$M$ is artinian or noetherian $\implies$ M has a composition series.

Question

Let $R$ be a semiprimary ring, that is let $R$ be a ring with its radical $J$ is nilpotent and $R/J$ is semisimple. Then for any $R$-module $M_R$ the following statements are equivalent:

$(1)$ $M$ is noetherian.

$(2)$ $M$ is artinian.

$(3)$ $M$ has a composition series.

proof: $(3) \implies (1)$ and $(3) \implies (2)$ are clear.

My questions are from $(1),(2) \implies (3)$:

Suppose that $M$ is noetherian or artinian and fix an integer $n$ such that $J^n=0$ and let $\bar{R}=R/J$. Consider the filtration $$M \supseteq MJ \supseteq MJ^2 \supseteq .... \supseteq MJ^n=0.$$ My fist question: The author says that it is enought to show that each filtration factor $MJ^i/MJ^{i+1}$ has a composition series.

How does this gaurantee that $M$ has a composition series?

My second question: each filtration factor $MJ^i/MJ^{i+1}$ can be view as an $\bar{R}$-module.

How should I define scaler multiplication on $MJ^i/MJ^{i+1}$ with the element of $\bar{R}$?

Answer 1

To your first question: You can find in some books or you can prove it yourself that if $M/N$ and $N$ have a composition series respectively, then $M$ has a composition series.

To your second question: You know that $M/JM$ is still a $R$-module, and since that $J_{.}M \subseteq JM $, then we can define naturally that $\overline{r}_{.} \overline{m}= \overline{rm} $ for $\overline{r} \in \overline{R}$ and $\overline{m} \in M/JM$. It make sense. Same for $J^iM/J^{i+1}M$.

Answer 2

The answer to my first question: If we will show that each filtration factor $MJ^i/MJ{i+1}$ has a composition series, then we can write that $MJ^{n-1}/MJ^n \cong MJ^{n-1}$ and $MJ^{n-2}/MJ^{n-1}$ has a composition series which imply that $MJ^{n-2}$ has a composition series. By using induction on n we can prove that $M$ has a composition series.

The answer to my second question: The scaler multiplication on $MJ^i/MJ^{i+1}$ with the element of $\bar{R}=R/J$ is defined as follows:

$$m.(J+r)=mr, \ \ where \ \ m \in MJ^i/MJ^{i+1} \ \ and \ \ (J+r) \in \bar{R}=R/J.$$ This scaler multplication is well defined since $MJ^i/MJ^{i+1}$ is annihilated by $J$.