The Stacks project

Proof. Assume that $\mathcal{A}_0$ is a finite type $\mathcal{O}_ S$-module. Choose an affine open $U = \mathop{\mathrm{Spec}}(R) \subset X$ such that $\mathcal{A}$ corresponds to a graded $R$-algebra $A$ with $A_0$ a finite $R$-module. Condition (1) means that (after possibly localizing further on $S$) that $A$ is a finite type $A_0$-algebra and condition (2) means that (after possibly localizing further on $S$) that $A$ is a finite type $R$-algebra. Thus these conditions imply each other by Algebra, Lemma 10.6.2.

A locally projective morphism is proper, see Morphisms, Lemma 29.43.5. Thus we may now assume that $S = \mathop{\mathrm{Spec}}(R)$ and $X = \text{Proj}(A)$ and that $A_0$ is finite over $R$ and $A$ of finite type over $R$. We will show that $X = \text{Proj}(A) \to \mathop{\mathrm{Spec}}(R)$ is projective. We urge the reader to prove this for themselves, by directly constructing a closed immersion of $X$ into a projective space over $R$, instead of reading the argument we give below.

By Lemma 31.30.2 we see that $X$ is of finite type over $\mathop{\mathrm{Spec}}(R)$. Constructions, Lemma 27.10.6 tells us that $\mathcal{O}_ X(d)$ is ample on $X$ for some $d \geq 1$ (see Properties, Section 28.26). Hence $X \to \mathop{\mathrm{Spec}}(R)$ is quasi-projective (by Morphisms, Definition 29.40.1). By Morphisms, Lemma 29.43.12 we conclude that $X$ is isomorphic to an open subscheme of a scheme projective over $\mathop{\mathrm{Spec}}(R)$. Therefore, to finish the proof, it suffices to show that $X \to \mathop{\mathrm{Spec}}(R)$ is universally closed (use Morphisms, Lemma 29.41.7). This follows from Lemma 31.30.3. $\square$