Let $H$ be a multiplicatively written, commutative monoid with identity $1\_H$, and let $\mathcal P(H)$ be the power set of $H$. If $X, Y \subseteq H$, we will set $$XY := \{xy: x \in X,\, y \in Y\}.$$
A *weak ideal system* on $H$ is a function $r: \mathcal P(H) \to \mathcal P(H)$ such that, for all $a \in H$ and $X, Y \subseteq H$, the following hold (given $V \subseteq H$, we will use $V\_r$ in place of $r(V)$):


1. $\emptyset = \emptyset\_r$ and $X \subseteq X\_r$.
2. $X \subseteq Y\_r$ implies $X\_r \subseteq Y\_r$.
3. $aH \subseteq \{a\}\_r$.
4. $a X\_r \subseteq (aX)\_r$.


In particular, $r$ is named an *ideal system* if the last condition holds as an equality.


Weak ideal systems are reminiscent of [closure operators](https://en.wikipedia.org/wiki/Closure_operator), and have a rich and far-reaching theory, which grew up from intuitions and pioneering work of W. Krull and H. Prüfer, and is well documented by M.D. Larsen and P.J. McCarthy's book *Multiplicative Theory of Ideals* and R. Gilmer's *Multiplicative Ideal Theory*. If asked for a neat and lucid treatment of the subject, I would go for F. Halter-Koch's 1998 monograph *Ideal Systems: An Introduction to Multiplicative Ideal Theory*, with the caveat that weak ideal systems are defined therein in a "slightly" different way than we're doing: In Halter-Koch's book, the theory of [weak] ideal systems is developed by assuming that monoids have an absorbing element $0\_H \ne 1\_H$, and the first condition in the above list is replaced with $X \cup \{0\_H\} \subseteq X\_r$.


With this in mind, let $r$ be a weak ideal system on $H$. We say that a set $I \subseteq H$ is an *$r$-ideal* if $I = I\_r$: If $H$ is the multiplicative monoid of a commutative unital ring $R$, then the function $$d: \mathcal P(H) \to \mathcal P(H): X \mapsto \{a\_1 x\_1 + \cdots + a\_n x\_n: a\_1, \ldots, a\_n \in H, \, x\_1, \ldots, x\_n \in X\}$$ is an ideal system on $H$, and the $d$-ideals of $H$ are nothing but the good, old, classical ideals of $R$.


We denote by $\mathcal I\_r(H)$ the set of $r$-ideals of $H$, which is made into a commutative, reduced monoid (this is simple, but not obvious) by the binary operation $$\cdot\_r: \mathcal I\_r(H) \times \mathcal I\_r(H) \to \mathcal I\_r(H): (I, J) \mapsto (IJ)\_r,$$
referred to as *$r$-multiplication*. Note that $H$ and $\emptyset$ are $r$-ideals, and they serve, respectively, as the identity and the absorbing element of $\mathcal I\_r(H)$.
An $r$-ideal $I$ of $H$ is termed a *prime $r$-ideal* if $I \ne H$ and $H \setminus I$ is a subsemigroup of $H$. In particular, the set of all prime $r$-ideals of $H$ is denoted by $r\text{-spec}(H)$ and called the *$r$-spectrum* of $H$.


If $I \in \mathcal I\_r(H)$, we take $\text{coht}\_r(I)$ to be the supremum of all $n \in \mathbf N^+$ for which there exist $\mathfrak p\_1, \ldots, \mathfrak p\_n \in r\text{-spec}(H)$ such that $\mathfrak p\_{i-1} \subsetneq \mathfrak p\_i$ for each $i \in [\![1, n]\!]$, where $\mathfrak p\_0 := I$ and $\sup \emptyset := 0$. We call $\text{coht}\_r(I)$ the *$r$-coheight* of $I$. It is straightforward that $\text{coht}\_r(H) = 0$ and $\text{coht}\_r(I) \le \text{coht}\_r(J)$ for all $I, J \in \mathcal I\_r(H)$ with $J \subseteq I$, and my question is:



> 
> **Q.** Is there any (non-trivial) characterization of the pairs $(H, r)$ which verify the triangle inequality, i.e., such that $\text{coht}\_r(I \cdot\_r J) \le 1 + \text{coht}\_r(I) + \text{coht}\_r(J)$ for all $I, J \in \mathcal I\_r(H) \setminus \{H\}$?
> 
> 
> 


I don't know of a *single* example in which the inequality is *not* satisfied, but I must also confess that I've not tried too hard to find one. On a positive note, it's not too difficult to prove that the inequality holds (even without the "1 + " on the right-hand side) in the (very) extremal cases when:


* $H$ is arbitrary and $r$ is the *discrete weak ideal system* on $H$ (that is, the unique weak ideal system for which $X\_r = H$ for every non-empty $X \subseteq H$);
* $H$ is a [null monoid](https://en.wikipedia.org/wiki/Null_semigroup) (i.e., there is a distinguished element $0\_H \in H$ such that $xy = 0\_H$ for all $x, y \in H \setminus \{1\_H\}$) and $r$ is the fundamental ideal system determined by taking $X\_r = XH$ for all $X \subseteq H$ (this is often called the *$s$-system*, and the corresponding ideals are precisely the ideals of $H$ in the usual sense of semigroup theory).


In addition, I *suspect* that the inequality is true when $H$ is the multiplicative monoid of a PID (which is, in a way, another extremal case), and can prove it in some special instances.




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*Edit 1.* After discussing the question with Paolo Leonetti, I changed my mind and added a "1 + " on the right-hand side of the triangular inequality. In hindsight, this has a "natural justification", as we'd like to have a function $\mathcal I\_r(H) \to \mathbf N \cup \{\infty\}$ which "separates" $H$ (namely, the only unit of $\mathcal I\_r(H)$) from the $r$-prime ideals. But that wouldn't be (always) the case with the $r$-coheight, as confirmed by the following example (due to Paolo): Consider the monoid $(\mathbf N, +)$ with the unique ideal system $r$ for which $X\_r = \mathbf N\_{\ge \min X}$ for every non-empty $X \subseteq \mathbf N$. Then observe that the $r$-ideals are all and only the subsets of $\mathbf N$ of the form $\mathbf N\_{\ge \kappa}$ with $\kappa \in \mathbf N \cup \{\infty\}$, and the unique prime $r$-ideals are $\emptyset$ and $\mathbf N^+$. Accordingly, ${\rm coht}\_r(\mathbf N^+) = 0$, and this implies that the inequality without the "1 + " is false, by taking $I = J = \mathbf N^+$.




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*Edit 2.* Here is a more conceptual counterexample to the inequality without the "1 + ": Let $H$ be a non-trivial, Dedekind-finite, commutative, reduced monoid with a non-empty set of atoms (e.g., $H$ may be the multiplicative monoid of the integers), take $r$ to be the $s$-ideal system mentioned in the above (i.e., $X\_r := XH$ for every $X \subseteq H$), and set $I := H \setminus \{1\_H\}$. We claim that $I$ is a prime $r$-ideal of $H$: Indeed, $IH \subsetneq H$, otherwise there would exist $x, y \in I$ such that $xy = 1\_H$, which is impossible by an [observation of Benjamin Steinberg](https://mathoverflow.net/a/262001/16537) and the assumption that $H$ is reduced. It follows that $I$ is an $r$-ideal of $H$, and is actually a prime $r$-ideal, because its complement in $H$ is the trivial subgroup of $H$. So $\text{coht}\_r(I) = 0$, while ${\rm coht}\_r(I^2) \ge 1$, since $I^2 \subsetneq I$ by the fact that $H$ has no non-trivial unit and the set of atoms of $H$ is non-empty.




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*Notes.* A monoid $H$ is *Dedekind-finite* if $xy=1\_H$, for some $x, y \in H$, implies $yx=1\_H$, and *reduced* if the only unit (or invertible element) of $H$ is the identity $1\_H$. Finally, an *atom* of $H$ is a non-unit element $a \in H$ for which there don't exist two other non-unit elements $x, y \in H$ such that $a = xy$.



Weak ideal systems $r$ for which the $r$-coheight satisfies a kind of triangle inequality
The inequality in the OP is true, and even in a stronger form: I'm indebted to Andrea Gagna for his [enlightening answer](https://mathoverflow.net/questions/267943/), which has greatly inspired the following proofs.


**Proposition.** Assume $H$ is a commutative monoid and $r$ a weak ideal system on $H$, and let $I, J \in \mathcal I\_r(H)$ and $\mathfrak p \in r\text{-spec}\_r(H)$. The following hold:


$\phantom{iii}$(i) If $I \cdot\_r J \subseteq \mathfrak p$, then $I \subseteq \mathfrak p$ or $J \subseteq \mathfrak p$.


$\phantom{iii}$(ii) $\text{coht}\_r(I \cdot\_r J) \le 1 + \max(\text{coht}\_r(I), \text{coht}\_r(J))$.


*Proof.* (i) Suppose $I \cdot\_r J \subseteq \mathfrak p$. By the very definition of the $r$-multiplication and condition 1 in the axiomatic definition of a weak ideal system, we have 
$$IJ \subseteq (IJ)\_r = I \cdot\_r J \subseteq \mathfrak p.$$
If $I \subseteq \mathfrak p$, we are done. Otherwise, pick $x \in I \setminus \mathfrak p$. Then $xy \in \mathfrak p$ for every $y \in J$, and this implies $J \subseteq \mathfrak p$, because $H \setminus \mathfrak p$ is a subsemigroup of $H$ (by the assumption that $\mathfrak p$ is a prime $r$-ideal), so the existence of an element $y \in J \setminus \mathfrak p$ would yield $xy \in H \setminus \mathfrak p$ (since $x \in H \setminus \mathfrak p$), a contradiction. 


(ii) The claim is obvious if the $r$-coheight of $I \cdot\_r J$ is zero. Otherwise, let $\mathfrak p\_1, \ldots, \mathfrak p\_n$ be $r$-prime ideals such that $\mathfrak p\_{i-1} \subsetneq \mathfrak p\_i$ for each $i \in [\![1, n]\!]$, where $\mathfrak p\_0 := I \cdot\_r J$ for convenience. Then (i) implies (by symmetry) $I \subseteq \mathfrak p\_1$. This shows that $n-1 \le \text{coht}\_r(I)$, and the rest is clear. []


