I know that if $A$ is a PID and $S\subset A^*$ is a multiplicative set. Then $S^{-1}A$ is a PID since ideals of $S^{-1}A$ is of the form $S^{-1}I$ where $I$ is an ideal of $A$. We have to find a counterexample for the converse.
I was thinking about $R=\Bbb{Z}[1/p]/\Bbb{Z}$ which is a PID. And $R=\frac{S^{-1}\Bbb{Z}}{\Bbb{Z}}$ where $S=\{p^n:\ n\ge0\}$. We know that for a ring $A$ and ideal $I$, $\frac{S^{-1}A}{S^{-1}I}\cong \bar{S}^{-1}(A/I)$. But there is no such ideal $I$ of $\Bbb{Z}$ such that $S^{-1}I=\Bbb{Z}$.
Can anyone give me any way-out or hint to construct such example? Thanks for your help in advance.
For example, $R=\mathbb Z[X]$ works.
It's a UFD, therefore a Krull domain. If you localize at a height $1$ prime, say $(X)$, you get a discrete valuation domain. DVR's are PIDs that are not fields. $R$ is not a PID because PIDs have Krull dimension $1$ and $R$ has Krull dimension $2$.