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Let $H$ be a commutative monoid (written multiplicatively). We say that a set $I \subseteq H$ is an ideal of $H$ if $IH = I$. The set $\mathcal I(H)$ of all ideals of $H$ is made into a (commutative) monoid by the binary operation $$ \mathcal I(H) \times \mathcal I(H) \to \mathcal I(H): (I, J) \mapsto IJ := \{xy: x \in I, \, y \in J\}. $$ The identity is $H$, and $\mathcal I(H)$ is actually a non-trivial, reduced monoid with an absorbing element (namely, the empty set). (In this context, "reduced" means that there is no other unit besides the identity.) Then my question, which comes, at least in part, out of curiosity, is the following:

Q1. Given a non-trivial, commutative, reduced monoid $K$ with an absorbing element, does there exist a commutative monoid $H$ such that $\mathcal I(H)$ is isomorphic to $K$? Has this realization problem been ever considered?

Edit 1. As an illustrative example, let us consider the case when $K$ is the (canonical) zeroification of the additive monoid of non-negative integers, that is, $K \cong_{\sf Mon} (\mathbf N \cup \{\infty\}, +)$. We want to find a commutative, reduced monoid $H$ such that $\mathcal I(H)$ is isomorphic to $K$. What could we try first? The most naif idea may be to start with $(\mathbf N, +)$. So, let's try first with this.

By definition, a non-empty set $I\subseteq \mathbf N$ is an ideal of $H = (\mathbf N, +)$ iff $$I = I+\mathbf N := \{x+n: x \in I, \, n \in \mathbf N\},$$ and this is possible iff $I$ is of the form $\mathbf N_{\ge k}$ for some $k \in \mathbf N$ (if there exists $n \in \mathbf N \setminus I$ such that $n \ge \min I$, then $\min I + (n - \min I) \in I + \mathbf N$, a contradiction). Consequently, we obtain $$\mathcal I(H) = \bigl\{\mathbf N_{\ge \kappa}: \kappa \in \mathbf N \cup \{\infty\}\bigr\},$$ and it is now trivial to check that $\mathcal I(H)$ and $(\mathbf N \cup \{\infty\}, +)$ are isomorphic monoids.

Edit 2. Benjamin Steinberg notes that the answer to Q1 is no, by observing that $\mathcal I(H)$ cannot realize any reduced, commutative monoid with an absorbing element embedding a non-trivial, finite group as a subsemigroup (e.g., the zeroification of the unconditional unitization of a non-trivial, finite, abelian group), because $I^{k+1} \subseteq I^k$ for all $I \in \mathcal I(H)$ and $k \in \mathbf N$. This leads to a natural refinement of Q1.

To wit, let me say that a monoid $H$ is weakly aperiodic if the period of every $x \in H$ is either $1$ or $\infty$ (the period of $x$ is the infimum of the set of all $r \in \mathbf N^+$ for which there exists $m \in \mathbf N^+$ with $x^m = x^{m+r}$, with the convention that $\inf \emptyset := \infty$). Then we may ask the following:

Q2. Given a non-trivial, weakly aperiodic, commutative, reduced monoid $K$ with an absorbing element, does there exist a commutative monoid $H$ such that $\mathcal I(H)$ is isomorphic to $K$? Has this realization problem been ever considered?

It's quite possible that further restrictions are needed for a realization result along these lines to be possible, but I don't have clear ideas about it right now.

Edit 3 (Apr 27, 2017). Also Q2 has a negative answer, as it is seen from the comments to Benjamin Steinberg's answer.

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  • $\begingroup$ Hi Salvo. As a newcomer to this thread, it's very hard to follow, because of the edits to the question in response to a commentary discussion on Benjamin Steinberg's answer. If there are further things you want to know, I suggest you ask them as a separate question. $\endgroup$
    – David White
    Commented Apr 27, 2017 at 15:08
  • $\begingroup$ Hi David. I get your point, and yes, I could (and should) have arranged things in a different way: I'll edit for one last time to make it clearer, I hope, that both questions in the OP have been answered in the negative. $\endgroup$
    – Salvo Tringali
    Commented Apr 27, 2017 at 15:27

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The answer is no. The problem is that $I^k\subseteq I^{k-1}$ so no element can generate a cyclic group of order 2 as a subsemigroup. So take $Z/2$ and adjoin a zero and a new identity and you can't realize it.

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  • $\begingroup$ Right! And this suggests an obvious refinement of the original question, so let me edit the OP. $\endgroup$
    – Salvo Tringali
    Commented Apr 25, 2017 at 11:20
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    $\begingroup$ There are other restrictions. Your monoid admits a compatible partial order with the identity the biggest element. This forces divisibility to be a partial order. In fact your monoid has arbitrary joins in the order and the product distributes over the join. So you have a commutative quantale with identity as the top. To be the multiplicative reductive of such a thing is very special. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 25, 2017 at 11:35
  • $\begingroup$ I've just noted your comment (I was editing the OP while you were posting it). Let me try to think if your further remarks provide a negative answer also to Q2. Btw, do you know of any conventional name for what I'm provisionally calling a "totally acyclic monoid" in the OP? $\endgroup$
    – Salvo Tringali
    Commented Apr 25, 2017 at 11:50
  • $\begingroup$ On a second thought, I've just realized that $\mathcal I(H)$ is strongly unit-cancellative (mathoverflow.net/a/259790/16537), and so is, much more in general, the set $\mathcal I_r(H)$ of all $r$-ideals of $H$, equipped with the operation $\cdot_r$ of $r$-multiplication, where $r$ is a weak ideal system on $H$ (mathoverflow.net/questions/267943): It is enough to observe that $I\cdot_r J \subseteq I \cap J$ for all $I,J \in \mathcal I_r(H)$. $\endgroup$
    – Salvo Tringali
    Commented Apr 25, 2017 at 12:15
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    $\begingroup$ A commutative monoid can have idempotent ideals so why must it be unit-cancellative? Take a meet semilattice. Every ideal is idempotent so satisfies $I^2=I$. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 25, 2017 at 13:55

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