Let $R$ be a unital ring. In the following modules will be left modules.

We will say that $R$ has the free submodule property if for any free module $F$ over $R$ and any submodule $F'\subseteq F$ we have that $F'$ is also free. It is well known, that if $R$ is a PID, then $R$ has the free submodule property. One can ask whether the converse is also true? We will try to answer this question.

Proposition. If $R$ is a commutative ring, which is not a PID, then $R$ does not have the free submodule property.

Proof. Assume that $R$ is not a PID. Then there are two possibilities: either $R$ is not a domain or there is an ideal $I\subseteq R$ which is not principal. Assume that $R$ is not a domain and let $a,b\in R$ be two zero divisors, i.e. $a\neq 0$ , $b\neq 0$ and $a\cdot b=0$ . Let $(b)\subseteq R$ be an ideal generated by $b$ . Then obviously $(b)$ is a submodule of $R$ (regarded as a $R$ -module). Assume that $(b)$ is free. In particular there exists $m \in (b)$ , $m\neq 0$ such that $r\cdot m=0$ if and only if $r=0$ . But $m$ is of the form $\lambda\cdot b$ and because $R$ is commutative we have $$a\cdot m=a\cdot (\lambda\cdot b)=\lambda\cdot (a\cdot b)=0.$$ Contradiction, because $a\neq 0$ . Thus $(b)$ is not free although $(b)$ is a submodule of a free module $R$ .

Assume now that there is an ideal $I\subseteq R$ which is not principal and assume that $I$ is free as a $R$ -module. Since $I$ is not principal, then there exist $a,b\in I$ such that $\{a,b\}$ is linearly independent. On the other hand $a,b\in R$ and $1$ is a free generator of $R$ . Thus $\{1,a\}$ is linearly dependent, so $$\lambda\cdot 1+ \alpha\cdot a=0$$ for some nonzero $\lambda, \alpha\in R$ (note that in this case both $\lambda,\alpha$ are nonzero, more precisely $\lambda=a$ and $\alpha=-1$ ). Multiply the equation by $b$ . Thus we have $$\lambda\cdot b + (\alpha\cdot b)\cdot a=0.$$ Note that here we used commutativity of $R$ . Since $\{a,b\}$ is linearly independend (in $I$ ), then the last equation implies that $\lambda=0$ . Contradiction. $\square$

Corollary. Commutative ring is a PID if and only if it has the free submodule property.