The Stacks project

Lemma 71.2.9. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $U \to X$ is an ├ętale morphism such that $\text{WeakAss}(\mathcal{F}) \subset \mathop{\mathrm{Im}}(|U| \to |X|)$, then $\Gamma (X, \mathcal{F}) \to \Gamma (U, \mathcal{F})$ is injective.

Proof. Let $s \in \Gamma (X, \mathcal{F})$ be a section which restricts to zero on $U$. Let $\mathcal{F}' \subset \mathcal{F}$ be the image of the map $\mathcal{O}_ X \to \mathcal{F}$ defined by $s$. Then $\mathcal{F}'|_ U = 0$. This implies that $\text{WeakAss}(\mathcal{F}') \cap \mathop{\mathrm{Im}}(|U| \to |X|) = \emptyset $ (by the definition of weakly associated points). On the other hand, $\text{WeakAss}(\mathcal{F}') \subset \text{WeakAss}(\mathcal{F})$ by Lemma 71.2.4. We conclude $\text{WeakAss}(\mathcal{F}') = \emptyset $. Hence $\mathcal{F}' = 0$ by Lemma 71.2.5. $\square$

Comments (2)

Comment #4071 by Matthieu Romagny on

Typo: replace Ass by WeakAss in the end of the proof.

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CUP. Beware of the difference between the letter 'O' and the digit '0'.