Unique Laurent decomposition for the Jones–Vassiliev basis

Let $Q=R[X]\mathrm{/}(G)$. By [F1] and [F2], $R[X]$ is a commutative polynomial ring, and [F3] makes $(G)$ a two-sided ideal and $Q$ a quotient ring. Since $G$ is monic of degree two, [F4] gives every class in $Q$ a representative $a(p)+b(p)X$ of degree at most one in $X$. If two such representatives define the same class, their difference is a multiple of $G$ and has degree less than two; uniqueness in [F4] forces the difference to be zero. Thus this representative is unique.

Define $\varphi :R[X]\to \mathbb{Z}[x,x^{-1}]$ on finite polynomials by $\varphi ({\sum }_j{c}_j(p)X^j)={\sum }_j{c}_j(x-x^{-1})x^j$. Substitution and finite convolution make $\varphi$ a ring homomorphism, and $\varphi (G)=x^2-(x-x^{-1})x-1=0$, so $\varphi$ is constant on cosets of $(G)$ and induces $\bar{\varphi }:Q\to \mathbb{Z}[x,x^{-1}]$. In $Q$, $X(X-p)=1$, so $X$ is a unit with inverse $X-p$. Define $\psi :\mathbb{Z}[x,x^{-1}]\to Q$ by $\psi ({\sum }_k{n}_k{x}^k)={\sum }_{k\ge 0}{n}_k{X}^k+{\sum }_{k<0}{n}_k(X-p{)}^{-k}$; because $X-p=X^{-1}$, finite Laurent multiplication is preserved. The composite $\bar{\varphi }\psi$ fixes $x$, while $\psi \bar{\varphi }$ fixes $X$ and $p$; since $Q$ is generated by $X$ and $p$, the maps are inverse. The unique representatives from step 1.1 therefore give the asserted unique Laurent decomposition.