How could I construct a module $M$ that has exactly $n$ composition series?

I can't seem to find a series of submodules where each have exactly $n \in \mathbb{N}$ composition series. I don't know if there is something like a classic example of this.

  • 1
    $\begingroup$ If $n=k!$ for some $k$, you can take the direct sum $M=L_1\oplus L_2\oplus \ldots \oplus L_k$ where $L_1,L_2,\ldots,L_k$ are non-isomorphic simple modules. It would be quite tricky if you require that $M$ be indecomposable. $\endgroup$ – user593746 Nov 20 '18 at 18:16
  • $\begingroup$ What do you mean "$n$ composition series"? You don't mean "composition length $n$ for a given $n$", right? You actually want $n$ distinct (in some sense) composition series? $\endgroup$ – rschwieb Nov 20 '18 at 18:55
  • $\begingroup$ Ooops, that was wrong. The $\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}$ gives $n=5$, not $n=4$. Old comment deleted. $\endgroup$ – user593746 Nov 21 '18 at 11:27

Consider the $\mathbb{Z}$-module $M=(\mathbb{Z}/2^{n-1}\mathbb{Z})\oplus(\mathbb{Z}/3\mathbb{Z})$. There are exactly $n$ distinct composition series for $M$. The composition series are given by $$0=M_0^i\subsetneq M_1^i \subsetneq M_2^i\subsetneq \ldots \subsetneq M_{n-1}^i\subsetneq M_n^i=M,$$ where $i=1,2,\ldots,n$, and $$M_j^i=\begin{cases}\big\langle (2^{n-1-j},0)\big\rangle&\text{for }j=0,1,2,\ldots,i-1,\\ \big\langle (2^{n-j},0),(0,1)\big\rangle&\text{for }j=i,i+1,i+2,\ldots,n.\end{cases}$$ Hence, for any integer $n\geq 0$, there exist a ring $R$ and a left $R$-module $M$ of finite length such that $M$ has exactly $n$ distinct composition series.

It would be an interesting question to fix the ring $R$ or demand that $M$ be indecomposable. For example, if $R$ is a finite field of order $q$, then due to this lecture note, the only possible values of $n$ are of the form $$n=[k]_q!=\prod_{i=1}^k\frac{q^i-1}{q-1}=\prod_{i=1}^k(1+q+q^2+\ldots+q^{i-1}).$$ If $R=\mathbb{Z}$ or $R$ is a semisimple ring, and $M$ is an indecomposable left $R$-module of finite length, then $M$ has exactly one composition series.

It is of course possible to have a module of finite length with infinitely many composition series. Take for example $R=\mathbb{Q}$ and $M=\mathbb{Q}^2$. Then, for every $r\in\mathbb{Q}$, $$0\subsetneq \operatorname{span}\big\{(1,r)\big\}\subsetneq M$$ is a composition series of $M$ (and along with $0\subsetneq \operatorname{span}\big\{(0,1)\big\}\subsetneq M$, these are all composition series of $M$).


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