Timeline for What are the main structure theorems on finitely generated commutative monoids?
Current License: CC BY-SA 4.0
35 events
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| Jan 26, 2022 at 15:20 | comment | added | Salvo Tringali | @JohnBaez An idempotent monoid has no atoms per "my" definition (some people would say "is antimatter"); in particular, this is the case of a unital semilattice. The order-theoretic notion of "atom" is rather generalized, on the level of monoids, by the notion of "quark": Given a preorder $\preceq$ on (the underlying set of) a monoid $H$, we let a $\preceq$-unit be an element $u \in H$ s.t. $u\preceq 1_H\preceq u$; and a $\preceq$-quark be a $\preceq$-non-unit s.t. there is no $\preceq$-non-unit $b$ with $b \prec a$. Now let $\preceq$ be the divisibility preorder on $H$ and you get a quark. | |
| Jan 26, 2022 at 11:50 | answer | added | Salvo Tringali | timeline score: 6 | |
| Jan 25, 2022 at 20:15 | comment | added | deaton.dg | @JohnBaez Take the free commutative monoid on a single generator, but with an extra half of two. By that, I mean $\mathbb{N}\{1,x\} / \langle 2x=2\rangle$. Every element has a canonical representative $n+bx$ where $n \in \mathbb{N}$ and $b\in\{0,1\}$. I believe this satisfies all the listed conditions, but it cannot be embedded in a free abelian group because then $2(f(1)-f(x))=0$ and so $f(x)=f(1)$. | |
| Jan 25, 2022 at 19:56 | comment | added | John Baez | Thanks! From logic I'm familiar with the concept of 'atom' in a lattice or semilattice, which is a special case of the definition you gave. But I'd never thought about it for more general commutative monoids. So I don't yet have much of a feeling for your theorem, but it's interesting - so if you feel like posting an answer containing it, please do! | |
| Jan 25, 2022 at 19:47 | comment | added | Salvo Tringali | @JohnBaez A monoid is atomic if every non-unit factors into a (finite) product of atoms. And an atom is a non-unit that doesn't factor as the product of two non-units. (The terminology goes back to P.M. Cohn's work on factorization in the 1960s.) | |
| Jan 25, 2022 at 19:46 | comment | added | John Baez | @SalvoTringali - I forget what an 'atomic' commutative monoid is. | |
| Jan 25, 2022 at 19:45 | comment | added | John Baez | @deaton.dg - I had said any finitely generated cancellative commutative monoid with $a + \cdots + a = 0 \implies a = 0$ embeds in a finitely generated free abelian group. Do you know a counterexample? Regardless, your definition of 'torsion-free' seems to be the standard one so I've changed mine to that. | |
| Jan 25, 2022 at 19:34 | comment | added | Salvo Tringali | @JohnBaez Does "If a cancellative, commutative monoid is finitely generated up to units, then it is atomic" count as an interesting structure theorem to you? If so, then I'll post an answer to provide further details (and discuss some generalizations). | |
| Jan 25, 2022 at 19:24 | history | edited | John Baez | CC BY-SA 4.0 |
fixed definition of "torsion-free" commutative monoid
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| Jan 24, 2022 at 4:41 | comment | added | deaton.dg | I believe the 4th bullet point is not quite correct. The definition of torsion-free should be that if $n$ is a positive integer and $na=nb$ then $a=b$. Otherwise, I can create such a monoid which does not embed into a finitely generated free abelian group. | |
| Feb 26, 2018 at 17:01 | answer | added | Tim Campion | timeline score: 14 | |
| Aug 12, 2013 at 14:53 | answer | added | arsmath | timeline score: 7 | |
| Aug 12, 2013 at 12:56 | answer | added | J.-E. Pin | timeline score: 9 | |
| May 20, 2013 at 23:53 | comment | added | John Baez | In case anyone reads this in the distant future, my question is now my last sentence, though it wasn't when Andres Caicedo asked. | |
| May 20, 2013 at 11:47 | comment | added | Benjamin Steinberg | @Yemon, I didn't see your comment the other day. Sorry to duplicate in my comment. | |
| May 20, 2013 at 10:30 | answer | added | Benjamin Steinberg | timeline score: 15 | |
| May 20, 2013 at 9:14 | comment | added | user6976 | @Noah: Submonoids of $\mathbb{N}$ differ from arithmetic progressions $k\mathbb{N}$ by finite sets. But the "by finite sets" part makes complete classification up to isomorphism hard. I do not know such a classification. In fact the lattice of subsemigroups of $\mathbb{N}$ contains every finite lattice as a sublattice (Repnitskii). On the other hand, the isomorphism problem for commutative semigroups is decidable which follows from a result of Taiclin and a result of Grunewald and Segal. | |
| May 20, 2013 at 4:22 | comment | added | Theo Johnson-Freyd | @Noah S: I imagine that there are certain versions of the problem "classify numerical monoids" that really mean "understand the primes". Certainly there are "classification" problems among prime numbers that are not known. | |
| May 20, 2013 at 3:28 | comment | added | Noah Schweber | One quick question: you said numerical monoids have not 'been "classified" in any useful sense.' Could you explain what a useful classification would entail? I mean, we know exactly what the numerical monoids are, up to isomorphism. (This is probably a naive question.) | |
| May 20, 2013 at 0:50 | history | edited | John Baez | CC BY-SA 3.0 |
added information about numerical monoids
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| May 20, 2013 at 0:44 | comment | added | John Baez | Benjamin wrote: "A finite commutative semigroup has a grading by a semilattice such that the homogeneous components are nilpotent extensions of abelian groups." Great! That sounds like the kind of thing I want to know about. | |
| May 20, 2013 at 0:43 | comment | added | John Baez | Benjamin wrote: "Another big result is that the first order theory is decidable. I can't recall the reference but Mark Sapir knows it." Two references to this - papers by M. A. Taiclin - are in the link I provided in my second item. This link is a long review article by Mark Sapir and a coauthor. The references are numbers 386 and 387. I added some words to clarify that indeed the whole elementary theory is decidable. | |
| May 20, 2013 at 0:39 | history | edited | John Baez | CC BY-SA 3.0 |
clarified the question
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| May 20, 2013 at 0:33 | history | edited | John Baez | CC BY-SA 3.0 |
added 86 characters in body
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| May 19, 2013 at 22:12 | comment | added | John Baez | Andres wrote: "is your question the last sentence?" No, it's the title: what are the main structure theorems on finitely generated commutative monoids? | |
| May 19, 2013 at 22:11 | comment | added | John Baez | Qiaochu wrote: "This sounds quite hard." I'm not expecting a full classification, just theorems that help us classify certain restricted classes of finitely generated commutative monoids, or at least describe their structure. For example, knowing that every cancellative one embeds in $\mathbb{Z}^n$ is worth something. Locally compact Hausdorff topological abelian groups is another category where we'll never get a full classification, but there are beautiful partial results. | |
| May 19, 2013 at 12:40 | comment | added | Benjamin Steinberg | Probably not what you are looking for but context-free subsets of commutative monoids are semilinear, so definable in pressburger arithmetic. They have decidable membership by integer programming. In particular integer programming decides membership in submonoids so the generalized word problem is decidable. | |
| May 19, 2013 at 12:27 | comment | added | Benjamin Steinberg | 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. | |
| May 19, 2013 at 12:20 | comment | added | Benjamin Steinberg | 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. | |
| May 19, 2013 at 12:17 | comment | added | Benjamin Steinberg | Another big result is that the first order theory is decidable. I can't recall the reference but Mark Sapir knows it. Also finitely generated commutative monoids are residually finite. A lot more is known form numerical and affine semigroups, eg, subsemigroups of N and of Z^m. | |
| May 19, 2013 at 6:59 | comment | added | Yemon Choi | Howie's book(s) on semigroup theory have, IIRC, a short discussion of the broad-brush structure theorem for decomposing a commutative semigroup as "a semilattice of archimedean subsemigroups". I'm away from my copy so can't give precise ref. | |
| May 19, 2013 at 5:53 | comment | added | Qiaochu Yuan | This sounds quite hard. Isn't the category of finitely generated commutative idempotent monoids equivalent to the category of finite lattices? | |
| May 19, 2013 at 1:46 | comment | added | user9072 | Pierre Grillet's Commutative Semigroups (2001) seems like (another) good place to start. | |
| May 19, 2013 at 1:34 | comment | added | Andrés E. Caicedo | John, sorry, is your question the last sentence? | |
| May 19, 2013 at 0:50 | history | asked | John Baez | CC BY-SA 3.0 |