Lemma 70.9.8. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $U \subset X$ be a quasi-compact open. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{G} \subset \mathcal{F}|_ U$ be a quasi-coherent $\mathcal{O}_ U$-submodule which is of finite type. Then there exists a quasi-coherent submodule $\mathcal{G}' \subset \mathcal{F}$ which is of finite type such that $\mathcal{G}'|_ U = \mathcal{G}$.

**Proof.**
Denote $j : U \to X$ the inclusion morphism. As $X$ is quasi-separated and $U$ quasi-compact, the morphism $j$ is quasi-compact. Hence $j_*\mathcal{G} \subset j_*\mathcal{F}|_ U$ are quasi-coherent modules on $X$ (Morphisms of Spaces, Lemma 67.11.2). Let $\mathcal{H} = \mathop{\mathrm{Ker}}(j_*\mathcal{G} \oplus \mathcal{F} \to j_*\mathcal{F}|_ U)$. Then $\mathcal{H}|_ U = \mathcal{G}$. By Lemma 70.9.2 we can find a finite type quasi-coherent submodule $\mathcal{H}' \subset \mathcal{H}$ such that $\mathcal{H}'|_ U = \mathcal{H}|_ U = \mathcal{G}$. Set $\mathcal{G}' = \mathop{\mathrm{Im}}(\mathcal{H}' \to \mathcal{F})$ to conclude.
$\square$

## 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Your email address will not be published.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)