Proving $\sqrt {P^ n}=P$ when $P$ is a prime ideal I am trying some assignments and I am struck on that.
If $R$ is a ring, and $A$ an ideal then $\sqrt{A}$ means  $\{a\in R| a^k \in A \text{ for some integer } k > 0\}$.
I need to show that if $P$ is a prime ideal in $R$ then $\sqrt{P^n} = P $.
I am sorry but I am unable to prove any of the inclusion.
$x\in \sqrt{P^n}$ implies that $\{x\in R| x^n \in P^n\}$. but I am unable to see how to use that it is a prime ideal to  prove x $\in  P$.
 A: Let $x\in P$, then $x^n\in P^n$ and so  $x\in \sqrt{P^n}$ whih proves the first inclusion.
If $x\in\sqrt{P^n}$ then there is some $m\in\mathbb{N}$ such that $x^m\in P^n$. But, by definition of product ideal, since $x^m\in P^n$ it is of the form
$x^m=p_{11}\dots p_{n 1}+\dots+p_{1 l}\dots p_{n l}$, where $p_{ij}\in P$ and $l\in\mathbb{N}$, so $x^m\in P$.  But $x^{m-1}x\in P$ implies (since $P$ is prime) that either $x\in P$ or $x^{m-1}\in P$. If $x\in P$ we are done, so we may assume $x^{m-1}\in P$. By induction on $m$ it is easy to check that $x\in P$, and so the inclusion is proved.
A: 
$x\in \sqrt{P^n}$ implies that $\{x\in R| x^n \in P^n\}$.

Well, that's a valiant start. But what does it mean to "imply a set"? That is not what you mean.
The real implication of $x\in \sqrt{P^n}$ is, according to the definition, that $x^k\in P^n$ for some integer $k>0$.  But $P^n\subseteq P$... so then what?
Conversely, discussed in the comments, $x\in P$ implies $x^n\in P^n$, so that direction is already clear.  All that remains is for you to connect the dots in the previous hint I gave.
