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I can prove the following result without too much trouble:

Let $f: X \to S$ be a proper flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then the following are equivalent,

(1) the generic fiber $X_\xi \to \mathrm{Spec}{\kappa(\xi)}$ is smooth

(2) there exists a smooth fiber $X_s \to \mathrm{Spec}(\kappa(s))$ for some $s \in S$

(3) there exists a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth.

A possible reference is [EGA-IV-4-12.2.4] then use properness to conclude that the image of the nonsmooth locus is closed.

However, I am interested in the case that $f$ is not proper. It is easy to show that (2) does not imply (3) without properness. E.g. take $$ X = \mathrm{Spec}{k[x, y, z]/(y(xz - 1))} \to \mathrm{Spec}{k[z]} = \mathbb{A}^1_k $$ which has only one smooth fiber (over $z = 0$). However, is it true that (1) still implies (3) without properness? Explicitly, is the following true:

Let $f: X \to S$ be a flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then if $X_\xi \to \mathrm{Spec}(\kappa(\xi))$ is smooth then there is a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth?

If this is true can somebody point me to a reference? If it is false can somebody provide a counterexample?

Many thanks.

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    $\begingroup$ You probably want to use Chevalley's theorem on the constructibility of the pushforward of constructible sets (EGA IV, 1.8.4.) here. $\endgroup$
    – Will Sawin
    Mar 27 '20 at 17:17
  • $\begingroup$ It looks like that does the trick. Thanks $\endgroup$
    – Ben C
    Mar 27 '20 at 18:20
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    $\begingroup$ I think you might find it helpful to look at "spreading out" in Bjorn Poonen's book on Rational points. For example, you can replace "smooth" by "proper", "unramified" "etale" "separated" "flat" etc. $\endgroup$ Mar 28 '20 at 10:55
  • $\begingroup$ Wow, this "spreading out" theorem is much stronger than I expected. Thanks for the reference. $\endgroup$
    – Ben C
    Mar 28 '20 at 16:50
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As Will Sawin said in his comment, this is true using Chevalley's theorem. A sketch goes as follows.

Suppose that $X_\xi \to \mathrm{Spec}(\kappa(\xi))$ is smooth. Let $N \subset X$ be the nonsmooth locus. For a morphism of finite presentation, the smooth locus is retrocompact open so $N$ is constructible. Thus by Chevalley's theorem $f(N)$ is constructible so $S \setminus f(N)$ is constructible. However, $\xi \in S \setminus f(N)$ because the generic fiber is smooth so $S \setminus f(N)$ contains a dense open $U \subset S \setminus f(N)$. Then $X_U \to U$ is smooth because $f$ is flat + finite presentation and each fiber above $s \in U$ is smooth since $U$ does not intersect $f(N)$ meaning the fiber over $s \in U$ does not intersect $N$.

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