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 $x^r$.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.