Let $Ra$ be the left ideal of a ring $R$ generated by an element $a \in R$.
Show that $Ra$ is a projective left $R$-module if and only if the left annihilator of $a$, $\{r \in R \mid ra = 0\}$ is of the form $Re$ for some idempotent element $e \in R$.
Note: I know that for an idempotent e, $Re$ is a projective left R-module, since $R \cong Re \bigoplus R(1-e)$ shows that $Re$ is a direct summand of a free module. It seems that this fact may be useful here, but I am not sure how to proceed.
Hints:
On one hand, you have an exact sequence
$$ 0\to\ell(a)\to R\to Ra\to 0 $$
where $\ell(a)$ is the left annihilator of $a$, the first mapping is the inclusion map, and the last map is the obvious projection map. If $Ra$ is projective, this sequence splits. Use this to conclude that $\ell(a)$ is a summand of $R$ and has the form $Re$.
On the other hand, suppose $\ell(a)\oplus C=R$. $C$, being a summand of the free module $R$, is clearly projective. We still can look at the surjection $r\mapsto ra$. Looking at the kernel of this map, we can see that $(x+c)a=0$ if and only if $ca=0$ iff $c\in \ell(a)$ iff $c=0$. Use this to conclude $C\cong Ra$.