$k[x,y]$ 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.

Let $$I=\langle x^2,y^2, xy \rangle $$ $$J=\langle x^2 \rangle$$ $$K=\langle y^2 \rangle$$ Then
$$I+(J \cap K) = \langle x^2,y^2, xy \rangle + \langle x^2y^2 \rangle=\langle x^2,y^2, xy \rangle=I$$
$$I \cap J = J \qquad \qquad \qquad I \cap K = K$$
$$(I \cap J) + (I \cap K)= J+K=\langle x^2, y^2 \rangle \neq I$$
Indeed, using this argument you can prove a more general result:

In any Prufer domain you have the equality of ideals:$$\langle a, b \rangle^2 = \langle a^2, b^2 \rangle$$

Since in $k[x,y]$ this does not hold ($\langle x, y \rangle^2 = \langle x^2, y^2, xy \rangle$ ) this is not a Prufer domain.
