A left $R$-module $M$ is said to be co-semisimple, if every proper submodule of $M$ is an intersection of maximal submodules. Also, the radical of $M$ denoted by $Rad(M)$ is the intersection of all maximal submodule of $M$.
I want to prove that $M$ is co-semisimple if and only if $Rad(M)=(0)$. Can anyone help me?
My idea is to prove by assuming the the radical is not zero.
Thanks in advance.
Best Answer
We have $Rad(M)=\bigcap \{N\leq M$ $\vert$ $N$ is maximal$\}$
I will prove: $M$ is co-semisimple iff every nonzero quotient of $M$ has zero radical.
First, take $K\lt M$ a proper submodule.
Since $M$ is co-semisimple $K=\bigcap N_\alpha$, each of wich is maximal.
By the correspondence theorem $N_\alpha$ appears in $M/K$ in the form $N_\alpha /K$ (and they are maximal submodules of the quotient), then $Rad(M/K)=\bigcap\{N/K$$\vert$$N/K$ is maximal$\}=\bigcap (N_\alpha/K)=0$.
Take $K=0$, since $M/0$ is isomorphic to $M$, we have $Rad(M)=0$
For the other direction take $M/K\neq 0$, then $Rad(M/K)=0$ implies $K$ is the intersection of maximal submodules by definition of $Rad$.