Lemma 69.14.3. 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 66.24.2) 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 69.14.1 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 66.12.3. By Lemma 69.13.2 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 67.14.1. Hence we can apply Lemma 69.14.2. 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. $\square$

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