Is torsion of an elliptic curve determined by its reductions modulo primes?

Question

A friend of mine has posed the following question to me:$\newcommand{\Q}{\mathbb Q}\newcommand{\F}{\mathbb F}\newcommand{\Z}{\mathbb Z}$

Let $E$ be an elliptic curve over $\Q$. Is it the case that $|E(\Q)_{tors}|$ is equal to the greatest common divisor of $|E(\F_p)|$ over all primes $p>2$ of good reduction of $E$?

The former certainly divides the latter, since $E(\Q)_{tors}$ embeds in $E(\F_p)$ for all odd primes of good reduction. The answer to the above question is no, as illustrated by an elliptic curve $E:y^2=x^3+x$, which has $E(\Q)_{tors}\cong\Z/2$, yet $|E(\F_p)|$ is divisible by $4$ for all $p\geq 3$. However, if we know the structure of $E(\F_p)$ for all $p$, then we can still recover $E(\Q)_{tors}$ - for instance, $E(\F_3)\cong\Z/4$ and $E(\F_5)\cong\Z/2\times\Z/2$ and so $E(\Q)_{tors}$, which embeds into both of those, must be equal to $\Z/2$ (it can't be smaller due to the obvious point $(0,0)$). This is Example 4.6 from Silverman-Tate's book.

Based on this latter observation, the friend has further queried whether it is always necessarily the case, that is whether $E(\Q)_{tors}$ can be always determined to be the largest group which embeds into $E(\F_p)$ for all odd primes of good reduction. I would be quite surprised if that was the case, and I vaguely recall seeing a counterexample, but I had no luck finding it online. We are in particular interested in the following special case, which is the main question in this post:

Suppose $E(\Q)_{tors}$ is trivial. Does it follow that there is no nontrivial group which embeds into $E(\F_p)$ for all odd primes of good reduction of $E$?

Answer 1

$\newcommand{\Z}{\Bbb Z}$Suppose the $p$-torsion is a nontrivial extension of $\Z/p\Z$ by $\mu_p$, say defined over $\Bbb Q(\zeta_p, N^{1/p})$. The Galois group of this is the Frobenius group of order $p(p-1)$ acting via matrices which are upper triangular and have a $1$ in the lower right corner. Exercise: every element of this group acting on $E[p]$ has a fixed vector. Thus at some prime of good reduction $q$, the Galois action (which factors through Frobenius) also has a fixed point so there will always be a $p$-torsion point mod $q$ even though there are no global points. There are examples for small p where the curve $E$ has this form. I guess an example which comes to mind is on of the three curves of conductor $11$ (the one without $5$-torsion! It’s the one which is neither $X_0(11)$ or $X_1(11)$ but sometimes called $X_2(11)$) which has trivial torsion.

Answer 2

I'm extremely late to the game here, but I think any answer to this question should mention the following theorem of Katz:

Theorem (Katz, '81): Let $E/K$ be an elliptic curve, and let $m\in \mathbb N$. Suppose that, for all but finitely many places $v$ of $K$, $$\#E(k_v) \equiv 0\pmod m,$$ where $k_v$ is the residue field. Then $E$ is isogenous over $K$ to an elliptic curve $E'$, such that $m\mid \#E(K)_{tors}$.

The proof makes crucial use of the bijection between:

• Elliptic curves $E'$ with a cyclic $\ell^n$-isogeny $E\to E'$.
• $G_K$-stable lattices inside the Galois representation $V_\ell(E) = T_\ell(E)\otimes {\mathbb Q}_\ell$.

In particular, if $\ell\mid \#E(k_v)$ for almost all $v$, then although $E$ may not contain a $\ell$-torsion point, it must be $\ell$-power isogenous over $K$ to an elliptic curve $E'$ that does. In particular, as a $G_K$ representation, $E[\ell]$ must be reducible.

Recall that, if $\rho\colon G\to \mathrm{GL}_2(\mathbb Q_\ell)$ is a Galois representation, then $\rho$ has a $G$-stable lattice $\Lambda\cong \mathbb Z_\ell^2\subset\mathbb Q_\ell^2$. With respect to this lattice, we have $\rho_\Lambda\colon G\to\mathrm{GL}_2(\mathbb Z_\ell)$, and we can define the residual representation $\overline\rho_\Lambda$ by composing with the map $\mathbb Z_\ell\to\mathbb F_\ell$. In general, different choices of lattices $\Lambda$ give non-isomorphic residual representations $\overline\rho_\Lambda$. However, the semisimplification $\overline\rho_\Lambda^{ss}$ does not depend on $\Lambda$.

There is a famous lemma, due to Ribet (Prop 2.1), which says:

Theorem (Ribet, '76): Let $\rho\colon G\to \mathrm{GL}_2(\mathbb Q_\ell)$ be an irreducible Galois representation, and let $\overline \rho^{ss}$ be the semisimplification of its reduction modulo $\ell$. Suppose that $\overline\rho^{ss} = \phi_1\oplus\phi_2$. Then there is a lattice $\Lambda\subset \mathbb Q_\ell^2$ such that $$\rho_\Lambda\simeq \begin{pmatrix}\phi_1&*\\0&\phi_2\end{pmatrix}$$ is reducible, but not semisimple.

Note that this theorem is symmetric: we are free to swap $\phi_1$ and $\phi_2$.

Now let's go back to the case of elliptic curves over $\mathbb Q$. Let $E/\mathbb Q$ be any elliptic curve such that $\ell\mid \#E(\mathbb F_p)$ for almost all $p$. Assume that $E$ is not CM, so that the Galois representation $\rho := V_\ell(E)$ is absolutely irreducible.

By Katz's theorem, $E$ is $\mathbb Q$-isogenous to an elliptic curve $E'$ with an $\ell$-torsion point. Hence, the Galois representation $E'[\ell]$ contains the trivial representation as a subrepresentation: a torsion point $P$ generates a line inside $E'[\ell]$ on which $G_\mathbb Q$ acts trivially. Since the determinant of $E'[\ell]$ is the cyclotomic character $\epsilon$, we see that $\overline \rho^{ss} = 1 \oplus \epsilon$.

By Ribet's Lemma, we can find a lattice $\Lambda\subset V_\ell(E')$ such that, with respect to this lattice, $$\overline\rho_\Lambda \simeq \begin{pmatrix} \epsilon & *\\0& 1\end{pmatrix}.$$

By the dictionary between lattices and isogenous curves, we have an elliptic curve $E''$, isogenous to $E$, such that $E''[\ell]\cong \overline\rho_\Lambda$.

Now, crucially, suppose that $\ell \ne 2$.

This elliptic curve $E''$ will typically give a counterexample to your question.

Indeed, since $\ell \ne 2$, $\epsilon$ is not the trivial character. Hence, $E''[\ell]$ does not contain the trivial character as a subrepresentation, so $E''[\ell](\mathbb Q) = 0$. But, by construction, $\ell\mid E''(\mathbb F_p)$ for almost all $p$!

The only thing that can go wrong is if $E$ had $\ell'$-torsion for a different prime $\ell'$. In particular, over $\mathbb Q$, we obtain a huge class of counterexamples (including the one in Mr. Basil's question), just by looking at elliptic curves with prime power isogenies.

For example, take any elliptic curve over $\mathbb Q$ with a $3$-isogeny, but no other isogeny. Then there will be two elliptic curves $E, E'$ in the isogeny class. One of them will have a $3$-torsion point. The other will not, and will give you a counterexample.