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, 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 (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.

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 $I = J = \mathbf N^+$.

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 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.

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 $a = xy$.


2 Answers 2


The case of the ideal system $d$ (on the multiplicative monoid of a commutative ring $A$) defined in the OP has a positive answer. In fact, the answer we get is rather strong and it is an immediate consequence of the following basic result.

Lemma. $-$ Let $A$ be a commutative unital ring, $\mathfrak{a}$ and $\mathfrak{b}$ two ideals and $\mathfrak{p}$ a prime ideal of $A$. If $\mathfrak{p}$ contains $\mathfrak{a}\cdot_d\mathfrak{b}$, then the prime ideal $\mathfrak{p}$ contains one among $\mathfrak{a}$ or $\mathfrak{b}$.

Proof. Suppose $\mathfrak{p}$ does not contain either $\mathfrak{a}$ or $\mathfrak{b}$. That is to say, there exist elements $x$ in $\mathfrak{a}$ and $y$ in $\mathfrak{b}$ such that $x$ and $y$ do not belong to $\mathfrak{p}$. Therefore $xy$ is not in $\mathfrak{p}$, as this is a prime ideal. Hence $\mathfrak{p}$ does not contain the product $\mathfrak{a}\cdot_d\mathfrak{b}$ and we get a contradiction.

Edit. Since I misread he definition of ideal in first place and it does not coincide with what I had in mind, I'd like to add some words about it and fix the proposition (see the comments).

Remarks. Notice that for any ideal $\mathfrak{a}$ of $A$, the value of $\text{coht}_d(\mathfrak{a})$ is precisely $\text{dim}(A/\mathfrak{a})$, the Krull dimension of $A/\mathfrak{a}$, when $\mathfrak{a}$ is a prime ideal and it is equal to $\text{dim}(A/\mathfrak{a}) + 1$ otherwise.

Proposition. $-$ Let $A$ be a commutative unital ring and $\mathfrak{a}$ and $\mathfrak{b}$ two ideals of $A$. Then $$\text{coht}_d(\mathfrak{a}\cdot_d\mathfrak{b}) \leq 1 + \text{max}\{\text{coht}_d(\mathfrak{a}), \text{coht}_d(\mathfrak{b})\}.$$

Proof. We can restrict to the case in which the $d$-coheights of the two ideals $\mathfrak{a}$ and $\mathfrak{b}$ are finite, the result being trivial otherwise. The previous lemma gives us the relation $$\text{dim}(A/\mathfrak{a}\cdot_d\mathfrak{b}) = \text{max}\{\text{dim}(A/\mathfrak{a}), \text{dim}(A/\mathfrak{b})\}.$$ Indeed, suppose wlog $n = \text{dim}(A/\mathfrak{a})\geq \text{dim}(A/\mathfrak{b})$ and consider a strict maximal chain of prime ideals $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \dots \subsetneq \mathfrak{p}_n$ of $A$, $n\geq 0$, such that $\mathfrak{a}\subseteq \mathfrak{p}_0$. If we had a prime ideal $\mathfrak{q}$ satisfying the relation $\mathfrak{a}\cdot_d \mathfrak{b} \subseteq \mathfrak{q} \subsetneq p_0$, the lemma would imply either $\mathfrak{a}\subseteq \mathfrak{q}$ or $\mathfrak{b}\subseteq \mathfrak{q}$, therefore producing a chain of primes over $\mathfrak{a}$ (resp. over $\mathfrak{b}$) of length $n+2$, which is impossible by assumption. The considerations of the above remark then immediately give the inequality regarding the $d$-coheights.

  • $\begingroup$ Doesn't the lemma in your answer prove only that $\text{coht}_d(\mathfrak a \cdot_d \mathfrak b) \le 1+ \max(\text{coht}_d(\mathfrak a), \text{coht}_d(\mathfrak b))$? The claim is obvious if the $d$-coheight of $\mathfrak a \cdot_d \mathfrak b$ is zero. Otherwise, let $\mathfrak p_1,\ldots,\mathfrak p_n$ be $d$-prime ideals s.t. $\mathfrak p_{i-1}\subsetneq\mathfrak p_i$ for each $i\in [\![1,n]\!]$, where $\mathfrak p_0:=\mathfrak a\cdot_d\mathfrak b$. Then the lemma in your answer implies, wlog, $\mathfrak a\subseteq\mathfrak p_1$: Note that we started with a strict inclusion, but (...) $\endgroup$ Apr 24, 2017 at 15:21
  • $\begingroup$ (...) we end up with a weak inclusion. This yields $n-1 \le \text{coht}_d(\mathfrak a)$, and we're done. [] (In any case, the main conclusion is still correct: The inequality in the OP holds true for the ideal system $d$ of the multiplicative monoid of any commutative, unital ring.) $\endgroup$ Apr 24, 2017 at 15:31
  • $\begingroup$ Btw, the inequality $\text{coht}_d(\mathfrak a\cdot_d \mathfrak b) \le 1+\max(\text{coht}_d(\mathfrak a),\text{coht}_d(\mathfrak b))$ cannot be improved, in general: In your notations, if $A$ is the ring of integers and $\mathfrak a = \mathfrak b = \{np: n \in \mathbf Z\}$ for some prime element of $\mathbf N^+$, then $\text{coht}_d(\mathfrak a)=\text{coht}_d(\mathfrak a)=0$, but $\text{coht}_d(\mathfrak a \cdot_d \mathfrak b)=1$. $\endgroup$ Apr 24, 2017 at 16:02

The inequality in the OP is true, and even in a stronger form: I'm indebted to Andrea Gagna for his enlightening answer, 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. []


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