The Stacks project

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 $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.69: Perfect complexes

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 066Z. Beware of the difference between the letter 'O' and the digit '0'.