Finite schemes over projective schemes are projective over a Noetherian affine base

Put $F=f_{\ast }{\mathcal{O}}_Y$. By [F1], on each affine $U=\mathrm{Spec}A\subseteq X$ the algebra $B=\mathrm{\Gamma }(f^{-1}(U),{\mathcal{O}}_Y)$ is a finite $A$-module. Since $X$ is projective over the Noetherian ring $R$, [F2] makes each such $A$ Noetherian, so $B$ is finitely presented over $A$ and $F$ is coherent by [F4]. The morphism $f$ is affine, and [F3] identifies $Y$ with ${\mathrm{Spec}}_{X}F$.

Define a graded quasi-coherent ${\mathcal{O}}_X$-algebra $\mathcal{C}$ by ${\mathcal{C}}_0={\mathcal{O}}_X$ and ${\mathcal{C}}_d=F$ for every $d\ge 1$. The degree-zero part acts on $F$ by its algebra structure, and the product of two positive-degree pieces is the multiplication $F\otimes F\to F$ placed in degree the sum of the degrees. Let $z\in \mathrm{\Gamma }(X,{\mathcal{C}}_1)$ be the unit section of $F$. On an affine $U=\mathrm{Spec}A$ where $B=0$, the positive-degree ideal of $\mathcal{C}$ is zero, so ${\mathrm{Proj}}_U\mathcal{C}{\mathrm{\mid }}_U=\mathrm{\varnothing }=\mathrm{Spec}B$. Otherwise let $z$ denote the unit section in degree one. For every affine $U$ and homogeneous $b\in {\mathcal{C}}_d(U)=F(U)$ with $d\ge 1$, the element $b$ in degree $d$ equals $z^{d-1}$ times the same section $b$ in degree one. For a degree-one section $b$, its square in ${\mathcal{C}}_2$ is $z$ times the section $b^2\in F$ placed in degree one. Thus a homogeneous prime containing $z$ contains every degree-one section $b$, and hence every positive-degree element, so is irrelevant; consequently $D_{+}(z)={\mathrm{Proj}}_X\mathcal{C}$. On $U=\mathrm{Spec}A$, multiplication by $z$ identifies the copies of $B$ in successive positive degrees, so ${\mathcal{C}}_{(z)}\cong B$ as an $A$-algebra. These canonical identifications commute with restriction, giving ${\mathrm{Proj}}_X\mathcal{C}\cong {\mathrm{Spec}}_{X}F\cong Y$.

Put $E={\mathcal{O}}_X\oplus F$. The degree-one map $E\to {\mathcal{C}}_1=F$, $(a,b)\mapsto a\cdot 1_F+b$, induces a graded algebra map $\mathrm{Sym}(E)\to \mathcal{C}$. It is surjective: in degree zero it is the identity on ${\mathcal{O}}_X$, and in every positive degree each local section $b\in F={\mathcal{C}}_d$ is the image of $1^{d-1}b$. By [F5], ${\mathrm{Proj}}_X\mathcal{C}$ is closed in ${\mathrm{Proj}}_X\mathrm{Sym}(E)$. Thus $Y$ has a closed immersion into this relative Proj.

Let $L$ be the very ample line bundle from a projective embedding $X\hookrightarrow {\mathbb{P}}_R^N$. By [F7], $E\otimes L^m$ is globally generated for some $m\ge 0$. Since $X$ is quasi-compact, finitely many generating sections give a surjection ${\mathcal{O}}_X^{q+1}\twoheadrightarrow E\otimes L^m$. It induces a graded quotient $\mathrm{Sym}({\mathcal{O}}_X^{q+1})\twoheadrightarrow \mathrm{Sym}(E\otimes L^m)$, and [F5] gives a closed immersion of the latter relative Proj into ${\mathbb{P}}_X^q$. Locally trivializing $L^m$ identifies ${\mathrm{Proj}}_X\mathrm{Sym}(E\otimes L^m)$ with ${\mathrm{Proj}}_X\mathrm{Sym}(E)$; the identifications differ on overlaps by the degree-one unit transition functions and glue. Composing with step 2.1 gives a closed immersion $Y\hookrightarrow {\mathbb{P}}_X^q$.

Base change of $X\hookrightarrow {\mathbb{P}}_R^N$ along ${\mathbb{P}}_R^q\to \mathrm{Spec}R$ gives a closed immersion ${\mathbb{P}}_X^q\hookrightarrow {\mathbb{P}}_R^q{\times }_R{\mathbb{P}}_R^N$, using [F8]. Composing $Y\hookrightarrow {\mathbb{P}}_X^q$ with this closed immersion and the Segre embedding of [F8] realizes $Y$ as a closed subscheme of ${\mathbb{P}}_R^{(q+1)(N+1)-1}$. Thus $Y$ is projective over $R$, proving the first two assertions.

For projective morphisms $Z\to Y\to X$, choose closed immersions $Y\hookrightarrow {\mathbb{P}}_X^m$ and $Z\hookrightarrow {\mathbb{P}}_Y^n$. Base change identifies ${\mathbb{P}}_Y^n$ with ${\mathbb{P}}_X^n{\times }_{X}Y$, which is closed in ${\mathbb{P}}_X^n{\times }_X{\mathbb{P}}_X^m$; the Segre embedding is a closed immersion into ${\mathbb{P}}_X^{(n+1)(m+1)-1}$. Their composite is a projective embedding of $Z$ over $X$. Iterating proves the finite-composition assertion.