Is the map which takes the radical of an ideal injective?

Question

Let $R$ be a commutative ring with unity. Let $I$ be an ideal of $R$.

Let $r(I)$ be the radical of $I$, defined by $r(I)=\{x\in R : x^n \in I \text{ for some }n\in\mathbb{N}\}$.

Let $A$ be the set of all ideals of $R$ and $P(R)$ be the set of all subsets of $R$.

Finally, define $\psi:A\to P(R)$ by $\psi(I)=r(I)$.

My question: is $\psi$ injective? As in, is it the case that $r(I)=r(J)\implies I=J$?

Answer 1

Short answer: no. A direct counterexample can be found by using two facts:

$(1)$ $r(I)$ is an ideal for any ideal $I$.

$(2)$ $r(r(I))=r(I)$

Since there are cases where $I\subsetneq r(I)$, we have that $\psi$ is not injective.