Let $k$ is a finite field, $G=\text{Gal}(\bar k/k)$ its absolute Galois group, and $l=\text{char}(k)$ is any prime.
Tate's Isogeny Theorem:
For all elliptic curves $E_1$, $E_2$ defined over $k$, the map $$ \phi: \text{Hom}_k(E_1,~E_2) \otimes_{\mathbb{Z}} \mathbb{Z}_l \to \text{Hom}_G(T_l(E_1),~T_l(E_2))$$ is isomorphic.
My question:
Why do we take tensor product by $\mathbb{Z}_l$ in the map ?
Why not simply $ \phi: \text{Hom}_k(E_1,~E_2) \to \text{Hom}_G(T_l(E_1),~T_l(E_2))$ ?
Is it because $\text{Hom}_G(T_l(E_1),~T_l(E_2))$ is a $\mathbb{Z}_l$-module of $G$-equivariant maps between $l$-adic Tate modules $T_l(E_1)$ and $T_l(E_2)$ ?
Kindly explain it
Best Answer
The map has no hope of being an isomorphism without tensoring the left hand side by $\mathbb{Z}_\ell$. Without tensoring with $\mathbb{Z}_\ell$, the left hand side is a finitely generated free abelian group and the right hand side is a finitely generated $\mathbb{Z}_\ell$ module. They don't even have the same cardinality.