TITLE: What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$?
QUESTION [5 upvotes]: 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?

REPLY [7 votes]: YCor's answer is good, but here's a slightly different way of seeing it. 
Given $x \in \mathbb Q$, let $[f(x)]$ be any lift of $f(x)$ from $\mathbb Z[1/p]/\mathbb Z$ to $ \mathbb Z[1/p]$. Then $p^n [ f(1/p^n)]$ is a $p$-adic Cauchy sequence because $p^n [f(1/p^n)] - p^m [ f(1/p^m)]$ is an integer multiple of $p^{\min (n,m)}$. Changing the lift also causes a difference by an integer multiple of $p^n$ and thus does not affect the limit. This defines for each $f$ an element of $\mathbb Q_p$.
To go the other way, given $\alpha \in \mathbb Q_p$, observe that multiplication by $\alpha$ defines a map $\mathbb Q$ to $\mathbb Q_p$. Compose this with the projection to $\mathbb Q_p / \mathbb Z_p \cong \mathbb Z[1/p] / \mathbb Z$ to obtain a map $\mathbb Q \to \mathbb Z[1/p]$.
It is straightforward to check that these are inverses. Starting with $\alpha$,  $[f(1/p^n)]$ is an element of $\mathbb Z_p$ plus $\alpha /p^n$ so $p^n [f(1/p^n)]$ is an approximation of $\alpha$. Starting with $f$, the limit $\alpha$ differs from $p^n [f(1/p^n)]$ by an element of $p^n \mathbb Z_p$, so $f(1/p^n)$ differs from $\alpha p^{-n}$ by an element of $\mathbb Z_p$, and then multiply these by elements of $\mathbb Z$ inverting every prime other than $p$ to get all rational numbers.