Let $H$ be a (multiplicatively written) commutative monoid with identity $1_H$. Given $X, Y \subseteq H$, we take $$XY := \{xy: x \in X,\, y \in Y\}.$$ We call a set $I \subseteq H$ an ideal of $H$ is $I = IH$. The set $\mathcal I^\ast(H)$ of all non-empty ideals of $H$ is made into a commutative, reduced monoid by endowing it with the binary operation $$\mathcal I^\ast(H) \times \mathcal I^\ast(H) \to \mathcal I^\ast(H): (I, J) \mapsto IJ.$$ Let $\mathcal R$ be the class of all monoids $K$ for which there exists a commutative monoid $H$ such that $K$ is isomorphic to $\mathcal I^\ast(H)$. A couple of days ago, I asked a question ultimately aimed at shedding light on the nature of the class $\mathcal R$, and an answer of Benjamin Steinberg resulted into a number of observations about the objects that sit in $\mathcal R$. But some of these observations were seriously flawed (and they were all mine).

In particular, I stated in the comments to the same answer that (a close relative of) $\mathcal I^\ast(H)$ is unit-cancellative (see the notes at the end of this post for terminology), as I overlooked a trivial case in the argument I pretended to use in the proof: What the argument does actually show is that, if $I, J \in \mathcal I(H)$ and $IJ = I$, then $J = I$ (and, hence, $I$ is an idempotent ideal). So, it seems plausible to me that $\mathcal I^\ast(H)$ will be unit-cancellative if $H$ is not "too wild", which leads to the following:

Q. Is it true that $\mathcal I(H)$ is unit-cancellative if so is $H$? If not, what about the (much more restrictive) case when $H$ is cancellative?

Notes. A commutative monoid $H$ with identity $1_H$ is unit-cancellative if $xy=x$, for some $x, y \in H$, implies $y \in H^\times$, where $H^\times$ is the set of units of $H$; reduced if $H^\times = \{1_H\}$; cancellative if the function $H \to H: x \mapsto ax$ is injective for every $a \in H$ (of course, if $H$ is cancellative, then it's also unit-cancellative).


1 Answer 1


Since $\mathcal I^*(H)$ is reduced, and $IJ = I$ implies $J = I$, the only question is whether $\mathcal I^*(H)$ has idempotents other than $H$. This is not even true if $H$ is cancellative and reduced:

Example. Let $H$ be the multiplicative monoid $\mathbb R_{\geq 1}$, and let $I = \mathbb R_{> 1}$. Then $I^2 = I$ since any $x > 1$ can be written as $x = \sqrt{x} \cdot \sqrt{x}$, with $\sqrt{x} > 1$. Moreover, $H$ is clearly cancellative, but $\mathcal I^*(H)$ is not unit-cancellative.

Similarly, you could use the additive monoids $\mathbb R_{\geq 0}$, $\mathbb Q_{\geq 0}$, or $\mathbb Z[\tfrac{1}{2}]_{\geq 0}$, with the ideal of positive elements being idempotent.

  • $\begingroup$ Right. In hindsight (and with a fresher mind) this is kind of obvious: All the monoids mentioned in your answer are divisibile, Dedekind-finite, commutative monoids where the group of units is a proper submonoid. In particular, "being Dedekind-finite & not a group" is a necessary and sufficient condition for the relative complement of the units to be a non-empty ideal (as already noted in mathoverflow.net/questions/267977), while "being Dedekind-finite & divisible" implies that the relative complement of the units is an idempotent ideal. $\endgroup$ Apr 28, 2017 at 6:09
  • $\begingroup$ Btw, I've just realized that I had essentially answered this question in the edit of April 25 at 23:55 to mathoverflow.net/questions/267977 (using the example of the monoid of non-negative rationals under addition). Maybe I should stop posting questions when it's so late in the night... $\endgroup$ Apr 28, 2017 at 6:18

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