Skip to main content
4 events
when toggle format what by license comment
Apr 1, 2018 at 15:56 history bounty ended CommunityBot
Apr 1, 2018 at 14:20 vote accept CommunityBot
Mar 30, 2018 at 17:40 comment added user111524 You don't need to go to Picard group ... $M \oplus R \cong R^2$ already implies $M$ is free of rank $1$. Also, that the sequence splits can be seen from the fact that $Ra+Rb$ is a finitely generated ideal , hence projective because the domain is Prufer.
Mar 29, 2018 at 9:33 history answered Laurent Moret-Bailly CC BY-SA 3.0