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](https://mathoverflow.net/a/268125/16537) 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?
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*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).



If $H$ is commutative and unit-cancellative, then so is the monoid of non-empty ideals of $H$
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.


