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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.

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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.

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  • $\begingroup$ 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 at 16:05
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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.

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    $\begingroup$ $J$ certainly preserves the lattice because it extends to an endomorphism over the complex numbers. So indeed $F$ must not. $\endgroup$ – Will Sawin Sep 22 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 at 16:09

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