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Excuse the possible naivete of this question. Since reading a nice survey article by Daniel Biss a few years ago, I'm always worried about what $P^1(R)$ is, for a ring $R$.

So that I stop worrying, I'm looking for an answer to the following question: For what (commutative, of course) rings $R$ is it true that $P^1(R)$ is naturally identifiable with the set of pairs $(a,b) \in R^2$ such that $(a,b)$ equals the unit ideal, modulo the natural action of $R^\times$?

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Similar discussion:… – Felipe Voloch Nov 23 '10 at 20:50
Thanks - I couldn't find that discussion when I searched earlier. – Marty Nov 23 '10 at 22:50
up vote 10 down vote accepted

This is equivalent to the property that every invertible (=rank-1 projective) $R$-module generated by two elements is free. Examples: semilocal rings, unique factorization domains, finite products of such rings.

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And in the Dedekind case, trivial class group is also necessary. – BCnrd Nov 23 '10 at 20:20
Thanks! That makes very good sense to me. – Marty Nov 23 '10 at 22:52
I think even for one-dimensional Noetherian domains, this property forces the Picard group to be trivial. – Hailong Dao Nov 24 '10 at 15:40

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