Take the tour ×  Let $f:X\to Y$ be a morphism of algebraic
varieties over an algebraically closed field $k$ (I am ready to
assume that $f$ is a smooth morphism, but that should not be
necessary). I want to check that the generic fiber of $f$ is
irreducible.  $\mathbf{Question:}$ Assume that the fibered
product $X\underset{Y}\times X$ is irreducible. Is it true that
the generic fiber of $f$ is irreducible?  share|improve this
question Suppose that $f$ is dominant. Let $\eta$ be the generic
point of $Y$. Since $X\times_Y X$ is irreducible, the generic
fibre of the morphism $X\times_Y X\to Y$ is also irreducible,
because the morphism is dominant. This generic fibre is equal to
$X_\eta\times_\eta X_\eta$ and this surjects onto $X_\eta$
(which is the generic fibre of $f$). Hence $X_\eta$ is
irreducible. –  Damian Rössler Feb 1 at 14:40 Pardon my
ignorance, but how do the notions of "fiber over the generic
point" and "general fiber" relate to one another? For example,
if we look at a 2:1 branched cover $\mathbb{P}^