The Stacks project

Proof. Pick $f \in S_ d$. Denote $s = \psi (f) \in \Gamma (Y, \mathcal{L})$. On the open set $Y_ s$ where $s$ does not vanish multiplication by $s$ induces an isomorphism $\mathcal{O}_{Y_ s} \to \mathcal{L}|_{Y_ s}$, see Modules, Lemma 17.25.10. We will denote the inverse of this map $x \mapsto x/s$, and similarly for powers of $\mathcal{L}$. Using this we define a ring map $\psi _{(f)} : S_{(f)} \to \Gamma (Y_ s, \mathcal{O})$ by mapping the fraction $a/f^ n$ to $\psi (a)/s^ n$. By Schemes, Lemma 26.6.4 this corresponds to a morphism $\varphi _ f : Y_ s \to \mathop{\mathrm{Spec}}(S_{(f)}) = D_{+}(f)$. We also introduce the isomorphism $\alpha _ f : \varphi _ f^*\mathcal{O}_{D_{+}(f)}(d) \to \mathcal{L}|_{Y_ s}$ which maps the pullback of the trivializing section $f$ over $D_{+}(f)$ to the trivializing section $s$ over $Y_ s$. With this choice the commutativity of the diagram in the lemma holds with $Y$ replaced by $Y_ s$, $\varphi $ replaced by $\varphi _ f$, and $\alpha $ replaced by $\alpha _ f$; verification omitted.

Suppose that $f' \in S_ d$ is a second element, and denote $s' = \psi (f') \in \Gamma (Y, \mathcal{L})$. Then $Y_ s \cap Y_{s'} = Y_{ss'}$ and similarly $D_{+}(f) \cap D_{+}(f') = D_{+}(ff')$. In Lemma 27.10.6 we saw that $D_{+}(f') \cap D_{+}(f)$ is the same as the set of points of $D_{+}(f)$ where the section of $\mathcal{O}_ X(d)$ defined by $f'$ does not vanish. Hence $\varphi _ f^{-1}(D_{+}(f') \cap D_{+}(f)) = Y_ s \cap Y_{s'} = \varphi _{f'}^{-1}(D_{+}(f') \cap D_{+}(f))$. On $D_{+}(f) \cap D_{+}(f')$ the fraction $f/f'$ is an invertible section of the structure sheaf with inverse $f'/f$. Note that $\psi _{(f')}(f/f') = \psi (f)/s' = s/s'$ and $\psi _{(f)}(f'/f) = \psi (f')/s = s'/s$. We claim there is a unique ring map $S_{(ff')} \to \Gamma (Y_{ss'}, \mathcal{O})$ making the following diagram commute

It exists because we may use the rule $x/(ff')^ n \mapsto \psi (x)/(ss')^ n$, which “works” by the formulas above. Uniqueness follows as $\text{Proj}(S)$ is separated, see Lemma 27.8.8 and its proof. This shows that the morphisms $\varphi _ f$ and $\varphi _{f'}$ agree over $Y_ s \cap Y_{s'}$. The restrictions of $\alpha _ f$ and $\alpha _{f'}$ agree over $Y_ s \cap Y_{s'}$ because the regular functions $s/s'$ and $\psi _{(f')}(f)$ agree. This proves that the morphisms $\psi _ f$ glue to a global morphism from $Y$ into $U_ d \subset X$, and that the maps $\alpha _ f$ glue to an isomorphism satisfying the conditions of the lemma.

We still have to show the pair $(\varphi , \alpha )$ is unique. Suppose $(\varphi ', \alpha ')$ is a second such pair. Let $f \in S_ d$. By the commutativity of the diagrams in the lemma we have that the inverse images of $D_{+}(f)$ under both $\varphi $ and $\varphi '$ are equal to $Y_{\psi (f)}$. Since the opens $D_{+}(f)$ are a basis for the topology on $X$, and since $X$ is a sober topological space (see Schemes, Lemma 26.11.1) this means the maps $\varphi $ and $\varphi '$ are the same on underlying topological spaces. Let us use $s = \psi (f)$ to trivialize the invertible sheaf $\mathcal{L}$ over $Y_{\psi (f)}$. By the commutativity of the diagrams we have that $\alpha ^{\otimes n}(\psi ^ d_{\varphi }(x)) = \psi (x) = (\alpha ')^{\otimes n}(\psi ^ d_{\varphi '}(x))$ for all $x \in S_{nd}$. By construction of $\psi ^ d_{\varphi }$ and $\psi ^ d_{\varphi '}$ we have $\psi ^ d_{\varphi }(x) = \varphi ^\sharp (x/f^ n) \psi ^ d_{\varphi }(f^ n)$ over $Y_{\psi (f)}$, and similarly for $\psi ^ d_{\varphi '}$. By the commutativity of the diagrams of the lemma we deduce that $\varphi ^\sharp (x/f^ n) = (\varphi ')^\sharp (x/f^ n)$. This proves that $\varphi $ and $\varphi '$ induce the same morphism from $Y_{\psi (f)}$ into the affine scheme $D_{+}(f) = \mathop{\mathrm{Spec}}(S_{(f)})$. Hence $\varphi $ and $\varphi '$ are the same as morphisms. Finally, it remains to show that the commutativity of the diagram of the lemma singles out, given $\varphi $, a unique $\alpha $. We omit the verification. $\square$