## 30.14 Coherent sheaves on Proj, I

In this section we discuss coherent sheaves on $\text{Proj}(A)$ where $A$ is a Noetherian graded ring generated by $A_1$ over $A_0$. In the next section we discuss what happens if $A$ is not generated by degree $1$ elements. First, we formulate an all-in-one result for projective space over a Noetherian ring.

Lemma 30.14.1. Let $R$ be a Noetherian ring. Let $n \geq 0$ be an integer. For every coherent sheaf $\mathcal{F}$ on $\mathbf{P}^ n_ R$ we have the following:

1. There exists an $r \geq 0$ and $d_1, \ldots , d_ r \in \mathbf{Z}$ and a surjection

$\bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_ R}(d_ j) \longrightarrow \mathcal{F}.$
2. We have $H^ i(\mathbf{P}^ n_ R, \mathcal{F}) = 0$ unless $0 \leq i \leq n$.

3. For any $i$ the cohomology group $H^ i(\mathbf{P}^ n_ R, \mathcal{F})$ is a finite $R$-module.

4. If $i > 0$, then $H^ i(\mathbf{P}^ n_ R, \mathcal{F}(d)) = 0$ for all $d$ large enough.

5. For any $k \in \mathbf{Z}$ the graded $R[T_0, \ldots , T_ n]$-module

$\bigoplus \nolimits _{d \geq k} H^0(\mathbf{P}^ n_ R, \mathcal{F}(d))$

is a finite $R[T_0, \ldots , T_ n]$-module.

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

$0 \to \mathcal{G} \to \bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_ R}(d_ j) \to \mathcal{F} \to 0$

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

$H^ i(\mathbf{P}^ n_ R, \bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_ R}(d_ j)) \to H^ i(\mathbf{P}^ n_ R, \mathcal{F}) \to H^{i + 1}(\mathbf{P}^ n_ R, \mathcal{G}).$

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

$0 \to \mathcal{G}(d) \to \bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_ R}(d_ j + d) \to \mathcal{F}(d) \to 0$

is short exact. The resulting cohomology sequence is

$H^ i(\mathbf{P}^ n_ R, \bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_ R}(d_ j + d)) \to H^ i(\mathbf{P}^ n_ R, \mathcal{F}(d)) \to H^{i + 1}(\mathbf{P}^ n_ R, \mathcal{G}(d)).$

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

$H^0(\mathbf{P}^ n_ R, \bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_ R}(d_ j + d)) \to H^0(\mathbf{P}^ n_ R, \mathcal{F}(d))$

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

$M_ k = \bigoplus \nolimits _{d \geq k} H^0(\mathbf{P}^ n_ R, \mathcal{F}(d))$

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

$N_ k = \bigoplus \nolimits _{d \geq k} H^0(\mathbf{P}^ n_ R, \bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_ R}(d_ j + d) )$

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

$N_ k = \bigoplus \nolimits _{j = 1, \ldots , r} R[T_0, \ldots , T_ n](d_ j)$

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

$\xymatrix{ N_{k_{-}} & N_{k_{+}} \ar[d] \ar[l] \\ M_ k & M_{k_{+}} \ar[l] }$

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. $\square$

Lemma 30.14.2. 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 $X = \text{Proj}(A)$. Then $X$ is a Noetherian scheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module.

1. There exists an $r \geq 0$ and $d_1, \ldots , d_ r \in \mathbf{Z}$ and a surjection

$\bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_ X(d_ j) \longrightarrow \mathcal{F}.$
2. For any $p$ the cohomology group $H^ p(X, \mathcal{F})$ is a finite $A_0$-module.

3. If $p > 0$, then $H^ p(X, \mathcal{F}(d)) = 0$ for all $d$ large enough.

4. For any $k \in \mathbf{Z}$ the graded $A$-module

$\bigoplus \nolimits _{d \geq k} H^0(X, \mathcal{F}(d))$

is a finite $A$-module.

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

$A_0[T_0, \ldots , T_ n] \longrightarrow A$

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

$\bigoplus \nolimits _{j = 1, \ldots , r} \mathcal{O}_{\mathbf{P}^ n_{A_0}}(d_ j) \longrightarrow i_*\mathcal{F}.$

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. $\square$

Lemma 30.14.3. Let $A$ be a graded ring such that $A_0$ is Noetherian and $A$ is generated by finitely many elements of $A_1$ over $A_0$. Let $M$ be a finite graded $A$-module. Set $X = \text{Proj}(A)$ and let $\widetilde{M}$ be the quasi-coherent $\mathcal{O}_ X$-module on $X$ associated to $M$. The maps

$M_ n \longrightarrow \Gamma (X, \widetilde{M}(n))$

from Constructions, Lemma 27.10.3 are isomorphisms for all sufficiently large $n$.

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

$\xymatrix{ M_ n \ar[d] \ar[r] & N_ n \ar[d] \ar@{=}[rd] \\ \Gamma (X, \widetilde{M}(n)) \ar[r] & \Gamma (X, \widetilde{N}(n)) \ar[r]^{\cong } & \Gamma (X, \widetilde{M}(n)) }$

is commutative. This means that the composition

$M_ n \to \Gamma (X, \widetilde{M}(n)) \to \Gamma (X, \widetilde{N}(n)) \to \Gamma (X, \widetilde{M}(n))$

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 $d$. $\square$

Let $A$ be a graded ring such that $A_0$ is Noetherian and $A$ is generated by finitely many elements of $A_1$ over $A_0$. Recall that $A_+ = \bigoplus _{n > 0} A_ n$ is the irrelevant ideal. Let $M$ be a graded $A$-module. Recall that $M$ is an $A_+$-power torsion module if for all $x \in M$ there is an $n \geq 1$ such that $(A_+)^ n x = 0$, see More on Algebra, Definition 15.88.1. If $M$ is finitely generated, then we see that this is equivalent to $M_ n = 0$ for $n \gg 0$. Sometimes $A_+$-power torsion modules are called torsion modules. Sometimes a graded $A$-module $M$ is called torsion free if $x \in M$ with $(A_+)^ n x = 0$, $n > 0$ implies $x = 0$. Denote $\text{Mod}_ A$ the category of graded $A$-modules, $\text{Mod}^{fg}_ A$ the full subcategory of finitely generated ones, and $\text{Mod}^{fg}_{A, torsion}$ the full subcategory of modules $M$ such that $M_ n = 0$ for $n \gg 0$.

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. $\square$

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).