Note that the last condition implies equality of the Weil polynomials thanks to
Exercise 19.6.2. Given this, the only issue is to prove that if the
\(L\)-polynomials coincide, then
\(A_1\) and
\(A_2\) are isogenous. Recall that the action of Frobenius on
\(T_\ell (A_i) \otimes_{\ZZ} \QQ\) is semisimple, and so is determined up to isomorphism by its characteristic polynomial. Consequently,
\(\Hom_{\ZZ_{\ell}[G_k]}(T_{\ell}(A_1),T_{\ell}(A_2))\) contains a map of full rank; by Tate’s theorem,
\(\Hom(A_1, A_2)\) must do so also.
Let us spell out this last point in more detail. If we start with an element of
\(\Hom_{\ZZ_{\ell}[G_k]}(T_{\ell}(A_1),T_{\ell}(A_2))\) of full rank, any sufficiently close
\(\ell\)-adic approximation
\(\varphi \in \Hom(A_1, A_2)\) also has full rank as an element of
\(\Hom_{\ZZ_{\ell}[G_k]}(T_{\ell}(A_1),T_{\ell}(A_2))\text{.}\) Now note that the connected part of
\(\ker(\varphi)\) has zero rational Tate module, and hence must be the zero abelian variety; similarly,
\(\coker(\varphi)\) must be the zero abelian variety. We conclude that
\(\varphi\) is an isogeny.