# Rational points on a special class of surfaces

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.

• Yes: $f=p_5(t)$. Apr 24 at 17:28
• I see that you have changed the question. The answer is still yes: $f=t(t^2(x^2+y^2)+1)$. Apr 25 at 9:37
• I changed it again. Apr 25 at 13:52
• The answer is still yes: $f=y^2 - t^3+t$. This example goes back to Fermat. Apr 25 at 14:11
• 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? Apr 25 at 18:01

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