Let $R$ be a normal Noetherian domain. Suppose there is $x \in R$ such that the principal ideal $xR$ has only one associated minimal prime ideal $\mathfrak{p}$. Assume there exists a multiplicative subset $A\subset R$ that does not contain zero and $A \cap \mathfrak{p} = \emptyset$, and such that in the localization $R_A$, $xR_A = \mathfrak{p} R_A$. Does it follow that $xR = \mathfrak{p}$?
Well, I know several basic facts: 1) $\mathrm{rad}(xR) = \mathfrak{p}$; 2) there exists $c \in R$ such that $\mathfrak{p} = (x:c)$. If $R$ were a UFD, the statement would be easy to verify.
I found an answer to this problem. It just required a few more facts.
So, assume on the contrary that there exists $y \in \mathfrak{p}\setminus xR$ but $x\cdot A^{-1}R = \mathfrak{p}\cdot A^{-1}R$. Then there exist $r \in R$, $a \in A$ such that $y = \frac{r}{a}x$ and $r \notin aR$. This implies that the ideal quotient $(a:r)$ is nontrivial, and all of its associated prime ideals are also associated prime ideals of $aR$. Since $R$ is normal and Noetherian, all the associated prime ideals of a principal ideal have height $1$; therefore, the same holds for $(a:r)$. From $y\cdot a = r \cdot x$, we see that $x \in (a:r)$, hence $x$ is in an associated prime ideal of $(a)$, which contradicts the assumption $\mathfrak{p}\cap A = \emptyset$. Thus the statement holds.