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

  • $\begingroup$ Is your question why we take the tensor product with $\mathbf{Z}_l$ at all, or why we in particular take it over $\mathbf{Z}$? $\endgroup$ Jul 15, 2020 at 15:08
  • $\begingroup$ @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))$ ? $\endgroup$
    – MAS
    Jul 15, 2020 at 15:27
  • 1
    $\begingroup$ 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 $\mathbb{Z}_l$. $\endgroup$ Jul 15, 2020 at 15:45
  • $\begingroup$ @AndreaMarino,thanks $\endgroup$
    – MAS
    Jul 15, 2020 at 16:00

1 Answer 1


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.


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