= Cubic nilpotency from a mixed chiral constraint
{title2=$X^2=XY=0\Longrightarrow Y^3=0$}
On the invertible-$F_X$ branch, use $y=(G\chi)/F_X-(GG)F_Y/(2F_X^2)$ and the two-component <Grassmann algebra> identity $(G\chi)^2=-\tfrac12(GG)(\chi\chi)$. Then $y^2=-(GG)(\chi\chi)/(2F_X^2)$, $y^3=y^2\chi=0$ and $y^2F_Y-y\chi\chi=0$. These are the components of $Y^3=0$. Generally $y^2$ is nonzero, so cubic nilpotency must not be replaced by quadratic nilpotency. The allowed analytic holomorphic monomials are $1,X,Y,Y^2$.
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