Is Prüfer group divisible over the ring of $p$-adic integers?

Question

Let $p$ be a prime number. Let $$\mathbb{Z}_{p}=\left\{\sum_{i=0}^\infty a_ip^i\mid a_i\in \{0,1,2,\dots,p-1\}\right\}~~and~~\mathbb{Z}_{p^\infty}=\left\{ \frac{a}{p^n}+\mathbb{Z}\mid a\in \mathbb{Z}\text{ and }n\in \mathbb{N}\right\}$$ denote the ring of $p$-adic integers and Prüfer $p$-group, respectively. Then, $\mathbb{Z}_{p^\infty}$ form a module over $\mathbb{Z}_{p}$ with respect to following external binary operation:

Let $\lambda=\sum_{i=0}^\infty a_ip^i\in \mathbb{Z}_{p}$ and $\frac{a}{p^n}+\mathbb{Z}\in \mathbb{Z}_{p^\infty}$ such that $o(\frac{a}{p^n}+\mathbb{Z})=n$. Then $(\sum_{i=0}^\infty a_ip^i)(\frac{a}{p^n}+\mathbb{Z})=(\sum_{i=0}^{n-1} a_ip^i)(\frac{a}{p^n})+\mathbb{Z}$. (see Is Prüfer group a module over the ring of $p$-adic integers? ). Now, I want to prove that this module is divisible.

My approach: Let $\lambda=\sum_{i=0}^\infty a_ip^i\in \mathbb{Z}_{p}$ and $\frac{a}{p^n}+\mathbb{Z}\in \mathbb{Z}_{p^\infty}$ such that $o(\frac{a}{p^n}+\mathbb{Z})=n$. Since $\mathbb{Z}_{p^\infty}$ is a divisible $\mathbb{Z}$-module, for any $\lambda_k= \sum_{i=0}^{k-1} a_ip^i\in \mathbb{Z}$ and $\frac{a}{p^n}+\mathbb{Z}\in \mathbb{Z}_{p^\infty}$, there exists $\frac{b}{p^r}+\mathbb{Z}\in \mathbb{Z}_{p^\infty}$ such that $o(\frac{b}{p^r}+\mathbb{Z})=r$ and $(\sum_{i=0}^{k-1} a_ip^i)(\frac{b}{p^r}+\mathbb{Z})=(\sum_{i=0}^{k-1} a_ip^i)(\frac{b}{p^r})+\mathbb{Z}=\frac{a}{p^n}+\mathbb{Z}$. There are following three cases:

1. If $r= k$, then $(\sum_{i=0}^{k-1} a_ip^i)(\frac{b}{p^r})+\mathbb{Z}=(\sum_{i=0}^{\infty} a_ip^i)(\frac{b}{p^r}+\mathbb{Z})$.
2. If $r>k$, take $a_i=0,\forall i\in \{k, k+1,\cdots,r-1\}$ and then $(\sum_{i=0}^{k-1} a_ip^i)(\frac{b}{p^r})+\mathbb{Z}=(\sum_{i=0}^{r-1} a_ip^i)(\frac{b}{p^r})+\mathbb{Z}=(\sum_{i=0}^{\infty} a_ip^i)(\frac{b}{p^r}+\mathbb{Z})$.
3. If $r<k$, then terms after $r-1$ will be absorbed by $\mathbb{Z})$. Hence, again we have $(\sum_{i=0}^{k-1} a_ip^i)(\frac{b}{p^r})+\mathbb{Z}=(\sum_{i=0}^{r-1} a_ip^i)(\frac{b}{p^r})+\mathbb{Z}=(\sum_{i=0}^{\infty} a_ip^i)(\frac{b}{p^r}+\mathbb{Z})$.

This shows that module is divisible. But, by direct calculation, I am not getting any conclusion.

Answer 1

$\Bbb Z_p$ is a DVR, i.e. a PID with a unique irreducible element up to units, in this case $p$ is a unique irreducible element up to a unit. By unique factorization in PIDs, we have that every non-zero element of $x \in \Bbb Z_p$ may be written as $x=p^n\cdot u$ for some $n \in \Bbb N_0$ and $u \in \Bbb Z_p$. From this, we can deduce the following criterion:

Let $M$ be a $\Bbb Z_p$-module, then $M$ is divisible if and only if multiplication by $p$ as an endomorphism of $M$ is surjective.

Proof Clearly if $M$ is divsible, then multiplication by $p$ is surjective.
Suppose that multiplication by $p$ is surjective. We need to show that multiplication by any non-zero scalar $x \in \Bbb Z_p$ is surjective. But we know that $x$ may be written as $x=p^n \cdot u$, so multiplication by $x$ may be written as a composition of multiplication by $p$ ($n$-times) and multiplication by $u$. Multiplication by a unit is an automorphism, just from the module axioms. As compositions of surjective maps are surjective, we are done.

With this criterion, we can easily show that $\Bbb Z_{p^\infty}$ is divisible: if $\frac{a}{p^n}+\Bbb Z \in \Bbb Z_{p^\infty}$, then we have $\frac{a}{p^n}+\Bbb Z=p(\frac{a}{p^{n+1}}+\Bbb Z)$.