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) 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).

• I don't understand what you wrote in parentheses. What quotient are you talking about exactly? What direct sum of copies of the semisimple ring? Why is the quotient semisimple? I did try to understand but I failed, sorry. All I thought: If $M$ is a non-zero $R$-module, in general, does an index set $I$ exist such that we have an epimorphism from $\bigoplus_{i\in I} R$ to $M$ ? Then we could quotient and $M$ would be a quotient of a direct sum of copies of the semisimple ring... – Guest Dec 17 '14 at 18:49
• (continued) Then I didn't know if the direct sum would be free, and if the quotient of the direct sum is the direct sum of quotients, etc. and if that would be useful. Essentially if you could elaborate then I could see what exactly you mean. – Guest Dec 17 '14 at 18:50
• @Guest Yes, there is always a free $R$-module $F$ and an epimorphism $F\to M$. (Is this clear?) Then recall that a direct sum of semisimple modules (and $R$ is a semisimple $R$-module by hypothesis) is also semisimple, so $F$ is semisimple. Moreover, from the fundamental theorem of isomorphism for modules there is $N\subset F$ a submodule such that $F/N\simeq M$. Now recall that a quotient module of a semisimple module is also semisimple. (More details can be found here, Corollary 2.6, 2.7.) – user26857 Dec 17 '14 at 18:58

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.