**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

\[ \bigoplus \nolimits _{dn + q \geq k} H^ p(X, \mathcal{F}(dn + q)) = \bigoplus \nolimits _{dn + q \geq k} H^ p(X', (i_*\mathcal{F}(q))(n) \]

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.
$\square$

## Comments (2)

Comment #5054 by Zhipu Zhao on

Comment #5273 by Johan on