The Stacks project

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. $\square$