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Suppose that $f\colon X\to Y$ is a proper (or even projective) morphism of (reduced) algebraic varieties over an algebraically closed field $k$. If fibers of $f$ over all closed points of $Y$ are reducible, does it imply that the generic geometric fiber of $f$ (that is, the pullback via $\mathrm{Spec}\,\overline{k(Y)}\to Y$) is also reducible? (I assume that $Y$ is irreducible.)

Thank you in advance,
Serge

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Geometric irreducibility is a constructible property (see EGA IV, §9, 9.7.7 (i)) so the answer is yes. –  Damian Rössler Aug 2 '13 at 10:37
    
Thanks a million! –  Serge Lvovski Aug 2 '13 at 10:55
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