Let $R$ be an integral domain. Let $f:X\rightarrow \mathrm{Spec}\,R$ be a flat proper morphism of schemes. Is it possible that $O_X(X)$ is not a flat $R$module?
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1$\begingroup$ 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. $\endgroup$ – gdb Mar 13 at 1:35

$\begingroup$ 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 $\epsilon\cdot \epsilon = 0$. $\endgroup$ – Jason Starr Mar 13 at 9:02