The Stacks project

Proof. Write $R = \Gamma (X, \mathcal{O}_ X)$. Consider the canonical morphism

of schemes, see Schemes, Lemma 26.6.4. Then the inverse image of the standard open $D(f)$ on the right hand side is $X_ f$ on the left hand side. Moreover, since $X$ is assumed quasi-compact and quasi-separated the morphism $\varphi $ is quasi-compact and quasi-separated, see Schemes, Lemma 26.19.2 and 26.21.13. Hence by Schemes, Lemma 26.24.1 we see that $\varphi _*\mathcal{F}$ is quasi-coherent. Hence we see that $\varphi _*\mathcal{F} = \widetilde M$ with $M = \Gamma (X, \mathcal{F})$ as an $R$-module. Thus we see that

which is exactly the content of the lemma. The first displayed isomorphism of the lemma follows by taking $\mathcal{F} = \mathcal{O}_ X$. $\square$

Proof. Assume $X$ is quasi-compact. Choose 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 $X_ s \cap U_ j = D(f_ j)$.

Proof of (1). Let $t/s^ n$ be an element in the kernel of (28.17.1.1). Then $t|_{X_ s} = 0$. Hence $(t|_{U_ j})|_{D(f_ j)} = 0$. By Lemma 28.17.1 we conclude that $f_ j^{e_ j} t|_{U_ j} = 0$ for some $e_ j \geq 0$. Let $e = \max (e_ j)$. Then we see that $t \otimes s^ e$ restricts to zero on $U_ j$ for all $j$, hence is zero. Since $t/s^ n$ is equal to $t \otimes s^ e/s^{n + e}$ in $\Gamma _*(X, \mathcal{L}, \mathcal{F})_{(s)}$ we conclude that $t/s^ n = 0$ as desired.

Proof of (2). Assume $X$ is quasi-compact and quasi-separated. Then $U_ j \cap U_{j'}$ is quasi-compact for all pairs $j, j'$, see Schemes, Lemma 26.21.6. By part (1) we know (28.17.1.1) is injective. Let $t' \in \Gamma (X_ s, \mathcal{F}|_{X_ s})$. For every $j$, there exist an integer $e_ j \geq 0$ and $t'_ j \in \Gamma (U_ j, \mathcal{F}|_{U_ j})$ such that $t'|_{D(f_ j)}$ corresponds to $t'_ j/f_ j^{e_ j}$ via the isomorphism of Lemma 28.17.1. Set $e = \max (e_ j)$ and

where $q_ j \in \Gamma (U_ j, \mathcal{L}|_{U_ j})$ is the trivializing section coming from the isomorphism $\mathcal{L}|_{U_ j} \cong \mathcal{O}_{U_ j}$. In particular we have $s|_{U_ j} = f_ j q_ j$. Using this a calculation shows that $t_ j|_{U_ j \cap U_{j'}}$ and $t_{j'}|_{U_ j \cap U_{j'}}$ map to the same section of $\mathcal{F}$ over $U_ j \cap U_{j'} \cap X_ s$. By quasi-compactness of $U_ j \cap U_{j'}$ and part (1) there exists an integer $e' \geq 0$ such that

as sections of $\mathcal{F} \otimes \mathcal{L}^{\otimes e + e'}$ over $U_ j \cap U_{j'}$. We may choose the same $e'$ to work for all pairs $j, j'$. Then the sheaf conditions implies there is a section $t \in \Gamma (X, \mathcal{F} \otimes \mathcal{L}^{\otimes e + e'})$ whose restriction to $U_ j$ is $t_ j \otimes s^{e'}|_{U_ j}$. A simple computation shows that $t/s^{e + e'}$ maps to $t'$ as desired. $\square$

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 $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$. $\square$