Because $\mathcal A_\bullet$ is coherent and locally generated by $\mathcal A_1$, there is locally a finite-rank coherent sheaf $\mathcal E$ and a surjection of graded algebras
$$
\operatorname{Sym}_{\mathcal O_X}(\mathcal E)\twoheadrightarrow\mathcal A_\bullet.
$$
The <Relative Proj construction> turns this into a <closed immersion>
$$
Y=\operatorname{Proj}_X\mathcal A_\bullet\hookrightarrow\mathbb P_X(\mathcal E).
$$
Consequently $Y\to X$ is a <projective morphism> and therefore a <proper morphism>. Since $X\to\operatorname{Spec}k$ is proper and proper morphisms are closed under composition, $Y\to\operatorname{Spec}k$ is proper.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 113 / 1 / c / Solution

{c}
{title2=$\operatorname{Proj}_X\mathcal A$}
{wiki=Proj_construction#Relative_Proj}

For a quasi-coherent graded $\mathcal O_X$-algebra $\mathcal A$, the relative Proj is obtained by gluing the schemes $\operatorname{Proj}\mathcal A(U)$ over affine open subsets $U\subseteq X$. If $\mathcal A$ is of finite type and generated in degree one, a surjection $\operatorname{Sym}(\mathcal E)\twoheadrightarrow\mathcal A$ realizes $\operatorname{Proj}_X\mathcal A$ as a closed subscheme of the projective bundle $\mathbb P_X(\mathcal E)$.


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 Incoming links: Relative Proj construction