Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 304 / 4 / b / Solution

In components the BRST transformation isThe last two equations immediately give $s^2\bar{c}=s^{2}B=0$. Applying the graded Leibniz rule to $s^{2}c$ produces a sum of three ghost monomials whose coefficient is the Jacobi identity $f^{e[ab}{f}^{c]de}$, so $s^{2}c=0$. Finally,the derivative terms cancel by anticommutation of the Faddeev-Popov ghost fields, and the remaining terms again cancel by the Jacobi identity. Hence $s^2=0$ on every field: $s$ is a nilpotent Grassmann-odd differential.