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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?

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    $\begingroup$ 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. $\endgroup$
    – Damian Rössler
    Commented Feb 1, 2013 at 14:40
  • $\begingroup$ 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 $\mathbb{P}^1\to \mathbb{P}^1$, then is it not the case that the general fiber is reducible (generally consisting of 2 points) but the fiber over the generic point (which is the general point of $\mathbb{P}^1$) is irreducible? (Even if this is the case, its possible Alexander cares about the fiber over the generic point instead of the fiber over a general point. $\endgroup$
    – Jack Huizenga
    Commented Feb 2, 2013 at 0:05

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It seems to me the answer should be yes, and is certainly yes in case $f$ is smooth. I'll sketch an argument in the smooth case; I think the details can probably be filled in more generally. (Note that if you just assume $X$ is nonsingular and the characteristic is 0, you can easily reduce to the smooth case by generic smoothness)

Put

$Z=\{ (x_1,x_2): f(x_1)=f(x_2)=y \textrm{ and } x_1,x_2 \textrm{ lie on the same component of } f^{-1}(y)\} \subset X \times_Y X$.

Check that $Z$ is a subvariety of $X\times_Y X$ with the same dimension as $X\times_Y X$ (this is easy if $f$ is smooth, and I think it should be true generally; in the general case, it is probably best to throw out some of the "bad" fibers of $f$ and take a closure to define $Z$). Then since $X\times_Y X$ is irreducible, we have $Z = X\times_Y X$, which gives the desired conclusion.

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  • $\begingroup$ (answering your comment above) I agree with you: In general (no pun intended...) the general fibre and the generic fibre are not related - as in your example above. Nevertheless, I you are interested in constructible propertieś like geometrically connected, geometrically irreducible etc. then there is a direct link. In this case, I really understood generic fibre and not general fibre. The distinction is important because as you noticed "irreducible" is not a constructible property. $\endgroup$
    – Damian Rössler
    Commented Feb 2, 2013 at 7:59

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