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Mar 13, 2019 at 9:02 comment added Jason Starr Welcome new contributor. As remarked by @gdb, this is false in general. Let $\mathcal{E}$ be any flat coherent $\mathcal{O}_X$-module such that $f_*\mathcal{E}$ is not flat. Now consider the new $R$-scheme structure on $X$ that replaces the structure sheaf $\mathcal{O}_X$ by the sheaf of $\mathcal{O}_X$-modules, $\mathcal{O}_X\oplus \left( \mathcal{E}\cdot \epsilon \right),$ with the unique structure of $\mathcal{O}_X$-algebra such that the "placeholder" $\epsilon$ satisfies $\epsilon\cdot \epsilon = 0$.
Mar 13, 2019 at 1:35 comment added gdb Two remarks: it is not possible if you assume that $R$ is normal (and noetherian) or if you assume that all fibres are geom. reduced (and $R$ is noetherian). It should be false in general, but I don't know an example on the top of my head.
Mar 12, 2019 at 21:33 history edited YCor
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Mar 12, 2019 at 20:52 history asked rori CC BY-SA 4.0