Let $E$ be an elliptic curve over $\mathbb{Q}$, and let $p$ and $\ell$ be two distinct primes of good reduction. Let $T_\ell = T_\ell(E) = \varprojlim E[\ell^n](\overline{\mathbb{Q}})$ be the $\ell$-adic Tate module, and let $F_p \in \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ be a Frobenius element at $p$. Then $F_p$ acts $\mathbb{Z}_\ell$-linearly on $T_\ell$, and this action depends only up to conjugation on the choice of $F_p$. In particular, its characteristic polynomial is well-defined. A basic result is that the coefficients of this characteristic polynomial are integers.

This last fact is usually proved by considering the reduction of $E$ modulo $p$, which does not change the $\ell$-adic Tate module, and using that we can realize the $F_p$-action in characteristic $p$ as coming from an actual morphism of elliptic curves, namely the Frobenius morphism $E \to E^{[p]}$. But I was wondering if it is possible to give a more direct proof, namely by constructing a $\mathbb{Z}$-lattice $\Lambda \subset T_\ell$ (by which I mean a rank 2 free $\mathbb{Z}$-module such that the map $\Lambda \otimes \mathbb{Z}_\ell \to T_\ell$ is an isomorphism) which is preserved by $F_p$ in the sense that $F_p(\Lambda) \subset \Lambda$ (note that one cannot expect equality here since the determinant of $F_p$ acting on $T_\ell$ is $p$). Certainly, if you already know that $F_p$ has integral characteristic polynomial, then you can easily construct such lattices: take any $t \in T_\ell \setminus \ell T_\ell$ that is not an eigenvector for $F_p$, then $\Lambda = t \mathbb{Z} + F_p(t)\mathbb{Z} \subset T_\ell$ is an $F_p$-invariant lattice. So there should be plenty such lattices. But the goal is to construct an $F_p$-invariant lattice without using that we already know that $F_p$ has integral characteristic polynomial.

One potential lattice can be constructed as follows. We choose a complex-analytic uniformization $E(\mathbb{C}) = \mathbb{C}/\Lambda_0$ for some lattice $\Lambda_0 \subset \mathbb{C}$. Then we define a map $\Lambda_0 \to T_\ell$ by sending $\lambda \in \Lambda_0$ to the sequence $(\ell^{-1} \lambda, \ell^{-2} \lambda, \ell^{-3}\lambda, \ldots) \in T_\ell$, which is well-defined because $\ell^{-n}\lambda \in E(\mathbb{C})[\ell^n] = E(\overline{\mathbb{Q}})[\ell^n]$. Let $\Lambda_\ell \subset T_\ell$ be the image of this map. It is not hard to prove that $\Lambda_\ell$ is free of rank 2 and that $\Lambda_\ell \otimes \mathbb{Z}_\ell \to T_\ell$ is an isomorphism. Also note that $\Lambda_\ell$ does not depend on the choice of the uniformization.

Question: Does $F_p(\Lambda_\ell) \subset \Lambda_\ell$ hold?

P.S. I've tried searching for results in this direction in various places, but did not find much. If someone has suggestions for references or keywords to search for, I would be much obliged.


2 Answers 2


There is a subtle problem with this idea, that causes serious problems. You observed that $\Lambda_\ell \otimes \mathbb Z_\ell = T_\ell$ but didn't find any other information for it. There is a reason for that.

Let $K$ be the field generated by the coordinates of the $\ell$-power torsion points of $E$. Given an $\ell$-power torsion point defined over $F$, to make Frobenius act on it, we need to know its reduction mod $p$, so we need to embed $F$ into the maximal unramified extension $\mathbb Q_p^{ur}$ of $\mathbb Q_p$.

Given a point in the homology of $E_{\mathbb C}$, to find the corresponding point of $F$, we need to express the coordinates as complex numbers, so we need to embed $F$ into $\mathbb C$.

Are these embeddings canonical? Well, if we define $F$ as the field generated by the complex coordinates of $\ell$-power torsion points, then the second embedding is canonical but the first isn't. If we define $F$ as the field generated by the $p$-adic coefficients of torsion points, then the first embedding is canonical but the second isn't. So regardless, there is some ambiguity - we can translate one of our embeddings by an automorphism of $F$ and get one that looks equally reasonable.

How bad is that ambiguity? Fixing an automorphism $\sigma \in \operatorname{Gal}(F/\mathbb Q)$ of $F$, making this change of embeddings corresponds exactly to translating your lattice by the action of $\sigma$ on $T_\ell(E)$. So the set of lattices we obtain your construction is a $\operatorname{Gal}(F/\mathbb Q)$-orbit in $T_\ell(E)$.

For $E$ generic, we have $\operatorname{Gal}(F/\mathbb Q) \cong GL_2(\mathbb Z_\ell)$, so the orbit is quite large. In fact every single lattice $\Lambda$ with $\Lambda \otimes \mathbb Z_\ell = T_\ell$ lies in this orbit, because we can find a matrix in $GL_2$ taking the basis of one such lattice to another. So there is no more information available about these lattices than your initial observation that $\Lambda \otimes \mathbb Z_\ell = T_\ell$!

Of course, there are examples of such $\Lambda$ stable under $F$ and examples not stable under $F$.

For any $E$ non-CM, the situation is the same, because the Galois group is an open subgroup of $GL_2(\mathbb Z_\ell)$ and these act transitively on the set $GL_2(\mathbb Z_\ell)/GL_2(\mathbb Z)$ of lattices $\Lambda$, since $GL_2(\mathbb Z)$ is dense in $GL_2(\mathbb Z_\ell)$.

For $E$ CM, the situation is different, as the Galois group is much smaller. If $p$ is a supersingular prime, then David Speyer's argument shows $\Lambda_\ell$ is never stable under Frobenius. Conversely, if $p$ is an ordinary prime, then the endomorphism $V =p /F$ lifts to an endomorphism of the curve over the CM field and thus an endomorphism of the curve over $\mathbb Q$, thus always preserves $\Lambda_\ell$, and because its determinant is $p$, $F= p/V$ necessarily preserves $\Lambda_\ell$ as well. So for CM curves, Frobenius preserves this lattice if and only if $p$ is ordinary.

  • $\begingroup$ Thanks! You are right, the question is subtly but seriously problematic. I actually mentioned in the question already that the action of $F_p$ is well-determined only up conjugation, but somehow I did not realize that this makes the question of whether $F_p(\Lambda_\ell) \subset \Lambda_\ell$ completely meaningless in most cases. $\endgroup$
    – Marc Paul
    Sep 22, 2020 at 16:05

Any construction along these lines is going to run into an obstruction pointed out by Serre. Consider the elliptic curve $E = \{ y^2 = x^3+x \}$ over $\mathbb{Z}[i]$, and let $p$ be a prime which is $3 \bmod 4$. Let $E/p$ be the reduction of $E$ modulo $p$ (which remains prime in $\mathbb{Z}[i]$). Then $E/p$ has the following endomorphisms:

  • The $p$-power Frobenius $F(x,y) = (x^p, y^p)$ and
  • The complex muliplication $J(x,y) = (-x, iy)$.

These maps obey $JF=-FJ$, $J^2 = -1$ and $F^2 = -p$.

There do not exist $2 \times 2$ integer matrices obeying these relations. (Proof below.) So there is no construction which associates a $\mathbb{Z}$-lattice to an elliptic curve and is functorial in charateristic $p$. So it is impossible that $J$ and $F$ both preserve your lattice. I didn't think about this in detail, but it seems much more likely that $J$ does than $F$.

Proof that there are not integer matrices obeying $J^2 = -1$, $JF = -FJ$ and $F^2 = -p$: Suppose otherwise. Using $J^2 = -1$, we can choose bases so that $J = \left[ \begin{smallmatrix} 0&-1 \\ 1&0 \end{smallmatrix} \right]$. The equation $JF=-FJ$ means that $F$ is of the form $F = \left[ \begin{smallmatrix} a&b \\ b&-a \end{smallmatrix} \right]$. Then $F^2 = (a^2+b^2) \mathrm{Id}$. There is no solution to $a^2+b^2 = -p$ in integers.

  • 1
    $\begingroup$ $J$ certainly preserves the lattice because it extends to an endomorphism over the complex numbers. So indeed $F$ must not. $\endgroup$
    – Will Sawin
    Sep 22, 2020 at 15:07
  • $\begingroup$ Thanks! This argument is quite famous and I have actually seen this argument before, so I should have known that what I was trying to do had no chance of ever working... $\endgroup$
    – Marc Paul
    Sep 22, 2020 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.