Let $R$ be a commutative ring with unity, $I$ be a primary ideal. Then $rad(I)=P$ is prime ideal.
Further if $P$ is finitely generated, then $P^m\subseteq I$ for some $m\geq 1$. This is the case for Noetherian rings.
Q. Is there a non-Noetherian ring $R$ with a primary ideal $I$ such that if $P$ is the radical of $I$ then no power of $P$ is in $I$?
(Here $P^m$ is defined to be ideal of $R$ generated by all sums $x_1x_2\cdots x_m$ with $x_i\in P$.)
Let $k$ be a field and consider the polynomial ring $k[x_1,x_2, \ldots]$ and the ideal $I=(x_1,x_2^2,x_3^3, \ldots)$. We have $P=\operatorname{rad}(I)=(x_1,x_2, \ldots)$ which is a maximal ideal, so $I$ is primary. For any $n \in \Bbb N$, we have $x_{n+1}^n \in P^n$, but $x_{n+1}^n \not \in I$, so no power of $P$ is contained in $I$.