Radical of an ideal - prove every prime ideal that contains $I$ also contains $\sqrt{I}$

Question

Let $R$ be a commutative ring with a unit and $I$ an ideal.
Please prove that every prime ideal that contains $I$ also contains $\sqrt{I}$.

I easily conclud that $I \subseteq \sqrt I$ but I couldn't find how to go on.

Answer 1

Suppose $\mathfrak a\subseteq \mathfrak p$. If $x^n\in\mathfrak a$ for some $n$; then $x^n\in\mathfrak p$, and since $\mathfrak p$ is prime...