Is $\mathbb{Z}_{p^\infty}$ a divisible group?

Question

Let $$\mathbb{Z}_{p^\infty}=\{x\in \mathbb{Q}/\mathbb{Z}: \ \exists n\in \mathbb{N} ,\ \ p^nx=0\}. $$

Is $\mathbb{Z}_{p^\infty}$ a divisible group ?

Answer 1

Yes, $\mathbb Z_{p^\infty}$ is a divisible group. To prove it, let us pick any positive integer $n$ and any element $x\in \mathbb Z_{p^\infty}$. We have to prove that there exists some $y\in \mathbb Z_{p^\infty}$ such that $x = ny$.

First, let us write $n=p^k m$, where $k$ is a non-negative integer and $m$ is an integer coprime to $p$, and represent $x$ by a rational number $\frac a{p^\ell}$, where $a$ is an integer coprime to $p$ and $\ell$ is a non-negative integer.

Now, $m$ is invertible in $\mathbb Z / p^\ell \mathbb Z$, that is, there exists an integer $m'$ such that $mm' \equiv 1$ modulo $p^\ell$. Set $b=am'$. The rational number $\frac b{p^\ell}$ represents an element $z$ of $\mathbb Z_{p^\infty}$ and one has $m\frac b{p^\ell} \equiv \frac a{p^\ell}$ modulo $1$, hence $mz=x$ in $\mathbb Z_{p^\infty}$. Finally, take the element $y\in \mathbb Z_{p^\infty}$ represented by $\frac b{p^{k+\ell}}$. We get $p^k y = z$, hence $$ ny = mp^k y = mz = x \, . $$ QED

EDIT (complement following comment by the author of the question): settting $Z:=\{[\frac a{p^n}]\in \mathbb Q /\mathbb Z \, : \ a,n \in \mathbb Z \text{ and } n\geq 0\}$, let us prove that $Z = \mathbb Z_{p^\infty}$. First, the inclusion $Z \subseteq \mathbb Z_{p^\infty}$ comes from the fact that $p^n [\frac a{p^n}] = 0$ in $\mathbb Q /\mathbb Z$ for all $a \in \mathbb Z$ and all $n \in \mathbb Z$ such that $n\geq 0$. Conversely, given $x\in \mathbb Z_{p^\infty}$,there exists $n\geq 0$ such that $p^n x = 0$ in $\mathbb Q /\mathbb Z$. Representing $x$ by a rational $\tilde x \in \mathbb Q$, this means that $p^n \tilde x$ is an integer. Hence, $\tilde x = \frac a{p^n}$ for some $a\in\mathbb Z$, which means that $x$ lies in $Z$. This proves the inclusion $Z \supseteq \mathbb Z_{p^\infty}$.