The Stacks project

Proposition 30.15.3. Let $A$ be a Noetherian graded ring. Set $X = \text{Proj}(A)$. The functor $M \mapsto \widetilde M$ induces an equivalence

\[ \text{Mod}^{fg}_ A/\text{Mod}^{fg}_{A, irrelevant} \longrightarrow \textit{Coh}(\mathcal{O}_ X) \]

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

Proof. We urge the reader to read the proof in the case where $A$ is generated in degree $1$ first, see Proposition 30.14.4. Let $f_1, \ldots , f_ r \in A$ be homogeneous elements of positive degree which generate $A$ over $A_0$. Let $d$ be the lcm of the degrees $d_ i$ of $f_ i$. Let $M$ be a finite $A$-module. Let us show that $\widetilde{M}$ is zero if and only if $M$ is an irrelevant graded $A$-module (as defined above the statement of the proposition). Namely, let $x \in M$ be a homogeneous element. Choose $k \in \mathbf{Z}$ sufficiently small and let $N \to N'$ and $M \to N'$ be as in Lemma 30.15.2. We may also pick $l$ sufficiently large such that $M_ n \to N_ n$ is an isomorphism for $n \geq l$. If $\widetilde{M}$ is zero, then $N = 0$. Thus for any $f \in A_+$ homogeneous with $\deg (f) + \deg (x) = nd$ and $nd > l$ we see that $fx$ is zero because $N_{nd} \to N'_{nd}$ and $M_{nd} \to N'_{nd}$ are isomorphisms. Hence $x$ is irrelevant. Conversely, assume $M$ is irrelevant. Then $M_{nd}$ is zero for $n \gg 0$ (see discussion above proposition). Clearly this implies that $M_{(f_ i)} = M_{(f_ i^{d/\deg (f_ i)})} = 0$, whence $\widetilde{M} = 0$ by construction.

It follows that the subcategory $\text{Mod}^{fg}_{A, irrelevant}$ is a Serre subcategory of $\text{Mod}^{fg}_ A$ as the kernel of the exact functor $M \mapsto \widetilde M$, see Homology, Lemma 12.10.4 and Constructions, Lemma 27.8.4. 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 irrelevant modules to $0$ which we have shown above.

By Lemma 30.15.1 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 $\text{Mod}^{fg}_ A \to \text{Mod}^{fg}_ A/\text{Mod}^{fg}_{A, irrelevant}$.

By Lemma 30.15.2 the composite left to right to left is isomorphic to the identity functor. Namely, let $M$ be a finite graded $A$-module and let $k \in \mathbf{Z}$ sufficiently small and let $N \to N'$ and $M \to N'$ be as in Lemma 30.15.2. Then the kernel and cokernel of $M \to N'$ are nonzero in only finitely many degrees, hence are irrelevant. Moreover, the kernel and cokernel of the map $N \to N'$ are zero in all sufficiently large degrees divisible by $d$, hence these are irrelevant modules too. Thus $M \to N'$ and $N \to N'$ are both isomorphisms in the quotient category, as desired.

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 0BXF. Beware of the difference between the letter 'O' and the digit '0'.