Factorization in monoids and rings (2005.01681v1)

Abstract: Let $H^\times$ be the group of units of a multiplicatively written monoid $H$. We say $H$ is acyclic if $xyz \ne y$ for all $x, y, z \in H$ with $x \notin H^\times$ or $z \notin H^\times$; unit-cancellative if $yx \ne x \ne xy$ for all $x, y \in H$ with $y \notin H^\times$; f.g.u. if there is a finite set $A \subseteq H$ such that every non-unit of $H$ is a finite product of elements of the form $uav$ with $u, v \in H^\times$ and $a \in A$; l.f.g.u. if, for each $x \in H$, the smallest divisor-closed submonoid of $H$ containing $x$ is f.g.u; and atomic if every non-unit can be written as a finite product of atoms, where an atom is a non-unit that does not factor into a product of two non-units. We generalize to l.f.g.u. or acyclic l.f.g.u. monoids a few results so far only known for unit-cancellative l.f.g.u. commutative monoids (cancellative monoids are unit-cancellative, and a commutative monoid is unit-cancellative if and only if it is acyclic). In particular, we prove the following: $\bullet$ If $H$ is an atomic l.f.g.u. monoid, then every non-unit has only finitely many factorizations (into atoms) that are "minimal" and "pairwise non-equivalent" (with respect to some naturally defined relations on the free monoid over the "alphabet" of atoms). $\bullet$ If $H$ is an acyclic l.f.g.u. monoid, then it is atomic; and moreover, each element has only finitely many "pairwise non-equivalent" factorizations if we additionally assume $H$ to be commutative.