- Let be $M$ a semi simple A-module and $f$ an homomorphism from $N$ to $M$. Then the Jacobson radical is contained in $Kerf$.
I’m studying semi simple algebras and im stuck with the the demonstration of the jacobson radical of $F[G]$.
So i wanna show that 1. My try:
We have $N/Kerf \cong M $ and then $N/Kerf$ is semisimple. So is a direct sum of two simple modules, $N/Kerf = N_1/Kerf \text{ }\oplus N_2/Kerf$. Then $Kerf$ is maximal of both $N_1$ and $N_2$. Is then maximal of $N_1 \oplus N_2 = N$?
Since $M$ is semisimple, it is a direct sum $\bigoplus_i S_i$ of simple modules. Thus, a morphism $f \colon N \to M$ is determined by its components $f_i \colon N \to S_i$. Moreover, $$ \ker f = \textstyle \bigcap_i \ker f_i. $$ Now use the following characterization of the Jacobson radical:
The Jacobson radical of a module $L$ is the intersection $\bigcap_g \ker g$, where $g$ runs over all morphisms from $L$ to a simple module.