Lemma 30.12.3. Let $X$ be a Noetherian scheme. 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 an integral closed subscheme $Z_ j \subset X$ 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

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

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

If $Z$ is not irreducible, then we can write $Z = Z_1 \cup Z_2$ with $Z_1, Z_2$ closed and strictly smaller than $Z$. Then we can apply Lemma 30.12.1 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 $Z$ 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 $Z$ is irreducible.

Suppose $Z$ is irreducible. Let $\mathcal{J}$ be the sheaf of ideals cutting out the reduced induced closed subscheme structure of $Z$, see Schemes, Lemma 26.12.4. By Lemma 30.10.2 we see there exists an $n \geq 0$ such that $\mathcal{J}^ n\mathcal{F} = 0$. Hence we obtain a filtration

$0 = \mathcal{J}^ n\mathcal{F} \subset \mathcal{J}^{n - 1}\mathcal{F} \subset \ldots \subset \mathcal{J}\mathcal{F} \subset \mathcal{F}$

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 $\mathcal{F}$.

In the case where $Z$ is irreducible and $\mathcal{J}\mathcal{F} = 0$ we can apply Lemma 30.12.2. This gives a short exact sequence

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

where $\mathcal{Q}$ is defined as the quotient. Since $\mathcal{Q}$ is zero in a neighbourhood of $\xi$ by the lemma just cited we see that the support of $\mathcal{Q}$ is strictly smaller than $Z$. Hence we see that $\mathcal{Q}$ has a filtration of the desired type by minimality of $Z$. But then clearly $\mathcal{F}$ does too, which is our final contradiction. $\square$

Comment #950 by correction_bot on

In replace the $\mathcal{I}$'s with $\mathcal{J}$'s.

Comment #2176 by Antoine Chambert-Loir on

In their recent paper, http://de.arxiv.org/pdf/1608.01919.pdf (Lemma 3.2.2), Burgos, Gubler, Jell, Künnemann and Martin note that the proof gives $\mathop{\rm supp} (I_j) = Z_j$.

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