Let $k$ be a field, $\overline{k}$ its algebraic closure, and $A$ an abelian variety defined over $k$, of dimension $g$. For each integer $m \ge 1$, let $A\_m$ denote the group of elements $a \in A(\overline{k})$ such that $ma = 0$. let $l$ be a prime number different from the characteristic of $k$, and let $T\_l(A)$ denote the projective limit of the groups $A\_{l^n}$ with respect to the maps $A\_{l^{n + 1}} \to A\_{l^n}$ which are induced by multiplication by $l$. It is well known that $T\_l(A)$ is a free module of rank $2g$ over the ring $\mathbb{Z}\_l$ of $l$-adic integers. The group $G = \text{Gal}(\overline{k}/k)$ operators on $T\_l(A)$.


Let $A'$ and $A^{\prime\prime}$ be abelian varieties defined over $k$. The group $\text{Hom}\_k(A', A^{\prime\prime})$ of homomorphisms of $A'$ into $A^{\prime\prime}$ defined over $k$ is $\mathbb{Z}$-free, and the canonical map$$\mathbb{Z}\_\ell \otimes \text{Hom}\_k(A', A^{\prime\prime}) \to \text{Hom}\_G(T\_l(A'), T\_l(A^{\prime\prime}))\tag\*{$(1)$}$$is injective.



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> **Theorem (Tate, 1966).** If $k$ is finite, the map $(1)$ is bijective.
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**Question.** Could anybody supply an outline of Tate's proof of this theorem and contribute their intuition as to why such a result is true?



On Tate's "Endomorphisms of Abelian Varieties over Finite Fields", sketch of proof of main result?
I don't have any contribution for the intuition beyond the fact that, I can't construct something outside the image of (1) so I hope it's surjective.


Here is a sketch of the central idea of Tate's proof. Consider $A = A' \times A''$ and try to find an endomorphism of $A$ from an endomorphism of its Tate module. The endomorphism of the Tate module gives you subgroups $G\_n$ of $A\_{\ell^n}$ for every $n$ and you can consider $A/G\_n$. Tate uses that there exists a moduli space of abelian varieties which is a variety over $k$ and thus has finitely many points over $k$ to show that the $A/G\_n$ have to repeat, i.e., for some $n, A/G\_n$ is isomorphic to some $A/G\_m, m \ne n$. This isomorphism produces an element of the endomorphism ring of $A$, tensored with $\mathbb{Q}$. This gives the required element on the right hand side of (1).


Incidentally, Faltings's proof that (1) is surjective for number fields follows the same strategy, except he shows that the height of $A/G\_n$ is bounded to get that they repeat. 


