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Oct 21, 2015 at 20:11 vote accept user46578
Oct 21, 2015 at 20:08 comment added Ariyan Javanpeykar @user46578 An open set could be empty. (In our situation, the morphism $f$ is generically flat and of finite presentation. The locus of flatness is therefore open (by Thm 11.3.1) and non-empty (by generic flatness). In particular, as $S$ is integral, the locus of flatness is dense open in $S$.)
Oct 21, 2015 at 20:06 history edited Ariyan Javanpeykar CC BY-SA 3.0
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Oct 21, 2015 at 20:06 comment added user46578 @LaurentMoret-Bailly But theorem $11.3.1$ does not mention non-emptyness of the flat locus.
Oct 21, 2015 at 19:58 comment added Ariyan Javanpeykar @LaurentMoret-Bailly Yes, thank you! (Thus, I should have written, "any morphism of schemes $f:X\to S$ with $S$ integral is generically flat. If $f$ is additionally of finite presentation, then $f$ is flat over a dense open of $S$.")
Oct 21, 2015 at 19:51 comment added Laurent Moret-Bailly @AriyanJavanpeykar : you probably need $f$ to be of finite presentation. In fact if $f$ is locally of finite presentation, then the flat locus in $X$ is open (EGA IV, (11.3.1)). If in addition $f$ is quasi-compact, your claim follows.
Oct 21, 2015 at 19:37 comment added Ariyan Javanpeykar Any morphism of schemes $f:X\to S$ with $S$ integral is flat over some dense open of $S$. As Matthieu Romagny mentions above, the question is more interesting when $S$ is only irreducible, e.g., $S$ is Spec $k[x]/x^n$.
Oct 21, 2015 at 19:34 comment added Ariyan Javanpeykar Right. "Everything is flat over a field."
Oct 21, 2015 at 19:32 comment added user46578 Wow. So I do not even need locally of finite-presentation or proper morphism, right?
Oct 21, 2015 at 19:29 history answered Ariyan Javanpeykar CC BY-SA 3.0