Question

Show that a principal right ideal aR in a ring R is projective as a right R-module iff annr(a) is of the form eR where e is an idempotent of R

Answer 1

Consider the exact sequence of right R-modules $0 \to \operatorname{ann}_r(a) \to R \xrightarrow{f} aR \to 0$ where $f$ is defined by left multiplication by $a$ . If $aR$ is projective, this sequence splits. Then $\operatorname{ann}_r(a)$ is a direct summand of $R_R$ , so $\operatorname{ann}_r(a) = eR$ for some $e^2 = e \in R$ . Conversely, if $\operatorname{ann}_r(a) = eR$ where $e^2 = e \in R$ , then the direct sum decomposition $R = eR \oplus (1 - e)R$ implies that $aR \cong R / eR \cong (1 - e)R$ , which is a projective right R-module.