$ab=0 \Rightarrow b$ is nilpotent in a Noetherian ring

Question

Let $A$ be a Noetherian ring and $a \in A$. Show that if $a$ is not contained in a minimal prime, $ab=0 \Rightarrow b$ is nilpotent.

I can't see a way to solve this. I tried to consider that in a Noetherian ring the nilradical is nilpotent, but I don't know how to use it.

Can anyone help me?

Answer 1

Hint:

The nilradical is the intersection of all minimal primes (whether $A$ is noetherian or not).