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. $\square$