$A$ is a reduced commutative ring with unit; $p$ is a minimal prime ideal. If $S = A \setminus{p}$ , I have to show that the ring $A_{p} = S^{-1}A$ is a field.
My thoughts: Since $p$ is a minimal prime, $S^{-1}p$ is the unique prime ideal of $A_{p}$, and is obviously also a maximal ideal. But how to use the fact that $A$ is reduced ?
As you already said, $p$ becomes the unique prime ideal of $A_p$, and hence the nilradical which consists precisely of all nilpotent elements of $A_p$. Since $A$ is reduced $A_p$ is reduced as well so the unique prime ideal of $A_p$ is the zero ideal.
Since every non-unit of a commutative ring lies in a (maximal) prime ideal, every non-zero element of $A_p$ must be a unit. Therefore $A_p$ is a field.