The Stacks project

Lemma 30.12.1. Let $X$ be a Noetherian scheme. Let $\mathcal{F}$ be a coherent sheaf on $X$. Suppose that $\text{Supp}(\mathcal{F}) = Z \cup Z'$ with $Z$, $Z'$ closed. Then there exists a short exact sequence of coherent sheaves

\[ 0 \to \mathcal{G}' \to \mathcal{F} \to \mathcal{G} \to 0 \]

with $\text{Supp}(\mathcal{G}') \subset Z'$ and $\text{Supp}(\mathcal{G}) \subset Z$.

Proof. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the sheaf of ideals defining the reduced induced closed subscheme structure on $Z$, see Schemes, Lemma 26.12.4. Consider the subsheaves $\mathcal{G}'_ n = \mathcal{I}^ n\mathcal{F}$ and the quotients $\mathcal{G}_ n = \mathcal{F}/\mathcal{I}^ n\mathcal{F}$. For each $n$ we have a short exact sequence

\[ 0 \to \mathcal{G}'_ n \to \mathcal{F} \to \mathcal{G}_ n \to 0 \]

For every point $x$ of $Z' \setminus Z$ we have $\mathcal{I}_ x = \mathcal{O}_{X, x}$ and hence $\mathcal{G}_{n, x} = 0$. Thus we see that $\text{Supp}(\mathcal{G}_ n) \subset Z$. Note that $X \setminus Z'$ is a Noetherian scheme. Hence by Lemma 30.10.2 there exists an $n$ such that $\mathcal{G}'_ n|_{X \setminus Z'} = \mathcal{I}^ n\mathcal{F}|_{X \setminus Z'} = 0$. For such an $n$ we see that $\text{Supp}(\mathcal{G}'_ n) \subset Z'$. Thus setting $\mathcal{G}' = \mathcal{G}'_ n$ and $\mathcal{G} = \mathcal{G}_ n$ works. $\square$

Comments (1)

Comment #949 by correction_bot on

Is it intentional that the first two lines of the proof are identical to the lines preceding the lemma?

There are also:

  • 5 comment(s) on Section 30.12: Devissage of coherent sheaves

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Your email address will not be published.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 01YD. Beware of the difference between the letter 'O' and the digit '0'.