Fpqc descent of properness components

Assume $f^{\mathrm{\prime }}$ is quasi-compact, and fix an affine open $U=\mathrm{Spec}A\subseteq S$. By [F4], $U$ is quasi-compact, so $p^{-1}(U)$ is a quasi-compact open of $S^{\mathrm{\prime }}$. Its inverse image under $f^{\mathrm{\prime }}$ is quasi-compact, and its projection onto $f^{-1}(U)$ is surjective because it is the base change of $p_U:p^{-1}(U)\to U$ along $f^{-1}(U)\to U$.

Suppose $f^{\mathrm{\prime }}$ is universally closed. For any $T\to S$, put $T^{\mathrm{\prime }}=T{\times }_S{S}^{\mathrm{\prime }}$ and let $q:T^{\mathrm{\prime }}\to T$ be the projection. The base change $X_{T^{\mathrm{\prime }}}\to X_T$ of $q$ is surjective by [F2]. For a closed subset $Z\subseteq X_T$, its inverse image $Z^{\mathrm{\prime }}\subseteq X_{T^{\mathrm{\prime }}}$ is closed by continuity [F15]. By [F25], $X_{T^{\mathrm{\prime }}}\to T^{\mathrm{\prime }}$ is a base change of $f^{\mathrm{\prime }}$, so [F16] makes the image of $Z^{\mathrm{\prime }}$ closed. The Cartesian square gives $f_{T^{\mathrm{\prime }}}(Z^{\mathrm{\prime }})\subseteq q^{-1}(f_T(Z))$. Conversely, if $t^{\mathrm{\prime }}\in q^{-1}(f_T(Z))$, choose $x\in Z$ above $q(t^{\mathrm{\prime }})$ and write $k=\kappa (q(t^{\mathrm{\prime }}))$. By AC extend $\{1\}\subset \kappa (x)$ to a $k$-basis and take the coordinate functional $\lambda :\kappa (x)\to k$ with $\lambda (1)=1$. Then $\lambda \otimes 1$ sends $1\otimes 1$ to $1$, so $\kappa (x){\otimes }_k\kappa (t^{\mathrm{\prime }})$ is nonzero. By [F22] it has a prime; [F20] gives a point of $X_{T^{\mathrm{\prime }}}$ over $(x,t^{\mathrm{\prime }})$. Thus $t^{\mathrm{\prime }}\in f_{T^{\mathrm{\prime }}}(Z^{\mathrm{\prime }})$, proving equality.

Assume $f^{\mathrm{\prime }}$ is of finite type. To descend this property, take a nonempty affine open $U=\mathrm{Spec}A\subseteq S$ and an affine open $V=\mathrm{Spec}C\subseteq f^{-1}(U)$. Since $p$ is quasi-compact, [F1, F3, F4] show $p^{-1}(U)$ is quasi-compact; choose a finite affine-open cover $U_i=\mathrm{Spec}{B}_i$. Since $p$ is surjective and $U$ is nonempty, discard empty members and the cover remains nonempty. For each $i$, $V_i=V{\times }_U{U}_i$ is affine with coordinate algebra $C{\otimes }_A{B}_i$, and its map to $U_i$ is finite type by [F6], because it is an affine-open restriction of the finite-type morphism $f^{\mathrm{\prime }}$.

Put $Y=X{\times }_{S}X$. Take the family of all pairs of affine opens $U_i=\mathrm{Spec}{B}_i\subseteq X$ and $V_i=\mathrm{Spec}{A}_i\subseteq S$ with $f(U_i)\subseteq V_i$; its source opens cover $X$ by [F4]. Let $Q_i={\mathrm{pr}}_1^{-1}(U_i)\cap {\mathrm{pr}}_2^{-1}(U_i)$. By [F9], $Q_i$ is open in $Y$ and represents $U_i{\times }_{V_i}{U}_i$, which [F8] identifies with $\mathrm{Spec}(B_i{\otimes }_{A_i}{B}_i)$. Set $Q={\bigcup }_i{Q}_i$. By [F26], ${\mathrm{\Delta }}_{X\mathrm{/}S}(X)\subseteq Q$ and ${\mathrm{\Delta }}_{X\mathrm{/}S}^{-1}(Q_i)=U_i$. On these affine charts the restricted diagonal corresponds to multiplication ${\mu }_i:B_i{\otimes }_{A_i}{B}_i\to B_i$, which is surjective because $b={\mu }_i(b\otimes 1)$. If $I_i=\ker {\mu }_i$, the explicit isomorphism $(B_i{\otimes }_{A_i}{B}_i)\mathrm{/}{I}_i\to B_i$, $r+I_i\mapsto {\mu }_i(r)$, identifies this chart map with a quotient map; [F17] and [F27] make each restriction a closed immersion. Since the $Q_i$ cover $Q$, [F28] makes ${\mathrm{\Delta }}_{X\mathrm{/}S}:X\to Q$ a closed immersion. The inclusion $Q\hookrightarrow Y$ is open, so [F21] now gives that ${\mathrm{\Delta }}_{X\mathrm{/}S}:X\to Y$ is an immersion.

A continuous image of a quasi-compact space is quasi-compact, so step 1.1 shows that $f^{-1}(U)$ is quasi-compact for every affine $U$. The affine-open criterion in [F3] proves that $f$ is quasi-compact. Conversely, if $f$ is quasi-compact, [F24] gives that $f^{\mathrm{\prime }}$ is quasi-compact.

By [F2], closedness of $q^{-1}(f_T(Z))$ implies that $f_T(Z)$ is closed. Since this holds for every $T\to S$ and every closed $Z\subseteq X_T$, [F16] proves that $f$ is universally closed. Conversely, universal closedness of $f$ implies that of $f^{\mathrm{\prime }}$ directly from [F16] and [F25].

Each $A\to B_i$ is flat by [F2]. Put $B={\prod }_i{B}_i$. The product-spectrum description [F12] and the cover of $p^{-1}(U)$ show that $\mathrm{Spec}B\to \mathrm{Spec}A$ is surjective; [F13] shows that $B$ is flat over $A$. Thus [F10] makes $A\to B$ faithfully flat.

Suppose $f^{\mathrm{\prime }}$ is separated, and write $Y=X{\times }_{S}X$. By [F19], its diagonal is the base change of ${\mathrm{\Delta }}_{X\mathrm{/}S}:X\to Y$ along the fpqc map $Y{\times }_S{S}^{\mathrm{\prime }}\to Y$, which is fpqc by [F1, F2]. The diagonal of $f^{\mathrm{\prime }}$ is a closed immersion by [F18], hence universally closed by [F16] and [F17]. The descent proved in steps 1.2 and 2.2, applied to ${\mathrm{\Delta }}_{X\mathrm{/}S}$ and this fpqc base change, shows that ${\mathrm{\Delta }}_{X\mathrm{/}S}$ is universally closed.

Choose finite $B_i$-algebra generators for each $C{\otimes }_A{B}_i$, and express each as a finite sum of pure tensors. Let $c_1,\dots ,c_r\in C$ be the finite list of all coefficients from $C$ appearing in those sums, and put $C_0=A[c_1,\dots ,c_r]\subseteq C$. The maps $B_i{\otimes }_A{C}_0\to B_i{\otimes }_{A}C$ are surjective because their images contain the chosen algebra generators. Since $B={\prod }_i{B}_i$ is a finite direct sum as an $A$-module, [F13] gives a surjection $B{\otimes }_A{C}_0\to B{\otimes }_{A}C$. By [F14], $B{\otimes }_A(C\mathrm{/}{C}_0)=0$; [F11] and faithful flatness of $B$ then imply $C\mathrm{/}{C}_0=0$. Hence $C$ is a finite-type $A$-algebra.

By [F16] and step 3.1, the image of ${\mathrm{\Delta }}_{X\mathrm{/}S}$ is closed: apply universal closedness to the closed whole source. Step 1.4 shows that this diagonal is an immersion. Therefore [F23] makes it a closed immersion, and [F18] gives that $f$ is separated. Conversely, if $f$ is separated, [F18] makes its diagonal a closed immersion; [F19] identifies the diagonal of $f^{\mathrm{\prime }}$ as a base change of it, so [F17] makes that diagonal a closed immersion and [F18] gives that $f^{\mathrm{\prime }}$ is separated.

Since $U$ and $V$ were arbitrary nonempty affine opens, [F5] gives that $f$ is locally of finite type. Step 2.1 gives quasi-compactness, hence $f$ is of finite type. Conversely, if $f$ is of finite type, [F24] gives that $f^{\mathrm{\prime }}$ is of finite type.

If $S$ or $X$ is empty, the assertions reduce to empty maps and hold directly; empty affine charts are omitted in the finite cover, and the zero algebra is generated by the empty list. If $p$ is the identity, each assertion is tautological. The finite generator lists in step 3.2 require only finite choice. AC is used exactly through [F2] for flat-chart testing, surjectivity and submersiveness, [F10] for the faithfully-flat criterion, [A1] for the linear functional, [F22] for prime existence in the residue-field tensor step, and [F17] for closed-immersion quotient and base change. The module-detection implication [F11] is used only after faithful flatness and requires no additional choice. The family of affine pairs in step 1.4 is the full family and uses no simultaneous pointwise choice. There are no endpoint parameters. The descent directions are proved for quasi-compactness in steps 1.1 and 2.1, finite type in steps 1.3, 2.3, 3.2 and 4.2, universal closedness in steps 1.2 and 2.2, and separatedness in steps 3.1, 1.4, and 4.1. The corresponding ascent directions are proved in steps 2.1, 4.2, 2.2 and 4.1. [A1, F2, F10, F11, F16, F17, F20, F22, step 1.1, step 1.2, step 1.3, step 1.4, step 2.1, step 2.2, step 2.3, step 3.1, step 3.2, step 4.1, step 4.2] $square$