$K$ a number field, $G_K$ its Galois group, $E_1, E_2$ two elliptic curves defined over $K$. The isogeny theorem says that if for some prime number $\ell$, The Tate modules (tensored with $\mathbb{Q}$) $V_{\ell}(E_1)$ is isomorphic to $V_{\ell}(E_2)$ as Galois modules. Then these two elliptic curves are isogenous.

My question is, when are these two curves isomorphic? Namely, what more invariants are needed to fully characterize the elliptic curve (besides the Tate modules). Thanks!

  • $\begingroup$ If you are interested in isomorphisms over the algebraic closure: $E_1$ and $E_2$ are isomorphic (over $\overline{K}$) iff their $j$-invariants are the same. Or are you interested in isomorphisms over $K$? $\endgroup$
    – felix
    Oct 12, 2010 at 19:12
  • $\begingroup$ I made a mistake in the original post, using $T_{\ell}$ of elliptic curves. It should use $V_{\ell}$. Thanks to Emerton for pointing out. $\endgroup$
    – natura
    Oct 13, 2010 at 1:11

1 Answer 1


If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by the twist by a locally free rank $1$ module over the endomorphism ring of one of them. This is true for all abelian varieties but for elliptic curves we only have two kinds of possibilities for the endomorphism ring; either $\mathbb Z$ or an order in an imaginary quadratic field. In the first case there is only one rank $1$ module so the curves are isomorphic. In the case of an order we get that the numbe of twists is a class number.

Addendum: Concretely, we have that $\mathrm{Hom}(E_1,E_2)$ is a rank $1$ projective module over $\mathrm{End}(E_1,E_1)$ (under the assumption that the Tate modules are isomorphic) and then $E_2$ is isomorphic to $\mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1)}E_1$ (the tensor product is defined by presenting $\mathrm{Hom}(E_1,E_2)$ as the kernel of an idempotent $n\times n$-matrix with entries in $\mathrm{End}(E_1)$ and $E_2$ is the kernel of the same matrix acting on $E_1^n$. Hence, given $E_1$ $E_2$ is determined by $\mathrm{Hom}(E_1,E_2)$ and every rank $1$ projective module appears in this way.

Addendum 1: Note that I was talking here about the $\mathbb Z_\ell$ (and not $\mathbb Q_\ell$ Tate modules. You can divide up the classification of elliptic curves in two stages: First you see if the $V_\ell$ are isomorphic (and there it is enough to look at a single $\ell$). If they are, then the curves are isogenous. Then the second step is to look within an isogeny class and try to classify those curves.

The way I am talking about here goes directly to looking at the $T_\ell$ for all $\ell$. If they are non-isomorphic (for even a single $\ell$ then the curves are not isomorphic and if they are isomorphic for all $\ell$ they still may or may not be isomorphic, the difference between them is given by a rank $1$ locally free module over the endomorphism ring. In any case they are certainly isogenous. These can be seen a priori as if all $T_\ell$ are isomorphic so are all the $V_\ell$ but also a posteriori essentially because a rank $1$ locally free module becomes free of rank $1$ when tensored with $\mathbb Q$.

Of course the a posteriori argument is in some sense cheating because the way you show that the curves differ by a twist by a rank $1$ locally free module is to use the precise form of the Tate conjecture: $$ \mathrm{Hom}(E_1,E_2)\bigotimes \mathbb Z_\ell = \mathrm{Hom}_{\mathcal G}(T_\ell(E_1),T_\ell(E_2)) $$ which for a single $\ell$ gives the isogeny.

Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection.

  • $\begingroup$ If I understand it correctly, in simpler terms, the above means: The two are isomorphic if and only if the kernel of the isogeny is a principal ideal in $\text{End} (E1,E2)$ (which is independent of which isogeny we take). Is this correct? $\endgroup$ Oct 12, 2010 at 21:27
  • $\begingroup$ Also, in the spirit of the question, do we need all Tate modules to be isomorphic? Or is it enough that for a positive density of $\ell$'s they are isomorphic? $\endgroup$ Oct 12, 2010 at 21:29
  • 2
    $\begingroup$ The kernel of the isogeny is finite and hence can not be an ideal in $\mathrm{End}(E_1)$. If $E_1$ and $E_2$ are just isogeneous, then their Tate modules are isomorphic for all but a finite number of $\ell$'s so density $1$ is not sufficient. $\endgroup$ Oct 12, 2010 at 21:33
  • $\begingroup$ Oops. Of course. The latter hints that there is actually a finite list of primes that need to be checked. I'll throw a guess once more: primes that divide conductor are enough? $\endgroup$ Oct 12, 2010 at 21:42
  • 1
    $\begingroup$ Dear Basic, Your assertion that "The Tate modules ... are isomorphic ... if [they are so] for just one prime" is not true. (I don't have Serre's book in front of me, but I presume he states this after tensoring with $\mathbb Q$.) E.g. if $E_1 \to E_2$ via a cyclic 5 isogeny (think about $X_1(11) \to X_0(11)$) then this will induce an isomorphism of $\ell$-adic Tate modules for every $\ell \neq 5$, but will not induce an isomorphism of $5$-adic Tate modules. (It will induce an embedding of one into the other, and the image will be of index 5. Furthermore, the $5$-adic Tate modules will ... $\endgroup$
    – Emerton
    Oct 12, 2010 at 23:46

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