The Stacks project

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. $\square$