In *Invent. math.* 116, 645-649 (1994) Dinakar Ramakrishnan proves a theorem which I understand to imply that the following statement (in light of the fact that elliptic curves over $\mathbb{Q}$ are modular):


Theorem: Let $E\_1 / \mathbb{Q}$ and $E\_2 / \mathbb{Q}$ be elliptic curves and suppose that the (mod $p$) reductions of $E\_1$ and $E\_2$ have the same number of points for a set of primes of Dirichlet density $> 7/8$. Then $E\_1$ and $E\_2$ are isogeneous.


Is $7/8$ expected to be sharp here? Could it be that $7/8$ can be replaced by any $a > 0$ with the conclusion in the above Theorem unaltered?


I'm interested in answers to these questions both in the level of generality in which Ramakrishnan works and in the special case that I give above.



Very strong multiplicity one for Hecke eigenforms
Let $E$ and $F$ be two elliptic curves. For convenience, suppose that they do not have CM. For each $p$, there is a Galois representation 


\[\rho: = \rho\_E + \rho\_F : G\_{\mathbb{Q}} \to \mathrm{GL}\_2(\mathbb{F}\_p) \times \mathrm{GL}\_2(\mathbb{F}\_p)\]


given by the action of Galois on the $p$-torsion of $E$ and $F$. For sufficiently large $p$, the Galois groups $\mathbb{Q}(E[p])$ and $\mathbb{Q}(F[p])$ are $\mathrm{SL}\_2(\mathbb{F}\_p)$-extensions of $\mathbb{Q}(\zeta\_p)$. If $p$ is big enough, $\mathrm{PSL}\_2(\mathbb{F}\_p)$ is simple. Thus, either:


1). These extensions are disjoint, and $\mathrm{im}(\rho) = \ker(\det(x)/\det(y))$.


2). These extensions intersect in a have a common $\mathrm{PSL}\_2(\mathbb{F}\_p)$ extension of $\mathbb{Q}(\zeta\_p)$, and hence a $\mathrm{PGL}\_2(\mathbb{F}\_p)$ extension of $\mathbb{Q}$. This implies that their projective representations are the same. This implies that they are equal up to twisting by a character. Since their determinants are also the same, this character is either trivial or quadratic.


If $E[p] = F[p]$ infinitely often then $E$ is isogenous to $F$, using Falting's proof of the Tate conjecture.


If Case 1) occurs, one can count that the density of pairs of elements $(\sigma,\tau)$ in $\mathrm{im}(\rho)$ such that $\mathrm{Tr}(\sigma) = \mathrm{Tr}(\rho)$ is $\sim 1/p$. By Cebotarev, this implies that the number of primes $\ell$ such that $a(E,\ell) = a(F,\ell) \pmod{p}$ has density at most $\sim 1/p$. Hence, if 1) occurs infinitely many times, then $a(E,\ell) = a(F,\ell)$ for set of density zero.


If $E[p]$ is a non-trivial quadratic twist of $F[p]$ for infinitely many primes, looking at the ramification of $E$ and $F$ it is easy to see that $E[p] = F[p] \otimes \chi$ for some fixed quadratic character $\chi$ for infinitely many $p$. Yet then using Faltings again, $E$ is isogenous to the twist of $F$ by $\chi$.


PS: This is a pretty standard argument.


PPS: The argument requires the existence of Galois representations (EDIT: with big image!) attached to $E$ (and $F$), which do not exist in general (EDIT: for example, weight one forms, see Toby's answer). The argument basically works for any pair of classical modular forms $e,f$ of weight at least $2$.


