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I want to understand the accepted answer to this question. The answer is supposed to work for the homotopy category of chain complexes of abelian groups too. (i.e. it shows that that category is not exact).

As mentioned in the comments the argument relies on the fact that the complex $$ \cdots \to 0 \to \mathbb{Z}/p^2\mathbb{Z} \to 0 \to \cdots $$
Is indecomposable in the homotopy category of chain complexes of $Ab$. I was not able to prove this myself.

How can this be shown?

Community
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Ward Beullens
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  • This question was moved from the mathoverflow site because it was considered off topic there. In a comment someone mentioned that any decomposition of this complex has a factor that has trivial homology. – Ward Beullens May 31 '16 at 11:44

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In any additive category, if an object $X$ is decomposable then its endomorphism ring has a non-trivial idempotent (projection onto one of the factors).

In the homotopy category of chain complexes of abelian groups, the complex $$\dots\to0\to\mathbb{Z}/p^2\mathbb{Z}\to0\to\dots$$ has endomorphism ring $\mathbb{Z}/p^2\mathbb{Z}$, which has no non-trivial idempotents.

Jeremy Rickard
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