Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed subset of $R$ then $S^{-1}R$ is also a principal ideal ring. Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed
  subset of $R$ then $S^{-1}R$ is also a principal ideal ring.


Thanks for any insight.
More generally, prove the following: Let $\phi : R \to R'$ be a ring homomorphism such that every element of $R'$ is associated to an element in $\phi(R)$. Note that localizations satisfy this property by construction, but also quotients. If $R$ is a principal ideal ring, then the same is true for $R'$. The proof is straight forward, since there is only one thing you can do: Use the assumption. Every ideal of $R'$ has some generating set, and by the assumption it may be choosen from $\phi(R)$ (since multiplication of units doesn't change an ideal). Take preimages in $R$, use that $R$ is a principal ideal ring, etc.
$S^{-1}R$ is a subring of a field (the field of fractions of $R$), and thus is an integral domain.

Let $I \subseteq S^{-1}R$ be an ideal. Let $D = \{a \in R \ |\ \frac{a}{b} \in I \}$.

claim that $D$ is an ideal of $R$.  

note that $\frac{0}{s} \in I$, so that $0 \in D$. 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 $a-b \in D$. 

let $r \in R$ and $a \in D$, with $d \in S$. Then $\frac{r}{d} \frac{a}{d} = \frac{ra}{d^2}$, so that $ra \in D$. Thus $D$ is an ideal of $R$, and thus we have $D = (\alpha)$. Let $d \in S$ 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,$S^{-1}R$ is a principal ideal domain.