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May 13, 2019 at 18:30 review Close votes
May 18, 2019 at 15:29
May 13, 2019 at 18:10 comment added R. van Dobben de Bruyn This doesn't use étale at all. If $\mathfrak q \subseteq B$ is a prime, then it has a preimage $\mathfrak p = f^{-1}(\mathfrak q) \subseteq A$. But then $\kappa(\mathfrak p) \otimes_A B$ maps to $\kappa(\mathfrak q)$, contradicting the hypothesis. We conclude that $\mathfrak q$ does not exist, i.e. $B$ is the zero ring. Geometrically, the assumption is that all fibres of $\operatorname{Spec} B \to \operatorname{Spec} A$ are empty, and the conclusion is that $\operatorname{Spec} B$ is empty.
May 13, 2019 at 17:56 history asked user223794 CC BY-SA 4.0