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Let $f:X\rightarrow Y$ be a morphism of schemes over $S$ (with possibly Noetherian conditions all over the place).

For every point $s\in S$, the morphism base changed to the fiber $f_s:X_s\rightarrow Y_s$ is faithfully flat.

Do we have that $f$ is flat? or even faithfully flat?

Edit: what if moreover both $X$ and $Y$ are flat over $S$?

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  • $\begingroup$ If $Y=S$ and $f:X\to S$ is onto, then for every $s\in S, f_s:X_s\to s$ is faithfully flat since $s$ is the spectrum of the field. But $f$ is obviously not necessarily flat. $\endgroup$ – Roland Jan 26 at 12:41
  • $\begingroup$ @Roland what if both of $X$ and $Y$ are flat over $S$? Then your example doesn't apply. $\endgroup$ – Lance Wu Jan 26 at 13:11

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