A line-bundle section cuts an affine open inside an affine scheme

Reduction. Write $U=\mathrm{Spec}R$ by [F2] and put $N=\mathrm{\Gamma }(U,L{\mathrm{\mid }}_U)$. By [F1] the $R$-module $N$ is locally free of rank one, so $N_{\mathfrak{p}}\cong R_{\mathfrak{p}}$ for every prime $\mathfrak{p}$; as the fibre of $L$ at the point $\mathfrak{p}$ is $N{\otimes }_R\kappa (\mathfrak{p})$, the point lies in $U\cap X_s$ exactly when the image of $s$ in $N{\otimes }_R\kappa (\mathfrak{p})$ is nonzero. Hence it suffices to prove that for a commutative ring $R$, a locally free rank-one $R$-module $N$ and an element $s\in N$, the set $V=\{\mathfrak{p}\in \mathrm{Spec}R:s\mathrm{\notin }\mathfrak{p}{N}_{\mathfrak{p}}\}$ is an affine open subscheme of $\mathrm{Spec}R$.

The algebra $B$. Let $T=T_R(N)={⨁}_{n\ge 0}{N}^{{\otimes }_{R}n}$ be the tensor algebra, and let $J\subseteq T$ be the two-sided ideal generated by all elements $m\otimes n-n\otimes m$ with $m,n\in N$ and by the element $s-1\in T$, where $s\in N=T_1$. Put $B=T\mathrm{/}J$. Then $B$ is a commutative $R$-algebra, because the degree-one generators of $T$ commute modulo $J$ and every element of $T$ is a sum of products of degree-one elements; it is generated as an $R$-algebra by the image of $N$. The relation $s=1$ need not preserve the tensor grading, and no grading on $B$ is used.

The tensor identity. Let $M$ be a second invertible sheaf and let $t\in \mathrm{\Gamma }(X,M)$; then $L\otimes M$ is invertible, and at every point $x$ the fibre of $L\otimes M$ is $(L\otimes \kappa (x)){\otimes }_{\kappa (x)}(M\otimes \kappa (x))$, a tensor product of one-dimensional vector spaces, in which the image of $s\otimes t$ is the product of the images of $s$ and of $t$. Hence the image of $s\otimes t$ is nonzero exactly when both are, so $X_{s\otimes t}=X_s\cap X_t$.

Base change of $B$. For every ring map $R\to R^{\mathrm{\prime }}$ the identifications of [F3] give a canonical isomorphism $B{\otimes }_R{R}^{\mathrm{\prime }}\cong T_{R^{\mathrm{\prime }}}(N{\otimes }_R{R}^{\mathrm{\prime }})$ modulo the ideal generated by the images of the generators of $J$, namely by the commutators and by the image of $s-1$; these identifications are compatible with a further base change. In the special case that $N=R^{\mathrm{\prime }}$ is free on the basis element $s$ one has $T_{R^{\mathrm{\prime }}}(N)=R^{\mathrm{\prime }}[s]$ and $B{\otimes }_R{R}^{\mathrm{\prime }}=R^{\mathrm{\prime }}[s]\mathrm{/}(s-1)\cong R^{\mathrm{\prime }}$.

Empty fibre where $s$ vanishes. Let $\mathfrak{p}\subseteq R$ be a prime with $s\in \mathfrak{p}{N}_{\mathfrak{p}}$, and put $\kappa =\kappa (\mathfrak{p})$, $N^{\mathrm{\prime }}=N{\otimes }_R\kappa$. Then $N^{\mathrm{\prime }}$ is one-dimensional over $\kappa$ by [F1], the image $s^{\mathrm{\prime }}$ of $s$ in $N^{\mathrm{\prime }}$ is zero, and step 2.1 gives $B{\otimes }_R\kappa =\kappa [u]\mathrm{/}(0-1)=0$, the quotient of the polynomial algebra $\kappa [u]$ by the unit relation $s^{\mathrm{\prime }}-1=-1$. Hence the fibre $\mathrm{Spec}(B{\otimes }_R\kappa (\mathfrak{p}))$ of $\mathrm{Spec}B\to \mathrm{Spec}R$ over $\mathfrak{p}$ is empty.

Nonempty fibre where $s$ generates. Let $\mathfrak{p}\subseteq R$ be a prime with $s\mathrm{\notin }\mathfrak{p}{N}_{\mathfrak{p}}$. Since $N_{\mathfrak{p}}$ is free of rank one over the local ring $R_{\mathfrak{p}}$, [F4] shows that $s$ is a basis of $N_{\mathfrak{p}}$; localising the isomorphism of step 2.1 at $\mathfrak{p}$ gives $B_{\mathfrak{p}}\cong R_{\mathfrak{p}}[s]\mathrm{/}(s-1)\cong R_{\mathfrak{p}}$, so $B_{\mathfrak{p}}\mathrm{\ne }0$ and $B{\otimes }_R\kappa (\mathfrak{p})\cong \kappa (\mathfrak{p})\mathrm{\ne }0$.

The morphism factors through $V$. By steps 3.1 and 3.2 the fibre of $\mathrm{Spec}B\to \mathrm{Spec}R$ is empty exactly over the primes outside $V$; in particular the set-theoretic image is contained in $V$, so by [F6] the affine morphism factors as a morphism $\mathrm{Spec}B\to V\subseteq \mathrm{Spec}R$ into the open subscheme $V$.

It is an isomorphism onto $V$. Let $\mathrm{Spec}{R}^{\mathrm{\prime }}\subseteq V$ be an affine open, and let $N^{\mathrm{\prime }}=N{\otimes }_R{R}^{\mathrm{\prime }}$. For every prime $\mathfrak{p}\subseteq R^{\mathrm{\prime }}$ the image of $s$ in $N_{\mathfrak{p}}$ generates, so $s$ generates $N^{\mathrm{\prime }}$ by [F5]; since $N^{\mathrm{\prime }}$ is locally free of rank one, $N^{\mathrm{\prime }}$ is free on the basis element $s$. Step 2.1 then gives $B{\otimes }_R{R}^{\mathrm{\prime }}\cong R^{\mathrm{\prime }}[s]\mathrm{/}(s-1)\cong R^{\mathrm{\prime }}$, that is, the base change of the morphism $\mathrm{Spec}B\to V$ of step 4.1 over the affine open $\mathrm{Spec}{R}^{\mathrm{\prime }}$ is an isomorphism. The affine opens $\mathrm{Spec}{R}^{\mathrm{\prime }}\subseteq V$ cover $V$, so by the local criterion of [F6] the morphism $\mathrm{Spec}B\to V$ is an isomorphism; in particular $V$ is affine, with coordinate ring $B$, and step 1.1 identifies $U\cap X_s$ with $V$.

Boundaries. If $s$ is nowhere vanishing then $X_s=X$ and $U\cap X_s=U$ is affine by hypothesis. If $s=0$ then $s\in \mathfrak{p}{N}_{\mathfrak{p}}$ for every prime, so $V=\mathrm{\varnothing }$ by step 3.1, which is affine because it is the spectrum of the zero ring by [F2], while step 1.3 gives $X_0\cap X_t=\mathrm{\varnothing }=X_{0\otimes t}$. Together with steps 1.1-5.1 this proves both claims of the Statement. [F2, step 1.1, step 1.3, step 5.1] \qed