Find a non-Noetherian ring that has an ideal which does not contain any power of its own radical

Question

Find a non-Noetherian ring and show that there exists some ideal $I$ where $\nexists n \in\mathbb{N}$ s.t. $[rad(I)]^n\subset I$

I have found some non Noetherian rings such as the ring of algebraic integers or the ring of continuous functions on $[a,b]$ but I cant find any ideal that does not contain any power of its radical.

Any discussion on that would be helpful.

Answer 1

Let $K$ be field, Let $R$ be the ring of polynomials in infinitely many variables $R:=K[x_{1},x_{2},\cdots] $, $R$ isn't Noetherian, this can be seen by considering the ascending chain of ideals $I_{k} :=<x_{1},\cdots, x_{k}>$. Consider the ideal $I=<x_{1}^{2},x_{2}^{3},\cdots, x_{n} ^{n+1},\cdots >$, then $Rad(I) =<x_{1},x_{2},\cdots, x_{n} ,\cdots >$, Now it is clear that $\forall n\in \mathbb{N}, Rad(I) ^{n} \nsubseteq I$