# A monoid where every element has finitely many divisors

Is there a special name or has there been any study of monoids of this form? This came up in considering the general construction of a multivariate power series algebra over a ring $R$; usually we take the set of functions $M\to R$ where $M$ is the set $[I]$ of finite multisets on the index set $I$ with pointwise addition and scalar multiplication, and define the multiplication as $(ab)_\alpha=\sum_{\beta\le\alpha} a_\beta b_{\alpha-\beta}$, where $\alpha,\beta\in[I]$, $\alpha-\beta$ is the multiset formed by removing all the elements of $\beta$ from $\alpha$ (with multiplicity), and $\beta\le\alpha$ means that all the elements in $\beta$ are in $\alpha$ with equal or greater multiplicity.

This definition makes all the monomials commute with each other, for example if $I=\{1,2\}$ then $x_1^2x_2=x_{\{1,1,2\}}=x_{\{2,1,1\}}=x_2x_1^2$ (where $x_{\{1,1,2\}}$ is notation for the coefficient function that is $1$ at $\{1,1,2\}$ and zero at every other $\alpha\in[I]$). This may not be desirable, so one can generalize to the situation where none of the variables commute, and in this case the monomials are given by strings like $x_1x_2x_1x_3$, which is to say we take $M$ to be the free monoid on $I$, with multiplication now being defined by $(ab)_\alpha=\sum_{\beta\gamma=\alpha} a_\beta b_\gamma$ (where the sum is taken over all possible values of $\beta,\gamma\in M$ such that $\beta\gamma=\alpha$). This is well defined in both the free monoid and the earlier multiset monoid (which is actually the free commutative monoid) because the equation $\beta\gamma=\alpha$ has only finitely many solutions.

So to return to the question, has there been any study or a name for monoids $M$ such that $\{\beta\in M:\beta\mid\alpha\}$ is finite for each $\alpha$, or the power series algebras that result on such monoids (by repeating the above construction on an arbitrary such monoid instead of a free monoid or free commutative monoid)?

• Nice question. Another example is the monoid $(\mathbb{N},\times,1).$ – goblin Nov 7 '15 at 4:01
• Related. – goblin Nov 7 '15 at 4:11
• @goblin Wow, it's creepy how similar our questions and thought processes are here. I'd close this as a dupe if your question had an answer. – Mario Carneiro Nov 7 '15 at 6:11
• Great minds think alike :) – goblin Nov 7 '15 at 14:34
• There seems to be a connection here to bornological sets. If we adjoin to my "cofilter axioms" the requirement that every singleton set be small, or equivalently, that the small sets cover the monoid $M$, then we obtain the definition of a bornological set. My two "monoid axioms" seem to just be saying that the monoid operation is a "bounded map." I haven't checked the details so this might not be quite right. But anyway, this seems to suggest that the finiteness axiom is the only one with any real "teeth" or substance. – goblin Apr 18 '16 at 12:23

I suppose you define division in a monoid $M$ as follows: $v$ divides $u$ if there exist $x, y \in M$ such that $u = xvy$. In semigroup theory, this property is usually denoted $u \leqslant_\mathcal{J} v$ (note the inversion: $v \mid u$ is the same as $u \leqslant_\mathcal{J} v$). The relation $\leqslant_\mathcal{J}$ is one of the Green's preorders.
A monoid is said to be finite $\mathcal{J}$-above if, for each $u \in M$, the set $\{v \in M \mid u \leqslant_\mathcal{J} v\}$ is finite. I am not sure whether this was the earliest occurrence, but the term appeared in  and has been used several times in the literature since then.
• Actually, I define the divides relation so that $v$ divides $u$ if there exists $x$ such that $vx=u$. (It is an asymmetric operation, so you could also call it "left division".) I made a mistake in the above characterization, I thought that $vx=u$ has a unique solution $x$ if it exists, in which case if $u$ has finitely many divisors you could conclude that $vx=u$ has finitely many solutions $(v,x)$ (which is what is needed for the well-definedness of multiplication). In fact such an element $x=u/v$ is uniquely defined in both the free monoid and free comm. monoid, but not in general. – Mario Carneiro Oct 5 '15 at 17:30
• Then your relation is the Green's preorder $\leqslant_{\mathcal{R}}$... – J.-E. Pin Oct 5 '15 at 17:37