The Stacks project

Lemma 15.77.3. Let $R$ be a ring. Let $K^\bullet $ be a pseudo-coherent complex of $R$-modules. Consider the following conditions

  1. $K^\bullet $ is perfect,

  2. for every prime ideal $\mathfrak p$ the complex $K^\bullet \otimes _ R R_{\mathfrak p}$ is perfect,

  3. for every maximal ideal $\mathfrak m$ the complex $K^\bullet \otimes _ R R_{\mathfrak m}$ is perfect,

  4. for every prime $\mathfrak p$ we have $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p)) = 0$ for all $i \ll 0$,

  5. for every maximal ideal $\mathfrak m$ we have $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak m)) = 0$ for all $i \ll 0$.

We always have the implications

\[ (1) \Rightarrow (2) \Leftrightarrow (3) \Leftrightarrow (3) \Leftrightarrow (4) \Leftrightarrow (5) \]

If $K^\bullet $ is bounded below, then all conditions are equivalent.

Proof. By Lemma 15.74.9 we see that (1) implies (2). It is immediate that (2) $\Rightarrow $ (3). Since every prime $\mathfrak p$ is contained in a maximal ideal $\mathfrak m$, we can apply Lemma 15.74.9 to the map $R_\mathfrak m \to R_\mathfrak p$ to see that (3) implies (2). Applying Lemma 15.74.9 to the residue maps $R_\mathfrak p \to \kappa (\mathfrak p)$ and $R_\mathfrak m \to \kappa (\mathfrak m)$ we see that (2) implies (4) and (3) implies (5).

Assume $R$ is local with maximal ideal $\mathfrak m$ and residue field $\kappa $. We will show that if $H^ i(K^\bullet \otimes ^\mathbf {L} \kappa ) = 0$ for $i < a$ for some $a$, then $K$ is perfect. This will show that (4) implies (2) and (5) implies (3) whence the first part of the lemma. First we apply Lemma 15.76.4 with $i = a - 1$ to see that $K^\bullet = \tau _{\leq a - 1}K^\bullet \oplus \tau _{\geq a}K^\bullet $ in $D(R)$ with $\tau _{\geq a}K^\bullet $ perfect of tor-amplitude contained in $[a, \infty ]$. To finish we need to show that $\tau _{\leq a - 1}K$ is zero, i.e., that its cohomology groups are zero. If not let $i$ be the largest index such that $M = H^ i(\tau _{\leq a - 1}K)$ is not zero. Then $M$ is a finite $R$-module because $\tau _{\leq a - 1}K^\bullet $ is pseudo-coherent (Lemmas 15.64.3 and 15.64.8). Thus by Nakayama's lemma (Algebra, Lemma 10.20.1) we find that $M \otimes _ R \kappa $ is nonzero. This implies that

\[ H^ i((\tau _{\leq a - 1}K^\bullet ) \otimes _ R^\mathbf {L} \kappa ) = H^ i(K^\bullet \otimes _ R^\mathbf {L} \kappa ) \]

is nonzero which is a contradiction.

Assume the equivalent conditions (2) – (5) hold and that $K^\bullet $ is bounded below. Say $H^ i(K^\bullet ) = 0$ for $i < a$. Pick a maximal ideal $\mathfrak m$ of $R$. It suffices to show there exists an $f \in R$, $f \not\in \mathfrak m$ such that $K^\bullet \otimes _ R^\mathbf {L} R_ f$ is perfect (Lemma 15.74.12 and Algebra, Lemma 10.17.10). This follows from Lemma 15.77.1. $\square$

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