The Stacks project

Lemma 15.74.14. Let $R$ be a regular ring. 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. Any perfect $R$-module is finite by definition. Conversely, let $M$ be a finite $R$-module. Choose a resolution

\[ \ldots \to F_2 \xrightarrow {d_2} F_1 \xrightarrow {d_1} F_0 \to M \to 0 \]

with $F_ i$ finite free $R$-modules (Algebra, Lemma 10.71.1). Set $M_ i = \mathop{\mathrm{Ker}}(d_ i)$. Denote $U_ i \subset \mathop{\mathrm{Spec}}(R)$ the set of primes $\mathfrak p$ such that $M_{i, \mathfrak p}$ is free; $U_ i$ is open by Algebra, Lemma 10.79.3. We have a exact sequence $0 \to M_{i + 1} \to F_{i + 1} \to M_ i \to 0$. If $\mathfrak p \in U_ i$, then $0 \to M_{i + 1, \mathfrak p} \to F_{i + 1, \mathfrak p} \to M_{i, \mathfrak p} \to 0$ splits. Thus $M_{i + 1, \mathfrak p}$ is finite projective, hence free (Algebra, Lemma 10.78.2). This shows that $U_ i \subset U_{i + 1}$. We claim that $\mathop{\mathrm{Spec}}(R) = \bigcup U_ i$. Namely, for every prime ideal $\mathfrak p$ the regular local ring $R_\mathfrak p$ has finite global dimension by Algebra, Proposition 10.110.1. It follows that $M_{i, \mathfrak p}$ is finite projective (hence free) for $i \gg 0$ for example by Algebra, Lemma 10.109.3. Since the spectrum of $R$ is Noetherian (Algebra, Lemma 10.31.5) we conclude that $U_ n = \mathop{\mathrm{Spec}}(R)$ for some $n$. Then $M_ n$ is a projective $R$-module by Algebra, Lemma 10.78.2. Thus

\[ 0 \to M_ n \to F_ n \to \ldots \to F_1 \to M \to 0 \]

is a bounded resolution by finite projective modules and hence $M$ is perfect. 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 15.74.7 $\square$

Comments (2)

Comment #7210 by Torsten Wedhorn on

I think, the hypothesis that is of finite dimension is superfluous. Every given complex in is still perfect but the tor-amplitude of might not be bounded dependent only on the cohomological amplitude of , i.e., the such that .

Comment #7211 by on

Thanks Torsten! I've strengthened the result as you suggested. See here.

There are also:

  • 6 comment(s) on Section 15.74: 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'.