Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

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|cite|improve this question
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

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.