P is a projective module. Prove that there exists $F$ free such that $P \oplus F \cong F$

Question

P is a projective module. Prove that there exists $F$ free such that $P \oplus F \cong F$

I am trying the following idea:

Since $P$ is projective it is the direct summand of a free module $F_0 = P\oplus T$. Then take $F = \underset{i \in \mathbb{N}}{\oplus}F_0$. With this we have:

$$ P \oplus F = P \oplus(\underset{i \in \mathbb{N}}{\oplus}F_0) = P \underset{i \in \mathbb{N}} \oplus (P\oplus T ) =$$ $$ = P \underset{i \in \mathbb{N}}\oplus (T\oplus P) = \underset{i \in \mathbb{N}} \oplus (P\oplus T ) = \underset{i \in \mathbb{N}}{\oplus}F_0 =F$$

First is this completely correct?

Second is there a better way to do this?

Answer 1

The idea is good, but you're misusing the symbols.

Take $T$ such that $P\oplus T$ is free. Take $F=\bigoplus_{i\in\mathbb{N}}(P\oplus T)$. Then $F$ is free as well and $$ P\oplus F= P\oplus\bigoplus_{i\in\mathbb{N}}(P\oplus T)\cong P\oplus\bigoplus_{i\in\mathbb{N}}(T\oplus P)\cong \bigoplus_{i\in\mathbb{N}}(P\oplus T)=F $$ with obviously defined isomorphisms.

Answer 2

This is the "Eilenberg--Mazur swindle" of homological algebra. It was used by Bass to prove that "big" projective modules are free in this very readable paper. In there, you can also find a proof of what you want.