The Stacks project

Lemma 68.14.3. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{F}$ be a coherent sheaf on $X$. There exists a filtration

\[ 0 = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \ldots \subset \mathcal{F}_ m = \mathcal{F} \]

by coherent subsheaves such that for each $j = 1, \ldots , m$ there exists a reduced closed subspace $Z_ j \subset X$ with $|Z_ j|$ irreducible and a sheaf of ideals $\mathcal{I}_ j \subset \mathcal{O}_{Z_ j}$ such that

\[ \mathcal{F}_ j/\mathcal{F}_{j - 1} \cong (Z_ j \to X)_* \mathcal{I}_ j \]

Proof. Consider the collection

\[ \mathcal{T} = \left\{ \begin{matrix} T \subset |X| \text{ closed such that there exists a coherent sheaf } \mathcal{F} \\ \text{ with } \text{Supp}(\mathcal{F}) = T \text{ for which the lemma is wrong} \end{matrix} \right\} \]

We are trying to show that $\mathcal{T}$ is empty. If not, then because $|X|$ is Noetherian (Properties of Spaces, Lemma 65.24.2) we can choose a minimal element $T \in \mathcal{T}$. This means that there exists a coherent sheaf $\mathcal{F}$ on $X$ whose support is $T$ and for which the lemma does not hold. Clearly $T \not= \emptyset $ since the only sheaf whose support is empty is the zero sheaf for which the lemma does hold (with $m = 0$).

If $T$ is not irreducible, then we can write $T = Z_1 \cup Z_2$ with $Z_1, Z_2$ closed and strictly smaller than $T$. Then we can apply Lemma 68.14.1 to get a short exact sequence of coherent sheaves

\[ 0 \to \mathcal{G}_1 \to \mathcal{F} \to \mathcal{G}_2 \to 0 \]

with $\text{Supp}(\mathcal{G}_ i) \subset Z_ i$. By minimality of $T$ each of $\mathcal{G}_ i$ has a filtration as in the statement of the lemma. By considering the induced filtration on $\mathcal{F}$ we arrive at a contradiction. Hence we conclude that $T$ is irreducible.

Suppose $T$ is irreducible. Let $\mathcal{J}$ be the sheaf of ideals defining the reduced induced closed subspace structure on $T$, see Properties of Spaces, Lemma 65.12.3. By Lemma 68.13.2 we see there exists an $n \geq 0$ such that $\mathcal{J}^ n\mathcal{F} = 0$. Hence we obtain a filtration

\[ 0 = \mathcal{I}^ n\mathcal{F} \subset \mathcal{I}^{n - 1}\mathcal{F} \subset \ldots \subset \mathcal{I}\mathcal{F} \subset \mathcal{F} \]

each of whose successive subquotients is annihilated by $\mathcal{J}$. Hence if each of these subquotients has a filtration as in the statement of the lemma then also $\mathcal{F}$ does. In other words we may assume that $\mathcal{J}$ does annihilate $\mathcal{F}$.

Assume $T$ is irreducible and $\mathcal{J}\mathcal{F} = 0$ where $\mathcal{J}$ is as above. Then the scheme theoretic support of $\mathcal{F}$ is $T$, see Morphisms of Spaces, Lemma 66.14.1. Hence we can apply Lemma 68.14.2. This gives a short exact sequence

\[ 0 \to i_*(\mathcal{I}^{\oplus r}) \to \mathcal{F} \to \mathcal{Q} \to 0 \]

where the support of $\mathcal{Q}$ is a proper closed subset of $T$. Hence we see that $\mathcal{Q}$ has a filtration of the desired type by minimality of $T$. But then clearly $\mathcal{F}$ does too, which is our final contradiction. $\square$


Comments (0)


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 07UR. Beware of the difference between the letter 'O' and the digit '0'.