A special Artinian module is also Noetherian Problem:
Let $ R $ be a unital right Artinian ring and $ M $ be a unitary right $R$-module. Suppose that $ M_R $ is right Artinian. Prove that it's right Noetherian.
Since $ R $ is a right Artinian ring, its Wedderburn radical $ W $ is a nilpotent ideal of $ R $. Therefore, we can find a smallest positive integer $ s $ such that $ W^s=0 $ but $ W^{\left(s-1\right)} \neq 0 $
I want to prove this result by using induction on $ s $. When $ s=1 $, $ R $ is a semiprime right Artinian ring and thus $ M_R $ is completely reducible. Those imply that $ M_R $ is the sum of some irreducible right $R$-modules. The number of those irreducible module is finite since $ M_R $ is right Artinian. Now, notice that every irreducible module is Notherian, we then get that $ M_R $ is again Noetherian.    
However, I have no idea to do the inductive step. :(
 A: The main theorem in this context is that a right artinian ring is also right noetherian (Hopkins-Levitzki). The proof can be found in any book on ring theory.
Let $J$ be the Jacobson radical of $R$. If $J=\{0\}$, then $R$ is semisimple artinian, so any module is a direct sum of simple modules and the result follows.
Since $J$ is nilpotent, we can do induction on its nilpotency index $n$ (the minimal integer such that $J^n=0$), and we have just proved the base step of the induction.
Consider the exact sequence $0\to MJ^{n-1}\to M\to M/MJ^{n-1}\to0$. By the induction hypothesis, $M/MJ^{n-1}$ is noetherian, being an artinian module over $R/J^{n-1}$ and by the induction hypothesis. So we are bound to prove that $L=MJ^{n-1}$ is noetherian.
Note that $LJ=0$, so $L$ is a module over the semisimple artinian ring $R/J$; since it is artinian, it is also noetherian, being semisimple.
Remark: if $I$ is an ideal of $R$, the structure of submodules of $M/MI$ is the same when we consider it either as a module over $R$ or over $R/I$; similarly, if $MI=0$, the structure of submodules of $M$ is the same when we consider it either as a module over $R$ or over $R/I$.
