Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a scheme which is integral, of finite type, flat and separated over $\mathbb{Z}$.

Let $D \subseteq X$ be a prime divisor on $X$ which is not flat over $\mathbb{Z}$.

Is it true that $D(\mathbb{F}_p) = \emptyset$ for all primes $p$, with at most one exception?

share|improve this question
2  
Remember that a morphism from an integral scheme to the spectrum of a Dedekind domain is flat if and only if it is dominant. –  user18119 May 16 '12 at 21:40
    
Hm, does this show that $D(\mathbb{F}_p) = \emptyset$ for all but finitely many $p$...? I must be missing something. –  Evariste May 16 '12 at 22:44
    
Aha, OK. The fact that $D$ is not $\mathbb{Z}$-flat shows that for only finitely many $p$, the fiber $D_{\mathbb{F}_p}$ is non-empty. But since $D$ is supposed to be irreducible, there can be at most one such $p$. Right? –  Evariste May 16 '12 at 22:54
    
Yes you are correct. –  user18119 May 16 '12 at 22:55
    
Thank you QiL! It wasn't so difficult after all, but I was looking at it the wrong way. –  Evariste May 16 '12 at 22:56
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.