The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BXD. Beware of the difference between the letter 'O' and the digit '0'.