We already knew that if $R$ is a Noetherian ring, then every ideal $I$ contains a power of its radical. Now suppose that $R$ is non-Noetherian, is there any example of $R$ such that one of its ideals, denoted by $I$, contains a power of $\sqrt{I}$, and the power is greater than or equal to $2$?
To make it easier, we can discuss with the ring $R=C^0(\mathbb{R},\mathbb{R})$, which is the ring of continuous map from $\mathbb{R}$ to $\mathbb{R}$. This ring is not Noetherian and I knew some of its maximal ideals. Is there any ideal $I$ such that $(\sqrt{I})^n=I$ for some $n \geq 2$?
Great to discuss with everyone for this.