The Stacks project

Proof. Let $\mathcal{J} \subset \mathcal{O}_ X$ be the ideal sheaf of $Z$. Let $\mathcal{F}' \subset \mathcal{F}$ be the subsheaf of local sections of $\mathcal{F}$ which are annihilated by $\mathcal{J}$. It is a quasi-coherent sheaf by Properties, Lemma 28.24.2. Moreover, $\mathcal{F}'_\xi = \mathcal{F}_\xi$ because $\mathcal{J}_\xi = \mathfrak m_\xi$ and part (3) of Properties, Lemma 28.24.2. By Lemma 30.9.5 we see that $\mathcal{F}' \to \mathcal{F}$ induces an isomorphism in a neighbourhood of $\xi$. Hence we may replace $\mathcal{F}$ by $\mathcal{F}'$ and assume that $\mathcal{F}$ is annihilated by $\mathcal{J}$.

Assume $\mathcal{J}\mathcal{F} = 0$. By Lemma 30.9.8 we can write $\mathcal{F} = i_*\mathcal{G}$ for some coherent sheaf $\mathcal{G}$ on $Z$. Suppose we can find a morphism $\mathcal{I}^{\oplus r} \to \mathcal{G}$ which is an isomorphism in a neighbourhood of the generic point $\xi$ of $Z$. Then applying $i_*$ (which is left exact) we get the result of the lemma. Hence we have reduced to the case $X = Z$.

Suppose $Z = X$ is an integral Noetherian scheme with generic point $\xi$. Note that $\mathcal{O}_{X, \xi } = \kappa (\xi )$ is the function field of $X$ in this case. Since $\mathcal{F}_\xi$ is a finite $\mathcal{O}_\xi$-module we see that $r = \dim _{\kappa (\xi )} \mathcal{F}_\xi$ is finite. Hence the sheaves $\mathcal{O}_ X^{\oplus r}$ and $\mathcal{F}$ have isomorphic stalks at $\xi$. By Lemma 30.9.6 there exists a nonempty open $U \subset X$ and a morphism $\psi : \mathcal{O}_ X^{\oplus r}|_ U \to \mathcal{F}|_ U$ which is an isomorphism at $\xi$, and hence an isomorphism in a neighbourhood of $\xi$ by Lemma 30.9.5. By Schemes, Lemma 26.12.4 there exists a quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ whose associated closed subscheme $Y \subset X$ is the complement of $U$. By Lemma 30.10.5 there exists an $n \geq 0$ and a morphism $\mathcal{I}^ n(\mathcal{O}_ X^{\oplus r}) \to \mathcal{F}$ which recovers our $\psi$ 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 because $X$ is integral and it is injective at the generic point of $X$ (easy proof omitted). $\square$