Jordan-Holder theorem for modules?

Question

Let $A$ be a finite dimensional algebra over some field $k$.

I think from the Jordan-Holder Theorem, one might be able to claim that every simple $A$-module occurs in the series (by this I mean it is isomorphic to the quotient of two successive submodules in the composition series).

I would be thankful if anyone could help me with the following questions.

1. Is my thought true (about the occurrence of simple modules)?

2. Is it possible to have two (possibly) different but isomorphic quotients in the composition series? In other words is it possible for a simple module to be isomorphic to more than one quotient of the successive submodules in the composition series.

I want to conclude from the above questions, whether there is a relation between the length of a composition series and the number of equivalences classes of simple $A$- modules?

**By modules I mean left $A$-modules and by a series I mean the composition series of $A$.

Answer 1

For 1

Yes, it's true. The trick is to remember that the simple modules of $A$ are the same as the simple modules of $A/J(A)$, where $J(A)$ is the Jacobson radical of $A$.

Since $A$ is a finite dimensional algebra, it is a right and left Artinian and Noetherian ring. As such, it has a composition as a left module over itself (and as a right module over itself too.) If we can show that every simple factor already appears in a composition series for $A/J(A)$, then we just link it up with a composition series for $J(A)$ and get a composition series for $A$ which contains copies of all isotypes of simple modules in its factors.

Now $A/J(A)$ is a semisimple ring, and all isotypes appear as factors in a composition series for $A/J(A)$. Can you see why this is?

A basic approach to prove this would be to remember that all isotypes of simple left $A/J(A)$ modules appear as minimal left ideals. That makes it clear that they appear in a decomposition of $A/J(A)$ into a direct sum of simple modules. By chopping off one summand at a time, you can produce a composition series displaying all the isotypes in its composition factors.

For 2

Sure, isotypes can occur multiple times, and you may have already noticed this if you have already carried out the last paragraph. Take a semisimple ring $R$ and decompose it into simple left $R$ modules: $R=S\oplus S'\oplus T$ where $S\cong S'$ are isomorphic simple left ideals and $T$ is another simple left ideal nonisomoprhic to $S$ and $S'$. Then this is a composition series:

$$ \{0\}\subseteq S\subseteq S\oplus S'\subseteq S\oplus S'\oplus T=R $$

The first two factors are isomorphic.

For the final question

Yes: your instinct is right. From this it follows that the composition length of $A$ is bounded from below by the number of distinct isotypes of simple $A$ modules. If you say a composition series $\{0\}=S_0\subseteq\ldots\subseteq S_n=A$ has length $n$, then $n$ is greater or equal to the number of distinct isoclasses.

Answer 2

Q1: Every simple $A$-module is of the form $A/m$ for some maximal ideal $m$ of $A$(proof is easy).Now we can write(as $A$ is noetherian and artinian) a composition series $A\supset m \supset \ldots \supset 0$ of $A$. So $A/m $ is occurring in at least one composition series as a factor .Then Jordan-Holder asserts that $A/m$ occurs in any composition series.