 Solution

We prove by induction on $n$ that every submodule $N\subseteq R^n$ is free. The case $n=0$ is immediate. Let $\pi :R^n\to R$ be projection onto the first coordinate. Since $R$ is a principal ideal domain,for some $d$. If $d=0$, then $N\subseteq R^{n-1}$ and induction applies. Otherwise choose $x\in N$ with $\pi (x)=d$. For every $y\in N$, write $\pi (y)=rd$; then $y-rx\in \ker (\pi {|}_N)$. Also $Rx\cap \ker (\pi {|}_N)=0$ because $R$ is a domain and $d\ne 0$. ThereforeThe kernel is a submodule of $R^{n-1}$ and is free by induction, while $Rx\cong R$. Hence $N$ is free. This is the submodule theorem for free modules over a principal ideal domain.

If $P$ is finitely generated and projective, choose a surjection $f:R^n\to P$. Projectivity supplies $h:P\to R^n$ with $f\circ h={\mathrm{id}}_P$, soThus $P\cong h(P)$ is a submodule of the finitely generated free module $R^n$, and the result just proved shows that $P$ is free.