Let $F$ be a field. Prove that in the ring of polynomials in $x$, $y$ and $z$ with coefficients in $F$ $:$
$\left (Y^2 Z^2, XYZ \right ) = \left (Y \right ) \cap \left (Z \right ) \cap \left (X,Y \right ).$
Any polynomial in $\left (Y^2 Z^2, XYZ \right )$ is of the form $p(x,y,z)y^2 z^2 + q(x,y,z) xyz$. Which can be written in three forms
(1) $y(p(x,y,z)yz^2 + q(x,y,z)xz)$ i.e. $yP(x,y.z)$ where $P(x,y,z) = p(x,y,z)yz^2 + q(x,y,z)xz.$
(2) $z(p(x,y,z)y^2 z + q(x,y,z)xy)$ i.e. $zQ(x,y,z)$ where $Q(x,y,z) = p(x,y,z)y^2 z + q(x,y,z)xy.$
(3) $y(p(x,y,z)yz^2) + x (q(x,y,z)yz)$ i.e. $xR(x,y,z) + yS(x,y,z)$ where $R(x,y,z) = y(p(x,y,z)yz^2)$ and $S(x,y,z) = x(q(x,y,z)yz).$
From (1), (2) and (3) it is clear that $\left (Y^2 Z^2, XYZ \right ) \subseteq \left (Y \right ) \cap \left (Z \right ) \cap \left (X,Y \right ).$
But how can I do the reverse part. Please help me.
Thank you in advance.
The ideal $(Y) \cap (Z) \cap (X,Y)$ simplifies to $(Y) \cap (Z)$, since $(Y) \subset (X,Y)$, and then further simplifies to $(YZ)$.
Thus, the reverse inclusion holds if and only if $YZ \in (Y^2Z^2,XYZ)$. \begin{align*} \text{But}\;\;&YZ \in (Y^2Z^2,XYZ)\\[4pt] \implies\;&YZ=aY^2Z^2+bXYZ,\;\text{for some}\;a,b \in F[X,Y,Z]\\[4pt] \implies\;&1 = aYZ + bX\\[4pt] \implies\;&1 = 0\qquad\text{[by substituting $X=Y=Z=0$]}\\[4pt] \end{align*} It follows that the reverse inclusion doesn't hold.