Lemma 68.14.1. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. 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 subspace structure on $Z$, see Properties of Spaces, Lemma 65.12.3. 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 geometric point $\overline{x}$ of $Z' \setminus Z$ we have $\mathcal{I}_{\overline{x}} = \mathcal{O}_{X, \overline{x}}$ and hence $\mathcal{G}_{n, \overline{x}} = 0$. Thus we see that $\text{Supp}(\mathcal{G}_ n) \subset Z$. Note that $X \setminus Z'$ is a Noetherian algebraic space. Hence by Lemma 68.13.2 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 (2)

Comment #7003 by Janos Kollar on

Comment #7225 by Johan on