Abelian group of finite rank Let given torsion free abelian group $A$ of finite rank. Let for prime number $p$, given that $\cap_i p^iA =\{0\}$. Is it true that for any $p$- torsion abelian group $B$, $\text{Hom}_{\mathbb{Z}}(A, B)$ is torsion $\mathbb{Z}$ module.
 A: Let $p$ be any prime. There exists a subgroup $A$ of $\mathbf{Z}[1/p]^2$ containing $\mathbf{Z}^2$ such that $\bigcap_n p^nA=\{0\}$ and $A/\mathbf{Z}^2$ is infinite (isomorphic to the quasi-cyclic group $P_p=\mathbf{Z}[1/p]/\mathbf{Z}$). Then $\mathrm{Hom}(A,P_p)$ contains $\mathrm{Hom}(P_p,P_p)\simeq\mathbf{Z}_p$ (the $p$-adics), so is not torsion.
A: Yes, i think it is true. Take $A\otimes_{\mathbb{Z}}\mathbf{Z}_p$ (tensor product with $p$-adic ring). If $e_1, ..., e_n\in A$ is basis of $A/pA$, then $A\otimes_{\mathbb{Z}}\mathbf{Z}_p = e_1\otimes_{\mathbb{Z}}\mathbf{Z}_p +... + e_n\otimes_{\mathbb{Z}}\mathbf{Z}_p$. Note that $\text{Hom}(A, B)\hookrightarrow\text{Hom}_{\mathbf{Z}_p}(A\otimes_{\mathbb{Z}}\mathbf{Z}_p, B\otimes_{\mathbb{Z}}\mathbf{Z}_p)$. For any $\phi\in\text{Hom}_{\mathbf{Z}_p}(A\otimes_{\mathbb{Z}}\mathbf{Z}_p, B\otimes_{\mathbb{Z}}\mathbf{Z}_p)$ and some $k$, $k\phi(e_i) = 0$, so $k\phi = 0$. So $\text{Hom}_{\mathbf{Z}_p}(A\otimes_{\mathbb{Z}}\mathbf{Z}_p, B\otimes_{\mathbb{Z}}\mathbf{Z}_p)$-torsion and so $\text{Hom}(A, B)$ - torsion. $\Box$
