Proving Projective Modules Have Free Modules as Direct Sums

[peteryellow]

Please help me to prove that for a projective module P there exists a free module F, such that P +F =F.

Here + denotes direct sum = denotes isomorphic.

Thanks

 

[gel]

Did you really mean to write P + F = F? It's a standard result that a module is projective if and only if there is a module Q and free module F such that P + Q = F. Can you prove that?

It does seem to follow that there exists a free module F' with P + F' = F', although I've never seen it stated like that before.

 

[PatF]

I have seen this called Eilenberg's trick. The idea is that Q+P=F1 where Q is projective and F1 is free. Now let F=F1+F1+F1.. a countable number of times.

Then, P+F is isomorphic to P+Q+P+Q+P+Q.. which is isomorphic to F.

 

[gel]

That's what I was thinking of, although I didn't know it was called Eilenberg's trick.