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?
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*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](https://mathoverflow.net/a/268125/16537) 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](https://en.wikipedia.org/wiki/Aperiodic_semigroup)* 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?
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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](https://mathoverflow.net/a/268125/16537).



Which monoids can be realized as the monoid of ideals of a commutative monoid?
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.


