I am studying the proof of this theorem (of T.Y.Lam). Namely: Let R be a ring for which rad R is nilpotent, and $R/radR$ is semisimple. Then for any R-module ${}_{R}M$, the following statements are equivalent:
1) M is notherian
2) M is artinian
3) M has a composition series
My question is (1) $\Rightarrow $ (3) and (2) $\Rightarrow $ (3). Here is the proof:
Assume M is either noetherian or artinian. For $J = rad R$, fix an integer n such that $J^n = 0$. Consider the filtration
$$M \supseteq JM \supseteq J^2M\supseteq...\supseteq J^nM = 0$$ It is enough to show that each filtration factor $J^iM/J^{i+1}M$ has a composition series.But $J^iM/J^{i+1}M$ is either noetherian or artinian, as a module over $R/rad R$. Since $R/rad R$ is semi simple, $J^iM/J^{i+1}M$ is a direct sum of simple $R/rad R$ - módules. The chain condition on $J^iM/J^{i+1}M$ implies that this direct sum must be finite, so $J^iM/J^{i+1}M$ does have a composition series as an $R/rad R$ - module. QED
My question is: why $J^iM/J^{i+1}M$ has a composition series as an R-module implies that $M \supseteq JM \supseteq J^2M\supseteq...\supseteq J^nM = 0$ is a composition series? Are each $J^iM/J^{i+1}M$ simple? why?
Any help would be great.
The claim is not that $M \supseteq JM \supseteq J^2M\supseteq...\supseteq J^nM = 0$ is a composition series, merely that it can be refined to one, which only requires that each of its factors has a composition series.
I should also note this is not the Hopkins-Levitzki Theorem, which states that a ring that is Artinian is Noetherian.