We are given a PID. Then we have to show that if F is a finitely generated free module of rank n, then a submodule M of F is free. This is fine. However, it then says to deduce that every finitely generated projective module is also free.
So, we have the result that F = L $\bigoplus$ $\alpha$(P), for some injective homomorphism $\alpha : P \to F$, F a free module and L a submodule of F. I think I'm missing something here because this suggests to me straight away that if P is not free, F cannot be free and we don't need to know about L. However, I'm assuming that we need to show F is in fact finitely generated as P is finitely generated and therefore L can taken as free and, from this, that we have P is therefore free as well? Would anyone be able to help?
Edit: As soon as I posted this, I realised the proof used the following. As M is a submodule, F = M $\bigoplus$ ker($\pi$) (a projection) and that ker($\pi$) is free. So, are we meant to deduce that P is ker($\pi$) and therefore free?
Let $A$ a PID. $P$ be a finitely generated, projective module over $A$. Finitely generated over $A$ means, there is some $n \in \mathbb{N}$ and some $A$-Module $K$, such that there's an exact sequence $$0 \rightarrow K \rightarrow A^n \rightarrow P \rightarrow 0$$ As $P$ is projective, this sequence is split exact. Or in other words, we can regard $P$ as a submodule of $A^n$ (which is by definition finitely generated free), as $A^n \cong K \oplus P$. Using the statement from the second sentence of your original post, we are done.