12
$\begingroup$

For each prime $p$, fix a place of $\bar{\mathbb{Q}}$ extending $p$ so we can reduce elements of $\bar{\mathbb{Q}}$ which are integral over $\mathbb Z$ and get elements of $\bar{\mathbb{F}}_p$. Now let $X$ be a smooth projective variety over $\bar{\mathbb{Q}}(t)$. For all but finitely many $p$ we can reduce the coefficients (of the coefficients in $t$) of the equations defining $X$ modulo $p$ and get $X_p$ defined over $\bar{\mathbb{F}}_p(t)$.

Is the following true/known: $X(\bar{\mathbb{Q}}(t)) = \emptyset$ if and only if $X_p(\bar{\mathbb{F}}_p(t)) = \emptyset$ for all but finitely many $p$?

Note that this is a somewhat non-standard local-global principle, as I am not using the places of $\bar{\mathbb{Q}}(t)/\bar{\mathbb{Q}}$.

I have an attempted proof that doesn't quite work. The set $\{P \in X(k(t)) \mid h(P) \le N\}$, for an algebraically closed field $k$, integer $N$ and some height function $h$ is a variety over $k$ and whether it is empty or not can be described by some polynomial identities, by the Nullstellensatz. So, this set is empty for $k=\bar{\mathbb{Q}}$ if and only if it is empty for $k=\bar{\mathbb{F}}_p$ for all but finitely many $p$. But that's not enough to solve my problem because I may have to apply it to $N$ depending on $p$. If, however, one has an a priori bound on $N$, then it works. For example, the above local-global principle works for non-isotrivial curves, since a result of Szpiro provides a bound on $N$.

Also, this will not work without the algebraic closures of the constant fields as counterexamples for the Hasse principle will work here too.

$\endgroup$
3
  • 1
    $\begingroup$ I would have thought one could produce counterexamples by considering isotrivial families of Fermat surfaces of large degree in $\mathbb{P}^3$. By Shioda, the reductions modulo infinitely many primes $p$ are (inseparably) rationally connected fibrations over a curve, but the degree $M$ of the unirational parameterization grows with $M$. $\endgroup$ – Jason Starr Oct 4 '16 at 9:23
  • 2
    $\begingroup$ Actually, now I see that there are some isotrivial families of Fermat hypersurfaces such that for all but finitely many $p$, the restriction of the family over $\overline{\mathbb{F}_p}(t)$ has no rational sections, even though they are inseparably rationally connected. This is one reason that inseparable rational connectedness is quite different from separable rational connectedness. $\endgroup$ – Jason Starr Oct 4 '16 at 11:04
  • 1
    $\begingroup$ Maybe ultraproducts can help? Haven't thought this through, just guessing. $\endgroup$ – Arne Smeets Oct 11 '16 at 6:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.