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In a previous version of this post, $H$ was an atomic commutative monoid such that the quotient $H/H^\times$ is finitely generated, and I was asking if such conditions were enough for $H$ to be BF. But the answer to that question is obviously no (as argued in the comments below). So I'm now posing an essentially different question.


Let $H$ be a multiplicatively written monoid with identity $1_H$, and denote by $\mathcal A(H)$ the set of atoms (or irreducible elements) of $H$, where $a \in H$ is called an atom if $a \notin H^\times$ (the group of units of $H)$ and $a$ doesn't split into the product of two non-units.

For every $x \in H \setminus \{1_H\}$, we write $\mathsf L_H(x)$ for the set of all $k \in \mathbf N^+$ for which there exist $a_1, \ldots, a_k \in \mathcal A(H)$ such that $x = a_1 \cdots a_k$. In addition, we take $\mathsf L_H(1_H) := \{0\} \subseteq \mathbf N$.

We say that $H$ is atomic (respectively, BF) if $\mathsf L_H(x)$ is non-empty (respectively, finite and non-empty) for every $x \in H \setminus H^\times$, and with these definitions in place, here comes my question:

Q. Assume that $H$ is atomic and unit-cancellative (i.e., $xy = x$ or $yx = x$, for some $x, y \in H$, only if $y \in H^\times$), and the quotient $\mathcal A(H)/\simeq_H$ is finite, where $\simeq_H$ is the equivalence relation on $H$ defined by $x \simeq_H y$ iff $x=uyv$ for some $u,v \in H^\times$ (namely, $x$ is associate to $y$). Is $H$ BF?

A couple of remarks. First, an atomic monoid $H$ is BF only if it's unit-cancellative. Second, the monoid $H := (\mathbf Q_{\ge 0}, +)$ is cancellative (and hence unit-cancellative), but is not atomic (and hence can't be BF), since $H^\times = \{0\}$ and $\mathcal A(H) = \emptyset$.

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  • $\begingroup$ On a second thought, the answer to the question as it stands is obviously in the negative: Just consider the monoid $H$ defined by the presentation $\langle a,b \mid ab^2=ab=ba \rangle$, so that $\mathcal A(H)=\{a,b\}$, $H$ is atomic and commutative, and $\mathsf L_H(ab)=\mathbf N_{\ge 2}$. Things get more interesting if $H$ is unit-cancellative (in the present version of the OP, it's only assumed that $H$ is atomic and commutative and $H/H^\times$ is f.g.). $\endgroup$ – Salvo Tringali Aug 19 '17 at 13:58

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