most recent 30 from http://mathoverflow.net 2013-06-20T09:39:31Z http://mathoverflow.net/feeds/question/120477 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120477/irreducibility-of-fibers-vs-irreducibility-of-fibered-product Alexander Braverman 2013-02-01T00:00:59Z 2013-02-01T02:06:15Z

<p>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.</p> <p>$\mathbf{Question:}$ Assume that the fibered product $X\underset{Y}\times X$ is irreducible. Is it true that the generic fiber of $f$ is irreducible?</p>

http://mathoverflow.net/questions/120477/irreducibility-of-fibers-vs-irreducibility-of-fibered-product/120481#120481 Jack Huizenga 2013-02-01T01:56:18Z 2013-02-01T02:06:15Z

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