TITLE: If $P$ is a prime ideal of $R$, $\sqrt{P^{n}}=P\ \forall n\in\mathbb{N}$?
QUESTION [7 upvotes]: Let $R$ be a commutative ring. If $P$ is a prime ideal of $R$, $\sqrt{P^{n}}=P\ \forall n\in\mathbb{N}$?

REPLY [3 votes]: Using the definition of $\sqrt{I}$ as "the intersection of prime ideals containing $I$," you immediately get that $\sqrt{P^n}\subseteq P$.
For the other containment, clearly any prime ideal containing $P^n$ must contain $P$, so the intersection of such ideals contains $P$.
Not positive what your definition is, but if you are not familiar with this one, it would be a good exercise to prove its equivalence (for commutative rings) to the other usual definition : $\sqrt{I}=\{x\in R\mid \exists n\in \Bbb N^+( x^n\in I)\}$.