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.
From: Yutong Zhang [view email]
[v1]
Wed, 20 May 2026 10:00:14 UTC (31 KB)
[v2]
Sun, 24 May 2026 07:38:18 UTC (31 KB)
[v3]
Wed, 27 May 2026 17:44:18 UTC (31 KB)
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。