= Solution
Prove the <submodule theorem for free modules over a principal ideal domain> by induction on the rank of $R^n$. For a submodule $N$, project onto its first coordinate. The image is an ideal $aR$. If $a=0$, then $N$ is a submodule of $R^{n-1}$ and induction applies. Otherwise choose $v\in N$ projecting to $a$. Every $w\in N$ has first coordinate $ra$ and hence $w-rv$ lies in the projection kernel. The domain property shows $Rv$ meets that kernel only at zero. Thus
$$
N=Rv\oplus\ker(\text{first-coordinate projection}|_N),
$$
and the kernel is a <free module> by induction. This proves that $N$ is free, of rank at most $n$.
To prove <free modules are projective>, choose a basis $(e_\lambda)$ of a free module $P$. For the given surjection $f$, choose $m_\lambda\in M$ with $f(m_\lambda)=g(e_\lambda)$ and extend $e_\lambda\mapsto m_\lambda$ linearly. Finite support of basis expansions makes the resulting <module homomorphism> well defined, and $f\circ h=g$.
Finally, take a finite generating set of a <projective module> $P$ to obtain a surjection $\pi:R^n\to P$. Projectivity applied to $\operatorname{id}_P$ gives a section $s:P\to R^n$ with $\pi s=\operatorname{id}_P$. Thus $s$ is injective and identifies $P$ with a submodule of $R^n$. The proved <submodule theorem for free modules over a principal ideal domain> shows that \b[every finitely generated projective module over a principal ideal domain is free].
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