Define $\mathbb{Z}_{(p)}$ = {$m/n \in \mathbb{Q} : p$ does not divide $n$}.
If $A \neq 0$ is an ideal of $\mathbb{Z}_{(p)}$ show that $A = <p^k>$ where $k \geq 0$ is the smallest integer such that $p^k \in A$.
Hint from the textbook: If $0 \neq m \in \mathbb{Z}$ then $m = p^rd$ where $r \geq 0$ and $p$ does not divide $d$.
I have literally no idea how to approach this problem. So far all I have is, from the hint: if $m/n \neq 0$ then $m/n = p^rd/n$ where $p$ does not divide $d$.
The previous part of the question asked us to show that $\mathbb{Z}_{(p)}$ is an integral domain, which I did. I'm not sure if that fact helps us here.
Fix an ideal $A$. As you've observed, every element $m\in A$ can be written as $p^rd$ with $r\ge 0$ and $d$ coprime to $p$.
In particular, there is an element $m_0\in A$ with $m_0 = p^{r_0}d_0$ with $r_0$ minimal. Can you show that $A = \langle p^{r_0}\rangle$?