Thread: Idempotents 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.

Originally Posted by robeuler

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.

Hint:

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.