The Stacks project

Proof. Any perfect $R$-module is finite by definition. Conversely, let $M$ be a finite $R$-module. Choose a resolution

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

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 $\square$