Are there cyclic, free modules where the generating element isn't a basis? Let $R$ be a ring, and $M$ a nontrivial cyclic, free $R$-module. Let $m$ generate $M$, so that $M = Rm$. Is it then the case that $m$ forms a basis for $M$, so that $\mbox{ann}_{R}(m) = (0)$?
I know that if $R$ is a domain or a commutative ring, it is easy to show that $m$ forms a basis. However, I am unsure as to whether or not it holds for general rings. Any insight would be appreciated. Thanks! 
 A: Let's take Georges proof and turn it into a counterexample (!)
Let $k$ be a field and $R$ the quotient of the free algebra $k\langle x,y, z\rangle$ by the ideal generated by $xy-1$ and $zy$. As a $k$-vector space, $R$ has a basis consisting of those non-commutative monomials which contain neither $xy$ nor $zy$ as subwords —this follows immediately from Bergman's Diamond Lemma, for example, or from a simple ad hoc argument (which surely will boil down to the Diamond lemma...)
Now $M=R$, viewed as a left $R$-module as usual, is generated by $m=y$, but of course $z\cdot m=0$, so $\{m\}$ is not a basis because $m$ has a non-trivial annihilator.
Notice that $M$ is of course free of rank $1$.
A: Yes, if $m$ generates $M$, it is a basis for $M$, if $R$ is commutative   .
Proof
Let $b$ be a basis of $M$, so that in particular $Ann(b)=0$.
Since  $m$ generates $M$ we can write  $b=rm$ for some $r\in R$.
On the other hand we can write $m=sb$ for some $s\in R$ since $b$, a basis, certainly generates $M$.
So we have $b=rm=rsb$, hence $(1-rs)b=0$ and thus $1-rs=0$ because $Ann(b)=0$.
We see that  $r,s\in R^*$ are  invertible and since $m=sb$ and $b$ is  a basis, $m$ is a basis too.  
Edit
I have used that a basis of a non-zero cyclic  free module has  just one element.
Since Isaac asks why in a comment, I'll give a proof.
I claim that if  $g$ is a generator of $M$, any two elements  on $M$ are linearly dependent (still assuming $R$ commutative !)
Indeed, if $u=ag$ and $v=bg$ are arbitrary   in $M$, we have a linear relation $bu-av=0$ and either this is  a nontrivial linear relation and $u,v$ are linearly dependent or  $a=b=0$ and then $u=v=0$ are  certainly linearly dependent in that case too.   
Important new edit
I had assumed in my proof that $R$ is commutative without saying so.   I have now made  this assumption explicit: all my apologies to all  and thanks to Mariano for calling my attention to this point. 
