38
$\begingroup$

I should read J. C. Rosales and P. A. García-Sánchez's book Finitely Generated Commutative Monoids and L. Redei's book The Theory of Finitely Generated Commutative Semigroups. I haven't. But here's what I've heard so far:

If we drop the property of being cancellative we get an enormous wilderness of finitely generated commutative monoids, so there shouldn't be any simple 'classification theorem'. But there still might be interesting structure theorems which help us understand this wilderness, just as there are for (say) compact topological abelian groups. What are they?

$\endgroup$
25
  • 6
    $\begingroup$ John, sorry, is your question the last sentence? $\endgroup$ May 19, 2013 at 1:34
  • 7
    $\begingroup$ A finite commutative semigroup has a grading by a semilattice such that the homogeneous components are nilpotent extensions of abelian groups. The buzzword is semilattice of Archimedean semigroups. I think Grillet will give the best results on such decompositions. $\endgroup$ May 19, 2013 at 12:20
  • 8
    $\begingroup$ In fact every commutative semigroup is a semilattice of Archimedean semigroups. The Archimedean components can be strange but if you have some extra conditions they will be cancellative and hence group embeddable. $\endgroup$ May 19, 2013 at 12:27
  • 6
    $\begingroup$ Andres wrote: "is your question the last sentence?" No, it's the title: what are the main structure theorems on finitely generated commutative monoids? $\endgroup$
    – John Baez
    May 19, 2013 at 22:12
  • 11
    $\begingroup$ In case anyone reads this in the distant future, my question is now my last sentence, though it wasn't when Andres Caicedo asked. $\endgroup$
    – John Baez
    May 20, 2013 at 23:53

5 Answers 5

13
$\begingroup$

The comments are getting a bit long so I'll put this as a partial answer. The case of von Neumann regular commutative semigroups was handled by Clifford in the 1940s. A semigroup is von Neumann regular if for all $a$, there exists $b$ with $aba=a$. Clifford proved a regular commutative semigroup is the same thing as a pair $(E,F)$ where $E$ is a poset with binary meets and $F$ is a presheaf of abelian groups on $E$. If the semigroup is a finitely generated monoid then $E$ will be a finite lattice.

For example, given such a pair, the underlying set of the semigroup is the disjoint union of the $F(e)$ with $e$ in $E$ (so the arrow set of the associated discrete fibration). The product of $a \in F(e)$ with $b \in F(e')$ is obtained by restricting both elements to the meet of $e$ and $e'$ and taking their product.

The more general semilattice decompositions in the comments are not as good as this.

$\endgroup$
2
  • $\begingroup$ This is what Howie calls a "strong semilattice" of abelian groups, right? (Presheaf over a semilattice is more informative terminology, I agree.) $\endgroup$
    – Yemon Choi
    May 26, 2013 at 23:37
  • $\begingroup$ @YC, yes but presheaf is a better known terminology outside semigroups. $\endgroup$ May 28, 2013 at 13:10
12
$\begingroup$

Have a look at Grillet's Commutative Semigroups. Let $C$ be a commutative semigroup. The outline of the structure theory is as follows (I'll include references to Grillet; see also V.5.7 for an outline of some structure theory for commutative semigroups):

  1. As arsmath says, $C$ decomposes as a semilattice of archimedean semigroups. The relevant semilattice is the universal semilattice $C_S = C / (2=1)$ on $C$. The decomposition is as follows: the the fibers of the universal map $C \to C_S$ are archimedean semigroups, called the archimedean components of $C$.
  • If $C$ is finitely-generated, then so is $C_S$, and hence $C_S$ is finite.
  1. $C$ is said to be complete if each archimedean component contains an idempotent, and subcomplete if each archimedean component embeds into a complete archimedean semigroup. An archimedean semigroup always contains at most one idempotent, so $C$ is complete iff the composite map $C^S \to C \to C_S$ (which is always injective (II.1.4)) is an isomorphism. Here $C^S = \{x \in C \mid 2x=x\}$ is the co-universal semilattice on $C$, i.e. the semilattice of idempotents in $C$.
  • If $C$ is finitely-generated, then $C$ is subcomplete (VII.1.1). The archimedean components of $C$ need not be finitely-generated ([VI.1.5, VI.2.5]), but they do have certain finiteness properties. Thus, arsmath's point that general archimedean commutative semigroups are complicated notwithstanding, for the finitely-generated case, we can focus on the more tractable class of subcomplete archimedean commutative semigroups.
  1. If $A$ is a complete archimedean semigroup, then $A$ is elementary (III.3.1), i.e. $A$ decomposes as an ideal extension $G \to A \to N$ where $G$ is a group and $N$ is a nilsemigroup, i.e. $N$ has an absorbing element -- an element $\infty \in N$ such that for every $x\in N$, there is $n \in \mathbb N$ such that for all $m \geq n$, $mx = \infty$. If $A$ is subcomplete archimedean, then it has a similar decomposition where $G$ is cancellative, i.e. $A$ is subelementary.
  • If $A$ is an archimedean component of a finitely-generated commutative semigroup $C$, then in the decomposition $G \to A \to N$, $G$ is finitely-generated (VI.2.4), but $N$ in general is not. Finitely-generated cancellative commutative semigroups are well-understood (they are products of finite groups and cones in $\mathbb Z^n$), so that part of the structure is comprehensible. But this nilsemigroup part is more mysterious, I think.

Finite generation of $C$ does imply some good properties of $N$, though -- the nilpotence degree has a uniform bound (VI.2.7), i.e. $N$ is a nilpotent semigroup (I find the terminology "nilpotent semigroup" vs. "nilsemigroup" to be confusing, but it's what Grillet uses; presumably it is standard in semigroup theory.). There's more to say here about control over $N$ (VI.3.3) but it involves more terminology which I do not have a great handle on.


Putting it all together, if you have a finitely-generated commutative semigroup $C$, then you can think of it as a finite lattice of a bunch of finite abelian groups equipped with certain positive cones, each of which has a nilpotent commutative semigroup on which it acts, with homomorphisms between these corresponding to the relations in the lattice.

$\endgroup$
5
  • 1
    $\begingroup$ This is a great answer, Tim - thanks! What does this mean? "$N$ is nilpotent, i.e. $N$ has an absorbing element $\infty$ such that for every $x\in N$, there is $n \in \mathbb N$ such that $mx = \infty$ for all $m \geq n$ and all $x \in N$." The typo leaves the order of quantifiers ambiguous. $\endgroup$
    – John Baez
    Apr 3, 2020 at 23:21
  • $\begingroup$ Glad to help! I've edited to try to clarify -- the quantifiers should go $\forall x \exists n \forall m \geq n (mx = \infty)$. $\endgroup$
    – Tim Campion
    Apr 4, 2020 at 0:51
  • $\begingroup$ Thanks! Good, that's what I'd call nilpotent. If there existed an $n$ that worked for all $x$ I guess I'd call it "uniformly nilpotent" or something. $\endgroup$
    – John Baez
    Apr 4, 2020 at 1:26
  • $\begingroup$ @JohnBaez Coming back to this today, I noticed that in the case where $C$ is finitely-generated, Grillet shows that the nilpotent part $N$ actually is "uniformly nilpotent" (or in his terminology, $N$ is a "nilpotent semigroup") and not just nilpotent (or in his terminology, a "nilsemigroup"). I've edited my answer to reflect this. $\endgroup$
    – Tim Campion
    Feb 1 at 18:47
  • 1
    $\begingroup$ The names nil semigroup and nilpotent semigroup come from ring theory where people talk about nil ideals versus nilpotent ideals or if they are brave enough not to require identities then nil rings and nilpotent rings. See en.wikipedia.org/wiki/Nil_ideal. This is consistent with the usage of idempotent ideal. In other words ideals in a ring form a multiplicative semigroup and nilpotent ideal means nilpotent in this semigroup. The same idea applies to semigroups. $\endgroup$ Feb 1 at 20:56
9
$\begingroup$

Let me add to the previous answers some important properties of the rational subsets. The rational subsets of a monoid $M$ form the smallest class of subsets of $M$ containing the singletons and closed under finite union, product and star (star = submonoid generated). By construction, rational sets are closed under finite union, product and star, but are not in general closed under complement.

However, if $M$ is a finitely generated commutative monoid:

(1) Every congruence on $M$ is a rational subset of $M \times M$.

(2) The rational subsets of $M$ are closed under Boolean operations (finite union, finite intersection and complement).

(3) Every rational subset of $M$ is unambiguously rational.

[1] S. Eilenberg and M.P. Schützenberger, Rational sets in commutative monoids. J. Algebra 13 (1969) 173-191. doi:10.1016/0021-8693(69)90070-2

P.S. Unambiguously rational subsets: same definition as for rational subsets, but only unambiguous versions of the three operations are allowed.
(a) Unambiguous union = disjoint union.
(b) Unambiguous product $XY$: if $x_1, x_2 \in X$, $y_1, y_2 \in Y$ and $x_1x_2 = y_1y_2$, then $x_1 = x_2$ and $y_1 = y_2$.
(c) Unambiguous star $X^*$: the monoid $X^*$ is free with base $X$.

$\endgroup$
2
  • 1
    $\begingroup$ What does "unambiguously rational" mean? $\endgroup$ Aug 12, 2013 at 13:18
  • $\begingroup$ @james-cranch Just added the definition in my answer. $\endgroup$
    – J.-E. Pin
    Aug 12, 2013 at 15:09
6
$\begingroup$

As a couple of people have mentioned, commutative semigroups can be decomposed as lattices of archimedean semigroups. My impression is that there is no general classification result for archimedean semigroups, but there is a reasonably strong result by Tamura:

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pja/1195522174

$\endgroup$
6
$\begingroup$

[In a comment to the OP, I asked whether a statement along the lines of Corollary 1 below would count as an "interesting structure theorem", and this post expands on John Baez's yes to that question.]

Let $H$ be a multiplicatively written monoid; unless stated otherwise, $H$ need not be commutative, cancellative, or whatever. An atom of $H$ is a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$. We write $H^\times$ and $\mathscr A(H)$, resp., for the group of units and the set of atoms of $H$.

Denote by $\mathscr F(X)$ the free monoid on a basis $X$; we'll refer to the elements of $\mathscr F(X)$ as $X$-words. Given a non-unit $x \in H$, we define $\mathcal Z_H(x)$ as the set of all $\mathscr A(H)$-words $\mathfrak a$ such that $x = \pi_H(\mathfrak a)$, where $\pi_H$ is the canonical (monoid) epimorphism $\mathscr F(H) \to H$ (see the section "Pills of factorization theory" in this answer if something in the terminology or in the notation doesn't look familiar). The $\mathscr A(H)$-words in $\mathcal Z_H(x)$ are the atomic factorizations of $x$ (relative to $H$).

We write $\preceq_H$ for the "shuffling preorder" induced on $\mathscr F(\mathscr A(H))$ by the divisibility preorder${}^{(1)}$ $\mid_H$ on $H$; roughly speaking, this means that an $\mathscr A(H)$-word $\mathfrak a$ is $\preceq_H$-smaller than an $\mathscr A(H)$-word $\mathfrak b$ if and only if $\mathfrak a$ is a subword of a permutation of $\mathfrak b$ modulo $\mid_H$-equivalence${}^{(2)}$ on the letters.

We call $H$ atomic (resp., an FF-monoid) if $\mathcal Z_H(x)$ is non-empty (resp., finite and non-empty modulo $\preceq_H$-equivalence) for every non-unit $x \in H$; and an FmF-monoid if, for each non-unit $x \in H$, the set of $\preceq_H$-minimal $\mathscr A(H)$-words in $\mathcal Z_H(x)$ is finite and non-empty modulo $\preceq_H$-equivalence.

Examples. (i) If $H$ is the multiplicative monoid of the ring of integers, then $\mathscr A(H)$ is the set of (positive or negative) primes, and the $H$-words $(-2) \ast 3 \ast (-5)$ and $(-3) \ast 5 \ast 2$ are $\preceq_H$-equivalent atomic factorizations of $-30$.

(ii) If $p \in \mathbf N^+$ is a prime, $n$ is an integer $\ge 2$, and $H$ is the multiplicative monoid of the ring of integers modulo $p^n$, then the $H$-word $\bar{p}^{\ast (n+1)}$ is an atomic factorization of $\bar{0}$ but is not $\preceq_H$-minimal (here, $\bar{x}$ is the residue class mod $p^n$ of an integer $x$).

We say that $H$ is finitely generated modulo units (or shortly, an f.g.u. monoid) if there is a finite subset $A \subseteq H$ such that $H \setminus H^\times$ is contained in the subsemigroup generated by the set $H^\times A H^\times$; and is locally finitely generated modulo units (or shortly, an l.f.g.u. monoid) if, for each $a \in H$, the smallest divisor-closed submonoid${}^{(3)}$ of $H$ containing $a$ is an f.g.u. monoid.

Examples. (iii) The monoid $H$ considered in Example (i) is not atomic: $0$ is a non-unit of $H$ with no atomic factorizations. Note, though, that $0$ is an irreducible of $H$ (see the definition before Theorem 2 below), and $H$ has the weaker property that every non-unit is a product of irreducibles.

(iv) The monoid $H$ considered in Example (ii) is FmF, but not FF; in particular, if $k$ an integer $\ge n$ and $u_1, \ldots, u_k \in H$ are units, then the $H$-word $\bar{p}u_1 \ast \cdots \bar{p}u_k$ is an atomic factorization of $\bar{0}$. (Note that two $\mathscr A(H)$-words are $\preceq_H$-equivalent only if they have the same number of letters.)

With these premises in place, we have:

Theorem 1. Every acyclic${}^{(4)}$, l.f.g.u. monoid is FmF.

Most notably, this implies the following:

Corollary 1. If $H$ is a unit-cancellative${}^{(5)}$, commutative monoid and the quotient monoid $H/H^\times$ is finitely generated, then $H$ is FF.

The corollary is essentially due to A. Geroldinger and F. Halter-Koch, cf. Proposition 2.7.8.4 in their monograph:

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

These results are basically a generalization of the existence part of the fundamental theorem of arithmetic and, at least to some extent, bring about the same kind of "structural content" (though atomic factorizations are, in general, far from being unique in any sensible way). In fact, they are part of a bigger picture; and though the OP asked only about finitely generated (commutative) monoids, it is perhaps worth seeing what can be done in a greater generality.

In particular, taking an irreducible of $H$ to be a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$ with $HxH \ne HaH \ne HyH$ leads to:

Theorem 2. If $H$ is a Dedekind-finite${}^{(6)}$ monoid satisfying the ACCP${}^{(7)}$, then every non-unit of $H$ factors as a product of irreducibles.

To my knowledge, a proof of Theorem 2 for commutative rings was first published by D.D. Anderson and S. Valdes-Leon in

  • Factorization in Commutative Rings with Zero Divisors, Rocky Mountain J. Math. 26 (1996), No. 2, 439-480;

their proof (see loc. cit., Theorem 3.2) carries over verbatim to commutative monoids (and is, in some sense, straightforward from the definitions). In a different direction, we have:

Theorem 3. The conditions below are equivalent:

  1. $H$ is acyclic and satisfies the ACCP.
  2. $H$ is unit-cancellative and satisfies the ACCPR and the ACCPL${}^{(8)}$.

Moreover, each of these conditions implies that $H$ is atomic.

Note that every atom is irreducible, and the converse is true if $H$ is acyclic. Insofar as I'm aware, the terms "atom" and "atomic" were first coined by P.M. Cohn, whom I believe deserves credit also for the statement and the proof of the following (see Question 395980 for discussion on this "controversial point"):

Corollary 2. If $H$ is cancellative and satisfies the ACCPR and the ACCPL, then it is atomic.

There is more to the story. For further details and generalizations, I can only address the interested reader to an article of mine on monoids and preorders (here) and to a couple of related papers with Austin A. Antoniou (here) and Laura Cossu (here).

Notes.

(1) $a \mid_H b$, for some $a, b \in H$, if and only if $b \in HaH$.

(2) Given a preorder $\preceq$ on a set $X$, we say that two elements $x, y \in H$ are $\preceq$-equivalent if $x \preceq y \preceq x$.

(3) A submonoid $K$ of $H$ is divisor-closed if $x \mid_H y$ for some $x \in H$ and $y \in K$ implies that $x \in K$ (among other things, this guarantees that $H$ and $K$ have the same units).

(4) $H$ is acyclic if $xyz \ne y$ for all $x, y, z \in H$ such that either $x$ or $z$ is not a unit.

(5) $H$ is unit-cancellative if $xy \ne x \ne yx$ for all $x, y \in H$ such that $y$ is a non-unit. Obviously, every cancellative monoid is unit-cancellative; moreover, every acyclic monoid is unit-cancellative and the two notions coincide on the level of commutative monoids. On the other hand, Adian's embedding theorem (that is, the special case of Guba's embedding theorem where the left and right graphs of a finite semigroup presentation are cycle-free) implies at once the existence of a finitely generated, cancellative monoid with trivial group of units that is not acyclic.

(6) $H$ is Dedekind-finite if every left- or right-invertible element is a unit (or equivalently, if $xy = 1_H$ for some $x, y \in H$ implies $yx = 1_H$). It is fairly clear that every unit-cancellative or commutative monoid is Dedekind-finite.

(7) $H$ satisfies the ACCP if there is no (infinite) sequence $a_1, a_2, \ldots$ of elements of $H$ such that $Ha_nH \subsetneq Ha_{n+1}H$ for each $n \in \mathbf N^+$ (a set of the form $HaH$ with $a \in H$ is called a principal ideal of $H$).

(8) $H$ satisfies the ACCPR if there is no (infinite) sequence $a_1, a_2, \ldots$ of elements of $H$ such that $a_nH \subsetneq a_{n+1}H$ for each $n \in \mathbf N^+$ (a set of the form $aH$ with $a \in H$ is called a principal right ideal of $H$); and it satisfies the ACCPL if the opposite monoid of $H$ satisfies the ACCPR.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.