The ring $R$ is commutative with unit. An ideal $I$ is called primary, if it stands the following:
If $ab \in I$ then $a \in I$ or $b^n \in I$, for a natural number $n$.
Show that if $I$ is a primitive ideal of $R$, then $Rad(I)$ is a prime ideal of it.
Could you give me a hint how we could show this?
EDIT:
That's what I have tried:
$Rad(I)=\{ x \in R| \exists n \in \mathbb{N} \text{ such that } x^n \in I \}$
$P$ is a prime ideal iff $a,b \in R$, $a \cdot b \in \mathbb{P}$, then $a \in P$ or $b \in P$
Let $a \cdot b \in Rad(I) \Rightarrow \exists m \in \mathbb{N}$ such that $(a \cdot b)^m \in I \Rightarrow a^m \cdot b^m \in I \Rightarrow a^n \in I \text{ or } (b^{m})^n \in I \Rightarrow a \in Rad(I) \text{ or } b^{m \cdot n } \in I \Rightarrow a \in Rad(I) \text{ or } b \in Rad(I)$
Could you tell me if it is right?
$ab\in\sqrt I\Rightarrow\exists m\ (ab)^m\in I$; $(ab)^m=a^mb^m\in I\Rightarrow a^m\in I$ or $\exists n\ (b^m)^n\in I$ $\Rightarrow\cdots$