Let $R$ be an integral domain. Show that $R$ is a PID is equivalent to every submodule of a free $R$-module of finite rank is free of finite rank.
(First question: why do we require $R$ being an integral domain? Is it because "rank" doesn't make sense for $R$ not an integral domain?)
One direction is a theorem, I am trying to show the other direction.
$R$ itself is a free $R$-module of rank 1. Let $I\neq 0$ be an ideal. It is a submodule of $R$, so by assumption free of rank $n$, $I\cong R^n$ as $R$-modules. We want to show that $n=1$. Can anybody give a hint to this one?
First, if $R$ is not an integral domain, there are elements $a,b \in R \setminus \{0\}$ such that $ab = 0$. Suppose now that the $R$-module generated by $a$ were free. By definition, it then has a basis (i.e., a generating system such that linear combinations are unique). However, if $e$ is an element of such a basis, $e$ is a multiple of $a$, such that $b \cdot e = 0 \cdot e = 0$ are two different linear combinations of $0$.
To your other question, assume that $n > 1$. Then, through the isomorphism $I \cong R^n$, you will find non-zero elements $a,b \in I$, such that $ab = 0$, a contradiction (since $R$ is a domain). (Or one could say that if $n > 1$, the ideal $I$ has a basis consisting of $\geq 2$ elements $r_1, ..., r_n$. However then, $r_1r_2 - r_2r_1 = 0$ is a non-trivial linear combination of $0$.)