TITLE: Algebraic analogue of the Moebius bundle over the circle
QUESTION [15 upvotes]: Let $R$ be the ring $R[X,Y]/(X^2+Y^2−1)$. The space of $\mathbb{R}$-rational points of the affine scheme associated to $R$ is the topological circle $S^1$.
An algebraic vector bundle over $R$ is an $R$-algebra $A$ with certain properties or equivalently a finitely generated projective $R$-module $A$.
My first question is: What is the explicit projective $R$-module $A$ corresponding to the topological Moebius bundle over $S^1$? (By 'corresponding' I mean in particular that the $\mathbb{R}$-rational points should induce the topological Moebius bundle and that it has rank one.)
My second question is motivated by the fact that topologically there are only two non-isomorphic rank one bundles over $S^1$: the trivial bundle and the Moebius bundle. What is the analogous algebraic situation? Are there more than two non-isomorphic and rank one projective modules over $R$?
Thank you!
(I apologize that I first asked this question on math.stackexchange, I deleted it there.)

REPLY [16 votes]: The internal note "Le ruban de Moebius comme représentation d'un idéal non principal" (Moebius band, as a non-principal ideal), by Daniel Ferrand, contains more or less the material you want. Daniel is retired and doesn't have a website, but, thank you Google!,
the note is downloadable from http://www.math.unibas.ch/~giordano/Moebius.pdf
The non-principal ideal you want is the ideal $(1+x,y)$ in the ring $R[x,y]/(x^2+y^2-1)$.

REPLY [7 votes]: If you want to describe the projective module as a direct summand of a free module, you can use the 2 by 2 projection matrix
$\frac{1+X}{2}$ $\frac{Y}{2}$
$\frac{Y}{2}$ $\frac{1-X}{2}$
Since this ring is a Dedekind domain, the projective modules of rank one correspond to the ideal classes. (And see Pete Clark's answer for what that works out to be. I think it's a fun exercise.)