Global Dimension 1, right annihilator

Question

Let $R$ be a ring with right global dimension 1.

Then I am trying to show that for any $a\in R$ if we define the right annihlator $r(a)=\{x\in R|ax=0\}$ then we have that $\exists e\in R$ such that $r(a)=eR$ and $e^2=e$

Thanks for any help

Answer 1

Another name for rings with (right) global dimension 1 are (right) hereditary rings. These are the rings in which every (right) ideal is projective.

Notice that there is a short exact sequence $$0 \to r(a) \overset{i}{\to} R \overset{p}{\to} aR \to 0,$$ where $i$ is the inclusion map, $aR = \{ar \mid r \in R\}$, and $p(r) = ar$. Because the $aR$ is a right ideal of $R$, it is projective. Thus this short exact sequence splits, and the right ideal $r(a)$ is a direct summand of $R$, and it is well-known that any right ideal direct summand is of the form $eR$ for some $e = e^2 \in R$.