Lemma 15.77.5. Let $A \to B$ be a ring map. Let $a, b \in \mathbf{Z}$. Let $d \geq 0$. Let $K^\bullet $ be a complex of $B$-modules. Assume

the ring map $A \to B$ is flat,

for every prime $\mathfrak p \subset A$ the ring $B \otimes _ A \kappa (\mathfrak p)$ has finite global dimension $\leq d$,

$K^\bullet $ is pseudo-coherent as a complex of $B$-modules, and

$K^\bullet $ has tor amplitude in $[a, b]$ as a complex of $A$-modules.

Then $K^\bullet $ is perfect as a complex of $B$-modules with tor amplitude in $[a - d, b]$.

**Proof.**
We may assume that $K^\bullet $ is a bounded above complex of finite free $B$-modules. In particular, $K^\bullet $ is flat as a complex of $A$-modules and $K^\bullet \otimes _ A M = K^\bullet \otimes _ A^{\mathbf{L}} M$ for any $A$-module $M$. For every prime $\mathfrak p$ of $A$ the complex

\[ K^\bullet \otimes _ A \kappa (\mathfrak p) \]

is a bounded above complex of finite free modules over $B \otimes _ A \kappa (\mathfrak p)$ with vanishing $H^ i$ except for $i \in [a, b]$. As $B \otimes _ A \kappa (\mathfrak p)$ has global dimension $d$ we see from Lemma 15.66.19 that $K^\bullet \otimes _ A \kappa (\mathfrak p)$ has tor amplitude in $[a - d, b]$. Let $\mathfrak q$ be a prime of $B$ lying over $\mathfrak p$. Since $K^\bullet \otimes _ A \kappa (\mathfrak p)$ is a bounded above complex of free $B \otimes _ A \kappa (\mathfrak p)$-modules we see that

\begin{align*} K^\bullet \otimes _ B^{\mathbf{L}} \kappa (\mathfrak q) & = K^\bullet \otimes _ B \kappa (\mathfrak q) \\ & = (K^\bullet \otimes _ A \kappa (\mathfrak p)) \otimes _{B \otimes _ A \kappa (\mathfrak p)} \kappa (\mathfrak q) \\ & = (K^\bullet \otimes _ A \kappa (\mathfrak p)) \otimes ^{\mathbf{L}}_{B \otimes _ A \kappa (\mathfrak p)} \kappa (\mathfrak q) \end{align*}

Hence the arguments above imply that $H^ i(K^\bullet \otimes _ B^{\mathbf{L}} \kappa (\mathfrak q)) = 0$ for $i \not\in [a - d, b]$. We conclude by Lemma 15.77.2.
$\square$

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