8
$\begingroup$

Is there a Structure Theorem for finitely generated commutative cancellative monoids?

Of course they can be densely embedded into a finitely generated abelian group, whose structure is known. Also, in the book of J. C. Rosales and P. A. García-Sánchez there are some special embedding theorems: If the monoid is torsionfree, it even embeds to some free abelian group, and if the monoid is also reduced, it embeds in some free commutative monoid.

But I want to know if it is possible to give a complete classification (for example, in terms of generators and relations, as in the case of groups).

$\endgroup$
1
  • $\begingroup$ Numerical monoids are probably the most famous example of a class of commutative cancellative monoids that are not really understood. $\endgroup$ Jan 22, 2015 at 3:34

3 Answers 3

6
$\begingroup$

I don't know of any structure theorem, and I imagine any classification would be quite complicated. If you restrict your view to finitely generated cancellative monoids that are also saturated ($ka \in M$ for any $k>1, a \in M^{gp}$ implies $a \in M$) and sharp (the only unit is zero), then the question amounts to classifying finitely generated cones in $\mathbb{Z}^n$ up to $GL_n(\mathbb{Z})$-equivalence, and there doesn't seem to be an explicit answer to even this special case. Passing to the positive real or rational span removes a large amount of arithmetic information.

On the positive side, you have the fact that any finitely generated commutative monoid $M$ is finitely presented, i.e., there is a coequalizer diagram $P_1 \rightrightarrows P_0 \to M$ with $P_1$ and $P_0$ free commutative and finitely generated. The question of determining when two presentations yield isomorphic monoids is algorithmically decidable but seems rather difficult.

$\endgroup$
5
$\begingroup$

To me, that such a monoid embeds in a finitely generated abelian group is a pretty good structure theorem. I would be surprised if one could get any sensible general statement much beyond that.

Some thoughts:

  • Factoring out the largest subgroup of the monoid reduces the problem to the classification of monoids having no nontrivial invertible element.

  • One can get quite complicated structures even when the group generated is cyclic times $\mathbb{Z}$.

  • Any antichain in $(\mathbf{N}\cup\{0\})^n$ under the partial order $(a_1,\ldots,a_n)\le(b_1,\ldots,b_n)$ if all $a_i\le b_i$ generates an example: take all $(b_1,\ldots,b_n)$ satisfying $(b_1,\ldots,b_n)\ge(a_1,\ldots,a_n)$ for some $(a_1,\ldots,a_n)$ in the antichain.

  • Insider $\mathbb{Z}^n$ one can get examples whose "positive real span" is an arbitrary rational cone, but each nontrivial rational cone arises in infinitely many ways. So the classification is more complicated that classifying rational cones up to $\mathrm{GL}_n(\mathbb{Z})$-equivalence.

  • and so on.

$\endgroup$
4
$\begingroup$

Finitely generated commutative semigroups have decidable elementary theories and the isomorphism problem for them is decidable. This means that some sort of description does exist. For example, with every pair of finitely generated commutative monoids, one can associate two tuples of integer matrices such that the monoids are isomorphic iff the tuples are conjugate (Taiclin), and then one can use a result of Grunewald and Segal or Sarkisyan. For references look at the survey http://www.math.vanderbilt.edu/~msapir/ftp/pub/survey/survey.pdf (the part about semigroups there).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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