Flat finite-presentation morphisms are open

We first prove the affine claim: if $A\to B$ is a flat homomorphism of finite presentation, then $\mathrm{Spec}B\to \mathrm{Spec}A$ is open. Fix $b\in B$ and let $E_b\subseteq \mathrm{Spec}A$ be the image of the distinguished open $D(b)\subseteq \mathrm{Spec}B$ under the map of spectra.

Now let $f:X\to S$ be flat and locally of finite presentation. Cover $X$ by affine opens $U_i=\mathrm{Spec}{B}_i$ such that $f(U_i)\subseteq V_i$ for affine opens $V_i=\mathrm{Spec}{A}_i\subseteq S$ with $A_i\to B_i$ of finite presentation, as [F2] allows; by [F4] and flatness of $f$ each $A_i\to B_i$ is flat.

The homomorphism $A\to B_b$ is flat: $A\to B$ is flat by hypothesis, the localisation $B\to B_b$ is flat by [F8], and a composite of flat ring homomorphisms is flat by [F8].

By [F5] the set $E_b$ is constructible in $\mathrm{Spec}A$.

The set $E_b$ is stable under generalisation. Let $\mathfrak{p}\in E_b$ and let ${\mathfrak{p}}^{\mathrm{\prime }}\subseteq \mathfrak{p}$ be a prime of $A$. Choose $\mathfrak{q}\in D(b)$ with $\mathfrak{q}\cap A=\mathfrak{p}$. Since $b\mathrm{\notin }\mathfrak{q}$, the prime $\mathfrak{q}$ determines a prime ${\mathfrak{q}}_2\in \mathrm{Spec}(B_b)$ contracting to $\mathfrak{p}$, and [F7] applied to the flat homomorphism $A\to B_b$ of step 2.1 produces ${\mathfrak{q}}_1\subseteq {\mathfrak{q}}_2$ with ${\mathfrak{q}}_1\cap A={\mathfrak{p}}^{\mathrm{\prime }}$; as $b\mathrm{\notin }{\mathfrak{q}}_1$ the prime ${\mathfrak{q}}_1$ defines a point of $D(b)$ lying over ${\mathfrak{p}}^{\mathrm{\prime }}$, so ${\mathfrak{p}}^{\mathrm{\prime }}\in E_b$.

By step 2.2 the set $E_b$ is constructible and by step 3.1 it is stable under generalisation, so [F6] shows that $E_b$ is open in $\mathrm{Spec}A$. Since $b\in B$ was arbitrary, the image of every distinguished open of $\mathrm{Spec}B$ is open.

Every open subset of $\mathrm{Spec}B$ is a union of distinguished opens by [F11], and the image of a union is the union of the images, so by step 4.1 the image of every open subset of $\mathrm{Spec}B$ is open in $\mathrm{Spec}A$. This proves the affine claim that $\mathrm{Spec}B\to \mathrm{Spec}A$ is open.

For each $i$ the affine claim of step 5.1 makes the restriction $f{\mathrm{\mid }}_{U_i}:U_i\to V_i$ open, hence also the map $U_i\to S$ is open in the sense of [F3], since $V_i$ is open in $S$.

Consequently $f$ is open: for an open $W\subseteq X$ one has $W={\bigcup }_i(W\cap U_i)$, so $f(W)={\bigcup }_{i}f(W\cap U_i)$, and each $f(W\cap U_i)$ is open in $S$ by step 6.1 applied to the open subset $W\cap U_i\subseteq U_i$, which is exactly the openness required of $f$ by [F3].

Finally let $T\to S$ be an arbitrary morphism and form the base change $f_T:X{\times }_{S}T\to T$ of [F10]. By [F9] the morphism $f_T$ is again flat and locally of finite presentation, so step 7.1 applies to $f_T$ and shows that $f_T$ is open. As $T\to S$ was arbitrary, $f$ is universally open by the definition recorded in [F3]. The Axiom of Choice [F12] is used exactly through [F5], [F6] and [F7], each invoked above. [F3, F9, F10, F12, step 7.1] $\mathrm{\square }$