compute $\text{Hom}(\mathbb Z_p, \mathbb Z)$ (p-adic integers) How can I compute $\text{Hom}(\mathbb Z_p, \mathbb Z)$ or $\text{Hom}(\mathbb Z_p, \mathbb Q)$? (By $\mathbb Z_p$ I mean p-adic integers.) 
 A: Not sure this a fully satisfactory answer, but I want to share the following construction proving, at least when assuming a suitable set theory, the existence of very many homomorphisms of additive groups from $\Bbb{Z}_p$ to $\Bbb{Q}$, and also to provide a certain kind of census of them.
Consider the field of $p$-adic numbers $\Bbb{Q}_p$. It is an extension of the field $\Bbb{Q}$, so it is a vector space over $\Bbb{Q}$. If we accept the Axiom of Choice this implies that there exists a $\Bbb{Q}$-basis $\mathcal{B}$ of $\Bbb{Q}_p$. Furthermore, as $\Bbb{Q}_p$ is uncountable, so is $\mathcal{B}$.
We thus have a large supply of $\Bbb{Q}$-linear mappings $f:\Bbb{Q}_p\to\Bbb{Q}$. Namely, any function $\tilde{f}:\mathcal{B}\to\Bbb{Q}$ gives rise to a unique linear transformation $f$ in the usual way.
Composing any such $f$ with the inclusion mapping $i:\Bbb{Z}_p\to\Bbb{Q}_p$ gives a homomorphism $F:=f\circ i:\Bbb{Z}_p\to\Bbb{Q}$.
A few remarks are due:


*

*Any homomorphism $f$ of additive groups from $\Bbb{Q}_p\to\Bbb{Q}$ is actually a linear transformation. If $q=m/n\in\Bbb{Q}$ and $z\in\Bbb{Q}_p$ are arbitrary then $y:=f(qz)$ must satisfy the equation $$ny=nf(qz)=f(nqz)=f(mz)=mf(z).$$ This equation has a unique solution $y=qf(z)$ in $\Bbb{Q}$ proving linearity.

*Every element $z\in\mathcal{B}$ is of the form $z=x/p^{n}$ for some $x\in\Bbb{Z}_p$ and $n\in\Bbb{N}$. So without loss of generality we can assume that $\mathcal{B}\subset\Bbb{Z}_p$. Consequently different choices of $f$ give rise to different homomorphisms $F$. 

*Because $\Bbb{Q}$ is divisible, it is an injective object in the category of abelian groups. Therefore any homomorphism $F:\Bbb{Z}_p\to\Bbb{Q}$ can be lifted to a corresponding homomorphism $f:\Bbb{Q}_p\to\Bbb{Q}$ such that $F=f\circ i$. In other words, we have accounted for all the homomorphisms of addtive groups $\in Hom(\Bbb{Z}_p,\Bbb{Q}).$



The homomorphisms of additive groups $\Bbb{Z}_p\to\Bbb{Q}$ are the restrictions of $\Bbb{Q}$-linear transformations $\Bbb{Q}_p\to\Bbb{Q}$. Distinct transformations have distinct restrictions, so in this sense
  $\text{Hom}(\mathbb Z_p, \mathbb Q)$ is the dual space $\widehat{\Bbb{Q}_p}.$

