Show that the only nonzero ideals of R are the principal ideals $\langle p^e \rangle$

Question

Let $p$ be a prime number in $\mathbb{Z}$. Let $R = R_p = \{x \in \mathbb{Q}\ |\ \textrm{ord}_p(x)\geq0\}$, which is a subring of $\mathbb{Q}$.

Show that the only nonzero ideals of $R$ are the principal ideals $ \langle p^e \rangle$ for $e \geq 0$.

No idea on how to answer this. The definition is clear, but I do not think I have to use it.

Answer 1

Let $I=\langle r_1,r_2,\dots\rangle$ be an arbitrary ideal in $R$. Let $e=\max_i ord_p(r_i)$. Then for all $i$ $r_i=p^e s_i$ where $s_i\in R$. Thus $r_i\in\langle p^e\rangle$ for all $i$. Thus $I\subseteq \langle p^e\rangle$. Now let $i$ be such that $ord_p(r_i)=e$. Then $r_i=rp^e$ where $r\in\Bbb Q$ and $ord_p(r)=0$. Thus $r$ is a unit in $R$ because also $\frac1r\in R$. Thus $p^e=\frac1r r_i$. Thus $\langle p^e\rangle\subseteq\langle r_i\rangle\subseteq I$. And since we also showed $I\subseteq \langle p^e\rangle$ it follows that $\langle p^e\rangle=I$.

Thus all ideals are of the form $\langle p^e\rangle$.