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.

If Pete or someone else is still interested despite the fine answers already given, here is an analysis of what might be the simplest situation. Let $k$ be a field of characteristic $\neq 2$ and define $A=k[X,Y,Z]/(X^2+Y^2+Z^2-1)=k[x,y,z]$ where $X,Y,Z$ are indeterminates.

Proposition 1 (Nagata) The ring $A$ is a UFD iff $-1$ is not a square in $k$.

Sketch of proof: If $-1=i^2$ for some $i\in k$, the equality $x^2+y^2=(1+z)(1-z)$ shows that $A$ is not a UFD. In the other direction, if $-1$ is not a square, factoriality follows fairly easily from the following theorem of Nagata: if a noetherian domain $A$ has a prime element $t\in A$ such that $A[\frac{1}{t}]$ is a UFD, then $A$ was already a UFD.

Proposition 2 (Serre, Samuel) Consider the module $M=A\partial _x \oplus A\partial _y \oplus A\partial _z$ (the free module of rank 3 over $A$ ) and its quotient $P=M/(x\partial _X+ y\partial _y+ z\partial _z)$ . Then $P$ is projective over $A$. If moreover the field $k$ is formally real, then that module $P$ is not free over $A$.

Comments Formally real means that $-1$ is not a sum of squares in $k$. By Artin-Schreier theory this is equivalent to $k$ being orderable. Projectivity follows from an easy explicit calculation. Non-freeness over $\mathbb R$ follows , by the Serre-Swan correspondence, from the well known theorem in topology that the tangent bundle to the sphere is not trivial . In the general case of a formally real field one resorts to logical trickery of Tarski type to reduce to the case $k=\mathbb R$. Interestingly up to a few years ago, experts assured me that there was no purely algebraic proof in the case of $\mathbb R$. I don't know if one has been found since. Finally, if $k=\mathbb C$ Serre has shown that $P$ is free (however, by Proposition 1, the ring $A$ is not a UFD in the case $k=\mathbb C$ )

The simplest example I can think of is $\mathbb C[x_1, x_2, x_3, x_4, x_5]/(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - 1)$. The proof I am going to give is almost certainly an overkill, a simpler approach should be possible. Call $X$ the spectrum of the algebra above. This is the ring of regular functions of the complement of a smooth projective quadric of dimension 3 in a smooth projective quadric of dimension 4. By standard results on Chow groups of quadrics, the Chow group of $X$ has rank 1 in dimension 2; hence the K-theory rings, which is rationally isomorphic to the Chow ring, has rank at least 2, and this implies that there must be vector bundles on $X$ which are non-trivial. These correspond to non-free projective modules.