= Nilpotent chiral superfield
{title2=$X^2=0$}
For $X=x+\sqrt2\theta G+\theta^2F$, 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>.
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