Proposition 30.14.4. Let $A$ be a graded ring such that $A_0$ is Noetherian and $A$ is generated by finitely many elements of $A_1$ over $A_0$. Set $X = \text{Proj}(A)$. The functor $M \mapsto \widetilde M$ induces an equivalence

$\text{Mod}^{fg}_ A/\text{Mod}^{fg}_{A, torsion} \longrightarrow \textit{Coh}(\mathcal{O}_ X)$

whose quasi-inverse is given by $\mathcal{F} \longmapsto \bigoplus _{n \geq 0} \Gamma (X, \mathcal{F}(n))$.

Proof. The subcategory $\text{Mod}^{fg}_{A, torsion}$ is a Serre subcategory of $\text{Mod}^{fg}_ A$, see Homology, Definition 12.10.1. This is clear from the description of objects given above but it also follows from More on Algebra, Lemma 15.88.5. Hence the quotient category on the left of the arrow is defined in Homology, Lemma 12.10.6. To define the functor of the proposition, it suffices to show that the functor $M \mapsto \widetilde M$ sends torsion modules to $0$. This is clear because for any $f \in A_+$ homogeneous the module $M_ f$ is zero and hence the value $M_{(f)}$ of $\widetilde M$ on $D_+(f)$ is zero too.

By Lemma 30.14.2 the proposed quasi-inverse makes sense. Namely, the lemma shows that $\mathcal{F} \longmapsto \bigoplus _{n \geq 0} \Gamma (X, \mathcal{F}(n))$ is a functor $\textit{Coh}(\mathcal{O}_ X) \to \text{Mod}^{fg}_ A$ which we can compose with the quotient functor $\text{Mod}^{fg}_ A \to \text{Mod}^{fg}_ A/\text{Mod}^{fg}_{A, torsion}$.

By Lemma 30.14.3 the composite left to right to left is isomorphic to the identity functor.

Finally, let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Set $M = \bigoplus _{n \in \mathbf{Z}} \Gamma (X, \mathcal{F}(n))$ viewed as a graded $A$-module, so that our functor sends $\mathcal{F}$ to $M_{\geq 0} = \bigoplus _{n \geq 0} M_ n$. By Properties, Lemma 28.28.5 the canonical map $\widetilde M \to \mathcal{F}$ is an isomorphism. Since the inclusion map $M_{\geq 0} \to M$ defines an isomorphism $\widetilde{M_{\geq 0}} \to \widetilde M$ we conclude that the composite right to left to right is isomorphic to the identity functor as well. $\square$

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