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.

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- Sep 29th 2009, 02:06 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. - Sep 29th 2009, 02:18 AMNonCommAlg
- Apr 16th 2013, 07:48 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.