Let $R$ be a ring (not necessarily commutative).
$\operatorname{rad}R:=\bigcap \operatorname{ann}M$, where $M$ ranges over all the simple left $R$-modules.
My question is as follow:
$R$ and $R/\operatorname{rad}R$ have the same simple left modules.
The proposition is from T. Y. Lam - A first course in noncommutative rings.
What I have done:
By Correspondence theorem for modules, the simple left modules of $R/\operatorname{rad}R$ are is the form of $I/\operatorname{rad}R$ such that $I$ is the simple left module of $R$.
According to T. Y. Lam, by the definition of $\operatorname{rad}R$ above the proposition is clear. But I can not see.
$\endgroup$1 Answer
$\begingroup$First of all, if $f\colon R\to S$ is a ring homomorphism, then every (left) $S$-module $N$ becomes in a natural way also a (left) $R$-module, by defining, for $y\in N$ and $r\in R$, $ry=f(r)y$.
Also, if $M$ is a (left) $R$-module and $IM=0$, where $I=\ker f$, then $M$ becomes an $S$-module (provided $f$ is surjective): if $x\in M$ and $s=f(r)\in S$, then define $sx=rx$.
It is in this sense, with the homomorphism being the canonical one $R\to R/\operatorname{rad}R$, that the two rings share the simple modules.
Prove that if $M$ is a left $R$-module and $(\operatorname{rad}R)M=0$, then the submodules of $M$ as $R/\operatorname{rad}R$-module are the same as the submodules of $M$ as $R$-module.
Prove also that if $N$ is a left $R/\operatorname{rad}R$-module, then the submodules of $N$ as $R$-modules are the same as the submodules of $N$ as $R/\operatorname{rad}R$-module.
Prove that if $M$ is a simple $R$-module, then $(\operatorname{rad}R)M=0$.
