# Tate's Isogeny Theorem: Why do we take tensor product by $\mathbb{Z}_l$ over $\mathbb{Z}$ in the map?

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

• Is your question why we take the tensor product with $\mathbf{Z}_l$ at all, or why we in particular take it over $\mathbf{Z}$? Jul 15, 2020 at 15:08
• @LetGBeTheGraph, I have edited the question. My question why we are taking tensor product with $\mathbb{Z}_l$ ? Why not simply $\phi: \text{Hom}_k(E_1,~E_2) \to \text{Hom}_G(T_l(E_1),~T_l(E_2))$ ?
– MAS
Jul 15, 2020 at 15:27
• I don't know about this particular theorem, but if the right hand side is a $\mathbb{Z}_{l}$ module it would be weird to have an isomorphism on the spot, without first tensoring for $\mathbb{Z}_l$. Jul 15, 2020 at 15:45
• @AndreaMarino,thanks
– MAS
Jul 15, 2020 at 16:00

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.