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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.

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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).

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  • $\begingroup$ Thanks! If you don't mind me asking, how did you come up with this example? $\endgroup$
    – BHT
    Dec 9, 2023 at 22:32
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    $\begingroup$ 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 $\endgroup$
    – math54321
    Dec 10, 2023 at 2:54

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