Let e in R be idempotent. Show that the ideal Re is a projective R module.

I want to do this without explicitly lifting. But I am struggling to write Re + M = N such that N is free.

Printable View

- September 29th 2009, 03:06 AMrobeulerIdempotents and Projective R-Modules
Let e in R be idempotent. Show that the ideal Re is a projective R module.

I want to do this without explicitly lifting. But I am struggling to write Re + M = N such that N is free. - September 29th 2009, 03:18 AMNonCommAlg
- April 16th 2013, 08:48 AMswangRe: Idempotents and Projective R-Modules
Do this hint work? I think R(1-e) = R because 1-e is unit by (1-e)(1+e)=1 if e is idempotent.