Lemma 31.2.8. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $U \subset X$ is open and $\text{Ass}(\mathcal{F}) \subset U$, 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 $\text{Supp}(\mathcal{F}') \cap U = \emptyset$. On the other hand, $\text{Ass}(\mathcal{F}') \subset \text{Ass}(\mathcal{F})$ by Lemma 31.2.4. Since also $\text{Ass}(\mathcal{F}') \subset \text{Supp}(\mathcal{F}')$ (Lemma 31.2.3) we conclude $\text{Ass}(\mathcal{F}') = \emptyset$. Hence $\mathcal{F}' = 0$ by Lemma 31.2.6. $\square$

## Comments (2)

Comment #3570 by Laurent Moret-Bailly on

Statement of lemma: I suppose you mean "$U\subset X$ is open and $\mathrm{Ass}(\mathcal{F})\subset U$".

There are also:

• 2 comment(s) on Section 31.2: Associated points

## 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Your email address will not be published.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 0B3L. Beware of the difference between the letter 'O' and the digit '0'.