$\newcommand{\cts}{\mathrm{cts}}$Thanks for your reading. Let $A,B$ be two $\mathbb{Z}_p$-modules, where $\mathbb{Z}_p$ is the $p$-adic integer ring. I have two questions.

Is $\mathrm{Hom}_{\mathbb{Z}}(A,B)=\mathrm{Hom}_{\mathbb{Z}_p}(A,B)$? I think the answer may be no, but I can't find a counterexample. I don't have an idea about the difference between the two.

Is $\mathrm{Hom}_{\mathbb{Z}_p}(A,B)=\mathrm{Hom}_{\cts}(A,B)$, where $A,B$ have $p$-adic topology and $\mathrm{Hom}_{\cts}$ means continuous homomorphism between the two topological modules? I guess it is true, but I'm not sure how to prove it. And if $A,B$ have other linear topology, maybe it will not be true?

Let me explain my motivation for asking the second question, since I'm not sure the thought is right. I find that there are many cases in algebraic number theory using Pontryagin dual as an important tool. In fact, for a profinite group $G$, the dual is defined as $\mathrm{Hom}_{\cts}(G,\mathbb{R}/\mathbb{Z})$. Since its image should be finite, so $\mathrm{Hom}_{\cts}(G,\mathbb{R}/\mathbb{Z})=\mathrm{Hom}_{\cts}(G,\mathbb{Q}/\mathbb{Z})$. For $p$-adic Lie group $G$, since any integer $n \neq p$ is a unit in $\mathbb{Z}_p$, then for $\mathbb{Z}_p$-module $G$( I think there always exists a $\mathbb{Z}_p$-module structure on $G$? not 100% sure), so non-trivial maps in $\mathrm{Hom}_{\cts}(G,\mathbb{Q}/\mathbb{Z})$ should have image in $\mathbb{Q}/\mathbb{Z}(p) \cong \mathbb{Q}_p/\mathbb{Z}_p$. So the Pontryagin dual actually should be $\mathrm{Hom}_{\cts}(G,\mathbb{Q}_p/\mathbb{Z}_p)$. However I find in many cases it is also written as $\mathrm{Hom}_{\mathbb{Z}_p}(G,\mathbb{Q}_p/\mathbb{Z}_p)$. So I guess we have the general conclusion as the second one.

Thanks for any comments and answers.

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