Prove that $\sqrt{I} = \sqrt{J} \Leftrightarrow V(I) = V(J)$ 
Question: Let $I$ and $J$ are ideals of a commutative ring $R$. Prove that $\sqrt{I} = \sqrt{J} \Leftrightarrow V(I) = V(J)$, in which $V(I) = \{P \in \text{Spec}(R)\ |\ I \subseteq P\}$.

We know that $\sqrt{I} = \{r \in R\ |\ r^n \in I \text{ with } n \text{ a positive integer}\}$, so $\sqrt{I} = \sqrt{J} \Rightarrow r_1^n = r_2^m, r_1 \in I, r_2 \in J$. So how to prove this?? 
My attempt: Suppose $V(I) = V(J)$, i.e. all prime ideals contains $I$ and all prime ideals contains $J$ are equivalent. So from $V(I) = V(J) \Rightarrow$ all prime ideals contains both $I$ and $J$, or there are two family of prime ideals, one contains $I$ and one contains $J$? After that, $\sqrt{I} = \cap \{P, P \text{ prime ideal and } I \subseteq P\}$, $\sqrt{J} = \cap \{Q, Q \text{ prime ideal and } J \subseteq Q\}$ implies $\sqrt{I} = \sqrt{J}$ if $P = Q$??? Is that true???
 A: I will show one direction: assuming $\sqrt{I} = \sqrt{J}$ we derive $V(I) = V(J)$.
To show $V(I) \subseteq V(J)$, you must show that prime ideals of $R$ containing $I$ also contain $J$. So let $P$ be a prime ideal containing $I$ and let $r \in J$. Now since $J \subseteq \sqrt{J}$ and $\sqrt{I} = \sqrt{J}$ by assumption, $r \in \sqrt{I}$. So $r^n \in I$ for some $n \in \mathbb{Z}_+$; thus $r^n \in P$ as we assumed $I \subseteq P$. Now


*

*if $n = 1$, we immediately have $r \in P$.

*If $n = 2$, $r^2 = rr \in P$ so as $P$ is a prime ideal either $r \in P$ or $r \in P$.


$$\vdots$$
$$\text{so on and so forth}$$
$$\vdots$$
Basically by induction and using the prime ideal property, you always get $r \in P$. Hence $J \subseteq P$ and $V(I) \subseteq V(J)$ as required. By an analogous argument switching the roles of $I$ and $J$ above, you can also show $V(J) \subseteq V(I)$ and thus $V(I) = V(J)$.
Now can you complete the other direction (i.e. assuming $V(I) = V(J)$ to derive $\sqrt{I} = \sqrt{J}$)?
