(Eilenberg) Prove that every projective left R-module P has...

(Eilenberg) Prove that every projective left $R$-module $P$ has a free complement; that is, there exists a free left $R$-module $F$ such that $P \oplus F$ is free.
Hint. If $P \oplus Q$ is free, consider $Q \oplus P \oplus Q \oplus P \oplus \cdots$.

Key Concepts

Direct Sum Decomposition

The concept of direct sum decomposition involves splitting a module into two or more submodules whose intersection is trivial, such that the original module is isomorphic to the direct sum of these submodules. This concept is key in the study of projective modules since every projective module being a direct summand of a free module leads naturally to questions about whether the complementary summand can also be chosen to be free.

Projective Modules

Projective modules are those modules for which every surjective homomorphism onto them splits, meaning they are direct summands of free modules. This property is essential because it guarantees that for any free module containing a projective module as a summand, there exists a complement, facilitating many decompositions and extensions in module theory.

Free Modules

Free modules, which can be thought of as modules with a basis, are central because they allow every element to be uniquely expressed as an R-linear combination of basis elements. They serve as the building blocks in module theory, and working with free modules simplifies many proofs due to their well-understood structure.

Eilenberg's Trick

Eilenberg's trick is a clever technique which involves constructing an infinite direct sum by interlacing copies of given modules to obtain a free module. In this context, it is used to show that if a module is a direct summand of a free module, then by carefully reordering and forming larger direct sums, one can obtain a free complement. This method is essential for creating the necessary complement that, when added to the projective module, results in a free module.

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