I need to find all the ideals of $A=\mathbb{Z}[\frac{1}{a}]$ for an integral number $a\in\mathbb{Z}$. If I can prove that $A$ is principal, I can show that all ideals are generated by an integral number $m$, since $\frac{1}{a}$ is invertible in $A$. I thought of building a suitable stathm over $A$, but there are two interrelated parameters: the power of $a$ and the absolue value over $\mathbb{Z}$, and I'm not sure how to balance between these two... I know that $A$ is noetherian, since it's a $Z-$module of finite type over $\mathbb{Z}$.
How can I prove that $A$ is principal (since I have a strong feeling that it is)? Moreover, what is a generalisation (i.d. what are the principal rings of the same type as $A$)?
Recall that for non-zero $a\in\Bbb Z$, then $\Bbb Z\hookrightarrow\Bbb Z[1/a]$ is a ring localization. Consequently, there exists a bijection between ideals in $\Bbb Z[1/a]$ and ideals in $\Bbb Z$ which doesn't meet $\{a^n:n\in\Bbb N\}$. Thus if $\mathfrak b\subseteq\Bbb Z[1/a]$ is an ideal, then $$\mathfrak b=(\mathfrak b\cap\Bbb Z)\Bbb Z[1/a]$$ For, the inclusion $(\mathfrak b\cap\Bbb Z)\Bbb Z[1/a]\subseteq\mathfrak b$ follows directly from $\mathfrak b\cap\Bbb Z\subseteq\mathfrak b$, while the opposite inclusion $\mathfrak b\subseteq(\mathfrak b\cap\Bbb Z)\Bbb Z[1/a]$ follows from the fact that every element in $\mathfrak b$ can be written, by bring to a common denominator, in the form $x/a^n$ for suitable $x\in\mathfrak b\cap\Bbb Z$ and $n\in\Bbb N$. Since $\Bbb Z$ is a principal ideal domain, then $\mathfrak b\cap\Bbb Z=b\Bbb Z$ for some $b\in\Bbb Z$, hence $\mathfrak b=b\Bbb Z[1/a]$.