= Solution
Induct on $n$ to prove the stronger assertion that any <submodule> $M\subseteq R^n$ is a <free module> of <rank of a free module> at most $n$. For $n=0$ this is immediate. Project onto the first coordinate by $\pi:R^n\to R$. Its image $\pi(M)$ is an <ideal> of the <principal ideal domain>, so $\pi(M)=(a)$.
If $a=0$, then $M$ embeds in the remaining $n-1$ coordinates and the induction hypothesis applies. Otherwise choose $v\in M$ with $\pi(v)=a$. Put $N=M\cap\ker\pi$, a <submodule> of $R^{n-1}$, which is free by induction. For every $m\in M$, there is $b\in R$ with $\pi(m)=ba$, so $m-bv\in N$. If $bv\in N$, then $ba=0$; the <integral domain> property and $a\ne0$ give $b=0$. Therefore
$$
\boxed{M=Rv\oplus N\cong R\oplus N}.
$$
Appending $v$ to a <basis> of $N$ proves that $M$ is a <free module> of <rank of a free module> at most $n$. This proof of the <submodule theorem for free modules over a principal ideal domain> does not assume in advance that $M$ is a <finitely generated module>.
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