The Stacks project

Proof. We will use that $\mathcal{O}_{\mathbf{P}^ n_ R}(1)$ is an ample invertible sheaf on the scheme $\mathbf{P}^ n_ R$. This follows directly from the definition since $\mathbf{P}^ n_ R$ covered by the standard affine opens $D_{+}(T_ i)$. Hence by Properties, Proposition 28.26.13 every finite type quasi-coherent $\mathcal{O}_{\mathbf{P}^ n_ R}$-module is a quotient of a finite direct sum of tensor powers of $\mathcal{O}_{\mathbf{P}^ n_ R}(1)$. On the other hand coherent sheaves and finite type quasi-coherent sheaves are the same thing on projective space over $R$ by Lemma 30.9.1. Thus we see (1).

Projective $n$-space $\mathbf{P}^ n_ R$ is covered by $n + 1$ affines, namely the standard opens $D_{+}(T_ i)$, $i = 0, \ldots , n$, see Constructions, Lemma 27.13.3. Hence we see that for any quasi-coherent sheaf $\mathcal{F}$ on $\mathbf{P}^ n_ R$ we have $H^ i(\mathbf{P}^ n_ R, \mathcal{F}) = 0$ for $i \geq n + 1$, see Lemma 30.4.2. Hence (2) holds.

Let us prove (3) and (4) simultaneously for all coherent sheaves on $\mathbf{P}^ n_ R$ by descending induction on $i$. Clearly the result holds for $i \geq n + 1$ by (2). Suppose we know the result for $i + 1$ and we want to show the result for $i$. (If $i = 0$, then part (4) is vacuous.) Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^ n_ R$. Choose a surjection as in (1) and denote $\mathcal{G}$ the kernel so that we have a short exact sequence

By Lemma 30.9.2 we see that $\mathcal{G}$ is coherent. The long exact cohomology sequence gives an exact sequence

By induction assumption the right $R$-module is finite and by Lemma 30.8.1 the left $R$-module is finite. Since $R$ is Noetherian it follows immediately that $H^ i(\mathbf{P}^ n_ R, \mathcal{F})$ is a finite $R$-module. This proves the induction step for assertion (3). Since $\mathcal{O}_{\mathbf{P}^ n_ R}(d)$ is invertible we see that twisting on $\mathbf{P}^ n_ R$ is an exact functor (since you get it by tensoring with an invertible sheaf, see Constructions, Definition 27.10.1). This means that for all $d \in \mathbf{Z}$ the sequence

is short exact. The resulting cohomology sequence is

By induction assumption we see the module on the right is zero for $d \gg 0$ and by the computation in Lemma 30.8.1 the module on the left is zero as soon as $d \geq -\min \{ d_ j\} $ and $i \geq 1$. Hence the induction step for assertion (4). This concludes the proof of (3) and (4).

In order to prove (5) note that for all sufficiently large $d$ the map

is surjective by the vanishing of $H^1(\mathbf{P}^ n_ R, \mathcal{G}(d))$ we just proved. In other words, the module

is for $k$ large enough a quotient of the corresponding module

When $k$ is sufficiently small (e.g. $k < -d_ j$ for all $j$) then

by our computations in Section 30.8. In particular it is finitely generated. Suppose $k \in \mathbf{Z}$ is arbitrary. Choose $k_{-} \ll k \ll k_{+}$. Consider the diagram

where the vertical arrow is the surjective map above and the horizontal arrows are the obvious inclusion maps. By what was said above we see that $N_{k_{-}}$ is a finitely generated $R[T_0, \ldots , T_ n]$-module. Hence $N_{k_{+}}$ is a finitely generated $R[T_0, \ldots , T_ n]$-module because it is a submodule of a finitely generated module and the ring $R[T_0, \ldots , T_ n]$ is Noetherian. Since the vertical arrow is surjective we conclude that $M_{k_{+}}$ is a finitely generated $R[T_0, \ldots , T_ n]$-module. The quotient $M_ k/M_{k_{+}}$ is finite as an $R$-module since it is a finite direct sum of the finite $R$-modules $H^0(\mathbf{P}^ n_ R, \mathcal{F}(d))$ for $k \leq d < k_{+}$. Note that we use part (3) for $i = 0$ here. Hence $M_ k/M_{k_{+}}$ is a fortiori a finite $R[T_0, \ldots , T_ n]$-module. In other words, we have sandwiched $M_ k$ between two finite $R[T_0, \ldots , T_ n]$-modules and we win. $\square$

Proof. By assumption there exists a surjection of graded $A_0$-algebras

where $\deg (T_ j) = 1$ for $j = 0, \ldots , n$. By Constructions, Lemma 27.11.5 this defines a closed immersion $i : X \to \mathbf{P}^ n_{A_0}$ such that $i^*\mathcal{O}_{\mathbf{P}^ n_{A_0}}(1) = \mathcal{O}_ X(1)$. In particular, $X$ is Noetherian as a closed subscheme of the Noetherian scheme $\mathbf{P}^ n_{A_0}$. We claim that the results of the lemma for $\mathcal{F}$ follow from the corresponding results of Lemma 30.14.1 for the coherent sheaf $i_*\mathcal{F}$ (Lemma 30.9.8) on $\mathbf{P}^ n_{A_0}$. For example, by this lemma there exists a surjection

By adjunction this corresponds to a map $\bigoplus _{j = 1, \ldots , r} \mathcal{O}_ X(d_ j) \longrightarrow \mathcal{F}$ which is surjective as well. The statements on cohomology follow from the fact that $H^ p(X, \mathcal{F}(d)) = H^ p(\mathbf{P}^ n_{A_0}, i_*\mathcal{F}(d))$ by Lemma 30.2.4. $\square$

Proof. Because $M$ is a finite $A$-module we see that $\widetilde{M}$ is a finite type $\mathcal{O}_ X$-module, i.e., a coherent $\mathcal{O}_ X$-module. Set $N = \bigoplus _{n \in \mathbf{Z}} \Gamma (X, \widetilde{M}(n))$. We have to show that the map $M \to N$ of graded $A$-modules is an isomorphism in all sufficiently large degrees. By Properties, Lemma 28.28.5 we have a canonical isomorphism $\widetilde{N} \to \widetilde{M}$ such that the induced maps $N_ n \to N_ n = \Gamma (X, \widetilde{M}(n))$ are the identity maps. Thus we have maps $\widetilde{M} \to \widetilde{N} \to \widetilde{M}$ such that for all $n$ the diagram

is commutative. This means that the composition

is equal to the canonical map $M_ n \to \Gamma (X, \widetilde{M}(n))$. Clearly this implies that the composition $\widetilde{M} \to \widetilde{N} \to \widetilde{M}$ is the identity. Hence $\widetilde{M} \to \widetilde{N}$ is an isomorphism. Let $K = \mathop{\mathrm{Ker}}(M \to N)$ and $Q = \mathop{\mathrm{Coker}}(M \to N)$. Recall that the functor $M \mapsto \widetilde{M}$ is exact, see Constructions, Lemma 27.8.4. Hence we see that $\widetilde{K} = 0$ and $\widetilde{Q} = 0$. Recall that $A$ is a Noetherian ring, $M$ is a finitely generated $A$-module, and $N$ is a graded $A$-module such that $N' = \bigoplus _{n \geq 0} N_ n$ is finitely generated by the last part of Lemma 30.14.2. Hence $K' = \bigoplus _{n \geq 0} K_ n$ and $Q' = \bigoplus _{n \geq 0} Q_ n$ are finite $A$-modules. Observe that $\widetilde{Q} = \widetilde{Q'}$ and $\widetilde{K} = \widetilde{K'}$. Thus to finish the proof it suffices to show that a finite $A$-module $K$ with $\widetilde{K} = 0$ has only finitely many nonzero homogeneous parts $K_ d$ with $d \geq 0$. To do this, let $x_1, \ldots , x_ r \in K$ be homogeneous generators say sitting in degrees $d_1, \ldots , d_ r$. Let $f_1, \ldots , f_ n \in A_1$ be elements generating $A$ over $A_0$. For each $i$ and $j$ there exists an $n_{ij} \geq 0$ such that $f_ i^{n_{ij}} x_ j = 0$ in $K_{d_ j + n_{ij}}$: if not then $x_ i/f_ i^{d_ i} \in K_{(f_ i)}$ would not be zero, i.e., $\widetilde{K}$ would not be zero. Then we see that $K_ d$ is zero for $d > \max _ j(d_ j + \sum _ i n_{ij})$ as every element of $K_ d$ is a sum of terms where each term is a monomials in the $f_ i$ times one of the $x_ j$ of total degree $d$. $\square$

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. $\square$