algebra, algebraic geometry, number theory

An Application to Elliptic Curves

Let’s do an application of our theorems about finitely generated projective modules over Dedekind domains. This is another one of those things that seems to be quite well known to experts, but it is not written anywhere that I know of. Suppose {E} and {F} are elliptic curves defined over a number field {K} (this works in more generality, but this assumption will allow us to not break into lots of weird cases), and assume that the {\ell}-adic Tate modules are isomorphic for all {\ell}.

Recall briefly that the {\ell}-adic Tate module is just the limit over all the {\ell^n} torsion points, i.e. {\displaystyle T_\ell(E)=\lim_{\longleftarrow} E(\overline{K})[\ell^n]} as a {G=Gal(\overline{K}/K)}-module. We discussed this before in this post. An isogeny {\phi: E\rightarrow F} defined over {K} induces an action via pushforward {T_\ell (E)\rightarrow T_\ell (F)} which is Galois equivariant. In fact, if {\ell \nmid \text{deg}(\phi)}, then it induces an isomorphism of Tate modules.

First define {Hom(E,F)} to be the set of isogenies over {K} (this could get me in trouble and has been the main delay in this post). If {E} and {F} are isogenous, then the natural action of {End(E)} by composing turns {Hom(E,F)} into a rank {1} projective module over {End(E)}.

The question we want to ask ourselves is how much information do we get from the Tate module. It seems that surely this would not be enough information to recover the curve up to isomorphism, but recall that most elliptic curves do not have complex multiplication. Let’s start with that case. Suppose {E} is non-CM so that {End(E)\simeq \mathbb{Z}}. The only endomorphisms are the isogenies given by multiplication by an integer.

The Tate conjecture formally says that there is an isomorphism {Hom(E,F)\otimes_\mathbb{Z} \mathbb{Z}_\ell \stackrel{\sim}{\rightarrow}Hom_G(T_\ell(E), T_\ell(F))} (proved by Faltings in this case). This tells us that if you have some isomorphism {T_\ell(E)\simeq T_\ell(F)}, then there is an isogeny that induces it (maybe there is a less powerful tool to see this in this case). But we’ve now assumed that {End(E)} is {\mathbb{Z}}, so {Hom(E,F)} is not just a locally free module of rank {1}, but just plain free of rank {1}. All other isogenies are just composing this one with multiplication by an integer.

Let {\phi} be the generator of {Hom(E,F)}. If {deg(\phi)} is {n}, then all isogenies are divisible by {n}. Since we assume the Tate modules are isomorphic for all {\ell}, just pick some {\ell} that divides {n}. Since Tate says there is an isogeny inducing the isomorphism we get a contradiction unless {n=1}. Thus {deg(\phi)=1} and hence the generator is actually an isomorphism. This proves a fact I’ve seen stated, but haven’t seen written anywhere. If {E} and {F} are non-CM elliptic curves with isomorphic Tate modules for all {\ell}, then they must be isomorphic.

This should seem a little strange, because it basically says we can recover the curve up to isomorphism merely from knowing {H_1}. It turns out that weirder things can happen for CM curves, but we can use our structure theory from the last post to figure out what is going on. Suppose now that {End(E)} is the full ring of integers in a quadratic imaginary field (the only other possibility is that it is merely an order in such a field).

It turns out that if {E} and {F} have isomorphic Tate modules for all {\ell}, then we can’t just conclude they are isomorphic. Here is a good way to think about this. We have that {End(E)} is a Dedekind domain, and {Hom(E,F)} is a rank {1} projective module over it, so it is either generated by {1} element and hence free in which case the same type of argument will show {E} and {F} must be isomorphic. The reason we get no information in the case where it is generated by two things is that these degrees can be coprime. In fact, they must be or else the same argument gives an isomorphism again.

This recently came up in something I was working on, and I couldn’t believe that I couldn’t find this fact stated anywhere (but several number theorists confirmed that this was something they knew). It might be because introductory books don’t want to assume the Tate conjecture, and anything that does assume the Tate conjecture assumes you can figure this out for yourself.


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