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We know for any principal ideal domain, objects in the bounded derived category are all formal hence we can classify those objects with finitely generated cohomology using structure theorem for finitely generated modules.

Now consider the bounded derived category of $\mathbb C[x]/x^2$-modules, how to classify indecomposable objects with finitely generated cohomology in this category? Examples include $\mathbb C[x]/x^2 \overset{x}{\rightarrow} \mathbb C[x]/x^2 \overset{x}{\rightarrow}... \overset{x}{\rightarrow} \mathbb C[x]/x^2 $ and $\mathbb C$.

Note there exists non-formal object.

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Up to shifts, every indecomposable object is of one of the forms described in the question.

I don't know an explicit reference, but here's a sketch of a proof.

By induction on the length, it's not hard to prove that every bounded complex of finite rank free modules, such that the image of each differential is contained in the radical of its codomain, is a direct sum of complexes of the form $\mathbb C[x]/x^2 \overset{x}{\rightarrow} \mathbb C[x]/x^2 \overset{x}{\rightarrow}... \overset{x}{\rightarrow} \mathbb C[x]/x^2 $.

Now consider a minimal projective resolution of an object of the bounded derived category, truncated to the left of its homology.

By the way, this is an example of a "derived discrete" algebra, and there's a fair amount of literature about these.

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  • $\begingroup$ Thank you, I have a question: $\mathbb C$ is indecomposable but not perfect in the derived category, so we can't find a projective resolution in the bounded derived category. How to rule out such possibility? $\endgroup$
    – Zhiyu
    Commented Dec 8, 2019 at 18:31
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    $\begingroup$ @sawdada That's why I truncated the projective resolution to the left of its homology. If, in degrees $n$ and below, a projective resolution of an object looks like $P_n\stackrel{d_n}{\to} P_{n-1}\to P_{n-2}\to\dots$, and if the object has zero homology in degrees $n$ and above, then the object is quasiisomorphic to $\dots\to0\to\ker{d_n}\to P_n\to P_{n-1}\to P_{n-2}\to\dots$. $\endgroup$
    – Jeremy Rickard
    Commented Dec 8, 2019 at 21:23
  • $\begingroup$ OK, I see. Can we classify things over $\mathbb C[x]/x^n$ using your method? $\endgroup$
    – Zhiyu
    Commented Dec 8, 2019 at 23:52
  • $\begingroup$ @sawdada I don't think so. I'm not sure, but I suspect that for most $n$ this is a wild classification problem. $\endgroup$
    – Jeremy Rickard
    Commented Dec 9, 2019 at 11:02
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    $\begingroup$ By arxiv.org/pdf/1910.01494.pdf , it seems that K[x]/(x^n) is not derived tame for n>=3 (but it is for n=2). $\endgroup$
    – Mare
    Commented Dec 10, 2019 at 0:13

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