# Classification of long exact sequences

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.)

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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
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

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.
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
@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
@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