Does $I(J\cap K)=IJ\cap IK$ hold in a finitely generated polynomial $K$-algebra for $K$ a field?

Question

Let $K$ be a field and $R:=K[X_1,X_2,\dots, X_n]$ for a certain $n\in\mathbb N$. If $I,J,K$ are three ideals of $R$, can we conclude that $I(J\cap K)=IJ\cap IK$?

Answer 1

An integral domain $R$ is Prüfer iff for any non-zero ideals $I,J,K$ of $R$ the following holds: $I(J\cap K)=IJ\cap IK$ (see here.)

A polynomial ring over a field is Prüfer iff $n=1$.

A concrete conterexample: $R=K[X,Y]$, $I=(X,Y)$, $J=(X)$, and $K=(Y)$.

Answer 2

More generally, if $R$ is a commutative ring with unity and $A,I,J$ are three ideals, then $$ A (I \cap J) \subseteq AI \cap AJ $$ because any element of $A (I \cap J)$ can be simultaneously seen as an element of $AI$ and of $AJ$.

The reverse inclusion holds if $A$ and $X \cap Y$ are coprime because in that case $A$ is coprime with $X$ and $Y$, too, so $$ A (X \cap Y) = A \cap (X \cap Y) = (A \cap X) \cap (A \cap Y) = AX \cap AY $$ I don't think that it should hold in general, but right now I can't think of a counterexample.