The Stacks project

Proof. We first prove the lemma in case $X = \mathop{\mathrm{Spec}}(A)$ is affine and $\mathcal{L} = \mathcal{O}_ X$. In this case $s$ corresponds to an element $f \in A$. Say $\mathcal{F} = \widetilde{M}$ and $\mathcal{G} = \widetilde{N}$ for some $A$-modules $M$ and $N$. Then the lemma translates (via Lemmas 28.16.1 and 28.16.2) into the following algebra statements

1. If $M$ is a finite $A$-module and $\varphi : M \to N$ is an $A$-module map such that the induced map $M_ f \to N_ f$ is zero, then $f^ n\varphi = 0$ for some $n$.

2. If $M$ is a finitely presented $A$-module, then $\mathop{\mathrm{Hom}}\nolimits _ A(M, N)_ f = \mathop{\mathrm{Hom}}\nolimits _{A_ f}(M_ f, N_ f)$.

The second statement is Algebra, Lemma 10.10.2 and we omit the proof of the first statement.

Next, we prove (1) for general $X$. Assume $X$ is quasi-compact and hoose a finite affine open covering $X = U_1 \cup \ldots \cup U_ m$ with $U_ j$ affine and $\mathcal{L}|_{U_ j} \cong \mathcal{O}_{U_ j}$. Via this isomorphism, the image $s|_{U_ j}$ corresponds to some $f_ j \in \Gamma (U_ j, \mathcal{O}_{U_ j})$. Then $X_ s \cap U_ j = D(f_ j)$. Let $\alpha /s^ n$ be an element in the kernel of (28.17.2.1). Then $\alpha |_{X_ s} = 0$. Hence $(\alpha |_{U_ j})|_{D(f_ j)} = 0$. By the affine case treated above we conclude that $f_ j^{e_ j} \alpha |_{U_ j} = 0$ for some $e_ j \geq 0$. Let $e = \max (e_ j)$. Then we see that $\alpha \otimes s^ e$ restricts to zero on $U_ j$ for all $j$, hence is zero. Since $\alpha /s^ n$ is equal to $\alpha \otimes s^ e/s^{n + e}$ in $M_{(s)}$ we conclude that $\alpha /s^ n = 0$ as desired.

Proof of (2). Since $\mathcal{F}$ is of finite presentation, the sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is quasi-coherent, see Schemes, Section 26.24. Moreover, it is clear that

for all $n$. Hence in this case the statement follows from Lemma 28.17.2 applied to $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$. $\square$