Let $I, J$ be ideals in a commutative ring with identity $R$. Define the quotient ideal $(I : J)$ by $$(I : J)=\{x\in R : xJ\subseteq I\}.$$
Define the radical $r(A)$, of an ideal $A$ of $R$ by
$$r(A)=\{x\in R\;:\;x^n\in A, \;\;for\;some\;n>0\}.$$
My question is: There is some link between $r[(I : J)]$ and the two radicals $r(I)$ and $r(J)$?
We have the following relations: $$(I:r(J))\subseteq(I:J)\subseteq r(I:J)\subseteq(r(I):J)=(r(I):r(J)).$$ The first and the second of these inclusions are immediately clear. For the third one, consider $x\in r(I:J)$ and $y\in J$. There exists $n\in\mathbb{N}^*$ with $x^ny\in I$, implying $(xy)^n=x^ny\cdot y^{n-1}\in I$, hence $xy\in r(I)$ and therefore $x\in(r(I):J)$. Finally, for the equality it suffices to show that the left hand side is contained in the right hand side. So, let $x\in(r(I):J)$ and let $y\in r(J)$. There exists $n\in\mathbb{N}^*$ with $y^n\in J$, hence $xy^n\in r(I)$, and thus there exists $m\in\mathbb{N}^*$ with $x^my^{m+n}\in I$. It follows $(xy)^{m+n}=x^nx^my^{m+n}\in I$, therefore $xy\in r(I)$, and thus $x\in(r(I):J)$.
None of the above inclusions needs to be an equality. For a counterexample, consider $R=A[X,Y]/\langle X^3,Y^2\rangle$ for some reduced ring $A\neq 0$, and the ideals $I=0$ and $J=\langle \overline{X}^2\rangle$ of $R$ (where $\overline{X}$ denotes the canonical image of $X$ in $R$).
