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I apologize in advance if this question is too elementary. Let $f\colon X\to Y$ be a flat proper morphism of algebraic varieties over an algebraically closed field, and assume that $Y$ is irreducible (not all my assumptions might be necessary). Let $y\in Y$ be a closed point such that the fiber $f^{-1}(y)$ is a smooth irreducible variety.

Questions. 1) Is it true that the fiber over the generic point of $Y$ is irreducible and regular?

2) Is it true that the set of points of $Y$, whose fibers are irreducible and regular, is Zariski open?

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    $\begingroup$ The answer to question 1 is yes, and abx answered a question about this a day or two ago. The answer to question 2 is yes in characteristic $0$, but it is no in positive characteristic (it is always "yes" if you replace "regular" by "smooth"). $\endgroup$ – Jason Starr Mar 11 '17 at 13:11
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I am just giving a counterexample to Question 2 in positive characteristic. Let $\kappa$ be an algebraically closed field of characteristic $p$. Let $Y$ be $\text{Spec}\ \kappa[t,t^{-1}]$. Let $X$ be $\text{Spec}\ \kappa[s,s^{-1}]$. Let $f:X\to Y$ be the $\kappa$-morphism associated to the homomorphism of $\kappa$-algebras, $$f^*:\kappa[t,t^{-1}]\to \kappa[s,s^{-1}], \ t\mapsto s^p.$$ For every closed point of $Y$, the fiber of $f$ over that point is nonreduced, hence it is not regular. Since $X$ is irreducible and regular, the fiber of $f$ over the generic point $\text{Spec}\ \kappa(t)$ of $Y$ is also irreducible and regular. Thus, the set of points of $Y$ where the fiber is irreducible and regular is precisely the generic point of $Y$, and this is not an open subset.

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