Show that in a Noetherian ring $I$ and $J$ have the same radical if and only if there is a positive integer $N$ such that $I^N \subset J$ and $J^N \subset I$. [Hint: for the ``if'' direction, use a result you proved in the previous homework: $\textrm{rad}(I_1I_2) = \textrm{rad}(I_1) \cap \textrm{rad}(I_2)$; for the converse use the previous exercise and the observation that $I \subset \textrm{rad}(I)$.]
Proof:
Suppose $I$ is an ideal of Noetherian ring R. If $\textrm{rad}(I)\textrm{rad}(J)$ is finitely generated, then $(\sqrt I)^N \subseteq I$ and $(\sqrt J)^N \subseteq J$ for some integer $N\ge 1$. In HW 7 $\#4$ we proved that $\textrm{rad}(IJ)=\textrm{rad}(I\cap J)=\textrm{rad}(I)\cap \textrm{rad}(J)$. So if $\textrm{rad}(I)=R$ then $I\subset \textrm{rad}(I)$ and $1\in R$. So $1\in \textrm{rad}(I)$ and $1=1^n\in I$ and $I=R$ since $I,J\in R$. So $I^N=I=J^N=J$. Therefore, $I^N\subset J$ and $J^N \subset I$. $\square$ (please check if this answer works)
Lemma: $r(IJ) = r(I) \cap r(J)$.
Lemma: If $R$ is Noetherian and $I$ is an ideal in $R$, then there exists $N > 0$ such that $r(I)^N \subseteq I$.
Claim: If $R$ is a Noetherian ring, then $r(I) = r(J)$ if and only if there exists $N > 0$ such that $I^N \subseteq J$ and $J^N \subseteq I$.
Proof: Suppose $r(I) = r(J)$. By the lemma there exist $N_1, N_2 > 0$ such that $r(I)^{N_1} \subseteq I$ and $r(J)^{N_2} \subseteq J$. Choose $N = \max\{N_1, N_2\}$. Notice that $$J^N \subseteq r(J)^N = r(I)^N \subseteq r(I)^{N_1} \subseteq I$$ and $$I^N \subseteq r(I)^N = r(J)^N \subseteq r(J)^{N_2} \subseteq J.$$
Conversely, suppose that there exists $N > 0$ such that $I^N \subseteq J$ and $J^N \subseteq I$. Observe that, by monotonicity of radical and previous lemma, $$r(I) = \underbrace{r(I) \cap \cdots \cap r(I)}_{N \text{ times}} = r(I^N) \subseteq r(J)$$ and $$r(J) = \underbrace{r(J) \cap \cdots \cap r(J)}_{N \text{ times}} = r(J^N) \subseteq r(I).$$ Conclude that $r(I) = r(J)$.