$k[x,y]$ is not a Prüfer Domain? I know that if $R$ is a domain, the following conditions are equivalent:

*

*$R$ is a Prüfer Domain,

*If $I,J,K \subset R$ are nonzero ideals, $I + (J \cap K) = (I \cap J) + (I \cap K)$

*If $I,J,K \subset R$ are nonzero ideals, $I \cap (J + K) = (I + J) \cap (I + K)$
It is stated that $k[x,y]$ is not a Prüfer Domain where $k$ is a field in When do (multivariate) polynomial rings fail to be Prüfer rings?. However, I can not find any three nonzero ideals $I,J,K \subset k[x,y]$ such that $I + (J \cap K) \neq (I \cap J) + (I \cap K)$ or $I \cap (J + K) \neq (I + J) \cap (I + K)$.
Any ideas would be appreciated.

A simple counterexample is the following: $I=(x^2)$, $J=(x)$, and $K=(y)$. Then $$I+(J\cap K)=(x^2)+(x)\cap(y)=(x^2,xy)$$ while $$(I\cap J)+(I\cap K)=(x^2)\cap(x)+(x^2)\cap(y)=(x^2,x^2y)=(x^2).$$
