Lemma 36.13.3. Let $X$ be an affine scheme and let $U \subset X$ be a quasi-compact open subscheme. For any pseudo-coherent object $E$ of $D(\mathcal{O}_ U)$ there exists a bounded above complex of finite free $\mathcal{O}_ X$-modules whose restriction to $U$ is isomorphic to $E$.

**Proof.**
By Lemma 36.10.1 we see that $E$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ U)$. By Lemma 36.13.1 we may assume $E = E'|U$ for some object $E'$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$. Write $X = \mathop{\mathrm{Spec}}(A)$. By Lemma 36.3.5 we can find a complex $M^\bullet $ of $A$-modules whose associated complex of $\mathcal{O}_ X$-modules is a representative of $E'$.

Choose $f_1, \ldots , f_ r \in A$ such that $U = D(f_1) \cup \ldots \cup D(f_ r)$. By Lemma 36.10.2 the complexes $M^\bullet _{f_ j}$ are pseudo-coherent complexes of $A_{f_ j}$-modules. Let $n$ be an integer. Assume we have a map of complexes $\alpha : F^\bullet \to M^\bullet $ where $F^\bullet $ is bounded above, $F^ i = 0$ for $i < n$, each $F^ i$ is a finite free $R$-module, such that

is an isomorphism for $i > n$ and surjective for $i = n$. Picture

Since each $M^\bullet _{f_ j}$ has vanishing cohomology in large degrees we can find such a map for $n \gg 0$. By induction on $n$ we are going to extend this to a map of complexes $F^\bullet \to M^\bullet $ such that $H^ i(\alpha _{f_ j})$ is an isomorphism for all $i$. The lemma will follow by taking $\widetilde{F^\bullet }$.

The induction step will be to extend the diagram above by adding $F^{n - 1}$. Let $C^\bullet $ be the cone on $\alpha $ (Derived Categories, Definition 13.9.1). The long exact sequence of cohomology shows that $H^ i(C^\bullet _{f_ j}) = 0$ for $i \geq n$. By More on Algebra, Lemma 15.63.2 we see that $C^\bullet _{f_ j}$ is $(n - 1)$-pseudo-coherent. By More on Algebra, Lemma 15.63.3 we see that $H^{-1}(C^\bullet _{f_ j})$ is a finite $A_{f_ j}$-module. Choose a finite free $A$-module $F^{n - 1}$ and an $A$-module $\beta : F^{n - 1} \to C^{-1}$ such that the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ is zero and such that $F^{n - 1}_{f_ j}$ surjects onto $H^{n - 1}(C^\bullet _{f_ j})$. (Some details omitted; hint: clear denominators.) Since $C^{n - 1} = M^{n - 1} \oplus F^ n$ we can write $\beta = (\alpha ^{n - 1}, -d^{n - 1})$. The vanishing of the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ implies these maps fit into a morphism of complexes

Moreover, these maps define a morphism of distinguished triangles

Hence our choice of $\beta $ implies that the map of complexes $(F^{-1} \to \ldots ) \to M^\bullet $ induces an isomorphism on cohomology localized at $f_ j$ in degrees $\geq n$ and a surjection in degree $-1$. This finishes the proof of the lemma. $\square$

## 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 (0)