On a proof that left artinian implies left noetherian 
Questions: [Refer to below] Could one elaborate on $\rm\color{#c00}{(a)}$, $\rm\color{#c00}{(b)}$ and $\rm\color{#c00}{(c)}$ ? My thoughts :
$\rm\color{#c00}{(a)}$ For $r+J\in R/J$ and $j+J^{i+1}\in J^i/J^{i+1}$, is $(r+J)(j+J^{i+1}):=rj+J^{i+1}$ ?
$\rm\color{#c00}{(b)}$ $J^i=Rj_1+Rj_2+\cdots+Rj_n\implies J^i/J^{i+1}=(R/J)j_1+(R/J)j_2+\cdots+(R/J)j_n$ ?
$\rm\color{#c00}{(c)}$ I guess it is a consequence of $R/J$ being semisimple, but I don't see it :(.

Thm: Let $R$ be a left artinian ring such that its Jacobson radical $J(R)$ is nilpotent. Then $R$ is left noetherian.
Proof: [Brief sketch] Let $J:=J(R)$ and
$$
R\supseteq J\supseteq J^2\supseteq \cdots\supseteq J^n=0.
$$
One can argue that it suffices to show $J^i/J^{i+1}$ is noetherian for all $i$.
$\rm\color{#c00}{(a)}$ $J^i/J^{i+1}$ is a $R/J$-module.
One can show that $R/J$ is semisimple and that $J^i$ is of finite type over $R$.
Hence,
$\rm\color{#c00}{(b)}$ $J^i/J^{i+1}$ is a $R/J$-module of finite type and $\rm\color{#c00}{(c)}$ $J^i/J^{i+1}$ is semisimple. 
Then $J^i/J^{i+1}$ is noetherian (by a known characterization of semisimple noetherian modules). $\blacksquare$
 A: (a) Yes.
(b) Yes.
(c) Over a semisimple ring every non-zero module is semisimple (as a quotient of a free module which is a direct sum of copies of your semisimple ring, hence semisimple).
A: Let $A$ and $B$ be ideals of the ring $R$. Then $A/BA$ is a module over $R/B$ in a natural way. Indeed, for $b\in B$ and $a\in A$, the product $ba\in BA$, so
$$
b(a+AB)=0+AB
$$
in the module $A/AB$. Thus $\operatorname{Ann}(A/BA)\supseteq B$. If $M$ is an $R$-module and $\operatorname{Ann}(M)\supseteq B$, then $M$ is a module over $R/B$ by defining
$$
(r+B)x=rx
$$
Specialize the case to $A=J^i$ and $B=J$.
If, with the above notation, $M$ is finitely generated as $R$-module, then it's clearly finitely generated also as $R/B$-module.
If $N$ is a module over $R/B$, then it's also a module over $R$ by defining $ry=(r+B)y$. The lattice of submodules of $N$ is the same when considered either a module over $R/B$ or over $R$. In particular, semisimplicity is preserved (it's a lattice property, that is, being complemented). So in your case you get the chain
$$
J^k=\{0\}\subseteq J^{k-1}\subseteq\dots\subseteq J\subseteq J^0=R
$$
where the quotients $J^{i-1}/J^{i}$ are noetherian as $R$-modules.
