## Rings of fractions over a PID are PIDs

Prove that if $R$ is a principal ideal domain and $D \subseteq R$ a set of denominators, then the set $D^{-1}R$ of fractions with denominators in $D$ is a principal ideal domain.

$D^{-1}R$ is a subring of a field (the field of fractions of $R$), and thus is an integral domain.

Let $I \subseteq D^{-1}R$ be an ideal. Let $S = \{a \in R \ |\ \frac{a}{b} \in I \}$. We claim that $S$ is an ideal of $R$. To see this, note that $\frac{0}{d} \in I$, so that $0 \in S$. Now let $\frac{a}{d}, \frac{b}{e} \in I$; then $\frac{a}{d} - \frac{e}{d}\frac{b}{e} = \frac{a-b}{d}$, so that $a-b \in S$. Finally, let $r \in R$ and $a \in S$, with $d \in D$. Then $\frac{r}{d} \frac{a}{d} = \frac{ra}{d^2}$, so that $ra \in S$. Thus $S$ is an ideal of $R$, and thus we have $S = (\alpha)$. Let $d \in D$ such that $\frac{\alpha}{d} \in I$; then $(\frac{\alpha}{d}) \subseteq I$. Now let $\frac{a}{b} \in I$ with $a = c\alpha$; we have $\frac{a}{b} = \frac{cd}{b} \frac{\alpha}{d}$, so that $I = (\frac{\alpha}{d})$ is principal.

Thus $D^{-1}R$ is a principal ideal domain.