Example of a finitely generated flat module which is not free I couldn't come up with an example of a finitely generated flat module which is not free. I know that over local rings, freeness and flatness are equivalent. So the ring cannot be a local ring. 

Let $R$ be a semisimple ring which isn't a division ring, and take an idempotent $e\notin \{0,1\}$. Then we have that $R=eR\oplus(1-e)R$ is a nontrivial decomposition of $R$, and both pieces are cyclic and projective (since they are summands of $R$) hence flat.
Actually neither piece is a free module, but we'll argue here that at least one of them isn't free to simplify things. If they were both free, that would imply that $R\cong R^n$ for some $n>1$ as modules. (Each factor contributes at least one $R$, you see.) But since $R$ has the IBN property, this is impossible. 
Thus at least one of the pieces is flat but not free.

If you really want to be concrete, you can use a finite semisimple ring to make things obvious: let's try $R=\Bbb F_2\times\Bbb F_2$. $\Bbb F_2$ is denoting the field of two elements.
The module $I=\Bbb F_2\times \{0\}$ is a direct summand of $R$, but it can't be free: a free module would have to have at least four elements, and this module only has two elements!
