About radical of ideal

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Let $\alpha$ be an ideal $\ne$ 1 in a commutative ring $A$. Show that $\alpha=r(\alpha)\Leftrightarrow$ $\alpha$ is an intersection of prime ideals.

Here $r(\alpha)=\{x\in A|\exists n\in\mathbb{N}_{\ge 1}:x^n\in \alpha\}$ is radical of ideal $\alpha$

I know that $R_A=\{x\in A|\exists n\in\mathbb{N}_{\ge 1}:x^n=0\}$ is an intersection of all prime ideals in $A$, so it's easy to see that $R_A\in \alpha$, but this doesn't help ...

Please give me some suggestions, thanks in advance!