Maybe no one would care about this, but I just want to write a proof about this statement. Let $R$ be a commutative ring (with $1$) and $S$ be a multiplicatively subset of $R$ (we can assume that $0\notin S$).
If we consider the natural homomorphism $\phi \colon R\rightarrow S^{-1}R$ given by $a\mapsto \frac{a}{1}$, then for ideals $I$ of $R$ and $\mathscr{I}$ of $S^{-1}R$, $I^e=\langle \phi(I)\rangle$ is an ideal of $S^{-1}R$ called the extension of $I$, and $\mathscr{I}^c=\phi^{-1}(\mathscr{I})$ is an ideal of $R$ called the contraction of $\mathscr{I}$.
Proposition.- Let the situation be as in the previous paragraph. Then there exits a bijective map $\Phi$ between the sets $A=\{P\in \text{Spec}(R): P\cap S=\emptyset\}$ and $B=\text{Spec}(S^{-1}R)$ given by $$\Phi\colon A\rightarrow B$$ $$\;\;\;\;\;P\mapsto P^e.$$
Whose inverse is $\Psi\colon B\rightarrow A$ given by $\mathscr{P}\mapsto \mathscr{P}^c$. Moreover, both $\Phi$ and $\Psi$ preserve inclusions.
Proof: This is the corollary of theorem 5.32 given in Sharp's "Steps in Commutative Algebra".
Now, in the particular case $S=R\setminus P$, where $P$ is minimal prime ideal of $R$, we have for $P'\in \text{Spec}(R)$, $P'\cap (R\setminus P)=\emptyset$ if and only if $P'\subseteq P$, which implies that $P'=P$. So it follows that $P^e=PR_P$ is the only prime ideal of $R_P$. But it's a standar result that $\text{Nil}(R_P)$ is the intersection of the elements of $\text{Spec}(R_P)$ and therefore $\text{Nil}(R_P)=PR_P$. Hence, every element of $PR_P$ is nilpotent.
