The Kauffman bracket skein module of S³ is rank one

We prove by induction on the number of crossings of a regular diagram for a blackboard-framed link that its class is a scalar multiple of $[\mathrm{\varnothing }]$. If the diagram has a crossing, the crossing relation in [F1] expresses its class as the sum of the two smoothing classes with coefficients $A$ and $A^{-1}$. The skein convention gives each smoothing its blackboard framing, and each has one fewer crossing, so the induction applies. A crossingless diagram is a finite disjoint union of planar circles. Choose an innermost circle; its planar disk has no other component in its interior and is disjoint from the remaining link. Its blackboard framing is the zero framing, so [F1] removes it with factor $\delta$. Remove this component and repeat; the crossingless link reduces to a scalar multiple of the empty link. For a general framed link, [F2] supplies a diagram and [F3], applied finitely many times to its framing-correction integer, reduces its class to a scalar multiple of the corresponding blackboard-framed diagram. Hence every class lies in $R[\mathrm{\varnothing }]$.

Define $\varphi :R\to {\mathcal{K}}_A(S^3)$ by $\varphi (r)=r[\mathrm{\varnothing }]$. Step 1.1 makes $\varphi$ surjective. If $\varphi (r)=0$, $R$-linearity and $E([\mathrm{\varnothing }])=1$ give $r=E(\varphi (r))=E(0)=0$. Thus $\varphi$ is injective as well as surjective, so the empty class is a free $R$-module generator.

The trivial-circle relation in [F1], with $L=\mathrm{\varnothing }$, gives $[O]=\delta [\mathrm{\varnothing }]$, as claimed.