The Stacks project

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 66.12.3. 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

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 69.13.2 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. $\square$

Proof. By assumption there exists a coherent $\mathcal{O}_ Z$-module $\mathcal{G}$ with support $Z$ and $\mathcal{F} \cong i_*\mathcal{G}$, see Lemma 69.12.7. Hence it suffices to prove the lemma for the case $Z = X$ and $i = \text{id}$.

By Properties of Spaces, Proposition 66.13.3 there exists a dense open subspace $U \subset X$ which is a scheme. Note that $U$ is a Noetherian integral scheme. After shrinking $U$ we may assume that $\mathcal{F}|_ U \cong \mathcal{O}_ U^{\oplus r}$ (for example by Cohomology of Schemes, Lemma 30.12.2 or by a direct algebra argument). Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals whose associated closed subspace is the complement of $U$ in $X$ (see for example Properties of Spaces, Section 66.12). By Lemma 69.13.4 there exists an $n \geq 0$ and a morphism $\mathcal{I}^ n(\mathcal{O}_ X^{\oplus r}) \to \mathcal{F}$ which recovers our isomorphism over $U$. Since $\mathcal{I}^ n(\mathcal{O}_ X^{\oplus r}) = (\mathcal{I}^ n)^{\oplus r}$ we get a map as in the lemma. It is injective: namely, if $\sigma$ is a nonzero section of $\mathcal{I}^{\oplus r}$ over a scheme $W$ étale over $X$, then because $X$ hence $W$ is reduced the support of $\sigma$ contains a nonempty open of $W$. But the kernel of $(\mathcal{I}^ n)^{\oplus r} \to \mathcal{F}$ is zero over a dense open, hence $\sigma$ cannot be a section of the kernel. $\square$

Proof. Consider the collection

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

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

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

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. $\square$

Proof. First note that if $\mathcal{F}$ is a coherent sheaf with a filtration

by coherent subsheaves such that each of $\mathcal{F}_ i/\mathcal{F}_{i - 1}$ has property $\mathcal{P}$, then so does $\mathcal{F}$. This follows from the property (1) for $\mathcal{P}$. On the other hand, by Lemma 69.14.3 we can filter any $\mathcal{F}$ with successive subquotients as in (2). Hence the lemma follows. $\square$

Proof. Consider the collection

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.

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

with $\text{Supp}(\mathcal{G}_ i) \subset Z_ i$. By minimality of $T$ each of $\mathcal{G}_ i$ has $\mathcal{P}$. Hence $\mathcal{F}$ has property $\mathcal{P}$ by (1), a contradiction.

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

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 by (1). 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. Denote $i : Z \to X$ the closed subspace corresponding to $\mathcal{J}$. Then $\mathcal{F} = i_*\mathcal{H}$ for some coherent $\mathcal{O}_ Z$-module $\mathcal{H}$, see Morphisms of Spaces, Lemma 67.14.1 and Lemma 69.12.7. Let $\mathcal{G}$ be the coherent sheaf on $Z$ satisfying (3)(a) and (3)(b). We apply Lemma 69.14.2 to get injective maps

where the support of the cokernels are proper closed in $Z$. Hence we find an nonempty open $V \subset Z$ such that

Let $\mathcal{I} \subset \mathcal{O}_ Z$ be a quasi-coherent ideal sheaf cutting out $Z \setminus V$ we obtain (Lemma 69.13.4) a map

which is an isomorphism over $V$. The kernel is supported on $Z \setminus V$ hence annihilated by some power of $\mathcal{I}$, see Lemma 69.13.2. Thus after increasing $n$ we may assume the displayed map is injective, see Lemma 69.13.3. Applying (3)(b) we find $\mathcal{G}' \subset \mathcal{I}^ n\mathcal{G}$ such that

is injective with cokernel supported in a proper closed subset of $Z$ and such that property $\mathcal{P}$ holds for $i_*\mathcal{G}'$. By (1) property $\mathcal{P}$ holds for $(i_*\mathcal{G}')^{\oplus r_1}$. By (1) and minimality of $T = |Z|$ property $\mathcal{P}$ holds for $\mathcal{F}^{\oplus r_2}$. And finally by (2) property $\mathcal{P}$ holds for $\mathcal{F}$ which is the desired contradiction. $\square$

Proof. We will show that conditions (1) and (2) of Lemma 69.14.4 hold. This is clear for condition (1). To show that (2) holds, let

If $\mathcal{T}$ is nonempty, then since $X$ is Noetherian, we can find an $i : Z \to X$ which is minimal in $\mathcal{T}$. We will show that this leads to a contradiction.

Let $\mathcal{G}$ be the sheaf whose scheme theoretic support is $Z$ whose existence is assumed in assumption (3). Let $\varphi : i_*\mathcal{I}^{\oplus r} \to \mathcal{G}$ be as in Lemma 69.14.2. Let

be a filtration as in Lemma 69.14.3. By minimality of $Z$ and assumption (1) we see that $\mathop{\mathrm{Coker}}(\varphi )$ has property $\mathcal{P}$. As $\varphi$ is injective we conclude using assumption (1) once more that $i_*\mathcal{I}^{\oplus r}$ has property $\mathcal{P}$. Using assumption (2) we conclude that $i_*\mathcal{I}$ has property $\mathcal{P}$.

Finally, if $\mathcal{J} \subset \mathcal{O}_ Z$ is a second quasi-coherent sheaf of ideals, set $\mathcal{K} = \mathcal{I} \cap \mathcal{J}$ and consider the short exact sequences

Arguing as above, using the minimality of $Z$, we see that $i_*\mathcal{I}/\mathcal{K}$ and $i_*\mathcal{J}/\mathcal{K}$ satisfy $\mathcal{P}$. Hence by assumption (1) we conclude that $i_*\mathcal{K}$ and then $i_*\mathcal{J}$ satisfy $\mathcal{P}$. In other words, $Z$ is not an element of $\mathcal{T}$ which is the desired contradiction. $\square$