Viewing $\mathbb{Q}_p/\mathbb{Z}_p$ as a direct limit

Question

I have been told that taking the direct limit of the system $\mathbb{Z}/p^n\mathbb{Z}\rightarrow\mathbb{Z}/p^{n+1}\mathbb{Z}$ with respect to multiplication by $p$ maps gives us $\mathbb{Q}_p/\mathbb{Z}_p$. It's not clear to me why this is true. What is the best way to go about proving this? Can we define an explicit isomorphism, or should we use the universal property for direct limits? It's clear that in $\mathbb{Q}_p/\mathbb{Z}_p$ every element has order $p^n$ for some $n\geq 0$ since its elements are cosets of the form $\sum_{i=-n}^0 a_ip^i+\mathbb{Z}_p$. So, essentially, we'd like to show that $\lim_\rightarrow \mathbb{Z}/p^n$ consists of roots of unity, each with order a power of $p$. I'm not sure how to show the direct limit also has this property. If we're trying to set up an isomorphism explicitly, one option must be to look at the coefficients of cosets $\sum_{i=-n}^0 a_ip^i+\mathbb{Z}_p$ and define some element of the direct limit, but I think possibly the group operations in the direct limit are inhibiting my ability to see the 'obvious' (is it obvious??) choice for such map.

Answer 1

$\require{AMScd}$Consider the following commutative diagram of abelian groups:

which gives rise to a commutative diagram with exact rows: \begin{CD} \{0\}@>>>\Bbb Z/p^n\Bbb Z@>\varphi_n>>\Bbb Q_p/\Bbb Z_p@>\psi_n>>\Bbb Q_p/p^{-n}\Bbb Z_p@>>>\{0\}\\ @.@V\mu_nVV@|@VV\nu_nV\\ \{0\}@>>>\Bbb Z/p^{n+1}\Bbb Z@>>\varphi_{n+1}>\Bbb Q_p/\Bbb Z_p@>>\psi_{n+1}>\Bbb Q_p/p^{-n-1}\Bbb Z_p@>>>\{0\} \end{CD} By exactness of direct limits, we get an exact sequence $$\{0\}\to\varinjlim\Bbb Z/p^n\Bbb Z\xrightarrow\varphi\Bbb Q_p/\Bbb Z_p\to\varinjlim\Bbb Q_p/p^{-n}\Bbb Z_p\to\{0\}$$ Since $$\varinjlim\Bbb Q_p/p^{-n}\Bbb Z_p\cong\Bbb Q_p/\bigcup_{n\in\Bbb N}p^{-n}\Bbb Z_p=\{0\}$$ this proves $\varphi$ to be an isomorphism.