Definition 21.47.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{E}^\bullet $ be a complex of $\mathcal{O}$-modules. We say $\mathcal{E}^\bullet $ is perfect if for every object $U$ of $\mathcal{C}$ there exists a covering $\{ U_ i \to U\} $ such that for each $i$ there exists a morphism of complexes $\mathcal{E}_ i^\bullet \to \mathcal{E}^\bullet |_{U_ i}$ which is a quasi-isomorphism with $\mathcal{E}_ i^\bullet $ strictly perfect. An object $E$ of $D(\mathcal{O})$ is perfect if it can be represented by a perfect complex of $\mathcal{O}$-modules.
21.47 Perfect complexes
In this section we discuss properties of perfect complexes on ringed sites.
If $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is quasi-compact (Sites, Section 7.17), then a perfect object of $D(\mathcal{O})$ is in $D^ b(\mathcal{O})$. But this need not be the case otherwise.
Lemma 21.47.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $E$ be an object of $D(\mathcal{O})$.
If $\mathcal{C}$ has a final object $X$ and there exist a covering $\{ U_ i \to X\} $, strictly perfect complexes $\mathcal{E}_ i^\bullet $ of $\mathcal{O}_{U_ i}$-modules, and isomorphisms $\alpha _ i : \mathcal{E}_ i^\bullet \to E|_{U_ i}$ in $D(\mathcal{O}_{U_ i})$, then $E$ is perfect.
If $E$ is perfect, then any complex representing $E$ is perfect.
Proof. Identical to the proof of Lemma 21.45.2. $\square$
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. $\square$
Lemma 21.47.4. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $E$ be an object of $D(\mathcal{O})$. The following are equivalent
$E$ is perfect, and
$E$ is pseudo-coherent and locally has finite tor dimension.
Proof. Assume (1). Let $U$ be an object of $\mathcal{C}$. By definition there exists a covering $\{ U_ i \to U\} $ 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 21.45.2. Moreover, a direct summand of a finite free module is flat, hence $E|_{U_ i}$ has finite Tor dimension by Lemma 21.46.3. Thus (2) holds.
Assume (2). Let $U$ be an object of $\mathcal{C}$. After replacing $U$ by the members of a covering we may assume there exist integers $a \leq b$ such that $E|_ U$ has tor amplitude in $[a, b]$. Since $E|_ U$ is $m$-pseudo-coherent for all $m$ we conclude using Lemma 21.47.3. $\square$
Lemma 21.47.5. Let $(f, f^\sharp ) : (\mathcal{C}, \mathcal{O}_\mathcal {C}) \to (\mathcal{D}, \mathcal{O}_\mathcal {D})$ be a morphism of ringed sites. Let $E$ be an object of $D(\mathcal{O}_\mathcal {D})$. If $E$ is perfect in $D(\mathcal{O}_\mathcal {D})$, then $Lf^*E$ is perfect in $D(\mathcal{O}_\mathcal {C})$.
Proof. This follows from Lemma 21.47.4, 21.46.5, and 21.45.3. $\square$
Lemma 21.47.6. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(K, L, M, f, g, h)$ be a distinguished triangle in $D(\mathcal{O})$. If two out of three of $K, L, M$ are perfect then the third is also perfect.
Proof. First proof: Combine Lemmas 21.47.4, 21.45.4, and 21.46.6. Second proof (sketch): Say $K$ and $L$ are perfect. Let $U$ be an object of $\mathcal{C}$. After replacing $U$ by the members of a covering we may assume that $K|_ U$ and $L|_ U$ are represented by strictly perfect complexes $\mathcal{K}^\bullet $ and $\mathcal{L}^\bullet $. After replacing $U$ by the members of a covering we may assume the map $K|_ U \to L|_ U$ is given by a map of complexes $\alpha : \mathcal{K}^\bullet \to \mathcal{L}^\bullet $, see Lemma 21.44.8. Then $M|_ U$ is isomorphic to the cone of $\alpha $ which is strictly perfect by Lemma 21.44.2. $\square$
Lemma 21.47.7. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. If $K, L$ are perfect objects of $D(\mathcal{O})$, then so is $K \otimes _\mathcal {O}^\mathbf {L} L$.
Proof. Follows from Lemmas 21.47.4, 21.45.5, and 21.46.7. $\square$
Lemma 21.47.8. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. If $K \oplus L$ is a perfect object of $D(\mathcal{O})$, then so are $K$ and $L$.
Proof. Follows from Lemmas 21.47.4, 21.45.6, and 21.46.8. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)