# Existence of constrained pairs of elements in $\mathbb{C}[x,y,z]/(x^3,y^3,z^3)$ that multiply to zero

Question: Do there exist a pair of elements $$a,b\in \mathbb{C}[x,y,z]/(x^3,y^3,z^3)$$, where $$a\notin(x,y)\cup(y,z)$$ and $$b\notin (x,z)\cup (y,z)$$ but $$ab=0$$?

For context, I'm trying to show that for a certain pair of subvarieties $$X,Y\subseteq \mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^2$$ the cup product of their fundamental classes $$[X][Y]$$ is nonzero, but I only have limited information about $$X$$ and $$Y$$. I was able to show $$[X]$$ and $$[Y]$$ satisfy the properties of $$a,b$$ above and I'm hoping this is enough to deduce $$[X][Y]\neq 0$$.

Unless I made a mistake somewhere, by a long argument I was able to show that if such $$a,b$$ exist with $$ab=0$$ then we must have $$a,b\in (x,y,z)^2\setminus (x,y,z)^3$$, i.e. $$a$$ and $$b$$ both have minimal degree $$2$$. I don't know how to proceed from here without explicitly solving the massive system of quadratic equations defining the set of pairs $$a,b\in (x,y,z)^2$$ with $$ab=0$$; based on my limited understanding of computer algebra, I doubt it's even possible for a computer to solve this system equations.

Note that $$a \not \in (x,y) \iff a$$ has a term which is a power of $$z$$ (i.e. $$cz^k$$ for some $$c \in \mathbb{C}^\ast$$ nonzero and $$k \in \{0,1,2\}$$).
Let $$\rho$$ denote a square root of $$\frac{3}{2}$$, i.e. $$2 \rho^2 - 3 = 0$$. Taking $$a := x^2 + xy + y^2 + 2 \rho xz + 2 \rho yz + z^2 \\ b := x^2 - 2xy + y^2 + \rho xz + \rho yz - 4z^2$$ in the polynomial ring $$\mathbb{C}[x,y,z]$$, one has $$ab \in (x^3, y^3, z^3)$$ (it suffices to check that the coefficients of the 6 monomials $$x^2y^2, x^2yz, x^2z^2, xy^2z, xyz^2, y^2z^2$$ in the product all vanish).
• I actually did use computer algebra software: starting with 2 quadrics $a = x^2 + z^2 + a_1xy + a_2y^2 + a_3xz + a_4yz$, $b = x^2 + y^2 + b_1xy + b_2z^2 + b_3xz + b_4yz$, the condition $ab \in (x^3,y^3,z^3)$ becomes a system of $6$ equations in $8$ unknowns. After verifying the solution set is nonempty (i.e. the system doesn't generate the unit ideal, which already proves existence), I then tried adding specializations of the form $a_1 = 1, a_2 = 1$ etc. until (fortunately) happening to find one where the equations reduced to the simple constraint in the answer Dec 10, 2023 at 2:54