Lemma 27.28.5. Let $S$ be a graded ring such that $X = \text{Proj}(S)$ is quasi-compact. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Set $M = \bigoplus _{n \in \mathbf{Z}} \Gamma (X, \mathcal{F}(n))$ as a graded $S$-module, see Constructions, Section 26.10. The map

\[ \widetilde{M} \longrightarrow \mathcal{F} \]

of Constructions, Lemma 26.10.7 is an isomorphism. If $X$ is covered by standard opens $D_+(f)$ where $f$ has degree $1$, then the induced maps $M_ n \to \Gamma (X, \mathcal{F}(n))$ are the identity maps.

**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 26.10.4 and 26.10.5. 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 26.8.4. 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 26.8.8. By Lemma 27.17.2 there is a canonical isomorphism

\[ M_{(f_ i)} = \Gamma _*(X, \mathcal{L}, \mathcal{F})_{(s_ i)} \to \Gamma (X_{s_ i}, \mathcal{F}). \]

The construction of the map in Constructions, Lemma 26.10.7 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.
$\square$

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