MHB Semigroup Structure: Exploiting Non-Finitely Generated Monoids

[mathbalarka]

What are the best ways to exploit the structures of a given non-finitely generated cancellative commutative monoid?

This is not the best way to put together a question, I admit, as what I am asking is very general, but I think it is best to put in here rather than the Chat Room.

Also, is the rank of such a monoid particularly interesting?

Balarka
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[Turgul]

With such a vague question, it is difficult to give more than a vague answer, but here are some thoughts. Monoids are very similar to groups (in precisely the same way as commutative rings are a lot like fields). How do we study any difficult group? We study its representations. To my knowledge, this is the best approach to studying monoids as well. In terms of your question, one would hope that, with a given monoid in mind, one would have a description of objects the monoid acts on along with the associated actions.

More generally, most methods of attack one has for groups generalize to monoids, there might just be less one can say for the general case.

One can go further; a standard trick when dealing with monoids is to look at their groupification (Grothendieck group), understand the structure of the group and see what data can be reclaimed about monoid (similar to studying an integral domain by looking at its field of fractions).

In general, I can't imagine that the isomorphism type of the monoids you are looking at are all that interesting in the same sense that isomorphism types of abelian groups are not all that interesting. The interesting question is how they arise as parts of other structures (knowing the isomorphism type of a finitely generated abelian groups is not all that exciting, knowing which ones arise as the rational points on an elliptic curve, and HOW they arise is far more noteworthy).

 

[mathbalarka]

One can go further; a standard trick when dealing with monoids is to look at their groupification (Grothendieck group), understand the structure of the group and see what data can be reclaimed about monoid (similar to studying an integral domain by looking at its field of fractions).

As I have indicated, the monoid is not finitely generated, thus it's group-extensions doesn't help much, as far as my thoughts go. (Although it can be embedded in a group, yes, as it satisfies the conditions of Ore).

(knowing the isomorphism type of a finitely generated abelian groups is not all that exciting, knowing which ones arise as the rational points on an elliptic curve, and HOW they arise is far more noteworthy).

Yes, I agree on that. I am working on a specific monoid but since it is both non-torsion, non-finitely generated, and I don't think there are much of literature on such types, I am in the complete dark.

 

[Turgul]

I guess I'm really not sure what kind of advice you are looking for. If all you have is algebraic data, then the best you could possibly hope to find is the isomorphism type of the monoid. If all the algebraic data you have is that which you've provided, then there is an enormous set of isomorphism types satisfying your constraints (consider just the groups having these properties; any free abelian group on infinitely many generators works and you get one isomorphism type for every infinite cardinal--there are lots of ways to find a monoid in any of these groups).

If you're looking for something else, additional context is surely necessary.

 

[mathbalarka]

The best information of the Grothendiek group I have up until now is that it is endomorphic (could be generalized to isomorphism, I think) to $(\mathbb{Q}^+, \times)$ but I think one an do better, as the semigroup I am talking about is atomic, whereas in rationals under usual multiplication, primes doesn't interact at all.

Furthermore, it is my belief that the monoid cannot be embedded to a ring, although a proof is lacking.

 

[Turgul]

As you point out, by unique factorization of integers, $(\mathbb{Q}^+,\times)$ is the free abelian group on countably many generators. Certainly this group embeds into the ring $\mathbb{Q}$, so your monoid does as well if your map to $\mathbb{Q}^+$ ends up being injective (I don't think endomorphic is the word you want, as an endomorphism is a map from an object to itself).

In the long run, it's unlikely that I'll be able to be all that helpful other than to offer the obvious. Most people, before they try to study infinitely generated objects, spend some time understanding finitely generated objects. If this is a serious project and you want to know how people think about finitely generated torsion free abelian monoids in the hopes of extending to non-finite generation, I'll just say that the only people I know who seriously study such monoids are people who study toric geometry (toric geometry is roughly to finitely generated torsion-free abelian monoids as (classical) algebraic geometry is to finitely generated nilpotent-free $\mathbb{C}$-algebras).

 

[mathbalarka]

As you point out, by unique factorization of integers, $(\mathbb{Q}^+,\times)$ is the free abelian group on countably many generators. Certainly this group embeds into the ring \mathbb{Q}

I thought the base set has to be the same for the embedding?

 

[Turgul]

An embedding is merely an injective map which preserves the structure of interest. Certainly $\mathbb{Q}^+$ embeds into the multiplicative monoid of $\mathbb{Q}$ via the obvious inclusion (it even embeds into the multiplicative group of units of the field $\mathbb{Q}^\times$. In general, an embedding is supposed to be a way to see your object sitting inside some other object.

It is fair that you have to be careful what you mean when you say "embedding" because you have to be clear about what structure you are hoping is preserved beyond the map just being an injection; the map $\mathbb{Z} \rightarrow \mathbb{Z}$ sending $1 \mapsto 2$ is an embedding as additive groups but is not an embedding as unital rings.

When you want to embed a group into a ring, generally you want to embed a group into the group of units of the ring (in which case the group identity must go to 1), or into the additive group of the ring (in which case the identity must go to 0). The same seems natural for embedding a monoid into a ring.