What is the simplest example of a domain $R$ which is regular (in particular Noetherian) and factorial which admits a finitely generated projective module that is not free?


In fact I'll be at least somewhat happy with any example, since I can't think of one at the moment.


Some brief comments: $R$ needs to have Krull dimension greater than one or else it is a PID. The module in question needs to have rank greater than one, because the hypotheses force 
the Picard group to be equal to the divisor class group and the divisor class group to be trivial. And famously, by work of Quillen and Suslin, one cannot take $R$ to be a polynomial ring over a field. Oh yes, and of course $R$ can't be local (or even semilocal, I suppose). I'm already out of ideas... 


P.S.: If you can get an easier example by removing the hypothesis of finite generation, I'd be interested in that as well.



Nonfree projective module over a regular UFD?
Depending on what you consider simple, let $k$ be the complex numbers, or the integers, or the field with two elements (or any other commutative ring you're fond of). Let $R=k[a,b,c,x,y,z]/(ax+by+cz-1)$. Map $R^3$ to $R$ by $(f,g,h)\mapsto xf+yg+zh$. Let $P$ be the kernel of this map.


$P$ is the universal example of a rank 2 projective module over a $k$-algebra that becomes free after adding a free rank-one direct summand. (That is, if $A$ is any other $k$-algebra with such a projective module $Q$, then there is a map from $R$ to $A$ such that $Q=P\otimes\_RA$.) One could argue that this makes it the simplest example. In particular, Hugh Thomas's example arises from this example in this way.


To see that $P$ is not free, use the fact that Hugh Thomas's example is not free. Alternatively, invoke the much more general result of Mohan Kumar and Nori, which says that if $R=k[x\_1,...,x\_n,a\_1,...a\_n]/(\sum x\_ia\_i-1)$, then the kernel of the map defined by the $1\times n$ matrix $(x\_1^{m\_1},\ldots x\_n^{m\_n})$ cannot be free unless $m\_1\ldots m\_n$ is divisible by $(n-1)!$.


If you want an example where the UFD property is obvious, Mohan Kumar's paper "Stably Free Modules" gives a family of examples over rings of the form $A\_f$ where the rings $A$ are polynomial rings over fields. These examples all have the property that they become free after adding a free rank-one module.


