Lemma 31.30.3. Let $S$ be a scheme. Let $\mathcal{A}$ be a quasi-coherent graded $\mathcal{O}_ S$-algebra. Let $p : X = \underline{\text{Proj}}_ S(\mathcal{A}) \to S$ be the relative Proj of $\mathcal{A}$. If $\mathcal{O}_ S \to \mathcal{A}_0$ is an integral algebra map1 and $\mathcal{A}$ is of finite type as an $\mathcal{A}_0$-algebra, then $p$ is universally closed.

Proof. The question is local on the base. Thus we may assume that $X = \mathop{\mathrm{Spec}}(R)$ is affine. Let $\mathcal{A}$ be the quasi-coherent $\mathcal{O}_ X$-algebra associated to the graded $R$-algebra $A$. The assumption is that $R \to A_0$ is integral and $A$ is of finite type over $A_0$. Write $X \to \mathop{\mathrm{Spec}}(R)$ as the composition $X \to \mathop{\mathrm{Spec}}(A_0) \to \mathop{\mathrm{Spec}}(R)$. Since $R \to A_0$ is an integral ring map, we see that $\mathop{\mathrm{Spec}}(A_0) \to \mathop{\mathrm{Spec}}(R)$ is universally closed, see Morphisms, Lemma 29.44.7. The quasi-compact (see Constructions, Lemma 27.8.9) morphism

$X = \text{Proj}(A) \to \mathop{\mathrm{Spec}}(A_0)$

satisfies the existence part of the valuative criterion by Constructions, Lemma 27.8.11 and hence it is universally closed by Schemes, Proposition 26.20.6. Thus $X \to \mathop{\mathrm{Spec}}(R)$ is universally closed as a composition of universally closed morphisms. $\square$

[1] In other words, the integral closure of $\mathcal{O}_ S$ in $\mathcal{A}_0$, see Morphisms, Definition 29.53.2, equals $\mathcal{A}_0$.

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