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Let $f: X \to S$ be a proper morphism (not necessarily smooth!, but perhaps flat) and $S$ be an excellent, regular, affine, integral variety over a finite field (I do not know which of these properties are really necessary).

Does $\{s \in S : X_s := f^{-1}(s) \text{ is (geometrically) integral}\}$ contain an open subset [edit: this was: "non-empty open subset"]?

(The same question for $\{s \in S : X_s := f^{-1}(s) \text{ is (geometrically) connected}\}$ seems to have a positive answer by Stein factorisation.)

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    $\begingroup$ Is $X$ integral? What do you want, to exclude e.g. $S \coprod S \to S$? $\endgroup$ Mar 7, 2016 at 16:37
  • $\begingroup$ I deleted the "non-empty". $\endgroup$
    – user19475
    Mar 7, 2016 at 16:56
  • $\begingroup$ Then... it contains the empty open subset. Do you just want to ask if this locus is open? (Which is certainly my intuition; do you have a counterexample?) $\endgroup$ Mar 7, 2016 at 17:23

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If $f$ is proper and flat, the "geometrically integral locus" is open by EGA IV, (12.2.1)(x). This works for any morphism of schemes which is proper, flat and finitely presented.
The geometrically connected locus is not open in general: think of a ramified double cover.

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