The Stacks project

Lemma 28.17.2. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf on $X$. Let $s \in \Gamma (X, \mathcal{L})$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module.

  1. If $X$ is quasi-compact, then ( is injective, and

  2. if $X$ is quasi-compact and quasi-separated, then ( is an isomorphism.

In particular, the canonical map

\[ \Gamma _*(X, \mathcal{L})_{(s)} \longrightarrow \Gamma (X_ s, \mathcal{O}_ X),\quad a/s^ n \longmapsto a \otimes s^{-n} \]

is an isomorphism if $X$ is quasi-compact and quasi-separated.

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 ( 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 ( 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

\[ t_ j = f_ j^{e - e_ j} t'_ j \otimes q_ j^ e \in \Gamma (U_ j, (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes e})|_{U_ j}) \]

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

\[ t_ j|_{U_ j \cap U_{j'}} \otimes s^{e'}|_{U_ j \cap U_{j'}} = t_{j'}|_{U_ j \cap U_{j'}} \otimes s^{e'}|_{U_ j \cap U_{j'}} \]

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. $\square$

Comments (4)

Comment #2595 by Rogier Brussee on

Suggested slogan: Multiplication by a section in a line bundle is an isomorphism unless.

Comment #2623 by on

Hmmm... not sure what you wanted to say...

Comment #4719 by Manuel Hoff on

I think there is a mistake in the proof of (2):

If one defines the 's as in the proof, then they will map to the elements . These are the wrong elements and in general the 's won't glue.

Instead, I think, one wants to define , where is the trivializing section corresponding to under .

One can also avoid working with the different 's by noting that one can choose them all to be equal.

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