A question about radical ideals Let $R$ be a Noetherian ring. If $I$ is an ideal of $R$, then we know that for some integer $n>1$, $(\sqrt I)^n \subseteq I$. My question is: > In a Noetherian ring, is it true that for every non-zero, proper ideal $I$, $\exists n>1$ such that for every $x \in \sqrt I \setminus I$, $x^n \in I$ but $x^{n-1} \notin I$ ?

Certainly not.

Even for something as simple as $\mathbb Z/8\mathbb Z$ this isn't true. Now both $2$ and $4$ are in $\sqrt{\\{0\\}}$ for this ring, and $2^3=0$ but $2^2\
eq 0$, but $4^3=4^2=0$.

In fact, the _only_ case in which it is going to be true is if $\sqrt{I}^2\subseteq I$. For, if there exists an $n>2$ and an element such that $x^n\in I$ but $x^{n-1}\
otin I$, it must be true that $y=x^{n-1}$ satisfies $y^2\in I$, thwarting the condition.