1
$\begingroup$

Let $E$ be an elliptic curve over $\mathbb{Q}$. Then how to compute the $p$-torsion elements of $E$ over the $p$-adic field $\mathbb{Q}_p$ using SAGE or any other means ? At least can we say whether $E(\mathbb{Q}_p)[p]=0$ or not ?

$\endgroup$
9
$\begingroup$

Suppose $p>2$ and that $E$ has good reduction. If the reduction $\tilde E(\mathbb{F}_p)$ has no $p$-torsion then there is no $p$-torsion in $E(\mathbb{Q}_p)$. Otherwise look at the exact sequence $$ 0\to E(\mathbb{Q}_p)[p]\to \tilde E(\mathbb{F}_p)[p]\to \hat E(p\mathbb{Z}_p)/p\hat E(p\mathbb{Z}_p).$$ Here $\hat E$ is the formal group. So to check if a $p$-torsion point $\tilde P$ in the reduction lifts to a $p$-torsion point in $\mathbb{Q}_p$ do the following. Take any lift $P\in E(\mathbb{Q}_p)$. Then $Q=pP\in \hat E(p\mathbb{Z}_p)$ is the image of $\tilde P$ under the boundary map. Now check if $Q$ belongs to $p\hat E(p\mathbb{Z}_p) = \hat E(p^2\mathbb{Z}_p)$ by looking at the valuation at a parameter $t(Q)=-x(Q)/y(Q)$ of the formal group.

Refined versions of this will work for any type of reduction over any local field.

$\endgroup$
4
  • $\begingroup$ The Sage code given in the previous post (now deleted) was not working. Can you suggest some other code for computing the $p$-torsion over $\mathbb{Q}_p$ ? $\endgroup$ – Suman May 26 '16 at 10:07
  • $\begingroup$ Yes, it was working. And for small $p$ it is very good to use that code. Otherwise the above translates easily to code. But this site is not meant to answer that sort of questions. $\endgroup$ – Chris Wuthrich May 26 '16 at 10:13
  • $\begingroup$ I tried for $p=11$. May be that's why it was not working. $\endgroup$ – Suman May 27 '16 at 8:05
  • 2
    $\begingroup$ sage: E = EllipticCurve("57a1") sage: Ep = E.base_extend(Qp(11,100)) sage: f = Ep.torsion_polynomial(11) sage: r = f.roots() sage: len(r) 0 shows that there are no11-torsion on that curve over $\mathbb{Q}_{11}$. $\endgroup$ – Chris Wuthrich May 27 '16 at 9:17

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.