# Why $A/\operatorname{rad}A$ is generated by $e_a$?

Let $A$ be an algebra over an algebraically field $K$ and $(Q_A)_0$ be its ordinary quiver. Let $\{e_a \mid a \in (Q_A)_0\}$. Then $\{e_a \mid a \in (Q_A)_0\}$ is a complete set of primitive orthogonal idempotents of $A$. How to show that $\{e_a \mid a \in (Q_A)_0\}$ generates $A/\operatorname{rad}A$? Thank you very much. This question is from Line -13 of page 64 of the book Elements of representation theory of associative algebras, volume 1. I attached the page of the book.

• @YACP, thank you very much. – LJR Apr 6 '13 at 11:22

If $\{e_a\}$ form a complete set of primitive orthogonal idempotents, then by definition $A=\bigoplus_a e_a A$. Thus their residue classes (by abuse of notation also denoted $e_a$ satisfy: $$A/\operatorname{rad} A=\bigoplus_ae_aA/\operatorname{rad} A.$$ Hence they span $A/\operatorname{rad}A$.