If $Ra$ is free for $a\neq 0,$ is $a$ regular? Let $R$ be a commutative ring with unity, and $0\neq a\in R.$ We will say that an element $x\in R$ is linearly independent if $\{x\}$ is a linearly independent set. A non-zero element of $R$ is called regular if it is not a zero divisor.
$a$ is a regular element iff $\{a\}$ is a linearly independent set. So if $a$ is regular, then the principal ideal $Ra$ is a free $R$-module with basis $\{a\}.$ My question is about the converse.

Suppose $Ra$ is a free $R$-module. Is $a$ regular?

Suppose for a moment that $Ra$ has a one-element basis $\{b\}.$ Then $Ra$ is isomorphic to $R$ as an $R$-module. The $R$-module $R$ has the following property. 

$(\mathrm P)$ Every element $x\in R$ generating the whole module $R$ is linearly independent. 

This is because the elements generating the whole $R$ are exactly the units of $R$, and the linearly independent elements are exactly the regular elements of $R$ and $0$. Units are always regular.
$Ra$, being isomorphic to $R,$ also satisfies $(\mathrm P).$ Therefore, since $a$ generates the whole $Ra,$ we have that $a$ is linearly independent, and hence regular.
So, returning to my main question, if $Ra$ is a free $R$-module, and we want to show that $a$ is regular, it is enough to show that $Ra$ has a one-element basis, not necessarily $\{a\}.$
It is not difficult to prove that $Ra$ must have a finite basis. Suppose $\{x_i\}_{i\in I}$ is a basis of $Ra$. Then $a$ can be written as $$a=\sum_{j\in J}r_ix_j,$$ where $J$ is a finite subset of $I.$ Clearly, $\{x_j\}_{j\in J}$ is linearly independent and generates the whole $Ra.$
I don't see how to go from a finite basis to a one-element basis. However, I believe commutative rings should be "nice" enough for the answer to be "yes". Could an ideal of a commutative ring have greater dimension than the ring itself?
 A: Yes, if $0\neq I\subset R$ is an  ideal (not assumed principal !) which is a free module, then $I$ has dimension $1$.  
Proof
A basis of $I$ can have  only one element because any two different elements $0\neq a,b\in I$ are linked by the non-trivial linear relation   (watch carefully !)
$$b\cdot a-a\cdot b=0$$  
Remark So a posteriori $I$ turns out to be principal after all, even if you didn't assume it.
A: I don't think it requires that many words: it's pretty elementary.
If $aR\cong R$ via a homomorphism $f$, then there exists $b$ such that $f(ab)=1$. 
If $a$ were not regular, say $c\neq0$ and $ca=0$, then $0=f(cab)=cf(ab)=c1$: a contradiction. So, $a$ must be regular.
Another way is to realize $ann(aR)=ann(a)$. So the module $aR$ is faithful iff $a$ is regular.  Clearly all free modules are faithful.
A: Thanks for elaborating the question so much, but I think you are missing something simple. An element is regular if it is neither $0$ nor a zero divisor. Now assume $a$ is not regular. Then if we first consider the possibility $a=0$ we see that $aR=\{0\}$, which is a free $R$ module, so the answer to your question is strictly speaking"no, $aR$ free does not imply $a$ regular". But this module is not of rank $1$, so you probably want to include that in your question: "if $aR$ is a free $R$ module of rank $1$, then does it follow that $a$ is regular?". Now we have excluded $a=0$, so remains the case of a zero divisor: $ab=0$ with $a,b$ nonzero. But this relation immediately shows that $aR$ is not a free $R$ module, since mutliplication by $b$ kills all of $aR$, but it doesn't kill all of $R$. So to the modified question the answer is "yes, if $aR$ is a free $R$ module of rank $1$, then $a$ is regular".
