# How does one get that for elliptic curves $A$, $B$ over a number field, $a_\mathfrak{l}=b_\mathfrak{l}$ implies that the curves are isogenous?

Let $K$ be a number field, and let $A$ and $B$ be two elliptic curves over $K$. For a nonzero prime ideal $\mathfrak{l}$ of $K$, outside a finite set of primes, let $a_\mathfrak{l}$ and $b_\mathfrak{l}$ be the usual error terms $N(\mathfrak{l})+1-\bar A_\mathfrak{l}(\mathcal{O}/\mathfrak{l})$ and $N(\mathfrak{l})+1-\bar B_\mathfrak{l}(\mathcal{O}/\mathfrak{l})$ respectively (here $N$ is the absolute norm and $\bar X_\mathfrak{l}$ denotes the reduction of the curve $X$ at a good place $\mathfrak{l}$).

It is well known that the collection of these error terms determines the $K$-isogeny class of the curve considered. I am at the moment missing how the "correct" proof of this fact goes.

What I see so far is that, thanks to Faltings' Isogeny Theorem, it is enough to show that ${\rm Hom}_{G_K}(T_p(A),T_p(B))$ is non-zero, for some prime $p$. Now, our assumption that the packages of error terms of $A$ and $B$ coincide for almost all primes of $K$ ensures that the $G_K$-representations $T_p(A)\otimes\mathbf{Q}_p$ and $T_p(B)\otimes\mathbf{Q}_p$ are ${\it locally}$ isomorphic for almost all primes of $K$. How do we get then that they are ${\it globally}$ isomorphic, and hence conclude

${\rm Hom}_{G_K}(T_p(A),T_p(B))\neq 0$? If they came from an automorphic form, I see that one can use the strong multiplicity one result by J-L. But otherwise how can one complete the argument?

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Since the two representations $T_p(A) \otimes \mathbb{Q}_p$ and $T_p(B) \otimes \mathbb{Q}_p$ are locally isomorphic (and unramified) for almost all primes, the traces of the Frobenius elements are the same for both. Since those representations are continuous and the Frobeniuses are dense in the Galois group by Cebotarev's theorem, the traces (characters) of the two representations are the same. Since they are irreducible, they are isomorphic. – Joël Nov 26 '11 at 13:15
Of course! Now I see, many thanks. – Tommaso Centeleghe Nov 26 '11 at 13:40
Joël: In the CM case, the representations are not necessary irreducible, are they? In any case, Faltings also proved that these representations are semisimple, so equality of traces implies isomorphy. – ACL Nov 26 '11 at 13:54
@Tommaso: you have to be a little careful with your suggested automorphic argument in your last para, because the representations may be reducible in the CM case, and a non-cuspidal auto rep for $GL(2)$ is not in general determined (even up to isomorphism) by the isomorphism class of all but finitely many of the factors ;-) – Kevin Buzzard Nov 26 '11 at 14:08
In fact, for elliptic curves, semisimplicity was already proved by Serre, see Abelian l-adic representations and elliptic curves, IV-15, Proposition. – ACL Nov 26 '11 at 14:15