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I came across this theorem:

Theorem. Let $I$ be an ideal of a Noetherian ring $R$. Then there is $r \in \mathbb{N}$ with $\operatorname{rad}(I)^r \subseteq I$.

Is there a concise way to state this in term of primary or nil ideals? At first I tried comparing this with the assertion "$\operatorname{rad}(I)/I$ is a nil ideal of $R/I$", but the latter only says that whenever $x \in \operatorname{rad}(I)$, there is $r \in \mathbb{N}$ with $x^r \in I$. So it's a weaker assertion for two reasons: one, in the theorem $r$ is not allowed to depend on $x$, and two, the theorem says that any product $x_1 \ldots x_r$ is in $I$, not just elements of the form $x^r$.

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  • $\begingroup$ A nil ideal is one whose elements are nilpotent, which I strictly weaker than what you’re describing. $\endgroup$ – rschwieb Apr 1 at 23:08
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You can say that ‘ $\operatorname{rad}I/I$ is nilpotent in $R/I$ (or the radical of $R/I$ is nilpotent).

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