I want to show that in the polynomial ring $K[X,Y,Z,W]$ (where $K$ is a field) the equality $(X,Y)\cap(Z,W)=(XZ,XW,YZ,YW)$ holds.
Obviously RHS is contained in LHS. How to show that LHS is contained in RHS? I took $fX+gY=hZ+kW\in(X,Y)\cap(Z,W)$ and wanted to show that both $f$ and $g$ are divisible by either $Z$ or $W$. How to achieve this? Thank you for any hints.
More general question: in the polynomial ring $K[X_1,...,X_n]$, is it true that $I\cap J=IJ$ for all ideals $I,J$? If it's not, what about if we restrict to prime ideals $I,J$?
Let $fX+gY \in (Z,W)$, while we can assume that $f$ does not admit a monomial containing $Y$ (if so, we would just shift it to $g$).
We obtain $fX = 0$ in $K[X,Y,Z,W]/(Y,Z,W)=K[X]$, hence $f=0$ in $K[X,Y,Z,W]/(Y,Z,W)$, which means $f \in (Y,Z,W)$. By the assumption on $f$ we obtain $f \in (Z,W)$, the desired result for $f$.
We are left with $g$, but this is an easy one now: In particular we have already shown $fX \in (Z,W)$, so we obtain $gY \in (Z,W)$, which immediately implies $g \in (Z,W)$, since $Y$ is a non-zero divisor in $K[X,Y,Z,W]/(Z,W) = K[X,Y]$.
In general $IJ = I \cap J$ is false even for prime ideals in polynomial rings. It is even though true if one of them is maximal and not containing the other one. Because then $I+J=(1)$ holds.