# Are projective modules over a certain localised Laurent polynomial ring free?

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?

• 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. – R. van Dobben de Bruyn May 16 '20 at 22:24
• @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? – YCor May 16 '20 at 22:54
• @YCor: see Mohan's answer (vector bundles on regular schemes of dimension $\leq 2$ extend). – R. van Dobben de Bruyn May 17 '20 at 1:03
• @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 $\mathbb{Z}[t]$. – 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 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.
• Actually the result that every projective module over $A[t]$ or $A[t,u]$ for $A$ PID was proved in the Vaserstein-Suslin paper (attributed to Suslin in Bass' Bourbaki seminar), which precedes Quillen-Suslin by... few years. – YCor May 16 '20 at 22:58
• 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 $S \subset A$? – Anthony Conway May 17 '20 at 2:34
• @AnthonyConway: more precisely, $S^{-1}(M^{**}) = (S^{-1}M)^{**}$. But $S^{-1}M$ is finite projective, so in particular reflexive. – R. van Dobben de Bruyn May 17 '20 at 3:15
• @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. – Wilberd van der Kallen May 17 '20 at 7:06