Remark: Formal ideals and the BRST boundary

In $\mathbb{C}[[\hslash ]]$, $(\hslash )$ contains $\hslash 1$ but not $1$. A commutator proportional to $\hslash 1$ therefore does not alone prove that a formal quotient is the zero algebra. Inverting $\hslash$ changes the problem.

The supplied X-G sheet separately constructs a free causal BRST differential and excludes its specified exact source-preserving variational upgrade. It leaves the interacting quantum BRST construction unresolved CheungXG. Neither that upgrade nor an interacting identity is inferred from the formal-order proposition.