Free Module $R^{n}$ with zero divisors In Artin's Algebra he says that $R^{n}$, where $R$ is a ring, is a free $R$ module.   But if $R$ contains zero divisors then how can $R$ be free since 
there exists $r,r_{1}$ such that $rr_{1}=0$ with $r,r_{1} \ne 0$ ?
 A: A (left) module $M$ over a ring $R$ is free if there exists a set $\{m_i:i\in I\}\subseteq M$ such that every element of $M$ can be uniquely written as a sum $\sum_ir_im_i$ with $r_i=0$ for all but finitely many $i$. Applying this to the $R$-module $R$ and the set consisting of the identity element $1\in R$, we see that $R$ is free as a module over itself. Similarly, for $M=R^n$, the elements $e_i=(0,\ldots,1,\ldots,0)$ (with the $1$ in the $i$-th spot) clearly satisfy this condition).
It doesn't matter that $R$ may have zero-divisors. What is true is that if $r\in R$ is a zero divisor, then the set $\{r\}$ does not satisfy the conditions above (it is not a basis for the $R$-module $R$) because $0$ can be written as $0\cdot r$ and $r^\prime\cdot r$ for some non-zero $r^\prime$ (since this is what it means to be a zero-divisor).
I think you might be mixing the notion of a free module with the notion of a torsion-free module; but the latter notion sort of only makes sense for $R$ a domain (then the set of torsion elements of an $R$-module is a submodule). What is true is that a free module over a domain is also torsion-free.
For example, the ring $\mathbf{Z}/4\mathbf{Z}$ is free as a module over itself (as I pointed out above, the element $1+4\mathbf{Z}$ constitutes a free generating set, or basis), even though it has a zero divisor. Note however that as a $\mathbf{Z}$-module, $\mathbf{Z}/4\mathbf{Z}$ is not free. In fact it is a torsion $\mathbf{Z}$-module.
