$P$ is projective . Show that $P$ is a direct summand of a free $R-$ module

Question

$P$ is projective . Show that $P$ is a direct summand of a free $R-$ module

I am using the definition of a projective module as $P$ is projective if every exact sequence $M\rightarrow P\rightarrow 0$ splits

How to proceed?

Answer 1

Choose generators of $P$ to get a surjection from a free module $F$ onto $P$. This gives you a split (by your definition!) exact sequence $0 \to K \to F \to P \to 0$, showing $F = K \oplus P$.