Lemma 37.68.1. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

1. $f$ is universally open,

2. for every morphism $S' \to S$ which is locally of finite presentation the base change $X_{S'} \to S'$ is open, and

3. for every $n$ the morphism $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is open.

Proof. It is clear that (1) implies (2) and (2) implies (3). Let us prove that (3) implies (1). Suppose that the base change $X_ T \to T$ is not open for some morphism of schemes $g : T \to S$. Then we can find some affine opens $V \subset S$, $U \subset X$, $W \subset T$ with $f(U) \subset V$ and $g(W) \subset V$ such that $U \times _ V W \to W$ is not open. If we can show that this implies $\mathbf{A}^ n \times U \to \mathbf{A}^ n \times V$ is not open, then $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is not open and the proof is complete. This reduces us to the result proved in the next paragraph.

Let $A \to B$ be a ring map such that $A' \to B' = A' \otimes _ A B$ does not induce an open map of spectra for some $A$-algebra $A'$. As the principal opens give a basis for the topology of $\mathop{\mathrm{Spec}}(B')$ we conclude that the image of $D(g)$ in $\mathop{\mathrm{Spec}}(A')$ is not open for some $g \in B'$. Write $g = \sum _{i = 1, \ldots , n} a'_ i \otimes b_ i$ for some $n$, $a'_ i \in A'$, and $b_ i \in B$. Consider the element $h = \sum _{i = 1, \ldots , n} x_ i b_ i$ in $B[x_1, \ldots , x_ n]$. Assume that $D(h)$ maps to an open subset under the morphism

$\mathop{\mathrm{Spec}}(B[x_1, \ldots , x_ n]) \longrightarrow \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n])$

in order to get a contradiction. Then $D(h)$ would map surjectively onto a quasi-compact open $U \subset \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n])$. Let $A[x_1, \ldots , x_ n] \to A'$ be the $A$-algebra homomorphism sending $x_ i$ to $a'_ i$. This also induces a $B$-algebra homomorphism $B[x_1, \ldots , x_ n] \to B'$ sending $h$ to $g$. Since

$\xymatrix{ \mathop{\mathrm{Spec}}(B[x_1, \ldots , x_ n]) \ar[d] & \mathop{\mathrm{Spec}}(B') \ar[l] \ar[d] \\ \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n]) & \mathop{\mathrm{Spec}}(A') \ar[l] }$

is cartesian the image of $D(g)$ in $\mathop{\mathrm{Spec}}(A')$ is equal to the inverse image of $U$ in $\mathop{\mathrm{Spec}}(A')$ and hence open which is the desired contradiction. $\square$

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