2
$\begingroup$

Let $K$ be a number field, and let $C/K$ be a curve of genus $g \geq 2$ over $K$. Suppose that the set of $K$-rational points $C(K)$ on $C$ is non-empty. This enables one to define an Abel-Jacobi map $\sigma$ defined over $K$, with a given point $P \in C(K)$, which gives an embedding of $C$ into the Jacobian $J_C$, i.e., $\sigma : C \hookrightarrow J_C$.

Let $J_C(K)$ be the Mordell-Weil group of $J_C$ over $K$, and let $J_C(K)^{\text{tors}}$ be the set of points of finite order. What is the pre-image of $J_C(K)^{\text{tors}}$ in $C$ under $\sigma$?

Are there non-trivial examples (i.e., when $C(K) \ne \emptyset$ and $|J_C(K)^{\text{tors}}| > 1$) of when $C(K) = \sigma^{-1} \left(J_C(K)^{\text{tors}}\right)$?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Did you mean "i.e., when $C(K)\ne\emptyset$ and $|J_C(K)^{\text{tors}}|\ne1$."? $\endgroup$
    – Joe Silverman
    Commented Nov 6, 2020 at 13:28
  • $\begingroup$ @JoeSilverman yes I think that makes more grammatical sense. I'll change it $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 6, 2020 at 13:39
  • 1
    $\begingroup$ What kind of description of $\sigma^{-1}(J_C(K)^{\text{tors}})$ are you expecting? There are certainly examples where this set is $C(K)$, e.g. when $J_C(K)$ has rank zero. $\endgroup$
    – Jef
    Commented Nov 6, 2020 at 14:14

1 Answer 1

Reset to default
8
$\begingroup$

Trivially (though non-trivially according to your definition), we have that $C(K) = \sigma^{-1}(J_C(K)^{\text{tors}})$ whenever $J_C(K)$ is finite. There are lots of examples of this kind. But it also should not be very hard to come up with examples of, say, curves of genus 2 over $\mathbb Q$ whose Jacobian has positive rank (and non-trivial torsion), but such that no point of infinite order can be written as a difference of two rational points on the curve: Searching the LMFDB for genus 2 curves with a single rational point whose Jacobian has positive rank and nontrivial torsion turns up a fairly long list of examples, the first of which is this curve: $$ y^2 + y = x^5 + 4x^3 - 6x^2 + x - 15 $$ that has the unique point at infinity as its only rational point and Mordell-Weil group isomorphic to ${\mathbb Z} \times {\mathbb Z}/3{\mathbb Z}$.

There certainly will also be examples with more than one rational point on the curve (e.g., when the curve is hyperelliptic and all rational points are Weierstrass points).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for the answer. Do you have some idea as to how dense (say, with hyperelliptic curves of genus $g$ given by $y^2 = a_{2g+2}x^{2g+2} + \cdots + a_1 x + a_0, a_i \in \mathbb{Z}$, ordered by the height $\max |a_i| \leq X$) the curves with a rational point and such that $C(\mathbb{Q})$ embeds into $\text{Jac}_C(\mathbb{Q}^{\text{tors}}$ under the Abel-Jacobi map, and such that the rank of the Jacobian is positive? $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 8, 2020 at 0:15
  • $\begingroup$ (1) The density of curves with a rational point is expected to be zero and known to tend to zero as the genus grows (by work of Bhargava). (2) Restricting to curves with $a_{2g+2}=0$ to force the existence of a rational point, it is again expected that this is the only rational point outside a density zero set, and it is known that the density of curves with additional rational points tends to 0zero as the genus grows (by work of Poonen and myself; the definition of height is a bit different here, though). TBC... $\endgroup$
    – Michael Stoll
    Commented Nov 8, 2020 at 11:09
  • $\begingroup$ (3) I think one expects that 50% of all Jacobians of odd degree hyperelliptic curves of fixed genus have rank 0 and 50% have rank 1. So, combining (2) and (3), you'll get at least 50% heuristically for curves with a rational Weierstrass point. But almost all of them will have trivial torsion. It might be possible to prove something for curves with two rational Weierstrass points (forcing the rational torsion subgroup to have even order); also in this case, the heuristic expectation is that 100% of all curves have no additional rational points and 50% have positive rank. $\endgroup$
    – Michael Stoll
    Commented Nov 8, 2020 at 11:12

You must log in to answer this question.

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