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.