Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathcal C$ be the category of long exact sequences of finitely generated abelian groups almost all of whose entries vanish.

The category $\mathcal C$ is naturally additive as a subcategory of complexes of abelian groups.

Question: Can we write down a complete list of isomorphism classes (up to translation) of indecomposable objects of $\mathcal C$?

It is easy to see that the number of such isomorphism classes is countably infinite.

Here are some indecomposable objects: $$ \cdots\to 0\to\mathbb Z\overset{m}{\to}\mathbb Z\to\mathbb Z_m\to 0\to\cdots, $$

$$ \cdots\to 0\to\mathbb Z_{(m,n)}\to\mathbb Z_n\overset{m}{\to}\mathbb Z_n\to\mathbb Z _{(m,n)}\to 0\to\cdots, $$ where $m$ is a natural number, $n$ is a prime power and $(m,n)$ denotes the greatest common divisor.

But there is more; for instance, if $p$ is prime then the indecomposable object $$ \cdots\to0\to\mathbb Z_p\to\mathbb Z_{p^2}\overset{p}{\to}\mathbb Z_{p^2}\overset{p}{\to}\cdots\overset{p}{\to}\mathbb Z_{p^2}\overset{p}{\to}\mathbb Z_{p^2}\to\mathbb Z_p\to 0\to\cdots $$ can have any finite "length."

If this classification problem has been solved, a reference would be great. Otherwise I would very much appreciate any idea/hint towards a general solution.

(I've added the noncommutative-algebra tag because chain complexes can be considered as modules over a certain non-commutative ring. The question I am asking is a sub-problem of classifying all finitely generated indecomposables for this ring.)

share|improve this question
    
There's a mistake in your second exact sequence. I think the last group should be $\mathbb Z_{(m,n)}$. If $m$ and $n$ are relatively prime, then multiplication by $m$ is surjective, so the last group should be $0$, not $\mathbb Z_m$. –  Dustin Cartwright Aug 16 '12 at 15:59
    
@Dustin Cartwright: You are right. Thank you for catching this. I will correct the mistake. –  Rasmus Bentmann Aug 16 '12 at 16:40
    
Rasmus -- ?$(m.n)=n/(m,n)$ –  algori Aug 16 '12 at 17:22
    
@algori: But for $m=1$ this would give me $n$, though it should give $1$. –  Rasmus Bentmann Aug 16 '12 at 17:55
1  
Regarding the function $?(m,n)$: Take the alternating sum over the images of the terms in the Grothendieck group of the category of torsion $\mathbb Z$-modules (= the free abelian group generated by all primes). If a complex is exact this sum has to be zero. This implies that $?(m,n)$ must be equal to $(m,n)$. The embedding will be given by $n/(n,m)$. –  Florian Eisele Aug 16 '12 at 18:09
show 2 more comments

1 Answer

up vote 10 down vote accepted

In a series of recent papers, Schmidmeier and Ringel show than the classification of monomorphisms in the category of finitely generated $\mathbb Z/p^n$-modules is a wild problem of representation theory for $n>6$. Hence, your problem is also wild. There's no hope to get what you want.

share|improve this answer
    
What is the definition of "wild" in this context? Thanks in advance for your help in understanding the answer! –  tweetie-bird Aug 16 '12 at 19:26
1  
The classification for the case when each group is cyclic is very simple. Just take a sequence of numbers $(a_1,...,a_n)$ and consider the exact sequence: $0 \to \mathbb Z_{a_1} \to \mathbb Z_{a_1a_2} \to \mathbb Z_{a_2a_3} \to \dots \to \mathbb Z_{a_{n-1}a_n} \to \mathbb Z_{a_n} \to 0$ or alternately you can replace $0 \to \mathbb Z_{a_1} \to$ with $0 \to \mathbb Z \to^{a_1} \mathbb Z \to $. This answer would then imply that there must be non-cyclic indecomposables. Is it possible to give a reasonably easy description of some non-cyclic indecomposable exact sequence? –  Will Sawin Aug 16 '12 at 19:30
1  
@tweetie-bird: In arxiv.org/abs/math/0409417 Schmidmeier and Ringel show that the category of finitely generated $\mathbb Z/p^n$-submodule inclusions is controlled $\mathbb Z/p$-wild. This seems to mean roughly that the classification of its objects is at least as complication as the classification of finitely generated modules over the free $\mathbb Z/p$-algebra on two generators. –  Rasmus Bentmann Aug 16 '12 at 20:04
1  
Thank you. That is very helpful. –  tweetie-bird Aug 16 '12 at 23:12
1  
@Will Sawin: looking at Ringel-Schmidmeier's paper one should be able to work out an example of a non-cyclic two-step indecomposable complex of $\mathbb{Z}/128$-module given by an injection. Perhapes even using $\mathbb{Z}/64$-modules, but probably not $\mathbb{Z}/32$-modules. This makes me guess that it would take hours, maybe days for me to come up with such example. –  Fernando Muro Aug 17 '12 at 7:44
show 2 more comments

Your Answer

 
discard

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.