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 $H$.

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    $\begingroup$ The standard notation for these preorders uses the reverse conventions. We write in semigroup theory $s\leq_{\mathcal J} t$ if $s=utv$, we write $s\leq_{\mathcal R} t$ if $s=tu$ and $s\leq_{\mathcal L} t$ if $s=ut$. For us riight versus left has to do with left ideal versus right ideal rather than left/divisibility versus right divisiblity. $\endgroup$ Mar 3, 2021 at 13:18
  • $\begingroup$ Also, I suspect.Green is using the conventions that I am which means maybe the his mnimum condition is your maximum? $\endgroup$ Mar 3, 2021 at 13:20
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    $\begingroup$ Green’s minimum condition is different than what you are considering. He is considering the minimum condition on principal left/right/two-sided ideals and you are considering the reverse orders. $\endgroup$ Mar 3, 2021 at 13:25
  • $\begingroup$ You are look at noetherian rather than artinian from the semigroup view point. $\endgroup$ Mar 3, 2021 at 13:27
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    $\begingroup$ If you look at Baer-Levi semigroups they are simple on one-side (so have only one class on the two-sided and one of the one sided classes, but I think they have infinite chains on the other side but I have to think which way the chains go. $\endgroup$ Mar 3, 2021 at 13:33

1 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.

  • $\begingroup$ If I understand the idea correctly, you consider a monoid $H$ with the property that (i) every element is left-invertible and (ii) the ACC on principal right ideals is not satisfied: In the parlance of the OP, it follows from (i) that the divisibility preorder is artinian (equivalently, $H$ satisfies the ACC on principal ideals), and so also is the divides-from-the-right preorder (equivalently, $H$ satisfies the ACC on principal left ideals); while (ii) means that the divides-from-the-left preorder is not artinian. Nice! $\endgroup$ Mar 3, 2021 at 21:51
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    $\begingroup$ It's not a monoid. It's a semigroup. All elements right divide each other but it has infinite ascending and descending chains from any element for left divisibility. If you adjoin an identity you get a monoid with 2 right divisibility classes $\endgroup$ Mar 3, 2021 at 22:03
  • $\begingroup$ But you don't have left invertibility of any non unit. $\endgroup$ Mar 3, 2021 at 22:17
  • $\begingroup$ Whoops! Somehow, I had mindlessly translated the condition "for fixed $a$ and $b$, the equation $xa = b$ has at least one solution" to "$x = ba'$ with $a'$ a left inverse of $a$". $\endgroup$ Mar 3, 2021 at 22:39
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    $\begingroup$ In the transformation monoid K, Green's relations are different. The monoid K doesn't have any chain conditions on left ideals. H does since you are Just adding a maximum to all your posets. $\endgroup$ Mar 4, 2021 at 12:10

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