Finite morphisms are integral and universally closed

Work on the affine chart $U=\mathrm{Spec}A$ with $f^{-1}(U)=\mathrm{Spec}B$, so that $B$ is a finite $A$-module by [F1]. If $A=0$ then $U$ is empty and there is nothing to check. Let $A^{\mathrm{\prime }}=\mathrm{im}(A\to B)\subseteq B$. If $A^{\mathrm{\prime }}=0$, unitality forces $B=0$, and the map is integral vacuously. Otherwise, a finite $A$-module generating list for $B$ also generates $B$ over $A^{\mathrm{\prime }}$, because every coefficient acts through its image in $A^{\mathrm{\prime }}$. For any $b\in B$, the module $B$ is an $A^{\mathrm{\prime }}[b]$-module, finite over $A^{\mathrm{\prime }}$, and faithful over $A^{\mathrm{\prime }}[b]$: if $r\in A^{\mathrm{\prime }}[b]$ satisfies $rB=0$, then $r=r\cdot 1=0$. The inclusion version of [F2], applied to $A^{\mathrm{\prime }}\subseteq B$, makes $b$ integral over $A^{\mathrm{\prime }}$. Lift the coefficients of that monic relation along the surjection $A\to A^{\mathrm{\prime }}$; the same relation in $B$ proves that $b$ is integral over $A$. Since $b$ was arbitrary, the ring map $A\to B$ is integral by [F3].

Let $J\subseteq B$ be an ideal and let $Z=V(J)\subseteq \mathrm{Spec}B$ be the corresponding closed subset, every closed subset of $\mathrm{Spec}B$ being of this form by [F6]. The composite $A\to B\to B\mathrm{/}J$ is integral: an element of $B\mathrm{/}J$ is the image of some $b\in B$, and reducing a monic relation for $b$ over $A$ modulo $J$ exhibits a monic relation for its image, by [F3]. Put $I=\ker (A\to B\mathrm{/}J)$ and let $\phi :\mathrm{Spec}B\to \mathrm{Spec}A$ be the map induced by $A\to B$. The image $\phi (Z)$ equals $V(I)$: if $\mathfrak{q}\supseteq J$ then $\mathfrak{q}\cap A\supseteq I$; conversely, if $\mathfrak{p}\supseteq I$ then $I=\ker (A\to B\mathrm{/}J)\subseteq \mathfrak{p}$, so [F4] applied to the integral map $A\to B\mathrm{/}J$ produces a prime of $B\mathrm{/}J$ contracting to $\mathfrak{p}$, that is a prime $\mathfrak{q}\supseteq J$ of $B$ with $\mathfrak{q}\cap A=\mathfrak{p}$. By [F6] the set $V(I)$ is closed in $\mathrm{Spec}A$. Thus every affine chart of a finite morphism maps closed subsets onto closed subsets.

Consequently every finite morphism $g:Y\to S$ is closed. Let $Z\subseteq Y$ be closed and let $S={\bigcup }_i{U}_i$ be an affine open cover with $U_i=\mathrm{Spec}{A}_i$ and $g^{-1}(U_i)=\mathrm{Spec}{B}_i$ affine. For each $i$ the trace $Z\cap g^{-1}(U_i)$ is closed in the open subspace $g^{-1}(U_i)$, and $g(Z)\cap U_i=g(Z\cap g^{-1}(U_i))$ is closed in $U_i$ by step 2.1. A subset of $S$ whose trace in every member of an open cover is closed in that member is closed, since its complement has open trace in every $U_i$. Hence $g(Z)$ is closed in $S$.

Now let $T\to S$ be arbitrary. By [F5] the base change $f_T:X{\times }_{S}T\to T$ is again finite, so step 3.1 applied to the finite morphism $f_T$ shows that $f_T$ is a closed map. Since $T\to S$ was arbitrary, $f$ is universally closed.

Step 1.1 shows that the induced ring maps on all affine charts are integral, and step 4.1 shows that $f$ is universally closed, in particular that the image of $X{\times }_{S}T$ and the image of any closed subset are closed in $\mathrm{\mid }T\mathrm{\mid }$. The Axiom of Choice is used exactly through [F4], which supplies primes over primes for the integral map $A\to B\mathrm{/}J$, and [F5], which provides the base-changed finite morphism; no other selection occurs. If $B=0$ the chart is empty, the closed subset $Z$ is empty and its image is empty and closed, so the same argument applies through the zero ring case of step 1.1.