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.
$\square$

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Comment #949 by correction_bot on

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