The Stacks project

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 $d$. $\square$