I was asked to prove that every ideal of $\mathbb{Z}_{(p)}$ are of the form $(p^{k})$ where $\mathbb{Z}_{(p)}$ refers to the localization of $\mathbb{Z}$ at the prime ideal $(p)$ for a prime number $p$.
In general, for a multiplicative set $S$ of a ring $R$ I know that the ideals of $S^{-1}R$ are of the form $S^{-1}I=\left(\frac{a}{1}\mid a\in I\right)$ so in this specific problem, the ideals of $\mathbb{Z}_{(p)}$ are of the form $(n)$ for each $n\in\mathbb{N}$ since the ideals of $\mathbb{Z}$ are $n\mathbb{Z}$.
So is it true that for every $n\in\mathbb{N}$, there exists $k\in\mathbb{N}$ such that $(n)=(p^{k})$ in $\mathbb{Z}_{(p)}$?.
In general, for a multiplicative subset $S$ of a ring $R$, the ideals of $S^{-1}R$ are of the form $\{a/s : a \in I,\, s \in S\}$, for some ideal $I$ of $R$.
Then, for an ideal $J$ of $\Bbb{Z}_{(p)}$, there exists a non-negative integer $n$ such that $$ \begin{split} J &= \left\{ \frac as : a \in n\Bbb{Z}, s \in \Bbb{Z} \setminus p\Bbb{Z} \right\} \\ &= \left\{ \frac{nb}s : b \in \Bbb{Z}, s \in \Bbb{Z} \setminus p\Bbb{Z} \right\} = n\Bbb{Z}_{(p)}. \end{split} $$
Let $k$ be the non-negative integer such that $p^k$ is the greatest power of $p$ that divides $n$. Then $n = p^ku$ with $u \in \Bbb{Z} \setminus p\Bbb{Z}$, so $u$ is a unit of $\Bbb{Z}_{(p)}$ and hence $n\Bbb{Z}_{(p)} = p^k\Bbb{Z}_{(p)}$.