Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $R^\times$?

share|improve this question
Similar discussion: mathoverflow.net/questions/46116/… –  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
add comment

1 Answer

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.