4
$\begingroup$

Given an elliptic curve $E/\mathbb{Q}$, is it possible to determine whether or not $E$ has torsion points just by looking at it's Hasse-Weil L function $L(E,s)$? In general, what effects does an elliptic curve having torsion points have on its L function?

The effect of having free points is clearly seen in the Birch and Swinnerton Dyer Conjecture which relates the order of the zero at $s=1$ of $L(E,s)$ to the rank of $E$. The BSD conjecture also relates the value of $L(E,1)$ to the order of the torsion group, but looking at $L(E,1)$ is not enough to determine the order $E(\mathbb{Q})_{\mathrm{tor}}$ because of all of the other invariants involved in the expression.

$\endgroup$
5
  • 8
    $\begingroup$ The $L$-function does not change under an isogeny, but the torsion point can. So there is no way one can detect the torsion points from the $L$-function alone. $\endgroup$ Feb 6, 2021 at 23:49
  • $\begingroup$ @ChrisWuthrich The fact that the elliptic curve has torsion point is still preserved though, since isogenies are surjective, so my question still stands $\endgroup$
    – Milo Moses
    Feb 7, 2021 at 0:00
  • 4
    $\begingroup$ I am afraid that is incorrect. Typically there may well be a curve in the isogeny class without any torsion points. Isogenies are not surjective on $K$-rational points, only over algebarically closed fields. $\endgroup$ Feb 7, 2021 at 0:04
  • $\begingroup$ @ChrisWuthrich Thank you for correcting me. I will make sure to learn more about isogenies. $\endgroup$
    – Milo Moses
    Feb 7, 2021 at 0:06
  • 4
    $\begingroup$ Although you can't completely determine the torsion, it may still be possible to find out something about it. From the L-function you can determine all its coefficients, and hence all sizes $|E(\mathbb F_p)|$. If $E$ has good reduction at $p$, then size of torsion divides this size of reduction, so you can determine a bound on torsion by finding the gcd of the reduction sizes. I doubt there is much more than this that you can do. $\endgroup$
    – Wojowu
    Feb 7, 2021 at 0:26

1 Answer 1

6
$\begingroup$

This has been alluded to in one of the comments, but if $E(\mathbb Q)$ has an $\ell$-torsion point, then at every prime $p$ of good reduction we have $$ p+1-a_p = \#E(\mathbb F_p) \equiv 0 \pmod \ell, $$ so the local factor of the $L$-function at $p$ satisfies $$ L_p(T) = 1 - a_p T + p T^2 \equiv 1 - (p+1) T + p T^2= (1-T)(1-pT) \pmod\ell. $$ Of course, for $L(E/\mathbb Q,s)$, we need to evaluate this at $T=p^{-s}$, and then the meaning of this congruence gets a bit dicey, especially when we multiply them over all $p$ to get $L(E/\mathbb Q,s)$. On the other hand, if you just want to think about the reduction curve $\tilde E_p/\mathbb F_p$, then its zeta function is $$ Z(\tilde E_p/\mathbb F_p) = \frac{1-a_pT+pT^2}{(1-T)(1-pT)} \equiv 1 \pmod\ell, $$ i.e., the $\ell$-torsion point means that the local zeta function is congruent to 1 modulo $\ell$.

$\endgroup$
2
  • 2
    $\begingroup$ One can go a bit further, I believe. From the $L$-series, one gets the $\ell$-adic representations, i.e. the $\mathbb{Q}_{\ell}$-vector space with the action of the Galois group on them. That in turn allows us to list all $\mathbb{Z}_{\ell}$-lattices stable under the Galois group. Therefore, we would find all torsion subgroups as the curve varies in its isogeny class over $\mathbb{Q}$. $\endgroup$ Feb 8, 2021 at 19:47
  • 1
    $\begingroup$ @ChrisWuthrich Nice point. Although my recollection is that it gets trickier for higher dimensional abelian varieties. $\endgroup$ Feb 8, 2021 at 20:14

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.