Rings of Fractions of Principle Ideal Domains (8.2.8)

Dummit and Foote Abstract Algebra, section 8.2, exercise 8:

Let $R$ be a principle ideal domain, and let $D$ be a multiplicatively closed subset of $R$. Prove that $D^{-1}R$ is a principle ideal domain.

Proof: Let $I$ be a typical ideal of $D^{-1}R$, and $J$ be the subset of elements of $R$ appearing as numerators of elements of $I$. We claim $J$ is an ideal: For any $a,c∈J$, we have some $\dfrac{a}{b},\dfrac{c}{d}∈I$ by definition, so $\dfrac{a}{b} \cdot \dfrac{b^2}{b} = \dfrac{ae}{e}∈I$ for any $e∈D$ and similarly $\dfrac{ce}{e}∈I$. Now $(\dfrac{ae}{e}-\dfrac{ce}{e}) \cdot \dfrac{e}{e^2}=\dfrac{(a-c)e}{e} \cdot \dfrac{e}{e^2}=\dfrac{a-c}{e}∈I$ so that $a-c∈J$ and $J$ is closed under subtraction. For any $r∈R$, we have $\dfrac{re}{e} \cdot \dfrac{a}{b}=\dfrac{ra}{b}∈I$ so that $ra∈J$ and $J$ is an ideal. Due to the hypothesis, $J=(\alpha)$ for some $\alpha∈R$.

We now claim $I=(\dfrac{\alpha e}{e})$. Since $\alpha∈J$, as before we have $\dfrac{\alpha e}{e}∈I$ so that $(\dfrac{\alpha e}{e}) \subseteq I$. Now, observe arbitrary $\dfrac{x}{y}∈I$, and write $x=z\alpha$. We have $\dfrac{\alpha e}{e} \cdot \dfrac{z}{b} = \dfrac{a}{b}$, so that $I \subseteq (\dfrac{\alpha e}{e})$ and now $I=(\dfrac{\alpha e}{e})$ is a principle ideal, and $D^{-1}R$ is a principle ideal domain.$~\square$