Torsion subgroups in families of twists of elliptic curves

Question

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.)

Answer 1

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 all but finitely many twists.

Answer 2

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.

Answer 3

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 $\mathbb Q$.

Answer 4

"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.