Let $R=\mathbb{Z}[t^{\pm 1}]$ be the ring of Laurent polynomials, and let $S \subset R$ be the multiplicative subset generated by the polynomial $t-1$. I am interested in the ring $S^{-1}R=\mathbb{Z}[t^{\pm 1},(t-1)^{-1}]$ obtained by inverting $t-1$. More specifically, I know that finitely generated projective $R$-modules are free (e.g. by the Quillen-Suslin theorem) and I would like to know whether finitely generated projective $S^{-1}R$-modules are free? 



Are projective modules over a certain localised Laurent polynomial ring free?
The answer is yes. Given any projective module $P$ over $S^{-1}A$, where $A=\mathbb{Z}[t]$ (and works for many other rings too), it is the localization $S^{-1}M$ of a projective module over $A$. The reason is, you can always find such a finitely generated module $M$ with $S^{-1}M=P$, but you may replace $M$ with its double dual without affecting the localization, but any reflexive module over $A$ is projective (and thus free, by Seshadri's theorem, which precedes Quillen-Suslin by many years).


To answer your questions in the comments below, double dual of any finitely generated module over $A$ is reflexive. Since $P$ is projective (and hence reflexive), it follows that if $S^{-1}M=P$,then so is $S^{-1}(M^{\*\*})$. For your last question, for any Noetherian ring $A$ and $S\subset A$ a multiplicatively closed set, given any finitely generated module $P$ over $S^{-1}A$, there exists a finitely generated module $M$ over $A$ such that $S^{-1}M=P$. Further, if $P$ reflexive, then you may replace $M$ by $M^{\*\*}$ and thus assume it is reflexive.


