3
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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 $M$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.