Let $R=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ and $A=\{(1,0)\}$. Then, what is the free $R$-module $F(A)$? (Here the module action is mutiplying componentwise. )
I tried spanning $A$ by $R$. But in $RA=\{(0,0), (1,0)\}$, the element $(1,0)$ does not have a unique representation, since $(1,0)=(1,0)\cdot(1,0)=(1,1)\cdot(1,0)$. So that $RA$ is not free on $A$. How can I get $F(A)$?
You have to distinguish between the free module spanned by one basis element and the ideal generated by an element of $R$.
The free module spanned by an element $e$ is just $\{0,(1,0)e,(0,1)e,(1,1)e\}$ and is isomorphic to $R$ as a module.
The ideal $RA$ generated by your $A$ is indeed $\{0,(1,0)\}$ and is NOT a free module over $R$. Ideals are free if and only if they are principal and generated by a non-zero divisor. So $RA$ is not free since $(1,0)$ is a zero divisor.