Edit (Apr 24, 2017). I'm updating this post in the light of the latest developments of a related thread.

Let $H$ be a multiplicatively written, commutative monoid, and set $M := H \setminus H^\times$, where $H^\times$ is the group of units of $H$. Note that $M$ is an ideal of $H$ if and only if $H$ is Dedekind-finite, because $MH \ne H$ if and only if $xy \in H^\times$ for some $x \in H \setminus H^\times$ and $y \in H$, and we can use an observation of Benjamin Steinberg to conclude. Moreover, the following are equivalent:

  1. $M$ is an ideal of $H$.
  2. $M$ is a prime ideal of $H$.
  3. $M$ is the maximum element of the poset of proper ideals of $H$ ordered by $\subseteq$.

Given $k \in \mathbf N^+$, we denote by $\mu_k(H)$ the supremum of all $n \in \mathbf N^+$ for which there exist prime ideals $\mathfrak p_1, \ldots, \mathfrak p_n$ of $H$ such that $\mathfrak p_{i-1} \subsetneq \mathfrak p_i$ for each $i \in [\![1, n]\!]$, where $\sup \emptyset := 0$ and $\mathfrak p_0 := M^k := \{x_1 \cdots x_k: x_1, \ldots, x_k \in M\}$. To wrap it in buzzwords, $\mu_k(H)$ is the coheight of $M^k$ relative to the fundamental ideal system of $H$.

We get from elsewhere and a trivial induction that $\mu_k(H) \le k-1$ (the bound can be sharpened by a slightly more clever induction, but never mind). So my first question is the following:

Q1. What about the set $V(H) := \{\mu_k(H): k \in \mathbf N^+\} \subseteq \mathbf N^+ \cup \{\infty\}$? More specifically, is $V(H)$ a bounded subset of $\mathbf N^+$ (respectively, an interval) for every $H$?

On the other hand, it is noted below that $\mu_k(H) = 1$ for all $k \ge 2$ provided that $H$ is Dedekind-finite and atomic and $H \ne H^\times$. Thus, my second question is:

Q2. Does there exist a commutative monoid $H$ such that $H \ne H^\times$ and $\mu_k(H) \ne 1$ for some $k \ge 2$?

Edit (Apr 25, 2017). In hindsight, Q2 is trivial: If $H$ is the additive monoid of the non-negative rational numbers with addition, then $M = \mathbf Q^+$, so that $M^k = M$, and hence $\mu_k(H) = 0 \ne 1$, for all $k \in \mathbf N^+$. Here is a better question:

Q3. Is $\mu_k(H) \le 1$ for every monoid $H$ and all $k \in \mathbf N^+$?

Notes. We say that $H$ is reduced if the only unit (or invertible element) of $H$ is the identity, and Dedekind-finite if $xy = 1_H$, for some $x, y \in H$, implies $yx = 1_H$. A set $I \subseteq H$ is called an ideal if $IH = I$. Then a prime ideal of $H$ is an ideal $I \ne H$ such that $H \setminus I$ is a subsemigroup of $H$.


1 Answer 1


This has just come to my mind: It is not an answer to the questions in the OP, but might be helpful to deal with them, so I've thought to post it as an answer (instead of adding to the OP and making it longer than an Egyptian papyrus).

We'll work out the case when $H$ is the monoid of positive integers under multiplication: This is not a difficult case, but is illustrative of a much more general setup.

To begin, pick $n \in \mathbf N_{\ge 2}$. We'll show that $\text{coht}(M^n) = 1$, where $M := \mathbf N^+ \setminus \{1\}$. Since $H$ is a reduced, commutative, atomic monoid (but not a group), $M$ is a maximal prime ideal (as it follows from the remarks in the OP), and hence $\text{coht}(M) = 0$.

Building on these premises, let $\mathfrak p$ be a prime ideal of $H$ containing $M$, and suppose for a contradiction that there exists a prime $p \in \mathbf N^+$ such that $p \notin \mathfrak p$. Then $p \in H \setminus \mathfrak p$, and hence $p^n \in H \setminus \mathfrak p$, since $\mathfrak p$ is a prime ideale (that is, $H \setminus \mathfrak p$ is a subsemigroup of $H$). To wit, $p^n \notin \mathfrak p$. This is, however, impossible, because $p^n \in M^n \subseteq \mathfrak p$.

It follows that $\mathfrak p$ contains all the primes of $\mathbf N^+$. But, on the other hand, if $I$ is any ideal of $H$ and $n \in I$, then $nk \in I$ for all $k \in \mathbf N^+$, because $nk \in IH = I$ (by the very definition of an ideal). So putting it all together, we have that $\mathfrak p$ contains $M$, and hence $\mathfrak p = M$ (recall that prime ideals are proper ideals), which shows that $\text{coht}(M^n) = 1$. []

In retrospect, we have actually proved the following:

Proposition. Pick $n \in \mathbf N^+$, and let $H$ be an atomic, Dedekind-finite, commutative monoid. Then $M := H \setminus H^\times$ is a maximal prime ideal of $H$, with $\text{coht}(M^n) = 0$ if $n = 1$ or $H = H^\times$, and $\text{coht}(M^n) = 1$ otherwise.

Proof. Conceptually the same as the proof of the basic case worked out in the above (recall that a monoid $H$ is atomic if every non-unit element of $H$ is a finite product of atoms of $H$). []


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