If $\mathfrak{p}S^{-1}A$ is principal, $\mathfrak{p}$ is also principal? Let $A$ be a Noetherian domain and $S$ be a multiplicatively closed set of A. 
Take a minimal prime ideal $\mathfrak{p}$ in $SpecA\verb│\│\{(0)\}$ such that $\mathfrak{p}\cap S=\emptyset$.
If $\mathfrak{p}S^{-1}A$ is principal, $\mathfrak{p}$ is also principal?
If not, is it true when $S=\{1,x,x^2,\cdots\}$ where $x\in A$ is a non-zero prime element?
 A: New, better answer which does things correctly:
The answer to part 1 is no, as mentioned in the comments: $A=\Bbb Z[\sqrt{-5}]$, $p=(2,1+\sqrt{-5})$, $S=A\setminus p$. This ring $A$ is of dimension one, so $p$ is both a minimal and maximal prime.
For part 2, the answer is yes. We use the assumption that $p\cap S=\emptyset$. Recall that the localization map $A\to S^{-1}A=A[1/x]$ is injective since $S$ contains no zero divisors.
Assume $p$ is not principal with a minimal set of generators $f_1,\cdots,f_n$, $n>1$, but $pA[1/x]$ is principal. Pick an element $q\in pA[1/x]\cap A$ which generates $pA[1/x]$ and isn't divisible by $x$. Then there are elements $y_1,\cdots,y_n \in A[1/x]\setminus \{0\}$ so that $y_iq=f_i$. At least one of these equations must have $y_i\notin A\subset A[1/x]$ as otherwise all the equations would be valid in $A$ and $q$ would be a generator of $p$. After multiplying through by a suitable power of $x$, we get the equation $y_i'q=f_ix^m$ where $y_1'\notin (x)$ and $m>0$. But this is a contradiction, as $y_i',q\in A\setminus (x)$ but $y_i'q\in(x)$.
