Lemma 15.69.13. Let $R$ be a regular ring of finite dimension. Then

1. an $R$-module is perfect if and only if it is a finite $R$-module, and

2. a complex of $R$-modules $K^\bullet$ is perfect if and only if $K^\bullet \in D^ b(R)$ and each $H^ i(K^\bullet )$ is a finite $R$-module.

Proof. By Algebra, Lemma 10.109.8 the assumption on $R$ means that $R$ has finite global dimension. Hence every module has finite tor dimension, see Lemma 15.63.19. On the other hand, as $R$ is Noetherian, a module is pseudo-coherent if and only if it is finite, see Lemma 15.62.18. This proves part (1).

Let $K^\bullet$ be a complex of $R$-modules. If $K^\bullet$ is perfect, then it is in $D^ b(R)$ and it is quasi-isomorphic to a finite complex of finite projective $R$-modules so certainly each $H^ i(K^\bullet )$ is a finite $R$-module (as $R$ is Noetherian). Conversely, suppose that $K^\bullet$ is in $D^ b(R)$ and each $H^ i(K^\bullet )$ is a finite $R$-module. Then by (1) each $H^ i(K^\bullet )$ is a perfect $R$-module, whence $K^\bullet$ is perfect by Lemma 15.69.7 $\square$

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