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.



Is this lattice in the Tate module of an elliptic curve, coming from complex-analytic uniformization, stable under Frobenius?
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.


