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 $R$ is of finite dimension is superfluous. Every given complex $K$ in $D^b_{coh}(R)$ is still perfect but the tor-amplitude of $K$ might not be bounded dependent only on the cohomological amplitude of $K$, i.e., the $i$ such that $H^i(K) \ne 0$.

Comment #7211 by on

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

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• 6 comment(s) on Section 15.74: Perfect complexes

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