5
$\begingroup$

Let $C$ be a nonsingular projective curve defined over $\mathbb{Q}$, which does not admit a map of degree 1 or 2 to $\mathbb{P}^1$ or to an elliptic curve. It is then a consequence of Corollary 3 of [1] that $C$ possesses only finitely many quadratic points; i.e., the set

$$\Gamma_C := \left\{p \in C : [\mathbb{Q}(p) : \mathbb{Q}] = 2\right\}$$

is finite. In particular, there exists a bound $D_C$ such that, if $D$ is a squarefree integer satisfying $|D| > D_C$, then $C(\mathbb{Q}(\sqrt{D})) = C(\mathbb{Q})$.

It is natural to ask if there is an algorithm to effectively compute $D_C$, given a model of $C$. Going through the proof of the Harris-Silverman result does not suggest any such algorithm to me (unless the rank of the Jacobian of $C$ is 0).

Given a model of $C$, is there an algorithm to effectively compute $D_C$?

I would also like to ask if $D_C$ is known when $C$ is the modular curve $X_0(N)$; there are only 55 values of $N$ for which $X_0(N)$ does not have finitely many quadratic points (see Theorem 4.9 in [2] for this list). Let $S$ be this set of 55 integers.

Given $N \notin S$, is there an algorithm to effectively compute $D_{X_0(N)}$?

[1]: J. Harris, J.H. Silverman. Bielliptic Curves and Symmetric Products. Proc. Amer. Math. Soc. 112 (1991), 347-356

[2]: F. Bars. On Quadratic Points of Classical Modular Curves. Manuscript, 2012, available from the author's website.

$\endgroup$

2 Answers 2

5
$\begingroup$

Hi Barinder!

As far as I know there is not an algorithm to do so. See for instance the following paper of Harris and Silverman:

http://www.ams.org/journals/proc/1991-112-02/S0002-9939-1991-1055774-0/

Especially in the proof of Corollary 3 you can really get a clear picture of the argument. Basically if your curve is not hyperelliptic or bielliptic then you find that the symmetric product of $C$ with itself has only finitely many rational points by Faltings' Theorem. Then you have a finite-degree map of sets between the rational points of the symmetric product and your $\Gamma_C$.

If you had a way to compute $D_C$ for any curve, then you'd have something close to an effective version of Faltings for these symmetric product surfaces. Given that Faltings theorem isn't effective for curves I think we're quite far away from that.

For modular curves in particular, well that question is actually on my upcoming agenda. If you'd like to talk about this some time I would probably not require much persuasion.

$\endgroup$
1
  • $\begingroup$ Hi Jim! Thanks for the reference; your question about modular curves does sound very interesting, I've been thinking about this recently as well. Let's talk about this the next time we meet, or I'll send you an email. $\endgroup$ Commented May 24, 2013 at 9:38
4
$\begingroup$

I was going to point out that the specific result you cite for quadratic points is actually a theorem of Joe Harris and mine (Proc. Amer. Math. Soc. 112 (1991), 347-356), but I see that stankewicz has already given the reference, so I won't repeat the link. What Abramovich and Harris did later (and this is much harder) is generalize the result to points of degree $d$ for all $d\ge2$, although the obvious generalization turns out to be false, one can't merely assume that there are no maps of degree at most $d$ to an elliptic curve or to $\mathbb P^1$. Anyway, as far as I know, all finiteness results of this sort rely on a theorem of Faltings' (generalizing Vojta's proof of the Mordell conjecture) that is highly noneffective. However, for what it's worth, it is possible to give an effective upper bound for $\#D_C$. More generally, one can give effective constants $K_1$ and $K_2$ such that $\#D_C$ has at most $K_1$ points whose height is greater than $K_2$. Unfortunately, we can't get our hands on those putative $K_1$ points of large height, although one suspects that there aren't any such points.

$\endgroup$
1
  • $\begingroup$ Thanks Joe! I've made a reference to your paper with Joe Harris in the post. $\endgroup$ Commented May 24, 2013 at 9:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .