A one-variable BV algebra and a classical master solution

Every element is $f(x)+\theta g(x)$ with $f,g\in \mathbb{C}[x]$ and ${\theta }^2=0$; the polynomial quotient multiplication is associative and the assigned degrees make it graded commutative. For homogeneous factors, the ordinary product rule gives the Leibniz rule for ${\mathrm{\partial }}_x$; direct expansion gives ${\mathrm{\partial }}_{\theta }(f\theta j)=fj$, ${\mathrm{\partial }}_{\theta }(\theta gh)=gh$, and ${\mathrm{\partial }}_{\theta }((\theta g)(\theta h))=0=g(\theta h)-(\theta g)h$, so ${\mathrm{\partial }}_{\theta }$ is a degree-$+1$ graded derivation. Thus $\mathrm{\Delta }(f+\theta g)=g^{\mathrm{\prime }}(x)$ has degree $+1$, $\mathrm{\Delta }(1)=0$, and ${\mathrm{\Delta }}^2=0$.

For a homogeneous derivation $D$, $[D,m_a]=m_{D(a)}$; the composition identity gives $[\mathrm{\Delta },m_a]=m_{\mathrm{\Delta }a}+m_{{\mathrm{\partial }}_{\theta }a}{\mathrm{\partial }}_x+(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{m}_{{\mathrm{\partial }}_{x}a}{\mathrm{\partial }}_{\theta }$. Commuting once more yields $[[\mathrm{\Delta },m_a],m_b]=m_{{\mathrm{\partial }}_{\theta }a}{m}_{{\mathrm{\partial }}_{x}b}+(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{m}_{{\mathrm{\partial }}_{x}a}{m}_{{\mathrm{\partial }}_{\theta }b}$, a multiplication operator. Multiplication operators gradedly commute, so the third commutator with $m_c$ is zero and [F1]'s order condition holds.

Direct substitution in [F1] gives $\{f(x),\theta g(x)\}=f^{\mathrm{\prime }}(x)g(x)$ because $\mathrm{\Delta }(f\theta g)=(fg{)}^{\mathrm{\prime }}$, $\mathrm{\Delta }f=0$, and $\mathrm{\Delta }(\theta g)=g^{\mathrm{\prime }}$. Hence $\{S,\theta g\}=x^{2}g$ and $\{S,f\}=0$; also $S$ and $S^2$ depend only on $x$, so $\{S,S\}=0$ by the same formula. Thus $S$ satisfies the CME [F2].

Step 2.2 gives $d_S(f+\theta g)=x^{2}g$, which is independent of $\theta$ and is killed by $d_S$. Therefore $d_S^2=0$ directly, in agreement with [F3].