Let $\mathbb{F}$ be a field with $char~\mathbb{F}=0$, $A$ is a finite dimensional $F$-algebra. If $\theta$ is a representation of $A$, define character of $\theta$, to be $\chi_{\theta}(x)=tr~\theta(x)$. Since, two modules are isomorphic as $A$-modules implies that the characters are same, we associate $\chi_M$ for each equivalence class of modules.
I have proved that if $0\rightarrow N\rightarrow M\rightarrow P\rightarrow 0$, is a short exact sequence of finite dimensional $A$-modules then $\chi_M=\chi_N+\chi_P$ (extend a basis of $Im~N$ to that of $M$). Using this result or otherwise I want to prove that:
If $M$ is a finite dimensional, right $A$-module, then $\chi_M=\chi_N$ for some semisimple module $N$.
I have taken $N=rad~M$, then $P$ can be chosen to be $M/rad~M$, which is semisimple, but unable to show that $\chi_{rad~M}=0$.
Please provide a hint in this or some other way of solving this problem. Thank you.
P.S. This is an exercise in the book associative algebra by R. Pierce (Chapter 5, exercise 5.6.1(e))
