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In my research I have come across the following condition on a monoid.

Every element $x$ satisfies the following property: there exists a natural number $n$ such that for any $m \geq n$ and any decomposition $x = x_1 \cdot x_2 \cdot \cdots \cdot x_m$ of $x$ it must be that at least one of $x_i$ is an idempotent.

Free monoids obviously satisfy this property, and every monoid which has only idempotent elements. And of course natural numbers under multiplication due to prime factorization.

A non-example is any monoid with a unit which is not the identity and things like positive rationals or reals under addition.

Decompositions of monoid elements seem to be studied in certain areas such as language theory, however I have not found a reference for this exact notion of "finitely decomposable", so I would like to know whether such a condition (or a condition close to it) has been encountered before, and if so, when.

If it helps in answering the question, all the monoids I am considering are commutative.

Aleš Bizjak
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  • A more usual thing to ask is that any sufficiently long factorization of x, some subword has product an idempotent. This weaker statement is true for finite monoids and is essentially the pumping lemma from automata theory. – Benjamin Steinberg Apr 30 '17 at 14:53
  • @BenjaminSteinberg It does really depend. What I find strange about this request, is that the OP doesn't allow for the "free use" of units in the factorizations: It may reflect the kind of applications he has in mind, but it is unusual from the point of view of factorization theory in the way it has been developed so far, where units are just absorbed by the atoms and treated as a kind of "negligible factors". – Salvo Tringali Apr 30 '17 at 15:26
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    A non-example is any monoid with an element which has an inverse, but is not its own inverse,

    You probably meant "a unit which is not the identity element".

     –  Apr 30 '17 at 16:36
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    This will fail in any semigroup containing generalized inverses a,b with aba=a and bab=b with a,b not idempotents. This is quite rare for finite semigroups. – Benjamin Steinberg Apr 30 '17 at 16:45
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    Every element of finite order must be an idempotent. If x is finite order and not an idempotent it has arbitrarily long powers that are the same element – Benjamin Steinberg Apr 30 '17 at 19:57
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    @AlešBizjak I think you don't mean "A non-example is [something] and things like positive rationals or reals under multiplication", since the latter are groups, so they are already included in [something]. I think you mean the positive rationals (respectively, reals) under addition: More in general, any non-trivial, divisible, cancellative monoid is a non-example. – Salvo Tringali Apr 30 '17 at 20:54
  • Probably there is a way to describe this class of commutative monoids. Take an example $S$, and decompose it as a semilattice of archimedian commutative semigroups $S_\alpha$, $\alpha\in I$. Then each $S_\alpha$ may contain at most one idempotent, $e_i$. So I think the archimedian parts can be characterized. Then look at the semilattice structure. There should be some sort of Noetherian property satisfied by the semilattice, and the structure homomorphisms should be possible to characterize. –  May 01 '17 at 00:00
  • I've edited the question to address some of the points raised in the comments. – Aleš Bizjak May 01 '17 at 09:10
  • @AlešBizjak I think that, if you are only interested in monoids with many idempotents, this should be made clear in the OP: Some of your examples and non-examples give a completely different impression. I see your edits of the original post, and you didn't address this issue (raised by Mark Sapir's comments to my answer). So, I do actually presume that you're interested in any sort of monoids that satisfy the condition stated in the OP, though your motivation comes from applications where (infinitely) many non-trivial idempotents exist. Is my interpretation of your request correct? – Salvo Tringali May 02 '17 at 07:22

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Reduced BF-monoids fit the bill here.

If $H$ is a multiplicatively written monoid with identity $1_H$, then an atom of $H$ is an element $a \in H \setminus H^\times$ for which there do not exist $x, y \in H \setminus H^\times$ such that $a = xy$, where $H^\times$ is the group of units of $H$. In particular, $H$ is called atomic if every non-unit of $H$ is a (finite) product of atoms, and is a BF-monoid if the factorizations (into atoms) of a fixed element cannot get arbitrarily long. On the other hand, we say that $H$ is reduced if $H^\times = \{1_H\}$.

There is a vast literature on the factorization theory of atomic monoids, though most of it is centered on the commutative and cancellative setting, for which you may want to have a look to:

A. Geroldinger and F. Halter-Koch, Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory, Pure Appl. Math. 278, Chapman & Hall/CRC, Boca Raton (FL), 2006.

If, on the other hand, you are also interested in non-cancellative or non-commutative monoids, then it's a totally different story, and the best I can do is to address you to my own work with Yushuang Fan, where you will find some pointers to relevant literature (most notably, Smertnig's and Baeth and Smertnig's papers on cancellative categories) and an entire section devoted to basic aspects of the theory (namely, Sect. 2):

Y. Fan and S. Tringali, Power monoids: A bridge between Factorization Theory and Arithmetic Combinatorics, preprint (arXiv:1701.09152).

Lastly, if you find yourself wondering about sufficient conditions for a monoid to be BF, then there might be just the thing for you in another thread: Among many others, free monoids and free abelian monoids are BF-monoids (this is obvious), and so is the multiplicative monoid of non-zero elements of a Noetherian integral domain (this is less obvious). Moreover, if $H$ is a unit-cancellative BF-monoid ("unit-cancellative" means that $xy=x$ or $yx=x$, for some $x, y \in H$, implies $y \in H^\times$) and $M$ is a submonoid of $H$ with $M^\times = M \cap H^\times$, then $M$ is a BF-monoid too (Theorem 2.22(iv) + Corollary 2.23 in the above preprint). In point of fact, some of the examples mentioned in the OP are a special case of these.

Of course, reduced BF-monoids are not the only semigroups that fit your requests. But, to the best of my knowledge, they are the only class for which a systematic theory of factorization has been so far developed.

Edit. In the comments to this answer, Mark Sapir writes, "This does not seem to answer the question. The OP likes idempotents and does not like units." So, let me try to clarify why this is an answer to the question, though not the only one possible (I'll continue with the same notations used in the above).

Given $x \in H \setminus \{1_H\}$, we denote by $\mathsf L_H(x)$ the set of all $k \in \mathbf N^+$ for which there exist atoms $a_1, \ldots, a_k \in \mathbf N^+$ such that $x = a_1 \cdots a_k$. Moreover, we take $\mathsf L_H(1_H) := \{0\}$. It can be proved (this is not for free) that ${\sf L}_H(u) = \emptyset$ for every $u \in H^\times \setminus \{1_H\}$. In addition, it is straightforward that $$ \mathsf L_H(x) + \mathsf L_H(y) \subseteq \mathsf L_H(xy),\ \text{ for all }x, y \in H. $$ Lastly, $H$ is a BF-monoid iff $\mathsf L_H(x)$ is finite and non-empty for every $x \in H \setminus H^\times$.

With this in mind, assume $H$ is a reduced BF-monoid, so that $\mathsf L_H(x)$ is finite and non-empty for all $x \in H$. Accordingly, pick $\bar x \in H \setminus \{1_H\}$, and let $\rho(\bar x)$ be the maximum of $\mathsf L_H(\bar x)$. If $m$ is an integer $\ge 1 + \rho(\bar x)$ and $\bar x = x_1 \cdots x_m$ for some $x_1, \ldots, x_m \in H$, then $$ \mathsf L_H(x_1) + \cdots + \mathsf L_H(x_m) \subseteq \mathsf L_H(\bar x), $$ and hence $$ \max \mathsf L_H(x_1) + \cdots + \max \mathsf L_H(x_m) \le \max \mathsf L_H(\bar x) = \rho(\bar x), $$ which is possible only if $\mathsf L_H(x_i) = \{0\}$, and hence $x_i = 1_H$, for some $i \in [\![1,m]\!]$. []

Salvo Tringali
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  • This does not seem to answer the question. The OP likes idempotents and does not like units. –  Apr 30 '17 at 16:39
  • @MarkSapir I maintain it does. But I could have given more details: I did it now, you may want to give it a second try. In short, assume $H$ is a reduced BF-monoid: The basic point of this answer is that, for every fixed element $x \in H$, there exists $n_x \in \mathbf N^+$ s.t., if $m \ge n_x$ and $x=x_1\cdots x_m$ for some $x_1,\ldots,x_m \in H$, then at least $m-n_x$ of the factors $x_i$ must be equal to the identity, and the identity is an idempotent, so... – Salvo Tringali Apr 30 '17 at 20:23
  • The OP likes many idempotents and does not like units at all. I do not know the reason (it would be interesting to know) but it seems clear from the question. A typical example: a Rees quotient of the direct product of a reduced BF-monoid with trivial group of units and a semilattice. –  Apr 30 '17 at 23:52
  • @MarkSapir Every monoid with the property of the OP must be reduced, so units are not at all an issue. On the other hand, I'm not so sure about your statement that the OP "likes many idempotents": 2 out of 3 of the examples he mentions (namely, free monoids and the natural numbers under multiplication) are reduced BF-monoids. So, I sympathize with you that the OP would be better to clarify his requests, but this doesn't change that reduced BF-monoids are a legitimate answer to the question (in fact, they supply a non-trivial and wide class of interesting examples), in the form it stands. – Salvo Tringali May 01 '17 at 03:48
  • If $a^{-1}$ exists, $a\ne 1$, then $a=aa^{-1}aa^{-1}...a$, and none of the factors is an idempotent. Non-trivial units cannot exist. If there are no idempotents, then the monoid simply is a BF monoid without non-trivial units. But that is not what the OP wants apparently. –  May 01 '17 at 08:26
  • @MarkSapir The context in which these monoids would be useful is models of separation-like logics. There monoid elements represent "resources" and idempotent elements (called duplicable in the separation logic literature) are those resources which can be shared. Thus a lot of monoids which are used have very many idempotents, but typically have no units. In general the monoids are not cancellative either (e.g., an often used one is natural numbers under maximum). – Aleš Bizjak May 01 '17 at 09:07
  • @MarkSapir From my previous comment, "Every monoid with the property of the OP must be reduced, so units are not at all an issue." To wit, it's clear to me that there can't be non-trivial units, so "units are not at all an issue", in that we may assume since the outset that, for the purposes of this thread, monoids are reduced. As for the rest, I'll give up: You continue to repeat the refrain that the OP is (only) interested in monoids with many idempotents, and to ignore that (i) 2 out of 3 of the OP's examples are reduced BF-monoids and (ii) some of his non-examples have no non-trivial (...) – Salvo Tringali May 02 '17 at 09:53
  • (...) idempotent. Not to mention that the OP is seeking information about semigroups $H$ with the property that, for every $x \in H$, there exists $n_x \in \mathbf N^+$ s.t., if $m \ge n_x$ and $x=x_1\cdots x_m$, then $x_i^2=x_i$ for some $i \in [![1,m]!]$: He writes, "I would like to know whether such a condition (or a condition close to it) has been encountered before, and if so, when." So, the only question should be: Do reduced BF-monoids satisfy, among many others, the condition stated in the OP? – Salvo Tringali May 02 '17 at 12:03