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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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: $\displaystyle Re \oplus R(1-e)=R.$
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.
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