**Proof.**
By Algebra, Lemma 10.110.8 the assumption on $R$ means that $R$ has finite global dimension. Hence every module has finite tor dimension, see Lemma 15.65.19. On the other hand, as $R$ is Noetherian, a module is pseudo-coherent if and only if it is finite, see Lemma 15.63.17. 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 15.73.7
$\square$

## Comments (0)

There are also: