Here is the short version:

Fix an elliptic curve $E/\mathbb{Q}$. How does the torsion structure $E_d(\mathbb{Q})_{tors}$ vary, as $E_d$ runs through the quadratic twists of $E$?

Here is the longer version:

I have been playing with SAGE this morning. I inserted the elliptic curve ('11a1') $$E : y^2 + y = x^3 - x^2 - 10x - 20$$ which has rational torsion subgroup isomorphic to $\mathbb{Z}/5\mathbb{Z}$. I then computed its quadratic twist $E_d$ for all squarefree $d$ up to 2000, and observed $E_d(\mathbb{Q})_{tors}$ was always trivial.

Can it be that, in this particular family of quadratic twists, all but one of the curves have trivial torsion? Is this a general phenomenon?

(I ran this experiment for several other curves $E$ and got the same impression; that all but one of the curves in a family of twists have the same torsion structure.)

  • 3
    $\begingroup$ For non-$2$-torsion, you have $$E_d(\mathbb{Q})[n] \oplus E(\mathbb{Q})[n] = E(\mathbb{Q}(\sqrt{d})[n]$$ if $n$ is odd. So you can ask what new torsion points arise if you make a quadratic extension. For a fixed $n$, only at most three different $d$ can have that. In most cases (maybe all), it will be none or one. $\endgroup$ Sep 26, 2011 at 12:57
  • $\begingroup$ Can you (Chris Wuthrich) give a proof or some reference for this ? $\endgroup$
    – Suman
    May 2, 2013 at 10:25
  • $\begingroup$ The non-trivial Galois element $\sigma$ of $\mathbb{Q}(\sqrt{d})/\mathbb{Q}$ acts on the right hand side. The subgroup fixed by $\sigma$ is clearly the $n$-torsion over $\mathbb{Q}$. Now if $\sigma$ sends a point $P$ to $-P$, then its $y$-coordinate is of the form $z\cdot\sqrt{d}$ for some rational number $z$. Now writing out the equation between $z$ and the $x$-coordinate of $P$ gives you exactly the Weierstrass equation of the twist of $E$ by $d$. $\endgroup$ May 2, 2013 at 10:36
  • $\begingroup$ More generally, it's easy to define a map $$ E_d(\mathbb{Q}) \oplus E(\mathbb{Q}) \to E(\mathbb{Q}(\sqrt{d})) $$ whose kernel and co-kernel are $2$-groups. The result on $n$-torsion for odd $n$ is then immediate, and in addition one obtains the well-known and useful formula $$ \operatorname{rank} E_d(\mathbb{Q}) + \operatorname{rank} E(\mathbb{Q}) \ = \operatorname{rank} E(\mathbb{Q}(\sqrt{d})) $$ $\endgroup$ May 2, 2013 at 14:15

4 Answers 4


Theorem (originally due to Setzer?): Fix $E/\mathbb{Q}$ with $j(E)$ not 0 or 1728. Then for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ is isomorphic to $E[2](\mathbb{Q})$, so in particular $E_d(\mathbb{Q})_{tors}$ has order 1, 2, or 4. (Probably he also proved it for number fields.)

There's a paper of mine$^1$ with a much more general theorem using the theory of heights. I don't recall Setzer's proof except that it doesn't use heights.

Theorem: Let $K$ be a number field and let $A/K$ be an abelian variety with $\mu_n\subset {\rm Aut}(A)$. (This means we can twist $A$ by $n$'th roots of $d$.) Then every point $P\in A_d(K)$ satisfies one of the following two conditions:

  • $P$ is fixed by a non-trivial $\zeta\in\mu_n$.
  • $\hat h(P) \ge C_1(A)h^{(n)}(d) - C_2(A)$.

Here $\hat h$ is the canonical height relative to an ample symmetric divisor, and $h^{(n)}(d)$ is a sort of "$n$'th power free height," say equal to the minimum of $h(du^n)$ for $u\in K^*$. The constants depend on $A$, but are independent of $d$.

It follows from the theorem that after discarding finitely many $d \in K^*/{K^*}^n$, a point in $A_d(K)$ is either $1-\zeta$ torsion (hence $nP=O$), or its canonical height is positive, and hence it is nontorsion.

Of course, to describe more precisely what happens for the finitely many exceptional $d$ can be a delicate matter, as some of the other answers have indicated. I think it's interesting to see how one can approach the problem via heights or via representation theory.

$^1$ J.H. Silverman, Lower bounds for height functions, Duke Math. J. 51 (1984), 395-403.

EDIT: Fixed statement of first theorem. I'd originally written that "for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ has at most two elements." This is clearly false, since if $E$ has the form $E:y^2=(x-a)(x-b)(x-c)$ with $a,b,c\in\mathbb{Q}$, then $E_{d}[2](\mathbb{Q})$ has order 4 for every twist.

  • 2
    $\begingroup$ Thank you very much for your answer Joe, it's nice to see a result that works for abelian varieties generally. Also, if I may say, I found your two books on Elliptic Curves to be a great inspiration. $\endgroup$
    – Giuseppe
    Oct 8, 2011 at 22:30
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    $\begingroup$ Thanks for saying so. Authors always like to hear that their books are being read and enjoyed. $\endgroup$ Oct 10, 2011 at 13:16

Let me expand my comment above. While we believe that we expect this very very frequently, it is not always the case.

As I commented, we have $$ E_d(\mathbb{Q})[n]\oplus E(\mathbb{Q})[n] = E(\mathbb{Q}(\sqrt{d}))[n] $$ if $n$ is odd. So you can ask what new torsion points arise if you make a quadratic extension. For a fixed $n$, only at most three different d can have that. In most cases, this will be at most one $d$.

But for instance the curve 98a3 has no $3$-torsion defined over $\mathbb{Q}$ but its twist by $21$ and $-7$ both have. So that is an explicit counter-example to the intuition that only one $E_d$ will have $3$ torsion.

In terms of the Galois representation $E[p]$, it is the question if the image of $\bar\rho_p$ from the absolute Galois group of $\mathbb{Q}$ to $GL_2(\mathbb{F}_p)$ has a quotient isomorphic to two copies of $\mathbb{Z}/2\mathbb{Z}$. For $p=3$, as in the case above, this can happen if the image is for instance the group of diagonal matrices. But obviously that is very rarely the case.

In summary, as JSE said, a twist can have a $p$-torsion point only if $E[p]$ is reducible (i.e. it has at least one isogeny of degree $p$ from $E$ defined over $\mathbb{Q}$) and there is only one if there is a unique such isogeny.

Of course, the $2$-torsion is different. All twists will have a $2$-torsion point if one is present in $E(\mathbb{Q})$ as the Galois group of any quadratic extension will act trivially on it.

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    $\begingroup$ There's also the curious case of the curves of conductor $50$ where one twist has $3$-torsion and another twist as $5$-torsion. $\endgroup$ Sep 26, 2011 at 15:05
  • $\begingroup$ Thank you for your answer Chris. May I ask why "only at most three different d" can give us new torsion points? Is there a reference for this? $\endgroup$
    – Giuseppe
    Sep 26, 2011 at 22:30
  • $\begingroup$ Hmmm. I can't remember why I said three. The question is how many isogenies of odd prime degree are defined over Q. Maybe the maximum is two. It is certainly finite. If you need the exact number, I could start to think about it $\endgroup$ Sep 28, 2011 at 9:03

"Can it be that, in this particular family of quadratic twists, all but one of the curves have trivial torsion? Is this a general phenomenon?"

Indeed, it is very likely. The only way E_d[p] can have a rational point is if E[p] has a cyclic subgroup on which Galois acts through a quadratic character. For a "typical" elliptic curve, there will be so such p, since all the mod p Galois representations will be irreducible; for every non-Cm elliptic curve p, there are only finitely many p such that E[p] is reducible.

For each such p, the quadratic action of Gal(Q) on E[p] is what it is, and it tells you the unique d such that E_d[p] has a rational point.

So yeah, the phenomenon you're observing is very much to be expected.

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    $\begingroup$ See Chris's comment above for a correction to my assertion that there is always a unique d such that E_d[p] has a rational point. $\endgroup$
    – JSE
    Sep 27, 2011 at 3:45

The maximum number of quadratic twists with $n$-torsion (n odd) that an elliptic curve over a number field $K$ can have is 2, and here is an easy proof.

You can see this by asking how many twists with $n$-torsion can an elliptic curve which already has $n$-torsion have? If this number is $k$, then clearly an elliptic curve without $n$-torsion can have $k+1$ twists with $n$-torsion.

Suppose now that $E/K,\ E(K)\supset \mathbb Z/ n\mathbb Z$, has 2 twists $E_{d_1}$ and $E_{d_2}$ with $n$-torsion. It follows from what Chris wrote that $$E(K(\sqrt{d_1},\sqrt{d_2}))[n]\supset E(K)[n] \oplus E_{d_1}(K)[n] \oplus E_{d_2}(K)[n]\supset (\mathbb Z/ n\mathbb Z)^3,$$ which is clearly impossible.

Note that the answer depends a lot on the number field $K$ and the number $n$ you choose. The possibility of 2 twists exists only over number fields $K$ such that the complete $n$-tosion can be defined over a quadratic extension of $K$. For example, it follows that over $\mathbb Q$, an elliptic curve can have at most 2 twists with 3-torsion, 1 twist with 5 or 7-torison and 0 twists with $p$ torsion for all primes $p>7$.

So, Giuseppe, your curve $E$ with 5-torsion, as any other curve with 5-torsion, cannot have a twist with 5-torsion over $\mathbb Q$.


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