# Idempotents and Projective R-Modules

• Sep 29th 2009, 03:06 AM
robeuler
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.
• Sep 29th 2009, 03:18 AM
NonCommAlg
Quote:

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: $Re \oplus R(1-e)=R.$
• Apr 16th 2013, 08:48 AM
swang
Re: 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.