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Let $R$ be a commutative Noetherian local ring and let $D(R)$ be the derived category of $R$-modules. Recall that a chain complex $C_\bullet$ of modules over $R$ is called perfect if it is isomorphic in $D(A)$ to a bounded complex of free and finitely generated $R$-modules. My question is this: is the category of perfect complexes with finite length homology a thick subcategory of $D(A)$? Actually I only need it to be triangulated. So if the answer to the previous question is negative, then I would like to know if the category of perfect complexes with finite length homology is a triangulated subcategory of $D(A)$? Any references for these will be appreciated.

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    $\begingroup$ Isn't this trivial? If you have a map between two bounded complexes of free and finitely generated R-modules, consider its cone, by definition, it is also a bounded complex of free and finitely generated R-modules because every item of the complex is defined by direct sum. $\endgroup$ – Chen Jiang Oct 4 '14 at 3:35
  • $\begingroup$ I think you are referring to the second question, correct? Because the mapping cone will also have homology of finite length. Is that what you are saying? $\endgroup$ – Mahdi Majidi-Zolbanin Oct 4 '14 at 6:05
  • $\begingroup$ Yes. For the thickness, you may see here: ncatlab.org/nlab/show/perfect+chain+complex. Where says that perfect objects=compact objects. And compactness is a property stable under taking direct summand. $\endgroup$ – Chen Jiang Oct 4 '14 at 8:58
  • $\begingroup$ Modules of finite length are closed under homomorphic images, submodules, extensions, retracts... So yes. $\endgroup$ – Fernando Muro Oct 4 '14 at 16:24
  • $\begingroup$ Ok, I see it. Thanks. I made an error and thought that $R$ (regarded as a complex in $D(R)$) had homology of finite length. $\endgroup$ – Mahdi Majidi-Zolbanin Oct 4 '14 at 19:12

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