The Stacks project

Proof. Identical to the proof of Lemma 20.47.2. $\square$

Proof. This follows from Lemma 20.49.2 and the fact that on $X$ every direct summand of a finite free module is finite locally free. See Modules, Lemma 17.14.6. $\square$

Proof. After replacing $X$ by the members of an open covering 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 $\mathcal{O}_ X$-module. Applying $- \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}$ the assumption that $E$ has tor amplitude in $[a, b]$ implies $\mathcal{K} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{E}^{a - 1} \otimes _{\mathcal{O}_ X} \mathcal{F}$ has image $\mathop{\mathrm{Ker}}(\mathcal{E}^{a - 1} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{E}^ a \otimes _{\mathcal{O}_ X} \mathcal{F})$. It follows that $\text{Tor}_1^{\mathcal{O}_ X}(\mathcal{E}', \mathcal{F}) = 0$ where $\mathcal{E}' = \mathop{\mathrm{Coker}}(\mathcal{E}^{a - 1} \to \mathcal{E}^ a)$. Hence $\mathcal{E}'$ is flat (Lemma 20.26.16). Thus $\mathcal{E}'$ is locally a direct summand of a finite free module by Modules, Lemma 17.18.3. Thus locally the complex

is quasi-isomorphic to $E$ and $E$ is perfect. $\square$

Proof. Assume (1). By definition this means there exists an open covering $X = \bigcup U_ i$ such that $E|_{U_ i}$ is represented by a strictly perfect complex. Thus $E$ is pseudo-coherent (i.e., $m$-pseudo-coherent for all $m$) by Lemma 20.47.2. Moreover, a direct summand of a finite free module is flat, hence $E|_{U_ i}$ has finite Tor dimension by Lemma 20.48.3. Thus (2) holds.

Assume (2). After replacing $X$ by the members of an open covering we may assume there exist integers $a \leq b$ such that $E$ has tor amplitude in $[a, b]$. Since $E$ is $m$-pseudo-coherent for all $m$ we conclude using Lemma 20.49.4. $\square$

Proof. This follows from Lemma 20.49.5, 20.48.4, and 20.47.3. (An alternative proof is to copy the proof of Lemma 20.47.3.) $\square$

Proof. First proof: Combine Lemmas 20.49.5, 20.47.4, and 20.48.6. Second proof (sketch): Say $K$ and $L$ are perfect. After replacing $X$ by the members of an open covering we may assume that $K$ and $L$ are represented by strictly perfect complexes $\mathcal{K}^\bullet$ and $\mathcal{L}^\bullet$. After replacing $X$ by the members of an open covering we may assume the map $K \to L$ is given by a map of complexes $\alpha : \mathcal{K}^\bullet \to \mathcal{L}^\bullet$, see Lemma 20.46.8. Then $M$ is isomorphic to the cone of $\alpha$ which is strictly perfect by Lemma 20.46.2. $\square$

Proof. Follows from Lemmas 20.49.5, 20.47.5, and 20.48.7. $\square$

Proof. Follows from Lemmas 20.49.5, 20.47.6, and 20.48.8. $\square$

Proof. Being a perfect complex is local on $X$. Thus it suffices to check that $Rj_*E$ is perfect when restricted to $U$ and $V = X \setminus j(T)$. We have $Rj_*E|_ U = E$ which is perfect. We have $Rj_*E|_ V = 0$ because $E|_{U \setminus T} = 0$. $\square$

Proof. Note that if $V \subset X$ is some open such that $H^ i(E)|_ V$ is finite locally free for all $i \in \mathbf{Z}$ then $V \subset U$. Let $x \in U$. We will show that an open neighbourhood of $x$ is contained in $U$ and that $H^ i(E)$ is finite locally free on this neighbourhood for all $i$. This will finish the proof. During the proof we may (finitely many times) replace $X$ by an open neighbourhood of $x$. Hence we may assume $E$ is represented by a strictly perfect complex $\mathcal{E}^\bullet$. Say $\mathcal{E}^ i = 0$ for $i \not\in [a, b]$. We will prove the result by induction on $b - a$. The module $H^ b(E) = \mathop{\mathrm{Coker}}(d^{b - 1} : \mathcal{E}^{b - 1} \to \mathcal{E}^ b)$ is of finite presentation. Since $H^ b(E)_ x$ is finite free, we conclude $H^ b(E)$ is finite free in an open neighbourhood of $x$ by Modules, Lemma 17.11.6. Thus after replacing $X$ by a (possibly smaller) open neighbourhood we may assume we have a direct sum decomposition $\mathcal{E}^ b = \mathop{\mathrm{Im}}(d^{b - 1}) \oplus H^ b(E)$ and $H^ b(E)$ is finite free, see Lemma 20.46.5. Doing the same argument again, we see that we may assume $\mathcal{E}^{b - 1} = \mathop{\mathrm{Ker}}(d^{b - 1}) \oplus \mathop{\mathrm{Im}}(d^{b - 1})$. The complex $\mathcal{E}^ a \to \ldots \to \mathcal{E}^{b - 2} \to \mathop{\mathrm{Ker}}(d^{b - 1})$ is a strictly perfect complex representing a perfect object $E'$ with $H^ i(E) = H^ i(E')$ for $i \not= b$. Hence we conclude by our induction hypothesis. $\square$