 Nilpotent chiral superfield

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.