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Jul 17, 2012 at 14:33 vote accept Flyingpanda
Jul 17, 2012 at 12:21 comment added Jason Starr @HU Zhengyu: To spell it out, for every $k$, by Theorem III.12.8, there exists a proper Zariski closed subset $Z_k \subset C$ such that for every $p\in C\setminus Z_k$, '$h^0(X_p,\omega_{X_p}^{\otimes k})$' equals '$h^0(X_\eta, \omega_{X_\eta}^{\otimes k})$'. If your ground field is uncountable, then for every dense open subset of $C$, or even for every countable intersection $G$ of dense open subsets, there exists a point $p$ of $G$ which is no $Z_k$. So for this "general point", for every $k$, $h^0(X_p,\omega_{X_p}^{\otimes k})$ equals `$h^0(X_\eta, \omega_{X_\eta}^{\otimes k})$'.
Jul 17, 2012 at 12:14 comment added Jason Starr @Hu Zhengyu: I suppose this depends on what you mean by "general fibers". If you are working over $\overline{\mathbb{F}_p}$, then I agree that your definition of "general fibers" (whatever it is) might disagree with the standard definition. However, if you are working over an uncountable algebraically closed field, or over any "sufficiently large" algebraically closed field, then this follows by upper semicontinuity, cf. Theorem III.12.8 of Hartshorne.
Jul 17, 2012 at 9:03 comment added Flyingpanda I wonder how you can ensure the generic fibre has nonnegative Kodaira dimension?
Jul 17, 2012 at 2:36 history answered Jason Starr CC BY-SA 3.0