A question concerning a right $A/\operatorname{rad}A$-module.

Question

Let $\DeclareMathOperator\Top{top}\DeclareMathOperator\Rad{rad}M$ be a right $A$-module, let $\Top M=M/\Rad M$.

We see that, if $e$ is a primitive idempotent in $A$, then $eA$ is a indecomposable module in $A_A$.

Let $B=A/\Rad A$, then $\Top eA=eA/\Rad eA=eA/e\Rad A$ can be seen as a right $B$-module (Since $eA\cdot \Rad A\subset e\Rad A$).

Let $\bar{e}=e+\Rad A\in B$, then what confused me is why $\Top eA\cong\bar{e}B$?

Thanks to everyone!

Answer 1

You can define a natural map $\alpha: top\ eA \rightarrow \overline{e}B$ by the form $\alpha(\overline{ea})= e \overline{a}$, and it's easy to check it is an isomorphism.