0

An $R$-module $M$ is Hopfian (resp. co-Hopfian) if every surjective (resp. injective) $f:M \rightarrow M$ is an isomorphism. Do we have a name for a module $M$ that has both properties? Such a module would be a direct analogue of a finite dimensional vector space. If no name exists yet, might it be appropriate to refer to such an $M$ as a finite object in the category of $R$-modules?

Chris Leary
  • 2,993
  • Do you have an example which is not a finite dimensional vector space? – Surb Oct 30 '18 at 20:12
  • If $R$ is artinian, it should also be hopfian and cohopfian: any module map $R \rightarrow R$ is either multiplication by a unit or a zerodivisor, hence surjective implies injective and vice versa. – red_trumpet Oct 30 '18 at 20:49
  • For Surb's question, see my comments below. For red_trumpet's, I am well-aware of these facts. I'm floating a trial balloon here. If this proposed name is not appropriate, I would welcome suggestions, or the opinion that it is better to stick with the rather cumbersome both Hopfian and co-Hopfian. I am in no way trying to be didactic and say they should be called "finite objects." I don't have that level of standing in the field. – Chris Leary Oct 31 '18 at 23:06

1 Answers1

0

I don't know a name for such modules (as far as I can tell from a little literature search they are just called "Hopfian and co-Hopfian"), but I think "finite object" is a profoundly inappropriate name and thinking of such modules as a generalization of finite-dimensional vector spaces is unhelpful. In particular, a Hopfian and co-Hopfian module need not be finitely generated; a simple example is that $\mathbb{Q}$ is Hopfian and co-Hopfian as a $\mathbb{Z}$-module. In fact, there even exist Hopfian and co-Hopfian $\mathbb{Z}$-modules that are uncountable.

Basically, being Hopfian and co-Hopfian doesn't have very much to do with finiteness at all. It has more to do with being "rigid" so that it's impossible to construct nontrivial injections or surjections from the module to itself. The case of vector spaces is very special since all vector spaces are quite "flexible" (they have bases which makes it very easy to define maps out of them), and so the only way to have such rigidity is to be finite-dimensional.

If you want a brief and snappy term to mean "Hopfian and co-Hopfian", I would suggest "bi-Hopfian" (though as far as I know this term has never been used before). The prefix bi- is sometimes used in categorical contexts to say that both a condition and its dual condition holds (e.g., biproducts which are both products and coproducts).

(Being Hopfian and co-Hopfian follows from certain natural finiteness properties. Specifically, any Noetherian module is Hopfian and any Artian module is co-Hopfian, so a module which is both Noetherian and Artinian (also known as a finite length module) is both Hopfian and co-Hopfian. But the converses to these statements are not true in general, because typically most submodules of a module cannot be obtained as images or kernels of an endomorphism of the module.)

Eric Wofsey
  • 330,363
  • As always, I appreciate your thoughtful input. Although the example of $\mathbb{Q}$ is not finitely generated, it is not that hard to show that a torsion-free abelian group is co-Hopfian if and only if it is a finite dimensional $\mathbb{Q}$-vector space, and, as such, is Hopfian as well. So, there is a measure of finiteness lurking in the background. Similarly, the co-Hopfian divisible groups are characterized by the fact that the number of summands $\mathbb{Q}$ and $\mathbb{Z}(p^{\infty})$ must each be finite. – Chris Leary Oct 30 '18 at 20:57
  • (continuing) The one uncountable example I am familiar with, $\prod \mathbb{Z}(p),$ the product over the positive primes $p$ and $\mathbb{Z}(p)$ being the cyclic group of order $p,$ nonetheless has a co-Hopfian basic subgroup $\bigoplus \mathbb{Z}(p)$ which has each $p$-component of finite $p$-rank. Again, a notion of finiteness lurking. I am not wedded to the idea of finite object. I'm just looking for a concise way to say a module is both Hopfian and co-Hopfian. – Chris Leary Oct 30 '18 at 21:04
  • Sure, there are connections with finiteness properties. But "finite object"?? That sounds like a much more robust finiteness property, like being finitely generated or finitely presented or finite length. Similarly, thinking of a Hopfian and co-Hopfian module as "a direct analogue of a finite dimensional vector space" sounds totally ridiculous to me. Finite dimensional vector spaces are great and well-behaved and very nice and finite in pretty much every way you can imagine. Hopfian and co-Hopfian modules in general have pretty much none of that niceness. – Eric Wofsey Oct 30 '18 at 21:12
  • Thanks. I guess I'll just carry on with "both H and co-H" and try to note any "finite" features that I happen to dig up. – Chris Leary Oct 30 '18 at 22:10
  • I was looking through my old questions for a totally unrelated reason, and noticed that I never accepted your answer to this one. After having some time to think about it, I think your suggestion of "bi-Hopfian" is worth trying out. In hindsight, "a direct analogue of a finite dimensional vector space" is a pretty ridiculous statement, since sharing some properties does not mean the sharing of others. – Chris Leary Jan 12 '21 at 20:38