Is a module that is isomorphic to its dual necessarily free?

Question

Suppose $M$ is a finitely generated module over an integral domain $R$. If there is a bilinear form $\langle-,-\rangle:M\times M\rightarrow R$ which induces an isomorphism $M\rightarrow \text{Hom}_R(M,R)$ via $m\mapsto \langle m,-\rangle$, is it possible to conclude that $M$ is a free $R$-module?

Answer 1

You can consider the particular case $M$ an ideal of $I$ of $R$. Then the dual of $I$ is an $R$ submodule of $K$ (the field of fractions) $$(R:I) = \{ \alpha \in K \|\ \alpha I \subset R\}$$

Now, it is possible for an ideal $I$ not to be principal, but its square $I^2$ is principal. That should provide a counterexample. Also see example of a nontrivial class groups