Exercise 4.2, Atiyah Macdonald

Question

I'm having difficulties in showing that a radical ideal has no embedded primes. I know that a radical ideal $I\subseteq A$ ($A$ ring) is equal to the intersection of the minimal primes containing $I$. If there were finitely many of these minimal primes $p_1,\dots, p_n$, such that their intersection was still $I$, I would be done: I could suppose that $I=\cap_{i=1}^np_i$ was a minimal primary decomposition, and so the associated primes would be again $p_1,\dots, p_n$, that are all minimal. However $A$ is not necessarily Noetherian, so it doesn't seem that the assumption in italics is granted. So what should I do to overcome this problem? Thank you