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Every commutative monoid $M$ is naturally equipped with its divisibility preorder, defined as follows.

$$x \mid y \leftrightarrow \exists a(ax=y)$$

Is there a name for those commutative monoids such that the above preorder is antisymmetric? In other words, I'm interested in those commutative monoids satisfying the following quasi-identity:

$$\frac{ax=y\quad by=x}{x=y}$$

Motivation. The category of all such structures is probably a reflective subcategory of the category of all commutative monoids, with the left-adjoint to the inclusion functor being the functor $F$ such that $F(M)$ is the commutative monoid obtained by identifying elements $x,y \in M$ satisfying $x \mid y$ and $y \mid x$. Now given a commutative monoid $M$, we are often interested in meets and joins with respect to the divisibility order, but uniqueness issues rear their annoying heads. They can be remedied by working not in $M$, but in $F(M).$

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    $\begingroup$ Meets and joins are special cases of limits and colimits. It is a natural thing that these are not unique on the nose, but only up to unique isomorphism. In the case of preorders $x$ and $y$ are isomorphic when $x \leq y \leq x$. There is no need to make $x$ and $y$ artificially equal. Abstractly, the $2$-category of preorders is $2$-equivalent to the $2$-category of partial orders. By the way, for the same reason the official definition of a subobject (in a category) as an equivalence class of a monomorphism should be replaced simply by a monomorphism (in my opinion). $\endgroup$
    – Martin Brandenburg
    Commented Jul 6, 2014 at 13:34
  • $\begingroup$ In order words, there are various higher notions of uniqueness. Set theory corresponds to $0$-category theory and the corresponding notion of uniqueness is often much too strong to be useful (except for set theorists of course). In $1$-category theory the notion of uniqueness involves unique isomorphisms, in $2$-category theory it involves equivalences which are unique up to unique $2$-isomorphisms, etc. $\endgroup$
    – Martin Brandenburg
    Commented Jul 6, 2014 at 13:40
  • $\begingroup$ To say something specific about the question: I don't know if these monoids have a name, but observe that divisibility considerations often appear in the multiplicative monoids of commutative rings, and these don't have the property because the rings have units: Associate elements don't have to be equal (and this is a good thing), only their generated ideals are equal. By the way, this also offers a nice construction of $F(M)$ in general, as the set of principal ideals in $M$. $\endgroup$
    – Martin Brandenburg
    Commented Jul 6, 2014 at 13:42
  • $\begingroup$ I think you mean "poset" by "strict poset" and "preorder" by "weak preorder". I advocate the definition in which the subobjects of $X$ form a thin category (thus canonically equivalent to a preorder). Objects are monomorphisms $Y \hookrightarrow X$, morphisms are commutative diagrams. $\endgroup$
    – Martin Brandenburg
    Commented Jul 6, 2014 at 18:46

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At least two different terms are used in the literature for a commutative monoid in which division is a partial order: holoid and naturally partially ordered. Another possibility would be $\mathcal{H}$-trivial since a commutative semigroup has the required property if and only if the Green's relation $\mathcal{H}$ is the equality in this monoid. See Grillet's book Commutative Semigroups (2001), pages 120 and 201. I was able to trace back the term "holoid" as early as 1942, but it might have been introduced long before.

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