Let $R$ be a commutative ring with unity.
If $P\in \operatorname{Spec} R\:$ is such that $R_P$ is reduced, then is it necessarily true that $\exists f\in R\setminus P$, i.e., $f\in R$ such that $P\in D(f)$, such that $R_f$ is also reduced ? If this is not true in general, what if we assume $R$ is Noetherian ?
(Here by reduced ring we mean a ring with no non-zero nilpotent element.)
I explained in the comments why Noetherianity is necessary (I can replicate the argument here if you want), here's why it's enough :
if $R$ is Noetherian, then its nilradical $\mathrm{Nil}(R)$ is finitely generated, say by $x_1,...,x_n$. $R_P$ is reduced, meaning $x_1,...,x_n$ are $0$ in $R_P$, so in particular there are $s_1,...,s_n\notin P$ such that $s_ix_i=0$.
If you now invert $f=s_1...s_n$ (which obviously isn't in $P$) then of course you invert each $s_i$ and so $x_i = 0$ in $R_f$.
Moreover, if $e^k = 0$ in $R_f$, then for some $l$ we have $f^le^k=0$ in $R$. Therefore (multiply by enough $f$ or enough $e$) you get for some $r$, $(fe)^r=0$ in $R$.It follows that $fe=\sum_i \lambda_i x_i$ in $R$, thus if you invert $f$, since the $x_i$ die, you get $e=0$ in $R_f$: $R_f$ is reduced.