 Solution

Let $J$ be an ideal of $R^{\prime }=S^{-1}R$ and contract it toThis is an ideal of $R$, so $I=(d)$ because $R$ is a principal ideal domain. Every $x^n/1$ is a unit in a ring $R^{\prime }$. Consequently, if $r/x^n\in J$, then $r/1=(x^n/1)(r/x^n)\in J$, so $r\in (d)$ and $r/x^n\in (d/1)$. Conversely $d/1\in J$, hence every multiple of it belongs to $J$. Thusso every ideal of $R^{\prime }$ is principal. The localization remains an integral domain; hence $R^{\prime }$ is a principal ideal domain.