The Stacks project

Proof. By Proposition 28.26.13 there exists an integer $n_0$ such that $\mathcal{L}^{\otimes n}$ is globally generated for all $n \geq n_0$. Let $x \in X$ be a point. By the above we can find $a \in S_{n_0}$ and $b \in S_{n_0 + 1}$ such that $a$ and $b$ do not vanish at $x$. Hence $f(x) \in D_{+}(a) \cap D_{+}(b) = D_{+}(ab)$. By Constructions, Lemma 27.10.4 we see that $f(x) \in W_1$ as desired. By Constructions, Lemma 27.14.1 which was used in the construction of the map $f$ the maps $f^*\mathcal{O}_ Y(n_0) \to \mathcal{L}^{\otimes n_0}$ and $f^*\mathcal{O}_ Y(n_0 + 1) \to \mathcal{L}^{\otimes n_0 + 1}$ are isomorphisms in a neighbourhood of $x$. By compatibility with the algebra structure and the fact that $f$ maps into $W$ we conclude all the maps $f^*\mathcal{O}_ Y(n) \to \mathcal{L}^{\otimes n}$ are isomorphisms in a neighbourhood of $x$. Hence we win. $\square$

Proof. Let $s \in S_{+}$ be homogeneous such that $X_ s$ is affine open in $X$. Recall that $\widetilde{M}|_{D_{+}(s)}$ corresponds to the $S_{(s)}$-module $M_{(s)}$, see Constructions, Lemma 27.8.4. Recall that $f^{-1}(D_{+}(s)) = X_ s$. As $X$ carries an ample invertible sheaf it is quasi-compact and quasi-separated, see Section 28.26. By Lemma 28.17.2 there is a canonical isomorphism $M_{(s)} = \Gamma _*(X, \mathcal{L}, \mathcal{F})_{(s)} \to \Gamma (X_ s, \mathcal{F})$. Since $\mathcal{F}$ is quasi-coherent this leads to a canonical isomorphism

Since $\mathcal{L}$ is ample on $X$ we know that $X$ is covered by the affine opens of the form $X_ s$. Hence it suffices to prove that the displayed maps glue on overlaps. Proof of this is omitted. $\square$

Proof. Since $X$ is quasi-compact we can find homogeneous elements $f_1, \ldots , f_ n \in S$ of positive degrees such that $X = D_+(f_1) \cup \ldots \cup D_+(f_ n)$. Let $d$ be the least common multiple of the degrees of $f_1, \ldots , f_ n$. After replacing $f_ i$ by a power we may assume that each $f_ i$ has degree $d$. Then we see that $\mathcal{L} = \mathcal{O}_ X(d)$ is invertible, the multiplication maps $\mathcal{O}_ X(ad) \otimes \mathcal{O}_ X(bd) \to \mathcal{O}_ X((a + b)d)$ are isomorphisms, and each $f_ i$ determines a global section $s_ i$ of $\mathcal{L}$ such that $X_{s_ i} = D_+(f_ i)$, see Constructions, Lemmas 27.10.4 and 27.10.5. Thus $\Gamma (X, \mathcal{F}(ad)) = \Gamma (X, \mathcal{F} \otimes \mathcal{L}^{\otimes a})$. Recall that $\widetilde{M}|_{D_{+}(f_ i)}$ corresponds to the $S_{(f_ i)}$-module $M_{(f_ i)}$, see Constructions, Lemma 27.8.4. Since the degree of $f_ i$ is $d$, the isomorphism class of $M_{(f_ i)}$ depends only on the homogeneous summands of $M$ of degree divisible by $d$. More precisely, the isomorphism class of $M_{(f_ i)}$ depends only on the graded $\Gamma _*(X, \mathcal{L})$-module $\Gamma _*(X, \mathcal{L}, \mathcal{F})$ and the image $s_ i$ of $f_ i$ in $\Gamma _*(X, \mathcal{L})$. The scheme $X$ is quasi-compact by assumption and separated by Constructions, Lemma 27.8.8. By Lemma 28.17.2 there is a canonical isomorphism

The construction of the map in Constructions, Lemma 27.10.7 then shows that it is an isomorphism over $D_+(f_ i)$ hence an isomorphism as $X$ is covered by these opens. We omit the proof of the final statement. $\square$