What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?


I have a reference that says the group in question is $\mathbb{Q}\_p,$ the additive group of the quotient field of the $p$-adic integers. Can anyone provide a reasonable derivation of this result?



What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?
Note that this equality is as topological groups, where Hom is endowed with uniform convergence on compact subsets. 


Every such homomorphism is zero on $p^n$ for some $n$. Hence the given homomorphism group $G=\mathrm{Hom}(\mathbf{Q},\mathbf{Z}(p^{\infty}))$ can be written as $\bigcup G\_n$, where $G\_n$ is the set of homomorphisms that vanish on $p^n$. Note that $G\_n$ is open.


By definition,
$$G\_0=\mathrm{Hom}(\mathbf{Q}/\mathbf{Z},\mathbf{Z}(p^\infty))=\mathrm{Hom}(\bigoplus\_\ell\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=\mathrm{End}(\mathbf{Z}(p^\infty)),$$
because for prime $\ell\neq p$ we have $\mathrm{Hom}(\mathbf{Z}(\ell^\infty),\mathbf{Z}(p^\infty))=0$. It is classical that $\mathrm{End}(\mathbf{Z}(p^\infty))$ is isomorphic to $\mathbf{Z}\_p$ as a topological ring (details upon request).


Hence each $G\_n$ is isomorphic to $\mathbf{Z}\_p$. Multiplication by $p$ maps $G\_n$ into $G\_{n-1}$. Also, mapping $f$ to $x\mapsto f(p^{-1}x)$ maps $G\_{n-1}$ into $G\_n$ and is an inverse to multiplication by $p$. So indeed $pG\_n=G\_{n-1}$, and we deduce the result.




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Edit: consider $\mathbf{Q}\_p$ acting on itself as its own endomophism group (as topological group). Restricting to $\mathbf{Q}$ at the origin, and modding out by $\mathbf{Z}\_p$ at the target yields a topological group homomorphism $\phi:\mathbf{Q}\_p\to\mathrm{Hom}(\mathbf{Q}\_p/\mathbf{Z}\_p)$ (where the latter Hom is in topological groups). This homomorphism $\phi$ is injective: indeed, an element $f$ in the kernel of $\phi$ would, in restriction to $\mathbf{Q}$, satisfy $f(\mathbf{Q}\_p)\subset\mathbf{Z}\_p$. Since $\mathbf{Q}\_p$ is divisible it has no nontrivial finite quotient and hence no nonzero homomorphism into any profinite group. So $f=0$.


Keeping in mind that there is a canonical isomorphism $\mathbf{Z}(p^\infty)\to\mathbf{Q}\_p/\mathbf{Z}\_p$, since we previously obtained that $\mathrm{Hom}(\mathbf{Q}\_p/\mathbf{Z}\_p)\simeq\mathbf{Q}\_p$ as topological group, we deduce that $\phi$ is a isomorphism. 


I should also add that it's important to phrase the result in terms of topological groups. Indeed, as an abstract group, $\mathbf{Q}\_p$ is just isomorphic to $\mathbf{Q}^{(c)}$ (and also to $\mathbf{R}$).


