Projective-space projection is universally closed by finite graded pieces

Universal closedness of $\pi$ is checked after base change and locally on the target: given $T\to S$ and a closed $Z\subseteq {\mathbb{P}}_T^n$, it suffices to test closedness of the image of $Z$ over an affine open cover of $T$; since ${\mathbb{P}}_S^n{\times }_{S}T={\mathbb{P}}_T^n$ by [F1] and restriction to an affine open $T^{\mathrm{\prime }}\subseteq T$ gives ${\mathbb{P}}_{T^{\mathrm{\prime }}}^n$, we may assume throughout that $T=\mathrm{Spec}B$ is affine and that $Z\subseteq {\mathbb{P}}_B^n=\mathrm{Proj}B[x_0,\dots ,x_n]$ is closed, the identification being [F2].

For a field $k$ and a homogeneous ideal $J\subseteq k[x_0,\dots ,x_n]$ put $R=k[x]\mathrm{/}J$: then $V_{+}(J)=\mathrm{\varnothing }$ if and only if $R_m=0$ for some $m\ge 1$. Indeed, $R$ is generated in degree one by the images of the $x_i$, so $R_{d+1}=R_1{R}_d$ for every $d\ge 1$. If $R_m=0$, induction gives $R_d=0$ for all $d\ge m$. Every product of $m$ positive-degree homogeneous elements has degree at least $m$, whence $R_{+}^m=0$; thus every homogeneous prime contains $R_{+}$ and $\mathrm{Proj}R=\mathrm{\varnothing }$. Conversely, if some $x_i$ is not nilpotent in $R$, then a maximal homogeneous ideal $\mathfrak{q}$ of the graded ring $R$ avoiding all powers of $x_i$ (Zorn, AC) is prime, by the minimal-homogeneous-component argument: if $ab\in \mathfrak{q}$ with $a,b\mathrm{\notin }\mathfrak{q}$ and $a_i,b_j$ are homogeneous components of least degree not in $\mathfrak{q}$, all other components of degree $i+j$ of $ab$ lie in $\mathfrak{q}$, so $a_i{b}_j\in \mathfrak{q}$; then $\mathfrak{q}+(a_i)$ and $\mathfrak{q}+(b_j)$ are homogeneous ideals strictly containing $\mathfrak{q}$, so each meets the powers of $x_i$, and the product of such powers lies in $\mathfrak{q}+(a_i{b}_j)\subseteq \mathfrak{q}$, a contradiction; the contraction of $\mathfrak{q}$ is then a homogeneous prime of $k[x]$ containing $J$ but not $x_i$, a point of $V_{+}(J)$, contradiction. Hence every $x_i$ is nilpotent, say $x_i^{m_i}=0$ in $R$; with $m\ge (n+1){\max }_i{m}_i$, every monomial of degree $m$ has some exponent $\ge m_i$ and hence vanishes in $R$, so $R_m=0$.

By [F3] there is a homogeneous ideal $I\subseteq B[x_0,\dots ,x_n]$ with $Z=V_{+}(I)$; for a prime $\mathfrak{p}\subseteq B$ the fibre of $Z$ over $\mathfrak{p}$ is $V_{+}(I\kappa (\mathfrak{p}))\subseteq \mathrm{Proj}\kappa (\mathfrak{p})[x_0,\dots ,x_n]$, where $I\kappa (\mathfrak{p})$ is the image of $I$ under $B[x]\to \kappa (\mathfrak{p})[x]$, because base change of $V_{+}(I)$ along $\mathrm{Spec}\kappa (\mathfrak{p})\to \mathrm{Spec}B$ is $V_{+}$ of the extended ideal by the naturality in [F2] and the definition of $V_{+}$ in [F3].

Put $M_m:=(B[x]\mathrm{/}I{)}_m$ for $m\ge 0$; each $M_m$ is a finitely generated $B$-module by [F8], and $(\kappa (\mathfrak{p})[x]\mathrm{/}I\kappa (\mathfrak{p}){)}_m\cong M_m{\otimes }_B\kappa (\mathfrak{p})$ by [F7]. Combining with step 1.2, the fibre $Z_{\mathfrak{p}}$ is empty if and only if $M_m{\otimes }_B\kappa (\mathfrak{p})=0$ for some $m\ge 1$.

For a fixed $m\ge 1$ and a prime $\mathfrak{p}$, the vanishing $M_m{\otimes }_B\kappa (\mathfrak{p})=0$ is equivalent to $\mathfrak{p}(M_m{)}_{\mathfrak{p}}=(M_m{)}_{\mathfrak{p}}$; since $(M_m{)}_{\mathfrak{p}}$ is finitely generated over the local ring $B_{\mathfrak{p}}$ with maximal ideal $\mathfrak{p}{B}_{\mathfrak{p}}$ and $\mathfrak{p}{B}_{\mathfrak{p}}$ is the Jacobson radical of $B_{\mathfrak{p}}$, Nakayama [F5] gives $(M_m{)}_{\mathfrak{p}}=0$; then [F6] provides $b\mathrm{\notin }\mathfrak{p}$ with $(M_m{)}_b=0$, and for every prime $\mathfrak{q}\in D(b)$ we get $(M_m{)}_{\mathfrak{q}}=0$ and hence an empty fibre over $\mathfrak{q}$ by step 3.1.

Conversely if $(M_m{)}_b=0$ for some $m,b$ then for every $\mathfrak{q}\in D(b)$ one has $M_m{\otimes }_B\kappa (\mathfrak{q})=0$, so the fibre is empty over $D(b)$; therefore the set $U=\{\mathfrak{p}\in \mathrm{Spec}B:Z_{\mathfrak{p}}=\mathrm{\varnothing }\}$ equals ${\bigcup }_{m\ge 1}\bigcup \{D(b):(M_m{)}_b=0\}$, a union of basic open sets, hence is open in $\mathrm{Spec}B$.

The complement of $U$ in $\mathrm{Spec}B$ is exactly the image of the closed set $Z$ under ${\pi }_B$, so step 5.1 shows that this image is closed; since the argument applies to every affine base and, by step 1.1, to every base $T\to S$ after restriction to an affine cover, the projection $\pi$ is universally closed. The cases $n=0$ (where $\pi$ is an isomorphism $S\to S$) and $S=\mathrm{\varnothing }$ (where the source is empty and the condition is vacuous) are included in this argument through [F1]; the Axiom of Choice is used exactly in the Zorn argument of step 1.2 and through Nakayama [F5] and the finite-generation step [F6].