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Hi, I bet this is a very silly question. If I have $f:X\rightarrow Y$ a fibration (i.e a surjective morphism with connected fibers) with $Y$ a smooth proj variety and $X$ a normal variety, is the generic fiber of $f$ normal. I believe it should be by diemensional reason but I am not too sure.

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    $\begingroup$ Can't funny things happen in characteristic $p$? For example if $f$ is the Frobenius then the fibres will be non-reduced but connected, and so the generic fibre will also be non-reduced and hence singular. Or should your definition of fibration exclude this, i.e. you actually want a smooth morphism? $\endgroup$
    – Daniel Loughran
    Commented Nov 9, 2011 at 13:49
  • $\begingroup$ Daniel, even for the Frobenius, the generic fiber is still ok, right? Consider $F : R \to R$ the Frobenius acting on a domain, the generic fiber is just $K^{1/p}$ (where $K$ is the fraction field of $R$). You are right though in my answer, if I want to pass to other fibers I need the generic fiber to be geometrically normal. $\endgroup$
    – Karl Schwede
    Commented Nov 9, 2011 at 18:08

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Working locally on $X$ and $Y$, we may assume they are affine and so the map $f : X \to Y$ corresponds to a ring map $S \to R$ (an inclusion) with $S$ smooth over the base field and $R$ normal. Then the generic fiber is simply $(S \setminus 0)^{-1} R$. That's certainly normal since a multiplicative set times a normal ring is still normal.

EDIT: If additionally the generic fiber is geometrically normal (this is not free, and may fail as Daniel points out above), then it easily follows that an open set of the other fibers are also normal.

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  • $\begingroup$ @Karl Schwede Thank you for your nice answer! But I got stuck with " geometrically normal => the general fibre is normal". I feel that this is a requirement for most such Bertini type results on general fibres, but I never truly understand how to derive from "geometrical property" to general fibres, could you explain more on this part? $\endgroup$
    – Li Yutong
    Commented May 7, 2014 at 23:43
  • $\begingroup$ @LiYutong I guess you already figured it out long ago, but just in case: a reference for this kind of result is EGA $IV_{3}$, Theorem 12.2.4 $\endgroup$
    – Pedro
    Commented Jan 19, 2020 at 22:06
  • $\begingroup$ Thank you the reference! $\endgroup$
    – Li Yutong
    Commented Jan 27, 2020 at 1:19

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