MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anybody give me an example of a "naturally-occurring" algebraic category $C$ in which:

  1. $C$ has two non-isomorphic objects $A$ and $B$ which are bi-embeddable via monic maps; but

  2. $C$ does NOT have any infinite collection $A_{1}$, $A_{2}$, ... of objects which are pairwise bi-embeddable (via monic maps) and pairwise nonisomorphic?

Alternatively: can anybody give a reason why, under some reasonable hypothesis about the category $C$, property 1 should imply that 2 fails ("there is an infinite collection of pairwise bi-embeddable, pairwise nonisomorphic objects")?

share|cite|improve this question
up vote 4 down vote accepted

I don't know if this counts as a "naturally-occuring" algebraic category but as I scanned through my internal list of categories of modules over a ring, I spotted that the following has (1) and (2).

Take $G$ to be the "$p$-adic dihedral group" that is a semidirect product of the p-adic integers and a cyclic group of order $2$ where the latter acts by $-1$ on the former. Then form the completed group algebra (Iwasawa algebra) with coefficients in the field with $p$ elements, $R=F_{p}\left[\left[G\right]\right]$.

Now there are precisely two isomorphism classes of indecomposable projective modules [$P_{1}$] and [$P_{2}$] and $P_{1}$ and $P_{2}$ are bi-embeddable via monic maps (see section 9.6 of for the proof of both these claims).

It follows that if we take the category of all finitely generated projective modules over $R$ then it consists of objects that are direct sums of m copies of $P_{1}$ and $n$ copies of $P_{2}$, say. Two such objects will be bi-embeddable if and only if they have the same value of $m+n$ thus whilst there are arbitrarily large finite collections of pairwise bi-embeddable but pairwise nonisomorphic objects in this category there are no infinite ones.

I am sure there will be many more examples along these lines, I just happen to spend a lot of time thinking about Iwasawa algebras.

share|cite|improve this answer
Thanks! This is exactly the kind of example I was looking for. Abstracting a bit, it seems that if you ever have a "reasonable" category of algebraic objects with two "indecomposables" A and B which are bi-embeddable but nonisomorphic, then you ought to be able to take the subcategory of all objects "finitely generated by A and B" and get a similar thing satisfying both 1 and 2. I'm too lazy at the instant to think what's the optimal way to make this precise -- maybe it should be a statement about abelian categories? – John Goodrick Oct 15 '09 at 22:54
If by finitely generated you mean under direct sum in an additive category then I think that will be true. – Simon Wadsley Oct 16 '09 at 7:05

Natural example: the category of function fields of supersingular elliptic curves over a fixed algebraically closed field of characteristic $p$ (you pick $p$, say bigger than $13$ or so to ensure more than one isomorphism class). Note that this is -- up to translations -- the opposite category to the Brandt module category, in which the objects are the ss elliptic curves themselves and the morphisms are isogenies.

As $p$ varies, this gives a family of (essentially) finite categories such that any two objects are mutually embeddable and the number of isomorphism classes of objects tends to infinity with $p$.

If I may be so bold, I spent much of a paper talking about the relation between two function fields that each is embeddable in the other, which I called (borrowing from the theory of elliptic curves) "isogeny".

PLC, On elementary equivalence, isomorphism and isogeny.
J. Théor. Nombres Bordeaux 18 (2006), no. 1, 29--58.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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