1. Let $E_1$ and $E_2$ be Elliptic curves over the field $K$ and $\ell\neq\mathrm{char}(K)$ be a prime number. Let $T_\ell(E_i)$ is the Tate module of $E_i$, $i=1,2$ and $\mathrm{Hom}_K(T_\ell(E_1),T_\ell(E_2))$ is the group of $\mathbb{Z}_\ell$-linear maps from $T_\ell(E_1)$ to $T_\ell(E_2)$ that commute with the action of $G_{\bar{K}/K}$ as given by the $\ell$-adic representation. Then the natural map $$\mathrm{Hom}_K(E_1,E_2)\otimes\mathbb{Z}_\ell\longrightarrow\mathrm{Hom}_K(T_\ell(E_1),T_\ell(E_2))$$ is an isomorphism if:

i) $K$ is a finite field.

ii) $K$ is a number field.

  1. Let $K$ be a number field and $E/K$ be an Elliptic curve without complex multiplication. Let $$\rho_\ell:G_{\bar{K}/K}\longrightarrow\mathrm{Aut}(T_\ell(E))$$ be the $\ell$-adic representation of $G_{\bar{K}/K}$ associated to $E$. Then:

i) $\rho_\ell(G_{\bar{K}/K})$ is of finite index in $\mathrm{Aut}(T_\ell(E))$ for all primes $\ell\neq\mathrm{char}(K)$.

ii)$\rho_\ell(G_{\bar{K}/K})=\mathrm{Aut}(T_\ell(E))$ for all but finitely many primes $\ell$.

I have recently started studying about Elliptic curves from the book of Silverman (The Arithmetic of Elliptic Curves) and I am a proper beginner to the theory of Elliptic curves so I am looking for the proof of 1. I checked out the cited paper of Tate in the book which further directs me to an article which is written in German so I couldn't read up anything there. And, for 2 I couldn't really understand the proof from the source cited in the book.

So, if anyone could explain me the idea behind the proofs of these two theorems or maybe just how to visualise these two theorems geometrically or algebraically I would really appreciate it.

P.S. I am not looking for a proper proof for either of this theorems.

  • 6
    $\begingroup$ 1.i) is a relatively classical result of Tate, 1.ii) is a theorem which was first established as a byproduct of Faltings's proof of Mordell conjecture. Both of these are discussed in this document (see this webpage for more context and details about Faltings's proof). Those are very difficult results, especially the latter. I don't know much of the history of 2, but I believe they are of comparable difficulty. $\endgroup$
    – Wojowu
    May 17, 2022 at 19:19

1 Answer 1


Here is a rough idea of the proof of 1. I'll highlight (with italics) concepts that you may not know now, but will be useful to learn in number theory, which hopefully will be motivated by this question.

Note that $a \in \operatorname{Hom}_K( T_\ell(E_1), T_\ell(E_2))$ induces $a_n : E_1 [\ell^n] \to E_2[\ell^n]$ for each $n$. We want to show, given such an $a \neq 0$, that there exist one or more homomorphism $E_1 \to E_2$ whose induced actions on the Tate module, in some $\mathbb Z_\ell$-linear combination, give $a$.

It's important to note that given just a single $a_n : E_1 [\ell^n] \to E_2[\ell^n]$, we cannot reach the desired conclusion. In fact, there can be such an $a_n$ for $n$ very large but no nonconstant maps $E_1 \to E_2$ at all.

So we have to take advantage of the fact that $a_n$ exists for infinitely many $n$.

We will do this by contrasting this infinitude against something else we know to be finite. For $K$ finite, this finiteness is going to ultimately boil down to the finiteness of $K$. For $K$ a number field, it's going to be subtler, but ultimately boil down to the fact that there are finitely many elements of bounded size (at each infinite place) in the ring of integers of $K$.

For each $n$, we can make an abelian surface that remembers the data of $a_n$, by writing $A_n= E_1 \times E_2 / \{ (x,y) \in E_1[\ell^n] \times E_2[\ell^n ] \mid y = a_n(x) \} $.

A priori, this appear to be infinitely many different abelian surfaces. The key claim to prove is that there are actually only finitely many different surfaces $A_n$ up to isomorphism.

Once we prove this, we will immediately get a bunch of isomorphisms $A_n \to A_m $ which combined with the natural maps $E_1 \to A_n$ and $A_m\to E_2$ will give us a lot of maps $E_1 \to E_2$. We have to do some algebra to show that we can combine these maps to recover $a$.

But the main arithmetic step is proving the finiteness of isomorphism classes.

For $K$ finite, we can first think about how we would solve the problem if the $A_i$ were elliptic curves. Then they would each have a $j$-invariant in the field $K$. There are infinitely many curves and only finitely many different $j$-invariants, and it's not so hard to see there can be finitely many (in fact, at most $6$) curves over a finite field with a given $j$-invariant, up to isomorphism.

In higher dimensions, the role of the $j$-invariant is portrayed by the coordinates of the moduli space of abelian varieties, which similarly parameterize isomorphism classes of abelian varieties. Since there are finitely many values for each coordinate, the moduli space has finitely many points. However, the moduli space only parameterizes abelian varieties with a principal polarization, so we use Zarhin's trick, which involves checking that $A_n^4$ has a principal polarization and using finiteness statements for higher-dimensional abelian varieties.

For $K$ a number field, it is much more complicated, even for elliptic curves. To get finitely many $j$-invariants, we have to show the numerator and denominator of the $j$-invariant are bounded. An equivalent statement is that the height of the $j$-invariant is bounded. To run the argument for abelian surfaces, we have to define a suitable height function on the moduli space, and show it is bounded on the sequence $A_n$.

The correct geometric method for defining a height function for this problem is Arakelov theory, which allows constructing height functions with much more precise properties than the elementary Weil's height machine. Faltings defined a specific height function that behaves very well under isogenies, like the isogenies between the $A_n$, in particular because they arise from a $p$-divisible group, enabling the use of $p$-adic Hodge theory to control the change in local terms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.