Let $k$ be an algebraically closed field, $f:X \to Y$ be a surjective proper $k$-morphism locally of finite presentation between irreducible noetherian schemes. Assume that $Y$ is reduced. Under what additional condition on $f$ (other than flatness/ generic flatness) does there exist a non-empty open set $U$ of $X$ such that $f|_U$ is flat?

If I further assmume that $X$ is reduced, then does there exists a non-empty open subset $U$ of $Y$ such that $f|_{f^{-1}(U)}$ is flat?

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    $\begingroup$ Generic flatness implies it is non-empty. $\endgroup$ – Mohan Oct 21 '15 at 19:28
  • $\begingroup$ @Mohan Sorry, I slightly edited the question. $\endgroup$ – user46578 Oct 21 '15 at 19:33
  • $\begingroup$ The question becomes interesting when you do not assume that $Y$ is reduced... $\endgroup$ – Matthieu Romagny Oct 21 '15 at 19:34
  • $\begingroup$ @MatthieuRomagny I will be very interested to know what happens when $Y$ is non-reduced. May be you could elaborate a bit or suggest a reference for that case. $\endgroup$ – user46578 Oct 21 '15 at 19:38

Let $f:X\to S$ be a morphism of schemes.

Assume that $S$ is integral, and let $K$ be its function field. As "everything is flat over a field", the generic fibre $f_K:X_K\to \mathrm{Spec} \ K$ is a flat morphism.

In particular, the locus of flatness is non-empty.

As Laurent Moret-Bailly points out below, if $f$ is of finite presentation (and $S$ still integral), then $f$ is flat over a dense open of $S$.

  • $\begingroup$ Wow. So I do not even need locally of finite-presentation or proper morphism, right? $\endgroup$ – user46578 Oct 21 '15 at 19:32
  • $\begingroup$ Right. "Everything is flat over a field." $\endgroup$ – Ariyan Javanpeykar Oct 21 '15 at 19:34
  • $\begingroup$ 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 $k[x]/x^n$. $\endgroup$ – Ariyan Javanpeykar Oct 21 '15 at 19:37
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    $\begingroup$ @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. $\endgroup$ – Laurent Moret-Bailly Oct 21 '15 at 19:51
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    $\begingroup$ @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 $S$.) $\endgroup$ – Ariyan Javanpeykar Oct 21 '15 at 20:08

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