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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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    $\begingroup$ Similar discussion: mathoverflow.net/questions/46116/… $\endgroup$
    – Felipe Voloch
    Commented Nov 23, 2010 at 20:50
  • $\begingroup$ Thanks - I couldn't find that discussion when I searched earlier. $\endgroup$
    – Marty
    Commented Nov 23, 2010 at 22:50

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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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    $\begingroup$ And in the Dedekind case, trivial class group is also necessary. $\endgroup$
    – BCnrd
    Commented Nov 23, 2010 at 20:20
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    $\begingroup$ I think even for one-dimensional Noetherian domains, this property forces the Picard group to be trivial. $\endgroup$
    – Hailong Dao
    Commented Nov 24, 2010 at 15:40

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