10
$\begingroup$

Question: Let $C$ be an algebraic curve over some field (like the rationals) given by a plane projective model (possibly with singularities). Is there an easy way to see if this curve has a non-trivial rational map to an elliptic curve?

Criteria involving the Jacobian (something like $J(C)$ has a subgroup of codimension $1$) wouldn't help, as the curves I am interested in have a big genus ($>10$) and high degrees in both variables, so it is very unlikely that anything about their Jacobians can be computed.

Background: This question arose from an attempt to study rational points (over $\mathbb Q$) on certain curves which possibly are coverings of lower genus curves. In particular, if one could compute a covering map to an elliptic curve, one could efficiently look for rationals points.

Improve this question
$\endgroup$
3
  • 6
    $\begingroup$ Over a finite field there are good algorithms for computing the zeta function of $C$, from which, via the Tate conjecture, one can decide the existence of a morphism from $C$ to an elliptic curve. But computing such a morphism, once its existence is known, seems to be a hard problem. [Ref. : Handbook of elliptic and hyperelliptic curve cryptography.] So the "if" in your final sentence is not a small one, I believe. $\endgroup$
    – inkspot
    Commented Aug 30, 2012 at 16:39
  • 2
    $\begingroup$ I have two dubious ideas. If it's possible to find a map through $\mathbb P^1$ and classify the monodromy actions of the ramification groups at the ramified points, you can check if the map factors through an elliptic curve using group theory. If there is some kind of obvious "distinguished" $1$-form on this curve, which has an especially simple formula or good symmetry properties something, you can try to show it's the derivative of the map to an elliptic curve, but I'm not sure how you'd do that. $\endgroup$
    – Will Sawin
    Commented Aug 30, 2012 at 16:39
  • $\begingroup$ If the curve has a nontrivial finite group action, the induced action on the Jacobian can split off an elliptic curve. This will be the case if a group element has an eigenvalue 1 of multiplicity one on the tangent space to the Jacobian (which is the dual to the space of 1-forms on the curve). $\endgroup$
    – Evgeny Shinder
    Commented Jun 23, 2020 at 8:54

2 Answers 2

Reset to default
10
$\begingroup$

$C$ will admit a nontrivial rational map to an elliptic curve $E$ if and only if $E$ appears as an isogeny factor of the Jacobian $J(C)$. So in some sense it is impossible to avoid what you lament in your first paragraph.

However, we don't necessarily need to compute the Jacobian $J(C)$ in any sense: rather, we need to show that it does not have an elliptic curve as an isogeny factor. The conventional wisdom is that (if $C$ has genus greater than one, as you must of course intend) then it is very likely that $J(C)$ is geometrically simple and even that $\operatorname{End} J(C) = \mathbb{Z}$.

There are some sufficient conditions for that! The one I know by heart is the following beautiful theorem of Zarhin:

Let $K$ be a field of characteristic different from $2$, and let $C_{/K}$ be a hyperelliptic curve given by $y^2 = P_{2g+2}(x)$. Then if the Galois group of $P_{2g+2}(X)$ is $S_{2g+2}$ (as it usually is!!) or $A_{2g+2}$, then $\operatorname{End} J(C) = \mathbb{Z}$.

Unfortunately I have forgotten much of what I used to know about this sort of thing, but I am reasonably confident that there are further results along these lines. The name "Arsen Elkin" (whom I think was a student of Zarhin) is coming to mind.

It would be very interesting to have a criterion that one could apply to a fairly arbitrary plane curve $C$ that would be sufficient to force its Jacobian to have endomorphism ring $\mathbb{Z}$. This time I have no memories, however dim, of such a result, but it would certainly be nice...

Added: I looked up Elkin's work on MathSciNet. He has several papers in this area, but so far as I can see they all further refine the hyperelliptic case.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for the answer. It still does not do what I would like to see. In the meantime I believe that there is no criterion of a kind I was hoping for. $\endgroup$
    – Peter Mueller
    Commented Jul 29, 2013 at 13:44
  • $\begingroup$ $@$Peter: I agree that if your curves are not hyperelliptic (and you said that they have high degree in both $x$ and $y$) then my answer is not helpful: sorry about that. Nevertheless it is not clear to me that the situation is hopeless... $\endgroup$
    – Pete L. Clark
    Commented Jul 30, 2013 at 2:10
0
$\begingroup$

Simon has a paper related to this:

https://onlinelibrary.wiley.com/doi/abs/10.1002/mana.201200285

The paper is fairly elementary and should be readable. However it did not give any criterion for ruling out which curves that cannot be constructed explicitly. The problem is still open to me.

Improve this answer
$\endgroup$

You must log in to answer this question.

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