there is a positive…

It has been a well-known fact since Euclid's time that there exist infinitely many rational primes. Two natural questions arise: In which other rings, sufficiently similar to the integers, are there infinitely many irreducible elements? Is…

In this paper, we show that it is possible for a commutative ring with identity to be non-atomic (that is, there exist non-zero nonunits that cannot be factored into irreducibles) and yet have a strongly atomic polynomial extension. In…

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,…

We provide an irreducibility test and factoring algorithm (with some qualifications) for formal power series in the unique factorization domain $R[[X]]$, where $R$ is any principal ideal domain. We also classify all integral domains arising…

We generalize the theory of radical factorization from almost Dedekind domain to strongly discrete Pr\"ufer domains; we show that, for a fixed subset $X$ of maximal ideals, the finitely generated ideals with $\mathcal{V}(I)\subseteq X$ have…

An integral domain $D$ is a {\em valuation ideal factorization domain} (VIFD) if each nonzero principal ideal of $D$ can be written as a finite product of valuation ideals. Clearly, $\pi$-domains are VIFDs. We study the ring-theoretic…

A computable ring is a ring equipped with mechanical procedure to add and multiply elements. In most natural computable integral domains, there is a computational procedure to determine if a given element is prime/irreducible. However,…

Let $D$ be an integral domain. A nonzero nonunit $a$ of $D$ is called a valuation element if there is a valuation overring $V$ of $D$ such that $aV\cap D=aD$. We say that $D$ is a valuation factorization domain (VFD) if each nonzero nonunit…

A ring has bounded factorizations if every cancellative nonunit $a \in R$ can be written as a product of atoms and there is a bound $\lambda(a)$ on the lengths of such factorizations. The bounded factorization property is one of the most…

In a Dedekind domain $D$, every non-zero proper ideal $A$ factors as a product $A=P_1^{t_1}\cdots P_k^{t_k}$ of powers of distinct prime ideals $P_i$. For a Dedekind domain $D$, the $D$-modules $D/P_i^{t_i}$ are uniserial. We extend this…

In this paper, we address various aspects of divisibility by irreducibles in rings consisting of integer-valued polynomials. An integral domain is called atomic if every nonzero nonunit factors into irreducibles. Atomic domains that do not…

In this paper, we study factorizations in the additive monoids of positive algebraic valuations $\mathbb{N}_0[\alpha]$ of the semiring of polynomials $\mathbb{N}_0[X]$ using a methodology introduced by D. D. Anderson, D. F. Anderson, and M.…

A domain $R$ is said to have the finite factorization property if every nonzero non-unit element of $R$ has at least one and at most finitely many distinct factorizations up to multiplication of irreducible factors by central units. Let $k$…

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 $\text{Int}(D)=\{f\in K[x]\mid f(D)\subseteq D\}$,…

For an element $a$ of an integral domain D under an equivalence relation \tau, the \tau-factorization of a is defined as \lambda a_1 a_2... a_k, where \lambda is a unit in D and a_i \tau a_j for all i, j. An irreducible element has no…