I know that there are other answers to this question, but I want help with a specific approach. I want to use the contrapositive. $R$ is commutative and unital.
Here is what I have so far:
Proof. Suppose that $V$ is a free $R$-module. Recall also that a module is free if and only if it has a basis.
Recall that a principal ideal domain is an integral domain where all ideals are generated by one element.
We prove the contrapositive, that is, if $R$ is not a PID, then there exists a submodule of $V$ that is not free. Since $R$ is not a PID, that means that there exists an ideal $I$ of $R$ that is generated by at least two elements. Recall that $IV \leqslant V$ (it is a submodule). We claim that $IV$ is not a free module. It is sufficient to show that $IV$ does not have a basis. We denote
\begin{align*}
IV &= \left\{\sum_{i=1}^k a_iv_i : a_i\in I, v_i \in V, k \in \mathbb{N}\right\}\\
\end{align*}
I am stuck here. Is this an approach that will work or should I just try the direct proof?
Best Answer
Note: saying an ideal is “generated by at least two elements” does not, in and of itself, preclude the possibility that it it principal. I mean, $(2,3)$ is an ideal of $\mathbb{Z}$ generated by at least two elements; it’s even a minimal generating set, as no proper subset generated the ideal! But of course $\mathbb{Z}$ is a PID.
The key observation is this: in a domain, no two nonzero elements can be independent.
So, let $I$ be an ideal of $R$ that is not principal. We show it has no basis, even though $R$ (viewed as an $R$-module) is free. Indeed, let $X$ be a generating set for $I$. We prove that $X$ is not a basis.
Because $I$ is not principal, $X$ has more than one nonzero element. Let $a,b\in I$, $a\neq b$, both nonzero. Now define $$r_x = \left\{\begin{array}{ll} b&\text{if }x=a,\\ -a&\text{if }x=b,\\ 0&\text{otherwise.} \end{array}\right.$$ and consider $\sum_{x\in X}r_xx$.