We fix a binary form $F \in \mathbb{Z}[x,y]$ with non-zero discriminant and degree $d = 2g+2$, and consider the hyperelliptic curve

$$C_F: \displaystyle z^2 = F(x,y).$$

We say that a point $(x,y,z)$ is a quadratic point if there exists a quadratic extension $K/\mathbb{Q}$ over which $x,y,z$ are defined.

Then $C_F$ is easily seen to have infinitely many quadratic points: indeed, for any rational integers $x,y$ we obtain the quadratic point $(x,y, \sqrt{F(x,y)})$. These are the so-called obvious quadratic points. If we take the height of the point $(x,y,z) \in C_F$ to simply be the Weil height of the point $[x:y]$ on $\mathbb{P}^1$, then counting the obvious quadratic points of $C_F$ having bounded height $X$ is tantamount to counting the number of co-prime integers $x,y$ such $\max\{|x|, |y|\} \leq X$.

Do we expect this count to be dominant? That is, do the non-obvious quadratic points on $C_F$, i.e., those where $x,y,z$ are all in some quadratic extension and at least two of them not in $\mathbb{Q}$, more/less/equally dense than the obvious points?

  • 2
    $\begingroup$ Small comment related to behavior of non-obvious points and g. When $g\geq 4$, you only have finitely many non-obvious points by a result of Faltings. When $g = 3$, the curve could be bi-elliptic and if this elliptic curve has positive rank over $\mathbb{Q}$, this gives you another source of infinitely many obvious points. When $g = 2$ and the Jacobian of $J_F$ has positive rank over $\mathbb{Q}$, you get inifnitely many non-obvious points (c.f.~Stoll's comment on Sawin's answer from mathoverflow.net/questions/199191/…) $\endgroup$ – Jackson Morrow Feb 18 at 1:13

Yes, this is (provably!) dominant.

As Jackson Morrow pointed out, quadratic points on $C$ are the same as rational points on the symmetric square of $C$. The obvious points are the rational points on a $\mathbb P^1$. Contracting this $\mathbb P^1$, we get a surface in an abelian variety, hence by Falting's theorem (as Jackson Morrow also pointed out), contained in finitely many abelian subvarieties plus finitely many points.

For each abelian variety, the number of points of height at most $X$ will be $O ((\log X)^{r/2})$ where $r$ is the rank. On the other hand, $\mathbb P^1$ has $\approx X^2$ points of height $<X$.

So we can see that the obvious points are dense, as soon as we check that the heights in $\operatorname{Sym}^2 C$ and $C$ are comparable. But this is straightforward - we can map $\operatorname{Sym}^2 C \to \operatorname{Sym}^2 \mathbb P^1 \to \mathbb P^2$ by $( (x_1:y_1),(x_2:y_2)) \to ( x_1x_2: x_1 y_2+ x_2 y_1 : y_1 y_2)$ which makes it clear that if $(x_1:y_1)$ and $(x_2:y_2)$ are conjugate and have the same Weil height, their heights are at least the square root of the height of $( x_1x_2: x_1 y_2+ x_2 y_1 : y_1 y_2)$.

  • $\begingroup$ I guess you mean $y_1y_2$ instead of $x_2y_2$? $\endgroup$ – Michael Stoll Feb 18 at 20:53
  • $\begingroup$ @MichaelStoll Indeed, thanks. $\endgroup$ – Will Sawin Feb 18 at 21:42

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.