Lemma 27.26.4. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$. For any affine $U \subset X$ the intersection $U \cap X_ s$ is affine.

**Proof.**
This translates into the following algebra problem. Let $R$ be a ring. Let $N$ be an invertible $R$-module (i.e., locally free of rank 1). Let $s \in N$ be an element. Then $U = \{ \mathfrak p \mid s \not\in \mathfrak p N\} $ is an affine open subset of $\mathop{\mathrm{Spec}}(R)$. This you can see as follows. Think of $s$ as an $R$-module map $R \to N$. This gives rise to $R$-module maps $N^{\otimes k} \to N^{\otimes k + 1}$. Consider

with transition maps as above. Define an $R$-algebra structure on $R'$ by the rule $x \cdot y = x \otimes y \in N^{\otimes n + m}$ if $x \in N^{\otimes n}$ and $y \in N^{\otimes m}$. We claim that $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is an open immersion with image $U$.

To prove this is a local question on $\mathop{\mathrm{Spec}}(R)$. Let $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$. Pick $f \in R$, $f \not\in \mathfrak p$ such that $N_ f \cong R_ f$ as a module. Replacing $R$ by $R_ f$, $N$ by $N_ f$ and $R'$ by $R'_ f = \mathop{\mathrm{colim}}\nolimits N_ f^{\otimes n}$ we may assume that $N \cong R$. Say $N = R$. In this case $s$ is an element of $R$ and it is easy to see that $R' \cong R_ s$. Thus the lemma follows. $\square$

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## Comments (2)

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