[proofplan] We use two standard forms of Faltings's Tate theorem for abelian varieties over number fields as external inputs: the rational full-faithfulness statement after tensoring with $\mathbb{Q}_l$, and the integral saturation statement for the Hom lattice inside Tate-module homomorphisms. The rational theorem identifies the ambient $\mathbb{Q}_l$-space of $G_K$-equivariant maps, while the saturation lemma rules out hidden $l$-adic denominators and gives the integral isomorphism. Finally, we deduce the isogeny criterion by using $\mathbb{Q}$-density in $\mathbb{Q}_l$ to replace an $l$-adic rational isomorphism by an actual rational homomorphism whose Tate-module determinant is nonzero. [/proofplan]

[step:Declare the Tate modules and the natural comparison map] For each integer $n \geq 1$, let $A[l^n]$ and $B[l^n]$ denote the finite subgroup schemes of $l^n$-torsion points of $A$ and $B$ over a fixed [algebraic closure](/page/Algebraic%20Closure) $\overline{K}$ of $K$. Define the $l$-adic Tate modules \begin{align*} T_lA &= \varprojlim_{n} A[l^n](\overline{K}), & T_lB &= \varprojlim_{n} B[l^n](\overline{K}), \end{align*} where the transition maps are multiplication by $l$. These are free $\mathbb{Z}_l$-modules of ranks $2\dim A$ and $2\dim B$, equipped with continuous actions of the absolute [Galois group](/page/Galois%20Group) $G_K := \operatorname{Gal}(\overline{K}/K)$. Every $K$-morphism $f: A \to B$ sends $A[l^n]$ into $B[l^n]$ for every $n$, hence induces a compatible family of group homomorphisms $A[l^n](\overline{K}) \to B[l^n](\overline{K})$. Since $f$ is defined over $K$, these homomorphisms commute with the $G_K$-action. Passing to inverse limits gives a $\mathbb{Z}_l$-linear $G_K$-equivariant map $T_lf: T_lA \to T_lB$. Thus we obtain the natural $\mathbb{Z}_l$-[linear map](/page/Linear%20Map) \begin{align*} \Phi_l: \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l &\longrightarrow \operatorname{Hom}_{G_K}(T_lA,T_lB), \\ f\otimes a &\longmapsto a\,T_lf . \end{align*} [/step]

[step:Use Faltings's rational Tate theorem to identify the rational Hom space]Let \begin{align*} V_lA &:= T_lA\otimes_{\mathbb{Z}_l}\mathbb{Q}_l, & V_lB &:= T_lB\otimes_{\mathbb{Z}_l}\mathbb{Q}_l \end{align*} be the rational Tate modules, with the induced continuous $G_K$-actions. We invoke the external input [Faltings's rational Tate theorem for abelian varieties over number fields](/page/Faltings%20Tate%20Theorem): for abelian varieties $A$ and $B$ over the number field $K$, the natural map \begin{align*} \Phi_{l,\mathbb{Q}}: \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Q}_l &\longrightarrow \operatorname{Hom}_{G_K}(V_lA,V_lB) \end{align*} is an isomorphism of $\mathbb{Q}_l$-vector spaces. This is the rational full-faithfulness part of Faltings's theorem, not the integral statement being proved here. The hypotheses of that theorem are exactly satisfied here: $A$ and $B$ are abelian varieties over the number field $K$, and $l$ is a prime. Tensoring $\Phi_l$ with $\mathbb{Q}_l$ gives $\Phi_{l,\mathbb{Q}}$. Therefore $\Phi_l$ is injective, because a homomorphism between finite free $\mathbb{Z}_l$-modules whose scalar extension to $\mathbb{Q}_l$ is injective is itself injective.[/step]

[guided]The deep part of the theorem is the rational full-faithfulness statement proved by Faltings. It says that, once denominators in $l$ are allowed, no extra $G_K$-equivariant maps appear on Tate modules: every element of $\operatorname{Hom}_{G_K}(V_lA,V_lB)$ comes from a unique element of $\operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Q}_l$. We verify that this result applies. Its input consists of two abelian varieties over a number field and a prime $l$. The theorem statement supplies exactly these data: $A$ and $B$ are abelian varieties over the number field $K$, and $l$ is a prime. Hence the external rational full-faithfulness theorem gives an isomorphism \begin{align*} \Phi_{l,\mathbb{Q}}: \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Q}_l &\longrightarrow \operatorname{Hom}_{G_K}(V_lA,V_lB). \end{align*} The integral map $\Phi_l$ becomes this rational map after tensoring with $\mathbb{Q}_l$, because $V_lA = T_lA\otimes_{\mathbb{Z}_l}\mathbb{Q}_l$ and similarly for $B$. Since $\Phi_{l,\mathbb{Q}}$ is injective, an element in the kernel of $\Phi_l$ becomes zero after tensoring with $\mathbb{Q}_l$. The source $\operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l$ is a finite free $\mathbb{Z}_l$-module, so it has no $\mathbb{Z}_l$-torsion. Therefore the element was already zero, and $\Phi_l$ is injective.[/guided]

[step:Show the integral lattice is saturated]Let $u \in \operatorname{Hom}_{G_K}(T_lA,T_lB)$. Its scalar extension \begin{align*} u_{\mathbb{Q}}: V_lA &\to V_lB, \\ x\otimes c &\mapsto u(x)\otimes c \end{align*} is $\mathbb{Q}_l$-linear and $G_K$-equivariant. By the rational isomorphism in the previous step, there exists a unique element \begin{align*} \alpha \in \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Q}_l \end{align*} such that $\Phi_{l,\mathbb{Q}}(\alpha)=u_{\mathbb{Q}}$. It remains to prove that $\alpha$ lies in the $\mathbb{Z}_l$-lattice $\operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l$. Choose the smallest integer $m \geq 0$ for which \begin{align*} l^m\alpha \in \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l. \end{align*} Write $\beta := l^m\alpha$. If $m>0$, then \begin{align*} \Phi_l(\beta)=l^m u \end{align*} has image contained in $l^mT_lB$, hence in $lT_lB$. Therefore the induced map \begin{align*} \overline{\Phi_l(\beta)}: T_lA/lT_lA &\to T_lB/lT_lB \end{align*} is zero. The reduction of $\Phi_l(\beta)$ modulo $l$ is the action of $\beta$ on $A[l](\overline{K})$. Thus $\beta$ kills $A[l](\overline{K})$. We now invoke the external input [Tate module saturation for abelian-variety Hom lattices](/page/Tate%20Module%20Saturation): if $A$ and $B$ are abelian varieties over a field of characteristic $0$, then the image of \begin{align*} \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l \end{align*} in $\operatorname{Hom}_{\mathbb{Z}_l}(T_lA,T_lB)$ is $l$-saturated. Equivalently, whenever an element of $\operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l$ acts as zero on $A[l](\overline{K})$, it is divisible by $l$ in that $\mathbb{Z}_l$-module. The hypotheses of this saturation theorem are satisfied because $K$ is a number field, hence has characteristic $0$, and $A$ and $B$ are abelian varieties over $K$. Applying it to $\beta$ gives $\beta=l\gamma$ for some \begin{align*} \gamma \in \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l. \end{align*} Then \begin{align*} l^{m-1}\alpha = \gamma, \end{align*} contradicting the minimality of $m$. Therefore $m=0$, so $\alpha \in \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l$. Since $u=\Phi_l(\alpha)$, the map $\Phi_l$ is surjective.[/step]

[guided]We have already used Faltings's theorem to obtain a rational preimage of $u$. The only possible obstruction to surjectivity of the integral map is denominators: the rational preimage might a priori involve a negative power of $l$. We now rule this out. Extend $u$ to a map on rational Tate modules by defining \begin{align*} u_{\mathbb{Q}}: V_lA &\to V_lB, \\ x\otimes c &\mapsto u(x)\otimes c. \end{align*} This map is $\mathbb{Q}_l$-linear and $G_K$-equivariant because $u$ is $\mathbb{Z}_l$-linear and $G_K$-equivariant. Faltings's rational theorem gives a unique \begin{align*} \alpha \in \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Q}_l \end{align*} with $\Phi_{l,\mathbb{Q}}(\alpha)=u_{\mathbb{Q}}$. Choose the smallest integer $m \geq 0$ such that $l^m\alpha$ lies in the integral lattice $\operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l$, and define $\beta := l^m\alpha$. If $m>0$, then the induced Tate-module map satisfies \begin{align*} \Phi_l(\beta)=l^m u. \end{align*} Since $u(T_lA)\subseteq T_lB$, the image of $l^m u$ is contained in $l^mT_lB$, and because $m>0$ this is contained in $lT_lB$. Therefore the reduction modulo $l$ of $\Phi_l(\beta)$ is the zero map \begin{align*} \overline{\Phi_l(\beta)}: T_lA/lT_lA &\to T_lB/lT_lB. \end{align*} Now $T_lA/lT_lA$ identifies with $A[l](\overline{K})$, and the same is true for $B$. Under these identifications, the reduction of $\Phi_l(\beta)$ modulo $l$ is exactly the map induced by $\beta$ on $l$-torsion. Thus $\beta$ kills $A[l](\overline{K})$. We use the external input [Tate module saturation for abelian-variety Hom lattices](/page/Tate%20Module%20Saturation). The theorem says that, for abelian varieties over a field of characteristic $0$, the image of \begin{align*} \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l \end{align*} in $\operatorname{Hom}_{\mathbb{Z}_l}(T_lA,T_lB)$ is $l$-saturated. In the present form, this means: if an element of $\operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l$ acts as zero on $A[l](\overline{K})$, then it is divisible by $l$ in that $\mathbb{Z}_l$-module. The hypotheses apply because $K$ is a number field, so $\operatorname{char} K=0$, and $A$ and $B$ are abelian varieties over $K$. Applying the saturation theorem to $\beta$ gives \begin{align*} \beta = l\gamma \end{align*} for some $\gamma \in \operatorname{Hom}_K(A,B)\otimes_{\mathbb{Z}}\mathbb{Z}_l$. But then \begin{align*} l^{m-1}\alpha = \gamma, \end{align*} which contradicts the minimality of $m$. Hence $m=0$, so the rational preimage $\alpha$ was already integral. Therefore every $u \in \operatorname{Hom}_{G_K}(T_lA,T_lB)$ is in the image of $\Phi_l$.[/guided]

[step:Derive the isogeny criterion from the Hom isomorphism] Assume first that $A$ and $B$ are $K$-isogenous. Let $f: A \to B$ be a $K$-isogeny. Since $f$ is finite and surjective, its kernel is finite. Therefore, after tensoring with $\mathbb{Q}_l$, the induced map \begin{align*} V_lf: V_lA &\to V_lB \end{align*} is an isomorphism of $G_K$-representations. Conversely, assume that $V_lA \cong V_lB$ as $G_K$-representations. Let \begin{align*} \psi: V_lA &\to V_lB \end{align*} be a $G_K$-equivariant $\mathbb{Q}_l$-linear isomorphism. Then $\dim_{\mathbb{Q}_l}V_lA=\dim_{\mathbb{Q}_l}V_lB$, so $2\dim A=2\dim B$ and therefore $\dim A=\dim B$. Set $M:=\operatorname{Hom}_K(A,B)$. By Faltings's rational isomorphism, there exists \begin{align*} \alpha \in M\otimes_{\mathbb{Z}}\mathbb{Q}_l \end{align*} with $\Phi_{l,\mathbb{Q}}(\alpha)=\psi$. Choose $\mathbb{Q}_l$-bases of $V_lA$ and $V_lB$. With respect to these bases, define \begin{align*} D: M\otimes_{\mathbb{Z}}\mathbb{Q}_l &\to \mathbb{Q}_l \end{align*} by taking $D(\eta)$ to be the determinant of the matrix of $\Phi_{l,\mathbb{Q}}(\eta):V_lA\to V_lB$. This is a polynomial function on the finite-dimensional $\mathbb{Q}_l$-[vector space](/page/Vector%20Space) $M\otimes_{\mathbb{Z}}\mathbb{Q}_l$, and $D(\alpha)=\det(\psi)\neq 0$. The subspace $M\otimes_{\mathbb{Z}}\mathbb{Q}$ is dense in $M\otimes_{\mathbb{Z}}\mathbb{Q}_l$ for the $l$-adic topology, because $\mathbb{Q}$ is dense in $\mathbb{Q}_l$ after choosing a $\mathbb{Z}$-basis of the finitely generated torsion-free group $M$. Since the nonvanishing locus $\{\eta:D(\eta)\neq 0\}$ is open and contains $\alpha$, choose \begin{align*} \eta \in M\otimes_{\mathbb{Z}}\mathbb{Q} \end{align*} with $D(\eta)\neq 0$. Choose an integer $N\geq 1$ clearing the rational denominators of $\eta$, so that \begin{align*} f:=N\eta \in M=\operatorname{Hom}_K(A,B). \end{align*} Then $D(f)=N^{2\dim A}D(\eta)\neq 0$, so \begin{align*} V_lf:V_lA&\to V_lB \end{align*} is an isomorphism. Because $V_lf$ is injective, the kernel of $f:A\to B$ has no positive-dimensional abelian subvariety: otherwise that subvariety would contribute a nonzero rational Tate submodule killed by $V_lf$. Hence $\ker f$ is finite. The image $f(A)$ is an abelian subvariety of $B$ of dimension $\dim A$, and since $\dim A=\dim B$, it follows that $f(A)=B$. Thus $f$ is a surjective $K$-homomorphism with finite kernel, so $f$ is a $K$-isogeny. This proves both the isomorphism of Hom modules and the stated criterion for $K$-isogeny. [/step]