Lemma 28.26.4. 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 $\mathop{\mathrm{Spec}}(R)$.

Let $A = \bigoplus _{n \geq 0} A_ n$ be the symmetric algebra of $N$ (which is commutative) and view $s$ as an element of $A_1$. Set $B = A/(s - 1)A$. This is an $R$-algebra whose construction commutes with any base change $R \to R'$. Thus $B' = B \otimes _ R R'$ is the zero ring if $s$ maps to zero in $N' = N \otimes _ R R'$. It follows that if $x \in \mathop{\mathrm{Spec}}(R) \setminus U$, then $B \otimes _ R \kappa (x) = 0$. We conclude that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R)$ factors through $U$ as the fibres over $x \not\in U$ are empty. On the other hand, if $\mathop{\mathrm{Spec}}(R') \subset U$ is an affine open, then $s$ maps to a basis element of $N'$ and we see that $B' = R'[s]/(s - 1) \cong R'$. It follows that $\mathop{\mathrm{Spec}}(B) \to U$ is an isomorphism and $U$ is indeed affine. $\square$

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