Lemma: Field nilpotence

The field differential satisfies $s^2=0$ off shell. Its zeroth-order part is

Proof. For the odd vector field $c$, set $w^{\mu }=c^{\nu }{\partial }_{\nu }{c}^{\mu }$. The graded Lie identity is ${{\mathcal{L}}_c}^2={\mathcal{L}}_w$. Since $sq=\kappa {\mathcal{L}}_{c}q$ and $sc=\kappa w$,

The fixed background implies $s^{2}r=0$. The same identity on scalars gives $s^2\varphi =0$; direct expansion gives $s^{2}c=0$. The nonminimal pair is nilpotent by definition.