5
$\begingroup$

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?

  • 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.2012-05-16
  • 0
    Hm, does this show that $D(\mathbb{F}_p) = \emptyset$ for all but finitely many $p$...? I must be missing something.2012-05-16
  • 0
    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?2012-05-16
  • 0
    Yes you are correct.2012-05-16
  • 0
    Thank you QiL! It wasn't so difficult after all, but I was looking at it the wrong way.2012-05-16

0 Answers 0