The Stacks project

Proof. We will prove this by reducing the statement to Lemma 30.14.2. 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 $d$ be a sufficiently divisible integer. Set $A' = A^{(d)}$ with notation as in Algebra, Section 10.56. Then $A'$ is generated over $A'_0 = A_0$ by elements of degree $1$, see Algebra, Lemma 10.56.2. Thus Lemma 30.14.2 applies to $X' = \text{Proj}(A')$.

By Constructions, Lemma 27.11.8 there exist an isomorphism of schemes $i : X \to X'$ and isomorphisms $\mathcal{O}_ X(nd) \to i^*\mathcal{O}_{X'}(n)$ compatible with the map $A' \to A$ and the maps $A_ n \to H^0(X, \mathcal{O}_ X(n))$ and $A'_ n \to H^0(X', \mathcal{O}_{X'}(n))$. Thus Lemma 30.14.2 implies $X$ is Noetherian and that (1) and (2) hold. To see (3) and (4) we can use that for any fixed $k$, $p$, and $q$ we have

by the compatibilities above. If $p > 0$, we have the vanishing of the right hand side for $k$ depending on $q$ large enough by Lemma 30.14.2. Since there are only a finite number of congruence classes of integers modulo $d$, we see that (3) holds for $\mathcal{F}$ on $X$. If $p = 0$, then we have that the right hand side is a finite $A'$-module by Lemma 30.14.2. Using the finiteness of congruence classes once more, we find that $\bigoplus _{n \geq k} H^0(X, \mathcal{F}(n))$ is a finite $A'$-module too. Since the $A'$-module structure comes from the $A$-module structure (by the compatibilities mentioned above), we conclude it is finite as an $A$-module as well. $\square$

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. $\square$

Proof. We urge the reader to read the proof in the case where $A$ is generated in degree $1$ first, see Proposition 30.14.4. Let $f_1, \ldots , f_ r \in A$ be homogeneous elements of positive degree which generate $A$ over $A_0$. Let $d$ be the lcm of the degrees $d_ i$ of $f_ i$. Let $M$ be a finite $A$-module. Let us show that $\widetilde{M}$ is zero if and only if $M$ is an irrelevant graded $A$-module (as defined above the statement of the proposition). Namely, let $x \in M$ be a homogeneous element. Choose $k \in \mathbf{Z}$ sufficiently small and let $N \to N'$ and $M \to N'$ be as in Lemma 30.15.2. We may also pick $l$ sufficiently large such that $M_ n \to N_ n$ is an isomorphism for $n \geq l$. If $\widetilde{M}$ is zero, then $N = 0$. Thus for any $f \in A_+$ homogeneous with $\deg (f) + \deg (x) = nd$ and $nd > l$ we see that $fx$ is zero because $N_{nd} \to N'_{nd}$ and $M_{nd} \to N'_{nd}$ are isomorphisms. Hence $x$ is irrelevant. Conversely, assume $M$ is irrelevant. Then $M_{nd}$ is zero for $n \gg 0$ (see discussion above proposition). Clearly this implies that $M_{(f_ i)} = M_{(f_ i^{d/\deg (f_ i)})} = 0$, whence $\widetilde{M} = 0$ by construction.

It follows that the subcategory $\text{Mod}^{fg}_{A, irrelevant}$ is a Serre subcategory of $\text{Mod}^{fg}_ A$ as the kernel of the exact functor $M \mapsto \widetilde M$, see Homology, Lemma 12.10.4 and Constructions, Lemma 27.8.4. 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 irrelevant modules to $0$ which we have shown above.

By Lemma 30.15.1 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 $\text{Mod}^{fg}_ A \to \text{Mod}^{fg}_ A/\text{Mod}^{fg}_{A, irrelevant}$.

By Lemma 30.15.2 the composite left to right to left is isomorphic to the identity functor. Namely, let $M$ be a finite graded $A$-module and let $k \in \mathbf{Z}$ sufficiently small and let $N \to N'$ and $M \to N'$ be as in Lemma 30.15.2. Then the kernel and cokernel of $M \to N'$ are nonzero in only finitely many degrees, hence are irrelevant. Moreover, the kernel and cokernel of the map $N \to N'$ are zero in all sufficiently large degrees divisible by $d$, hence these are irrelevant modules too. Thus $M \to N'$ and $N \to N'$ are both isomorphisms in the quotient category, as desired.

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. $\square$