# Hyperelliptic curves with fixed genus and many rational points

It is a famous theorem of Faltings, previously a conjecture by Mordell, that any algebraic curve of genus at least $2$ defined over the rational numbers have at most finitely many rational points. A hyperelliptic curve is a special algebraic curve of the form

$$\displaystyle z^2 = f(x,y),$$

where $f(x,y) \in \mathbb{Z}[x,y]$ is a binary form of degree $2n+2$. The genus of the curve is equal to $n$. For genus $1$ and $2$, all algebraic curves are hyperelliptic.

What is unknown from Faltings' theorem is an effective bound for the height of rational points lying on a given curve. Therefore, even if one obtains an effective upper bound for the number of rational points lying on a given curve $C$, there is no way to check how close to being optimal the bound is.

A recent theorem of Bhargava asserts that most hyperelliptic curves, ordered by naive height of the coefficients of $f$, have no rational points as the genus $n$ tends to infinity. Indeed, there are $o(2^{-n})$ hyperelliptic curves of genus $n$ defined over $\mathbb{Q}$ with at least one rational point.

My question is, what is the quantity

$$\mathfrak{S}(n) = \displaystyle \sup_{\substack{C \text{ hyperelliptic} \\ \text{genus of } C = n}}\#\{(x,y,z) \in C: (x,y,z) \text{ rational; inequivalent}\}.$$

Here two points $(x, y,z), (x',y',z')$ are inequivalent if there does not exist a non-zero rational number $\lambda$ such that $(x,y,z) = (\lambda x', \lambda y', \lambda^{n+1} z')$. In other words, the points $(x,y,z), (x',y',z')$ are not in the same equivalence class in the weight projective space $\mathbb{P}(1,1,n+1)$.

Plainly, $\mathfrak{S}(n) \gg n$. Indeed, this can be seen by considering the curve

$$\displaystyle C: z^2 = x^{2n+2} + (y - x)(y - 2x) \cdots (y - (2n+2)x)$$

which has the pairwise inequivalent points $(1,1,1), (1,2,1), \cdots, (1,2n+2,1)$. It is not clear to me whether or not these are essentially all of the inequivalent rational points on $C$.

Is $\mathfrak{S}(n)$ bounded? If so, then can we give an upper bound in terms of $n$? If not, can one give an explicit family of curves $C_1, C_2, \cdots$ such that

$$\displaystyle \# C_1(\mathbb{Q}) < \#C_2(\mathbb{Q}) < \cdots ?$$

• For genus 0,1 there are infinitely many points. – joro Aug 19 '15 at 7:02
• Perhaps I'm misunderstanding the question, but I'm fairly confident that $\mathfrak{S}(n)$ is not known to be finite. – dhy Aug 19 '15 at 13:58
• ams.org/journals/jams/1997-10-01/S0894-0347-97-00195-1 – Felipe Voloch Aug 19 '15 at 14:35
• This is an open problem. As Felipe points out, the existence of a bound would follow from the weak Lang conjecture (rational points on varieties of general type are not Zariski dense). There are bounds when one also bounds the rank $r$ of the Mordell-Weil group (it has to be $\le n-3$), see my paper at mathe2.uni-bayreuth.de/stoll/schrift.html#AG48 and arxiv.org/abs/1504.00694 for a generalization to arbitrary curves... – Michael Stoll Aug 19 '15 at 14:59
• Another note: the current best lower bound for $\mathfrak S(2)$ is 642, see mathe2.uni-bayreuth.de/stoll/recordcurve.html . – Michael Stoll Aug 19 '15 at 15:01

There are bounds when one also bounds the rank $r$ of the Mordell-Weil group (it has to be $\le n−3$), see this paper for the case of hyperelliptic curves (with a bound $\ll (r+1)n$) and this one by Katz, Rabinoff and Zureick-Brown for a generalization to arbitrary curves.
There are (in my view) convincing heuristic arguments in favor of a bound depending on $n$ and $r$ only. So your question may be related to the (equally open) question whether Mordell-Weil ranks of Jacobians of (hyperelliptic) curves of fixed genus are bounded or not. If they are, then it seems very likely that the number of points is bounded, too.
Another note: the current best lower bound for ${\mathfrak S}(2)$ is 642, given by this curve.