Lemma 36.66.3. Let $A \to B$ be a ring map. Say $B$ is generated as an $A$-module by $b_1, \ldots , b_ d \in B$. Set $h = \sum x_ ib_ i \in B[x_1, \ldots , x_ d]$. Then $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is universally open if and only if the image of $D(h)$ in $\mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ d])$ is open.

**Proof.**
If $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is universally open, then of course the image of $D(h)$ is open. Conversely, assume the image $U$ of $D(h)$ is open. Let $A \to A'$ be a ring map. It suffices to show that the image of any principal open $D(g) \subset \mathop{\mathrm{Spec}}(A' \otimes _ A B)$ in $\mathop{\mathrm{Spec}}(A')$ is open. We may write $g = \sum _{i = 1, \ldots , d} a'_ i \otimes b_ i$ for some $a'_ i \in A'$. 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 $B[x_1, \ldots , x_ n] \to A' \otimes _ A B$ sending $h$ to $g$. Since

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. $\square$

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