Lemma 68.14.6. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $\mathcal{P}$ be a property of coherent sheaves on $X$. Assume

1. For any short exact sequence of coherent sheaves on $X$ if two out of three have property $\mathcal{P}$ so does the third.

2. If $\mathcal{P}$ holds for $\mathcal{F}^{\oplus r}$ for some $r \geq 1$, then it holds for $\mathcal{F}$.

3. For every reduced closed subspace $i : Z \to X$ with $|Z|$ irreducible there exists a coherent sheaf $\mathcal{G}$ on $X$ whose scheme theoretic support is $Z$ such that $\mathcal{P}$ holds for $\mathcal{G}$.

Then property $\mathcal{P}$ holds for every coherent sheaf on $X$.

Proof. We will show that conditions (1) and (2) of Lemma 68.14.4 hold. This is clear for condition (1). To show that (2) holds, let

$\mathcal{T} = \left\{ \begin{matrix} i : Z \to X \text{ reduced closed subspace with }|Z|\text{ irreducible such} \\ \text{ that }i_*\mathcal{I}\text{ does not have }\mathcal{P} \text{ for some quasi-coherent }\mathcal{I} \subset \mathcal{O}_ Z \end{matrix} \right\}$

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 68.14.2. Let

$0 = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \ldots \subset \mathcal{F}_ m = \mathop{\mathrm{Coker}}(\varphi )$

be a filtration as in Lemma 68.14.3. 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

$0 \to \mathcal{K} \to \mathcal{I} \to \mathcal{I}/\mathcal{K} \to 0 \quad \text{and} \quad 0 \to \mathcal{K} \to \mathcal{J} \to \mathcal{J}/\mathcal{K} \to 0$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).