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Is it possible to have an elliptic curve $E$ over $\Bbb Q$ such that $E( \overline{\Bbb Q})$ is a torsion abelian group?

I know that $E(\Bbb Q)$ can be a finite group. I know that $E(\Bbb C)$ is a complex torus, so it is never a torsion group. Finally I know that if $E(\overline{\Bbb Q})$ is torsion, then it is isomorphic to $(\Bbb Q/\Bbb Z)^2$ (notice that in particular it is not finitely generated as abelian group).

Thank you.

Alphonse
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  • Related: https://math.stackexchange.com/questions/1577405 – Watson Oct 23 '18 at 08:56
  • See corollary 19.11.7 in Vakil's notes http://math.stanford.edu/~vakil/216blog/FOAGjun1113public.pdf if the base field is uncountable and algebraically closed. – Watson Nov 23 '20 at 14:40
  • More general: https://mathoverflow.net/questions/55953/torsion-points-in-abelian-varieties-over-number-fields – Alphonse May 20 '21 at 08:35

1 Answers1

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The answer is no. Any elliptic curve has non torsion points in infinitely many quadratic extensions of $\mathbb{Q}$. Equivalently, it has infinitely many quadratic twist with rank $>0$ (and even with rank $>1$).

The proof of the first assertion is elementary: Suppose $E$ is given by an equation of the form $y^2=p(x)$, with $p(x)\in \mathbb{Q}[x]$ of degree 3 (and no double roots). Then, for any rational number $a\in \mathbb{Q}$, not a root of $p(x)$, consider the number $p(a)\in \mathbb{Q}^*$. Then the elliptic curve $E_{p(a)}$, quadratic twist of $E$ with respect of $p(a)$, so given by the equation $p(a) y^2=p(x)$, has the point $(x,y)=(a,1)$. This point is non-torsion for all $a$ outside a finite set, and you get infinitely many non-isomorphic between these curves (this is Silverman's Specialization Theorem (see Theorem 11.4 in his book on Advanced topics s in the Arithmetic of Elliptic Curves)

In fact Mestre showed that there are even infinitely many twists of rank $\ge 2$ in J-F. Mestre, Rang de courbes elliptiques d’invariant donné, C. R. Acad. Sci. Paris 314 (1992), 919–922

There is another related result: Frey and Jarden showed that elliptic curves have rank infinite over $\mathbb{Q}^{\operatorname{ab}}$. G. Frey and M. Jarden, Approximation theory and the rank of abelian varieties over large algebraic fields, Proc. London Math. Soc. (3) 28, 112–128 (1974).

So the answer is not only negative: they have always rank infinite over $\overline{\mathbb{Q}}$.

xarles
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  • Thank you, but do you have a reference for these results? I was not aware of those. – Alphonse Oct 19 '18 at 13:29
  • And you mean rank $>0$ (i.e. $\geq 1$), not $>1$, right? – Alphonse Oct 19 '18 at 13:30
  • I corrected misprint and added proofs and references. – xarles Oct 19 '18 at 16:15
  • Thank you very much, I learned a lot! Just two typos : 1) the polynomial $p$ should be named $f$ (it is written so in the sequel) 2) at the very end, I guess you want to say "they have always rank infinite over $\Bbb Q^{\text{ab}}$". – Alphonse Oct 19 '18 at 16:31
  • Thanks, typos corrected. – xarles Oct 19 '18 at 16:33
  • Dear @xarles, I would like to ask how to prove that $(a,1)$ is non-torsion on the curve $p(a) y^2 = p(x)$ ? I don't see how to apply Silverman's Specialization Theorem 11.4. – Alphonse Aug 25 '20 at 08:41