# Normal fibrations

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.

• 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? – Daniel Loughran Nov 9 '11 at 13:49
• 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. – Karl Schwede Nov 9 '11 at 18:08

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.
• @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 – Pedro Jan 19 '20 at 22:06