Lemma 15.74.3 (066Q)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.74: Perfect complexes
Lemma 15.74.3 (cite)

numberstags

Lemma 15.74.3. Let $M$ be a module over a ring $R$ . The following are equivalent

$M$ is a perfect module, and
there exists a resolution

$0 \to F_ d \to \ldots \to F_1 \to F_0 \to M \to 0$

with each $F_ i$ a finite projective $R$ -module.

Proof. Assume (2). Then the complex $E^\bullet$ with $E^{-i} = F_ i$ is quasi-isomorphic to $M[0]$ . Hence $M$ is perfect. Conversely, assume (1). By Lemmas 15.74.2 and 15.64.4 we can find resolution $E^\bullet \to M$ with $E^{-i}$ a finite free $R$ -module. By Lemma 15.66.2 we see that $F_ d = \mathop{\mathrm{Coker}}(E^{d - 1} \to E^ d)$ is flat for some $d$ sufficiently large. By Algebra, Lemma 10.78.2 we see that $F_ d$ is finite projective. Hence

$0 \to F_ d \to E^{-d+1} \to \ldots \to E^0 \to M \to 0$

is the desired resolution. $\square$

Comments (0)

There are also:

7 comment(s) on Section 15.74: Perfect complexes

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).

Comments (0)

Post a comment