The Stacks project

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 exist an integral closed subscheme $Z_ j \subset X$ and a nonzero coherent 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. $\square$

Comments (4)

Comment #950 by correction_bot on

In replace the 's with 's.

Comment #2176 by Antoine Chambert-Loir on

In their recent paper, (Lemma 3.2.2), Burgos, Gubler, Jell, K√ľnnemann and Martin note that the proof gives .

Comment #6815 by Yuto Masamura on

This lemma can be easily strengthened by adding the condition for each (the proof remains the same). This makes the proof of Lemma 33.45.2 a little simpler, as I will comment there.

Comment #6958 by on

For some reason I am reluctant to add either the statement that or the statement that . I would say the following: those statements are both true and follow immediately from the lemma as stated now, provided that is not zero. Thus I have instead added this as an outcome.

Unfortunately, the lemma (and a couple of other lemmas) forgot to state that the ideal sheaves are quasi-coherent or equivalently in this case, coherent. So I have added this as well.

The changes are here.

There are also:

  • 5 comment(s) on Section 30.12: Devissage of coherent sheaves

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