2
$\begingroup$

Consider a smooth surface of the following form $$ S = \{f(x,y,t) = p_0(t)x^2+p_1(t)xy+p_2(t)x+p_3(t)y^2+p_4(t)y+p_5(t) = 0\}\subset\mathbb{A}^3 $$ over $\mathbb{Q}$, and set $$ U_S = \{t' \in \mathbb{Q} : |\: f(x,y,t') = 0 \text{ for some } (x,y)\in\mathbb{Q}^2\}\subset\mathbb{Q}. $$ Is there any example of such a smooth and irreducible surface $S$ such that the projection $S\rightarrow\mathbb{A}^1_t$ is dominant, and $U_S$ is non empty and non Zariski dense in $\mathbb{Q}$?

Thank you.

$\endgroup$
6
  • 4
    $\begingroup$ Yes: $f=p_5(t)$. $\endgroup$ Apr 24 at 17:28
  • 2
    $\begingroup$ I see that you have changed the question. The answer is still yes: $f=t(t^2(x^2+y^2)+1)$. $\endgroup$ Apr 25 at 9:37
  • 2
    $\begingroup$ I changed it again. $\endgroup$
    – LaGra
    Apr 25 at 13:52
  • 2
    $\begingroup$ The answer is still yes: $f=y^2 - t^3+t$. This example goes back to Fermat. $\endgroup$ Apr 25 at 14:11
  • $\begingroup$ This example is really interesting, thank you. Do you know if there is such an example of the form $f = p_0(t)x^2+p_3(t)y^2+p_5(t) $ where the $p_i$ are all non constant? $\endgroup$
    – LaGra
    Apr 25 at 18:01

1 Answer 1

5
$\begingroup$

I am just posting my comments as an answer. Without a hypothesis that the geometric generic fiber of $\pi:S\to \mathbb{A}^1_t$ is irreducible, the result is false. For a smooth compactification of $S$ on which $\pi$ extends to a morphism to $\mathbb{P}^1_t$, the finite part of the Stein factorization of $\pi$ is either an isomorphism to $\mathbb{P}^1_t$ (precisely when the geometric generic fiber is irreducible) or it is a degree-$2$ cover, i.e., a hyperelliptic curve. For appropriate choices of the coefficient polynomials $p_i(t)$, this can be any hyperelliptic curve. If the genus is $\geq 2$, then this curve has only finitely many $\mathbb{Q}$-points (by Mordell's Conjecture / Falting's Theorem), so the image in $\mathbb{P}^1_t$ is also a finite set.

However, if the geometric generic fiber is irreducible, then the compactification over $S$ is a conic bundle over $\mathbb{P}^1_t$. After base change from $\mathbb{Q}$ to some number field, this surface is rational, i.e., the surface is geometrically rational. There is a conjecture (perhaps due to Colliot-Thélène) that the set of rational points on a geometrically rational variety over a number field is dense in the Brauer subset of the set of adelic points. Assuming this conjecture, once there is a single rational point (so that the Brauer subset is nonempty), the set of rational points is Zariski dense. Thus, the image in $\mathbb{P}^1_t$ is also Zariski dense.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer. Are you assuming that the single rational point is a smooth point of $S$ or could it be singular? $\endgroup$
    – LaGra
    Apr 30 at 12:00

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.