Minimal prime ideals consist of zerodivisors

Question

I don't find the proof for this little demonstration ...

Let $P$ be a minimal prime ideal of $A$. Show that $P$ is contained in the set of zero divisors of $A$.

Answer 1

Hints (assuming commutative Noetherian):

1) The only prime ideal of the localization $\;A_p\;$ is $\;pA_p\;$

2) We have that $\;x\in p\implies \frac x1\in pA_p\;$ is nilpotent