Finite factorization is detected by undermonoids

TL;DR

This paper proves that in cancellative commutative monoids, the finite factorization property is equivalent for all submonoids and undermonoids, confirming a conjecture by Gotti and Li.

Contribution

It establishes that the finite factorization property is hereditary from undermonoids to all submonoids in cancellative commutative monoids.

Findings

Every submonoid of a cancellative commutative monoid is an FFM if and only if every undermonoid is an FFM.

The proof uses fixed length factorizations and ideal enlargements to derive a contradiction.

The result confirms that finite factorization is a hereditary property in this context.

Abstract

Let $M$ be a cancellative commutative monoid and call a submonoid $S$ of $M$ an undermonoid if $\G(S)=\G(M)$ inside the Grothendieck group of $M$. Gotti and Li asked whether the finite factorization property is hereditary once it is known on all undermonoids: if every undermonoid of $M$ is a finite factorization monoid, must every submonoid of $M$ be a finite factorization monoid? We give an affirmative answer. Equivalently, for every cancellative commutative monoid $M$, the following two conditions coincide: every submonoid of $M$ is an FFM, and every undermonoid of $M$ is an FFM. The proof isolates a fixed length $\ell$ and an infinite set of length-$\ell$ factorizations of one element $b$. In the non-group case, a divisor-complement ideal $I=\{m\in M:m\nmid_M b\}$ enlarges the bad submonoid to a bad undermonoid while preserving the chosen length-$\ell$ factorizations. In the group case, a maximality argument over submonoids for which these factorizations survive is combined with a two-sided perturbation $S\mapsto S+\Nzero(2b+u)$. The key point is that the perturbation creates no new units and does not split any atom occurring in the fixed factorizations. This yields an undermonoid with infinitely many factorizations of $b$, contradicting the hypothesis.