# A nice way to phrase this theorem about ideals in Noetherian rings?

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 $$x^r$$.

• A nil ideal is one whose elements are nilpotent, which I strictly weaker than what you’re describing. – rschwieb Apr 1 at 23:08

You can say that ‘ $$\operatorname{rad}I/I$$ is nilpotent in $$R/I$$ (or the radical of $$R/I$$ is nilpotent).