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.

  • $\begingroup$ Can you show that $(1-e)a=a$? Have you tried to show that this leads to $Ra\cong R(1-e)$? I didn't check everything, so this may not work out, but it would be the first thing that comes to mind here... $\endgroup$ – Jyrki Lahtonen Nov 4 '13 at 17:28


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 $C\cong Ra$.

  • $\begingroup$ Thank you! I have been able to fill in all of the details, except for one point: In your first paragraph, we arrive at $R \cong l(a) \bigoplus Ra$. But I don't quite see how to show that $l(a)$ must have the form $Re$. $\endgroup$ – GoatsRule Nov 4 '13 at 19:22
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    $\begingroup$ Dear @GoatsRule : Consider it a lemma: every direct summand of $_RR$ is generated by an idempotent. Hint: try looking at $1=a+b\in A\oplus B=R$. It is a super useful lemma! $\endgroup$ – rschwieb Nov 4 '13 at 19:29

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