Nilpotent chiral superfield ($X^2=0$)

For $X=x+\sqrt{2}\theta G+{\theta }^{2}F$, the nilpotency condition gives $x^2=xG=0$ and $2xF=GG$. On the branch with invertible commuting part of $F$, $x=GG/(2F)$, while the goldstino $G$ and auxiliary field $F$ are independent. Products of three identical two-component odd spinor entries vanish, verifying the remaining constraints. The scalar is therefore composite. The branch hypothesis matters: $X=0$ also obeys the constraint and need not break supersymmetry.

With $X^2=0$ and invertible $F_X$, the additional chiral superfield $Y=y+\sqrt{2}\theta \chi +{\theta }^2{F}_Y$ satisfying $XY=0$ has $y=(G\chi )/F_X-(GG)F_Y/(2F_X^2)$. Its Weyl spinor $\chi$ and auxiliary field $F_Y$ remain independent. This eliminates the independent scalar. It does not force $Y^2=0$.

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 $(G\chi {)}^2=-\frac{1}{2}(GG)(\chi \chi )$. Then $y^2=-(GG)(\chi \chi )/(2F_X^2)$, $y^3=y^2\chi =0$ and $y^2{F}_Y-y\chi \chi =0$. 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 $1,X,Y,Y^2$.