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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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  • $\begingroup$ 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 $\mathbb Z_m$. $\endgroup$
    – Dustin Cartwright
    Commented Aug 16, 2012 at 15:59
  • $\begingroup$ @Dustin Cartwright: You are right. Thank you for catching this. I will correct the mistake. $\endgroup$
    – Rasmus
    Commented Aug 16, 2012 at 16:40
  • $\begingroup$ Rasmus -- ?$(m.n)=n/(m,n)$ $\endgroup$
    – algori
    Commented Aug 16, 2012 at 17:22
  • $\begingroup$ @algori: But for $m=1$ this would give me $n$, though it should give $1$. $\endgroup$
    – Rasmus
    Commented Aug 16, 2012 at 17:55
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    $\begingroup$ 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 $n/(n,m)$. $\endgroup$
    – Florian Eisele
    Commented Aug 16, 2012 at 18:09

1 Answer 1

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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.

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  • $\begingroup$ What is the definition of "wild" in this context? Thanks in advance for your help in understanding the answer! $\endgroup$
    – tweetie-bird
    Commented Aug 16, 2012 at 19:26
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    $\begingroup$ 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? $\endgroup$
    – Will Sawin
    Commented Aug 16, 2012 at 19:30
  • $\begingroup$ @tweetie-bird, wildness is a precise measure of intractability, take an introductory book on representation theory of algebras and you'll find explicit definitions and illustrative examples. $\endgroup$
    – Fernando Muro
    Commented Aug 16, 2012 at 19:58
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    $\begingroup$ @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. $\endgroup$
    – Rasmus
    Commented Aug 16, 2012 at 20:04
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    $\begingroup$ @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. $\endgroup$
    – Fernando Muro
    Commented Aug 17, 2012 at 7:44

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