Lemma 15.69.3. Let $M$ be a module over a ring $R$. The following are equivalent

1. $M$ is a perfect module, and

2. 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 15.69.2 and 15.62.4 we can find resolution $E^\bullet \to M$ with $E^{-i}$ a finite free $R$-module. By Lemma 15.63.2 we see that $F_ d = \mathop{\mathrm{Coker}}(E^{d - 1} \to E^ d)$ is flat for some $d$ sufficiently large. By Algebra, Lemma 10.77.2 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. $\square$

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