 Solution

False. In the polynomial ring $R=k[x,y]$ over a field, takeThe ideal $P$ is a maximal ideal, with quotient ring $R/P\cong k$. AlsoIn this last ring, an element with nonzero constant term is a unit, and an element with zero constant term is a multiple of $y$ and has square zero. Thus every zero divisor is a nilpotent element. This is equivalent to $I$ being a primary ideal: a zero product in the quotient ring with one nonzero factor forces the other factor to be nilpotent. Its radical of an ideal is $P$, so $I$ is $P$-primary.

However, $y\notin I$, so $I\ne P$. Moreover $x\in I$ while $x\notin P^2$, since every polynomial in $P^2$ has no term of total degree one. For every $n\ge 2$, $P^n\subseteq P^2$, so $I\ne P^n$. Hence