A question about projective modules and free modules.

Question

If $P$ is a projective $R$-module does there exist a $R$-module $Q$ such that $F=P \oplus Q$ is free?

I am studying "Module Theory" from N.S.Gopalakrishnan's "Commutative Algebra".In page 10 while proving the proposition 1.2.12 the author has made this assertion without proof.I am a new to this subject.If somebody helps me understanding it then I will be really benefitted by it.

Thank you in advance.

Answer 1

it is well known that one characterization of a projective module is that it is a direct summand of a free module.

to see this, let P be projective and choose a generating set $S=\{m_{\alpha}\}_{\alpha \in A}$

let $F$ be the free module on a set $X$ isomorphic to $S$ and look at the surjective map: $$ p:F \to P $$ defined by $p(x_{\alpha})=m_{\alpha}$

by the defining property of a projective module there is a map $q:P \to F$ (a lift of the identity map $P \to P$) which satisfies $q\circ p = id_P$

you now have: $$ F = ker(p) \oplus im(q) $$