Field prime to characteristic torsion and Tate module

By [F1] $A[{\mathrm{\ell }}^{\nu }]$ is finite locally free of rank ${\mathrm{\ell }}^{2g\nu }$; by [F2] it is etale over $F$, so its geometric points are separable and $\mathrm{\mid }A[{\mathrm{\ell }}^{\nu }](F^{\mathrm{s}\mathrm{e}\mathrm{p}})\mathrm{\mid }={\mathrm{\ell }}^{2g\nu }$. In particular $A[\mathrm{\ell }](F^{\mathrm{s}\mathrm{e}\mathrm{p}})$ has ${\mathrm{\ell }}^{2g}$ elements, and $H=A[{\mathrm{\ell }}^{\nu }](F^{\mathrm{s}\mathrm{e}\mathrm{p}})$ is a finite abelian $\mathrm{\ell }$-group whose $\mathrm{\ell }$-torsion has ${\mathrm{\ell }}^{2g}$ elements; by the elementary divisor classification [F3], $H\cong (\mathbb{Z}\mathrm{/}{\mathrm{\ell }}^{\nu }\mathbb{Z}{)}^{2g}$.

Multiplication by $\mathrm{\ell }$ maps $H_{\nu +1}=A[{\mathrm{\ell }}^{\nu +1}](F^{\mathrm{s}\mathrm{e}\mathrm{p}})$ into $H_{\nu }$ with kernel $H_1$ of order ${\mathrm{\ell }}^{2g}$. The order computation in step 1.1 makes its image have order ${\mathrm{\ell }}^{2g\nu }$, so it is surjective. Choose a basis of $H_1$ and recursively lift each basis vector through these maps. The lifted vectors form a basis of $H_{\nu +1}$ over $\mathbb{Z}\mathrm{/}{\mathrm{\ell }}^{\nu +1}\mathbb{Z}$: a relation, after applying $\mathrm{\ell }$, has all coefficients divisible by ${\mathrm{\ell }}^{\nu }$ by the basis property in $H_{\nu }$; multiplying the lifted vectors by ${\mathrm{\ell }}^{\nu }$ gives the original basis of $H_1$, so the remaining coefficients are zero modulo $\mathrm{\ell }$. Independence and the equal orders then give generation. DC (and hence the assumed AC) permits the countable recursive choice of compatible bases. These compatible bases identify the inverse system with the reductions of $({\mathbb{Z}}_{\mathrm{\ell }}{)}^{2g}$, and hence identify its inverse limit with that module.

The projections $T_{\mathrm{\ell }}(A)\to H_{\nu }$ are surjective by the compatible-basis construction. Their kernel is ${\mathrm{\ell }}^{\nu }{T}_{\mathrm{\ell }}(A)$. Indeed the inclusion from right to left follows because $H_{\nu }$ is killed by ${\mathrm{\ell }}^{\nu }$. Conversely, for a compatible sequence $(a_r)$ with $a_{\nu }=0$, put $b_r=a_{r+\nu }$. Compatibility gives ${\mathrm{\ell }}^r{b}_r=a_{\nu }=0$, so $b_r\in H_r$, and $\mathrm{\ell }{b}_{r+1}=b_r$, so $(b_r)\in T_{\mathrm{\ell }}(A)$. Also ${\mathrm{\ell }}^{\nu }{b}_r=a_r$, giving the reverse inclusion. This proves $T_{\mathrm{\ell }}(A)\mathrm{/}{\mathrm{\ell }}^{\nu }{T}_{\mathrm{\ell }}(A)\cong H_{\nu }$.

For (c), an element $\sigma$ of ${\mathrm{Aut}}_F(F^{\mathrm{s}\mathrm{e}\mathrm{p}})$ acts on $T_{\mathrm{\ell }}(A)$ coordinatewise and on each $A[{\mathrm{\ell }}^{\nu }](F^{\mathrm{s}\mathrm{e}\mathrm{p}})$ by functoriality; the quotient identifications of step 3.1 are $\sigma$-equivariant, so $\sigma$ acts trivially on $T_{\mathrm{\ell }}(A)$ if and only if it acts trivially on each quotient, i.e. on every finite torsion group. This is an elementary group argument on top of the multiplication supplier; no $H^1$ duality statement is asserted, and the choice assumptions of [F1] persist.