Q1. Is there any standard name for a (multiplicatively written) monoid $H$ with the property that, for all $x, y \in H \setminus H^\times$, there exist $m, n \in \mathbf N^+$ and $u, v \in H^\times$ such that $x^m = uy^n v$? Here, $H^\times$ is, as usual, the set of units (or invertible elements) of $H$.

A few examples of monoids with the above property: (a) groups; (b) numerical monoids, that is, submonoids of $(\mathbf N, +)$; (c) Puiseux monoids, i.e., submonoids of $(\mathbf Q_{\ge 0}, +)$.

Q2. What about other interesting examples from the literature?

Of course, Puiseux monoids are more general than numerical monoids, but there are some good reasons for distinguishing them.

Edit. Benjamin Steinberg points out that, if we require that, for all $x, y \in H$, there exist $m,n \in \mathbf N^+$ such that $x^m \mid_H y^n$ (i.e., if we drop the restriction about non-units), then what we get are the archimedean monoids. So, I should probably add that the reason for requiring $x, y \in H \setminus H^\times$ and $u,v \in H^\times$ in the above definitions, comes from the fact that I'm interested in the "structure" of the system of sets of lengths of $H$, that is, the family of sets $$ \mathscr{L}(H) := \{\mathsf L_H(x): x \in H\} \setminus \{\emptyset\} \subseteq \mathcal P(\mathbf N), $$ where for a fixed $x\in H\setminus\{1_H\}$ we denote by $\mathsf L_H(x)$ the set of all $k \in \mathbf N^+$ for which there exist atoms $a_1,\ldots,a_k\in H$ such that $x=a_1\cdots a_k$, while $\mathsf L_H(1_H):=\{0\}$. So, in this context, $H$ is interesting only if $\mathscr L(H)$ is "rich" (in particular, $H$ should have at least one atom).


1 Answer 1


If you ignore the bit about restricting to non-units you get what is called an archimedean semigroup.

  • $\begingroup$ At least in (classical) factorization theory, a (multiplicatively written) commutative monoid $H$ is called archimedean if $\bigcap_{n \ge 0} a^n H = \emptyset$ for all $a \in H \setminus H^\times$, cf., e.g., Halter-Koch's book Ideal Systems: An Introduction to Multiplicative Ideal Theory (Ch. 3, Exercise 6). This, however, seems quite different from the property stated in the OP. On the other hand, if I get your answer correctly, you take a semigroup $H$ to be archimedean if, for all $x, y \in H$, there exist $m,n \in \mathbf N^+$ with $x^m = y^n$: In particular, (...) $\endgroup$ Apr 8, 2017 at 18:54
  • $\begingroup$ (...) an archimedean monoid $H$ in the sense of this definition would have the property that, for all $x \in H$, there is $n \in \mathbf N^+$ with $x^n = 1_H$, and hence would be a group. But groups are not really interesting from the point of view of (classical) factorization theory. So I guess the restriction to non-units in the OP is what makes the difference here. $\endgroup$ Apr 8, 2017 at 18:54
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    $\begingroup$ People allow noncommutative archimedean semigroups and it is the same as what you write of you ignore the restriction about non units. Otherwise there is no name. $\endgroup$ Apr 8, 2017 at 18:58
  • $\begingroup$ Got it, thanks! I will wait to see if anyone can at least provide some interesting examples of monoids that satisfy the condition in the OP. If not, I will eventually accept your answer. $\endgroup$ Apr 8, 2017 at 19:01
  • $\begingroup$ I don't know interesting examples. For a finite monoids this means the complement of the group of units I'd a nilpotent extension of a completely simple semigroup. $\endgroup$ Apr 8, 2017 at 19:20

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