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

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.

• 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. – user593746 Nov 20 '18 at 18:16
• 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? – rschwieb Nov 20 '18 at 18:55
• 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. – user593746 Nov 21 '18 at 11:27

1 Answer

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 $$M$$).