MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.