Well-foundedness of divisibility vs well-foundedness of right- and left-divisibility

Question

Say that a preorder (i.e., a reflexive and transitive binary relation) $\preceq$ on a set $X$ is

• artinian if there is no sequence $(x_n)_{n \ge 1}$ of elements of $X$ with $x_{n+1} \prec x_n$ for each $n$, where $u \prec v$ means as usual that $u \preceq v$ and $v \not\preceq u$ (some authors prefer the term "well-founded", others the term "noetherian"; I'm going for the term "artinian" because it sounds natural in the light of certain applications);
• noetherian if its dual $\preceq^{\rm op}$ is artinian, where $x \preceq^{\rm op} y$ iff $y \preceq x$.

Next, let $H$ be a (commutative or non-commutative) monoid and denote

• by $\mid_H$ the divisibility preorder (on $H$), defined by $x \mid_H y$ iff $y = uxv$ for some $u, v \in H$;

• by $\dashv_H$ the "divides-from-the-right'' preorder, defined by $x \dashv_H y$ iff $y = ux$ for some $u \in H$;

• by $\vdash_H$ the "divides-from-the-left" preorder, that is, the "divides-from-the-right" preorder in the opposite monoid $H^{\rm op}$ of $H$.

My question is whether $\mid_H$ is artinian iff both $\dashv_H$ and $\vdash_H$ are artinian. I'm sure this is well known, but I haven't been able so far to find a reference. (By the way, is there a more standard (relational) symbol for the preorders I'm denoting by $\dashv_H$ and $\vdash_H$?)

The duals of these preorders were thoroughly studied in

J.A. Green, On the Structure of Semigroups, Annals of Math. 54 (1951) 163-172;

whence they are often referred to as the Green preorders. In particular, Theorem 4 in Green's paper implies that, if both $\dashv_H$ and $\vdash_H$ are artinian noetherian, then so also is $\mid_H$ (thanks to Benjamin Steinberg who made me notice in the comments below that I had misread Green's definitions and hence the conclusions of the theorem).

For the record, what I can prove is that the following are equivalent:

1. $H$ is acyclic (i.e., $uxv \ne x$ for all $u, v, x \in H$ with $u \notin H^\times$ or $v \notin H^\times$) and $\mid_H$ is artinian.
2. $H$ is unit-cancellative (i.e., $xy \ne x$ and $x \ne yx$ for all $x, y \in H$ with $y \notin H^\times$) and both $\dashv_H$ and $\vdash_H$ are artinian.

Here, $H^\times$ is the group of units of the monoid $H$.

Answer 1

I will give a semigroup example. You can adjoin an identity to get a monoid example.

I think your question (and also what Green had in mind, which is something different) is answered by Baer-Levi semigroups. Let $X$ be a countably infinite set and let $S$ be the semigroup of all one-to-one maps $f\colon X\to X$ with $X\setminus f(X)$ infinite. This is a left cancellative and left simple semigroup with no idempotents. A proof can be found in Clifford and Preston, Algebraic Theory of Semigroups, Volume 2 in Theorem 8.2, except they follow the convention of writing $xf$ instead of $f(x)$ and using right actions and hence say right simple and right cancellative.

So it follows that for any $f,g\in S$, there is $h\in S$ with $hf=g$ and so there is one right divisibility class and one two-sided divisibility class (or in semigroup parlance one $\mathcal L$-class and one $\mathcal J$-class). I claim there are infinite chains in both direction for the $\mathcal R$-order (what you call left divisibility). I have to confess I can never read papers talking about left and right divisibility because they are both switching left and right and also up and down with respect to the way I think.

Edit. Since the proof is short, I am adding a proof that $S$ is left simple. If $f,g\colon X\to X$ are injective with $X\setminus f(X)$ and $X\setminus g(X)$ infinite, choose an infinite subset $Y$ of $X\setminus g(X)$ with $(X\setminus g(X))\setminus Y$ infinite. Define $h\colon X\to X$ by $h(f(x)) =g(x)$ and defining $h$ on $X\setminus f(X)$ to be some arbitrary bijection between $X\setminus f(X)$ and $Y$. Then $hf=g$, $h$ is injective and $h\in S$ since $h(X) = g(X)\cup Y$ and so $X\setminus h(X) = (X\setminus g(X))\setminus Y$, which is infinite.

I claim that if $f,g\in S$, then $f=gh$ for some $h$ if and only if the range of $f$ is contained in the range of $g$ and $X\setminus g^{-1}(f(X))$ is infinite.

Obviously, if $f=gh$ with $h$ any function, then $f(X)\subseteq g(X)$ and since $g$ is one-to-one we must have $h=g^{-1}\circ f$ (which makes sense since $f(X)\subseteq g(X)$). For this to belong to $S$, we need $X\setminus g^{-1}(f(X))$ to be infinite.

To get an infinite descending chain of right ideals is now easy (and this is what Green was likely thinking of). One always has $fS\supsetneq f^2S\supsetneq\cdots$ since by left cancellativity, if $f^n=f^{n+1}g$, then since $f$ is injective $fg=1_X$, which contradicts $X\setminus f(X)$ being infinite. So this does left divisibility is not noetherian.

To go the other way, let $X$ be countably infinite and $f\colon X\to X$ any element of $S$. Let $Z$ be an infinite subset of $X\setminus f(X)$ with $X\setminus (f(X)\cup Z)$ is infinite. Now we can choose a partition of $X$ into two infinite sets $X_1,X_2$ and have $g$ send $X_1$ bijectively to $f(X)$ and send $X_2$ bijectively to $Z$. Then $fS\subsetneq gS$ by the criterion above and so we can build an infinite ascending chain as well by continuing this process. So left divisibility is not artinian.