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 $t1$. I am interested in the ring $S^{1}R=\mathbb{Z}[t^{\pm 1},(t1)^{1}]$ obtained by inverting $t1$. More specifically, I know that finitely generated projective $R$modules are free (e.g. by the QuillenSuslin theorem) and I would like to know whether finitely generated projective $S^{1}R$modules are free?

$\begingroup$ Out of curiosity: is there a reason you don't ask the stronger question about $\mathbf Z[t]$? That would be an even closer analogue to Quillen–Suslin. $\endgroup$ – R. van Dobben de Bruyn May 16 '20 at 22:24

$\begingroup$ @R.vanDobbendeBruyn since there's a single question in the post, I'm not sure what you mean by "the stronger question". Do you mean, for every $S$ instead of this specific one? $\endgroup$ – YCor May 16 '20 at 22:54

$\begingroup$ @YCor: see Mohan's answer (vector bundles on regular schemes of dimension $\leq 2$ extend). $\endgroup$ – R. van Dobben de Bruyn May 17 '20 at 1:03

$\begingroup$ @R.vanDobbendeBruyn : no sensible reason, other than it is a vice of my trade, knot theorists often work with $\mathbb{Z}[t^{\pm 1}]$ instead of $\mathbb{Z}[t]$. $\endgroup$ – Anthony Conway May 17 '20 at 2:38
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 QuillenSuslin 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.

2$\begingroup$ Actually the result that every projective module over $A[t]$ or $A[t,u]$ for $A$ PID was proved in the VasersteinSuslin paper (attributed to Suslin in Bass' Bourbaki seminar), which precedes QuillenSuslin by... few years. $\endgroup$ – YCor May 16 '20 at 22:58

$\begingroup$ Thank you for your answer. Do you mind expanding a bit please? From what I read, you are saying that $S^{1}M=S^{1}M^{**}$ for $M$ f.g. which then implies that $M$ is reflexive? I fear I don't understand either of these steps. Where are you using that $P$ is projective? Did you use anything about my specific localisation or are you working with any $A$ of dimension $\leq 2$ and any multiplicative subset $S \subset A$? $\endgroup$ – Anthony Conway May 17 '20 at 2:34

$\begingroup$ @AnthonyConway I have added a few more comments in my answer. $\endgroup$ – Mohan May 17 '20 at 3:00

$\begingroup$ @AnthonyConway: more precisely, $S^{1}(M^{**}) = (S^{1}M)^{**}$. But $S^{1}M$ is finite projective, so in particular reflexive. $\endgroup$ – R. van Dobben de Bruyn May 17 '20 at 3:15

1$\begingroup$ @Mohan You really should explain what you use about the ring. For instance, it would not work when $A$ is a polynomial ring in three variables over the reals. $\endgroup$ – Wilberd van der Kallen May 17 '20 at 7:06