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.
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1$\begingroup$ Would you define what you mean by "rank"? There are many possible non-equivalent meanings. $\endgroup$– YCorCommented Nov 18, 2015 at 17:19
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$\begingroup$ en.m.wikipedia.org/wiki/Rank_of_an_abelian_group $\endgroup$– solver6Commented Nov 18, 2015 at 18:12
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$\begingroup$ OK thanks. To avoid ambiguity it's easy to just say $\mathbb{Q}$-rank or torsion-free rank. (Well there's indeed less ambiguity as soon as the group is assumed torsion-free.) $\endgroup$– YCorCommented Nov 18, 2015 at 18:18
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2 Answers
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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.
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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. $\Box$
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$\begingroup$ The problem in this argument is that you lose the condition $\bigcap p^nA=0$ when you tensor with $\mathbf{Z}_p$ (this is not very visible since you don't explicitly use it... anyway if $A$ is in my example, $A\otimes\mathbf{Z}_p\simeq\mathbf{Q}_p^2$ and $\mathrm{Hom}(\mathbf{Q}_p^2,\mathbf{Q}_p/\mathbf{Z}_p)$ is not torsion. Or you seem to assume that $A\otimes\mathbf{Z}_p$ is finitely generated as $\mathbf{Z}_p$-module, which is not the case. $\endgroup$– YCorCommented Nov 18, 2015 at 23:25
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$\begingroup$ But $A\otimes\mathbf{Z}_p = e_1\otimes\mathbf{Z}_p+...+e_n\otimes\mathbf{Z}_p$ can't be isomorfic to $\mathbf{Q}_p^2$? $\endgroup$ Commented Nov 19, 2015 at 6:29
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$\begingroup$ I used that $A\otimes\mathbf{Z}_p\cong\varprojlim_i p^iA = e_1\otimes\mathbf{Z}_p+...+e_n\otimes\mathbf{Z}_p$ $\endgroup$ Commented Nov 19, 2015 at 6:39
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$\begingroup$ Your claim that lifts of generators mod $p$ generate at the $p$-adic level is false. $\endgroup$– YCorCommented Nov 19, 2015 at 12:33