Lemma 15.77.6. Let $A \to B$ be a local ring homomorphism. 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,

the ring $B/\mathfrak m_ AB$ is regular of dimension $d$,

$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, in fact it suffices if $H^ i(K^\bullet \otimes _ A^\mathbf {L} \kappa (\mathfrak m_ A))$ is nonzero only for $i \in [a, b]$.

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

**Proof.**
By (3) we may assume that $K^\bullet $ is a bounded above complex of finite free $B$-modules. We compute

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

The first equality because $K^\bullet $ is a bounded above complex of flat $B$-modules. The second equality follows from basic properties of the tensor product. The third equality holds because $K^\bullet \otimes _ A \kappa (\mathfrak m_ A) = K^\bullet / \mathfrak m_ A K^\bullet $ is a bounded above complex of flat $B/\mathfrak m_ A B$-modules. Since $K^\bullet $ is a bounded above complex of flat $A$-modules by (1), the cohomology modules $H^ i$ of the complex $K^\bullet \otimes _ A \kappa (\mathfrak m_ A)$ are nonzero only for $i \in [a, b]$ by assumption (4). Thus the spectral sequence of Example 15.62.1 and the fact that $B/\mathfrak m_ AB$ has finite global dimension $d$ (by (2) and Algebra, Proposition 10.110.1) shows that $H^ j(K^\bullet \otimes _ B^{\mathbf{L}} \kappa (\mathfrak m_ B))$ is zero for $j \not\in [a - d, b]$. This finishes the proof by Lemma 15.77.2.
$\square$

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