The Stacks project

Proof. The map $N \to N'$ in (3) comes from Constructions, Equation (27.10.1.5) by taking global sections.

By Constructions, Equation (27.10.1.6) there is a map of graded $A$-modules $M \to \bigoplus _{n \in \mathbf{Z}} H^0(X, \widetilde{M(n)})$. If the generators of $M$ sit in degrees $\geq k$, then the image is contained in the submodule $N' \subset \bigoplus _{n \in \mathbf{Z}} H^0(X, \widetilde{M(n)})$ and we get the map in (4).

By Algebra, Lemmas 10.58.2 and 10.58.1 the ring $A_0$ is Noetherian and $A$ is generated over $A_0$ by finitely many elements $f_1, \ldots , f_ r$ homogeneous of positive degree. Let $d = \text{lcm}(\deg (f_ i))$. Then we see that (6) holds for example by Constructions, Lemma 27.10.4.

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. Thus part (2) follows from Lemma 30.15.1.

We will deduce (1) from (2) using a trick. For $q \in \{ 0, \ldots , d - 1\} $ write

By part (2) these are finite $A$-modules. The Noetherian ring $A$ is finite over $A^{(d)} = \bigoplus _{n \geq 0} A_{dn}$, because it is generated by $f_ i$ over $A^{(d)}$ and $f_ i^ d \in A^{(d)}$. Hence ${}^ qN$ is a finite $A^{(d)}$-module. Moreover, $A^{(d)}$ is Noetherian (follows from Algebra, Lemma 10.57.9). It follows that the $A^{(d)}$-submodule ${}^ qN^{(d)} = \bigoplus _{n \in \mathbf{Z}} {}^ qN_{dn}$ is a finite module over $A^{(d)}$. Using the isomorphisms $\widetilde{M(dn + q)} = \widetilde{M(q)}(dn)$ we can write

Thus $N'$ is finite over $A^{(d)}$ and a fortiori finite over $A$. Thus (1) is true.

Let $K$ be a finite $A$-module such that $\widetilde{K} = 0$. We claim that $K_ n = 0$ for $d|n$ and $n \gg 0$. Arguing as above we see that $K^{(d)}$ is a finite $A^{(d)}$-module. Let $x_1, \ldots , x_ m \in K$ be homogeneous generators of $K^{(d)}$ over $A^{(d)}$, say sitting in degrees $d_1, \ldots , d_ m$ with $d | d_ j$. 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 $x_ j/f_ i^{d_ i/\deg (f_ i)} \in K_{(f_ i)}$ would not be zero, i.e., $\widetilde{K}$ would not be zero. Here we use that $\deg (f_ i) | d | d_ j$ for all $i, j$. We conclude that $K_ n$ is zero for $n$ with $d | n$ and $n > \max _ j (d_ j + \sum _ i n_{ij} \deg (f_ i))$ as every element of $K_ n$ is a sum of terms where each term is a monomials in the $f_ i$ times one of the $x_ j$ of total degree $n$.

To finish the proof, we have to show that $M \to N'$ is an isomorphism in all sufficiently large degrees. The map $N \to N'$ induces an isomorphism $\widetilde{N} \to \widetilde{N'}$ because on the affine opens $D_+(f_ i) = D_+(f_ i^ d)$ the corresponding modules are isomorphic: $N_{(f_ i)} \cong N_{(f_ i^ d)} \cong N'_{(f_ i^ d)} \cong N'_{(f_ i)}$ by property (6). By Properties, Lemma 28.28.5 we have a canonical isomorphism $\widetilde{N} \to \widetilde{M}$. The composition $\widetilde{N} \to \widetilde{M} \to \widetilde{N'}$ is the isomorphism above (proof omitted; hint: look on standard affine opens to check this). Thus the map $M \to N'$ induces an isomorphism $\widetilde{M} \to \widetilde{N'}$. 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 $\widetilde{Q} = 0$. By the result of the previous paragraph we see that $K_ n = 0$ and $Q_ n = 0$ for $d | n$ and $n \gg 0$. At this point we finally see the advantage of using $N'$ over $N$: the functor $M \leadsto N'$ is compatible with shifts (immediate from the construction). Thus, repeating the whole argument with $M$ replaced by $M(q)$ we find that $K_ n = 0$ and $Q_ n = 0$ for $n \equiv q \bmod d$ and $n \gg 0$. Since there are only finitely many congruence classes modulo $n$ the proof is finished. $\square$