Lemma 37.74.3. 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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