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 $A$.

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    $\begingroup$ In Q1 are you asking whether every simple module occurs in the composition series of the (left) module $A$? You didn't specify the module, and the answer clearly depends :-) $\endgroup$ Mar 7, 2014 at 12:55
  • $\begingroup$ math.berkeley.edu/~serganov/math252/notes9.pdf might be helpful? $\endgroup$
    – hmmmm
    Mar 7, 2014 at 12:57
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    $\begingroup$ Your edit failed to address the main part of Jyrki's question. Namely, the composition series of which module (probably $A$ as a left module over itself). $\endgroup$ Mar 7, 2014 at 13:00
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    $\begingroup$ The simplest answer to question 2 is: "Yes, this is possible. Any simple module $M$ occurs twice as a composition factor of the module $M\oplus M$." A more interesting question is, whether this can happen with an indecomposable module. Again the answer is affirmative, but that does not work for all modules and for all algebras. $\endgroup$ Mar 7, 2014 at 13:05
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    $\begingroup$ Ok. Sorry about being obtuse. What I meant is that if $A$ is not semisimple, then some simple modules may never appear with multiplicities higher than one as composition factors of indecomposable modules, but some other simple modules may do so. $\endgroup$ Mar 7, 2014 at 19:31

2 Answers 2


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.

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    $\begingroup$ what a nice answer :), thank you very much. Regarding your question in the answer, yes I see it (you have also described it later). $\endgroup$
    – Math137
    Mar 7, 2014 at 13:18
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    $\begingroup$ @math137 Thank you :) Glad it helped! One thing I did not go into detail on is that the simple $A$ modules are "the same as" the simple $A/J(A)$ modules. Maybe you've already worked this out, but if you haven't, it's worthwhile doing :) $\endgroup$
    – rschwieb
    Mar 7, 2014 at 13:22
  • $\begingroup$ I see why is that true, thank you very much. $\endgroup$
    – Math137
    Mar 7, 2014 at 13:23

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.


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