Let $p \in \mathbb N$ be a prime. Let

$$Q_p : = \left \{ x \in \mathbb Q : (\exists k \in \mathbb Z)\ \mathrm {and}\ (\exists n \in \mathbb N)\ \mathrm {such}\ \mathrm {that}\ x= \frac {k} {p^n} \right \}.$$

Show that $Q_p / \mathbb Z$ is divisible as $\mathbb Z$-module.

How should I proceed? Please help me.

Thank you in advance.


closed as off-topic by Saad, José Carlos Santos, Hanul Jeon, Shailesh, Henrik Apr 8 '18 at 10:17

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  • $\begingroup$ What are your thoughts about that? What have you tried so far? Where have you got stuck? $\endgroup$ – Taroccoesbrocco Apr 8 '18 at 9:22
  • $\begingroup$ @Taro see my comment below the answer of Lord Shark the Unknown. $\endgroup$ – A.Chattopadhyay Apr 8 '18 at 10:01

You need to prove that for $x=a/p^r$ and $n\in\Bbb N$ then there is a $y\in\Bbb Q_p$ with $ny\equiv x\pmod{\Bbb Z}$. You can reduce to the case where $n$ is prime. If $n=p$ that's easy: take $y=x/p$.

So, let $n\ne p$ also be a prime. Try $y=b/p^r$ with $b$ an integer. Then one needs $nb\equiv a\pmod{p^r}$. This congruence is soluble.

  • 1
    $\begingroup$ WLOG we may assume that $p \mid n$ and $p^r \nmid n$ for otherwise we have a solution to the given congruence. Let $m$ be the highest power of $p$ dividing $n$ where $0 <m <r$. Then $(n,p^r) = p^m$. But then $p^m \mid n$ and $p^m | nb - a \implies p^m\mid a$. Which proves that the given congruence equation has a solution modulo $p^m$. Isn't it @Lord Shark the Unknown? $\endgroup$ – A.Chattopadhyay Apr 8 '18 at 8:27

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