Does $IJ=IK\implies J=K$ always hold for integral domain and finitely generated nonzero ideal $I$?

Question

Let $R$ be a commutative integral domain, $I,J,K$ three ideals of $R$ with $I\neq (0)$ being finitely generated. Then does $IJ=IK$ imply $J=K$?

With Nakayama lemma, I can prove it if one of $J$ and $K$ equals to $R$. And I also know it holds when $R$ is a Prüfer domain or $I$ is singly generated.

Answer 1

I think taking $R:=F[X,Y]$, $I:=(X,Y)$, $J:=(X^2,Y^2)$ and $K:=(X^2,XY,Y^2)$ might be a more interesting counterexample, as user26857 noted in the comment.

Answer 2

It doesn't always hold. For example, Let $R:=\mathbb Z[\sqrt{-3}]$, take $I:=(2,1+\sqrt{-3})$, $J:=(2)$ and $K:=(1+\sqrt{-3})$, then $IJ=IK$.