Lemma 21.47.3. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $E$ be an object of $D(\mathcal{O})$. Let $a \leq b$ be integers. If $E$ has tor amplitude in $[a, b]$ and is $(a - 1)$-pseudo-coherent, then $E$ is perfect.

**Proof.**
Let $U$ be an object of $\mathcal{C}$. After replacing $U$ by the members of a covering and $\mathcal{C}$ by the localization $\mathcal{C}/U$ we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet $ and a map $\alpha : \mathcal{E}^\bullet \to E$ such that $H^ i(\alpha )$ is an isomorphism for $i \geq a$. We may and do replace $\mathcal{E}^\bullet $ by $\sigma _{\geq a - 1}\mathcal{E}^\bullet $. Choose a distinguished triangle

From the vanishing of cohomology sheaves of $E$ and $\mathcal{E}^\bullet $ and the assumption on $\alpha $ we obtain $C \cong \mathcal{K}[2 - a]$ with $\mathcal{K} = \mathop{\mathrm{Ker}}(\mathcal{E}^{a - 1} \to \mathcal{E}^ a)$. Let $\mathcal{F}$ be an $\mathcal{O}$-module. Applying $- \otimes _\mathcal {O}^\mathbf {L} \mathcal{F}$ the assumption that $E$ has tor amplitude in $[a, b]$ implies $\mathcal{K} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{E}^{a - 1} \otimes _\mathcal {O} \mathcal{F}$ has image $\mathop{\mathrm{Ker}}(\mathcal{E}^{a - 1} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{E}^ a \otimes _\mathcal {O} \mathcal{F})$. It follows that $\text{Tor}_1^\mathcal {O}(\mathcal{E}', \mathcal{F}) = 0$ where $\mathcal{E}' = \mathop{\mathrm{Coker}}(\mathcal{E}^{a - 1} \to \mathcal{E}^ a)$. Hence $\mathcal{E}'$ is flat (Lemma 21.17.15). Thus there exists a covering $\{ U_ i \to U\} $ such that $\mathcal{E}'|_{U_ i}$ is a direct summand of a finite free module by Modules on Sites, Lemma 18.29.3. Thus the complex

is quasi-isomorphic to $E|_{U_ i}$ and $E$ is perfect. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #8698 by Shizhang on

Comment #9374 by Stacks project on