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Let $A$ be an abelian variety over a field $K$. It is shown that its $p$-adic Tate module $T_p(A)= \varprojlim_{n} A[p^n](\overline{K}) \cong Hom(\mathbb{Q}_p/\mathbb{Z}_p,A(\overline{K})= \varprojlim Hom (\mathbb{Z}/p^n \mathbb{Z}, A(\overline{K}))$.

Now I consider the following projective system $[p]: A(\overline{K}) \rightarrow A(\overline{K}) $ and I want to compute

$B(A)= \varprojlim \left(A(\overline{K}) \overset{[p]}{\leftarrow} A(\overline{K}) \leftarrow \cdots \leftarrow A(\overline{K}) \leftarrow \cdots \right)$

We have the exact sequence $0 \rightarrow T_p A \rightarrow B(A) \rightarrow A(\overline{K}) \rightarrow 0$

My question is can I have any similar result that is $B(A) \cong Hom(?,A(\overline{K}))$ ?

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I think the group $(\mathbf{Q}_p/\mathbf{Z}_p)\oplus\mathbf{Z}[\tfrac1p]$ does the job, i.e. $$ \mathrm{Hom}((\mathbf{Q}_p/\mathbf{Z}_p)\oplus\mathbf{Z}[\tfrac1p],A(\bar K))\cong B(A). $$ Indeed, let $A(\bar K)_{\mathrm{tor}}$ be the torsion subgroup of $A(\bar K)$. We have a short exact sequence $$ 0\rightarrow A(\bar K)_{\mathrm{tor}}\rightarrow A(\bar K)\rightarrow A(\bar K)/A(\bar K)_{\mathrm{tor}}\rightarrow 0. $$ Since $A(\bar K)$ is a divisible group, the quotient group $A(\bar K)/A(\bar K)_{\mathrm{tor}}$ is a uniquely divisible abelian group, i.e., a $\mathbf{Q}$-vector space $V$. Since $A(\bar K)_{\mathrm{tor}}$ is also divisible, the sequence splits and one has an isomorphism $$ A(\bar K)\cong A(\bar K)_{\mathrm{tor}}\oplus V. $$ Denoting by $A(\bar K)[p^\infty]$ the subgroup of $A(\bar K)$ of all elements whose order is a power of $p$, and by $A(\bar K)[\not p]$ the subgroup of $A(\bar K)$ of all elements of order prime to $p$, one has $$ A(\bar K)_{\mathrm{tor}}=A(\bar K)[p^\infty]\oplus A(\bar K)[\not p]. $$ Hence, $$ A(\bar K)\cong A(\bar K)[p^\infty]\oplus W $$ with $W=A(\bar K)[\not p]\oplus V$ an abelian group for which multiplication-by-$p$ is a bijection. It follows that $$ B(A)\cong T_pA\oplus W. $$

Now, $$ \mathrm{Hom}(\mathbf{Z}[\tfrac1p],A(\bar K)[p^\infty])=0 $$ since all elements of $A(\bar K)[p^\infty]$ are of $p$-power torsion. On the other hand $$ \mathrm{Hom}(\mathbf{Z}[\tfrac1p],W)\cong W $$ since $W$ is uniquely $p$-divisible. It follows that $$ \mathrm{Hom}((\mathbf{Q}_p/\mathbf{Z}_p)\oplus\mathbf{Z}[\tfrac1p],A(\bar K))\cong T_pA\oplus W\cong B(A). $$

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    $\begingroup$ Thank you for your answer. Can you explain more about $B(A)\cong T_pA\oplus W$? It is really amazing for me! $\endgroup$
    – Desunkid
    Commented Jan 7, 2022 at 23:10
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    $\begingroup$ Limits commute with limits. In particular, $\lim(A(\bar K)[p^\infty]\times W)=(\lim A(\bar K)[p^\infty])\times (\lim W)$, and $\lim W=W$ since multiplication by $p$ is a bijection on $W$. $\endgroup$
    – Johannes Huisman
    Commented Jan 8, 2022 at 7:04
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    $\begingroup$ I think it''s stange for me because we have $0 \rightarrow T_p A \rightarrow B(A) \rightarrow A(\overline{K}) \rightarrow 0$, the map $B(A) \rightarrow A(\overline{K})$ send $(u_0,u_1,...) \mapsto u_0$. It follows that $B(A)=T_p(A) \times A(\overline{K})$ as groups? $\endgroup$
    – Desunkid
    Commented Jan 8, 2022 at 8:49
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    $\begingroup$ That exact sequence is not necessarily split. It is like $\lim \mu_{p^\infty}$, where $\mu_{p^\infty}$ is the group of roots of unity in $\mathbf C$ of order a power of $p$. That projective limit is torsion free. Hence, the surjective map from $\lim \mu_{p^\infty}$ to $\mu_{p^\infty}$ does not have a section. $\endgroup$
    – Johannes Huisman
    Commented Jan 8, 2022 at 11:19

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