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. $\square$

Comment #7003 by Janos Kollar on

May be worth adding: there is a unique sequence where G has no associated points in Z'. Then G' is the largest subsheaf supported on Z'.

Comment #7225 by on

Hi Janos, we have a discussion along these lines for Noetherian schemes in Section 51.15 and there we reference two of your papers! What we need to do is to write a section like that for coherent modules on (locally) Noetherian algebraic spaces as well. To everyone: email me if you are interested in doing this.

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