# What are the main structure theorems on finitely generated commutative monoids?

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 a commutative monoid is finitely generated it is finitely presented.

• If a finitely generated commutative monoid is cancellative ($a + b = a' + b \Rightarrow a = a'$) then it embeds in a finitely generated abelian group.

• If a finitely generated commutative monoid is cancellative and torsion-free ($a + a + \cdots + a = 0 \Rightarrow a = 0$) then it embeds in a finitely generated free abelian group. (This follows easily from the previous claim.)

• If a commutative monoid is a submonoid of $(\mathbb{N},+,0)$ it is called a numerical monoid and of course it is cancellative. A lot is known about numerical monoids, though I don't believe they have been "classified" in any useful sense.

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?

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John, sorry, is your question the last sentence? –  Andres Caicedo May 19 '13 at 1:34
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. –  Benjamin Steinberg May 19 '13 at 12:20
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. –  Benjamin Steinberg May 19 '13 at 12:27
Andres wrote: "is your question the last sentence?" No, it's the title: what are the main structure theorems on finitely generated commutative monoids? –  John Baez May 19 '13 at 22:12
In case anyone reads this in the distant future, my question is now my last sentence, though it wasn't when Andres Caicedo asked. –  John Baez May 20 '13 at 23:53

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.

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

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 $X$.

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What does "unambiguously rational" mean? –  James Cranch Aug 12 '13 at 13:18
@james-cranch Just added the definition in my answer. –  J.-E. Pin Aug 12 '13 at 15:09

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

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