# Commutative Algebra/Normal and composition series

## Normal series

Definition 12.1:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module. A finite sequence of submodules

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$

is called normal series of ${\displaystyle M}$.

Note that a normal series of a module is a normal series of the underlying group ${\displaystyle (M,+)}$; indeed, each subgroup of an abelian group is normal, hence each normal series in modules gives rise to a normal series of groups. The other direction is not true, since additive subgroups need not be closed under multiplication by elements of ${\displaystyle R}$.

Definition 12.2:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module, and let a normal series

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$

be given. This normal series is said to be without repetitions if each inclusion of modules is, in fact, strict.

## Refinements and composition series

Definition 12.3:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module, and let a normal series

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$

be given. A refinement of this normal series is another normal series

${\displaystyle M\geq L_{1}\geq L_{2}\geq \cdots \geq L_{m-1}\geq \langle 0\rangle }$

such that ${\displaystyle \{N_{1},\ldots ,N_{n-1}\}\subseteq \{L_{1},\ldots ,L_{m-1}\}}$.

Note that this implies ${\displaystyle m\geq n}$. Refinements arise from a normal series

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$

by inserting submodules ${\displaystyle L}$ between two modules of the composition series ${\displaystyle N_{k}}$ and ${\displaystyle N_{k+1}}$; that is, we start with two modules of a composition series ${\displaystyle N_{k}}$ and ${\displaystyle N_{k+1}}$, find a submodule ${\displaystyle L}$ of ${\displaystyle M}$ such that ${\displaystyle N_{k}\geq L\geq N_{k+1}}$, and then just insert this into the normal series.

Definition 12.4:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module. We say that ${\displaystyle M}$ is simple if and only if it has no proper submodules (i.e. no submodules of ${\displaystyle M}$ which are neither equal to ${\displaystyle M}$ nor equal to ${\displaystyle \langle 0\rangle }$).

Definition 12.5:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module. A composition series of ${\displaystyle M}$ is a normal series of ${\displaystyle M}$, say

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$,

such that

1. there are no repetitions, and
2. each so-called composition factor of that series, namely ${\displaystyle N_{k}/N_{k+1}}$ for ${\displaystyle k\in \{0,\ldots ,n-1\}}$, is simple (where we set ${\displaystyle N_{0}:=M}$ and ${\displaystyle N_{n}:=\langle 0\rangle }$).

Equivalently, a composition series is a normal series without repetitions, such that any proper refinement of it has repetitions.

To any module, we may associate a so-called length. This concept is justified by the following theorem:

Theorem 12.6 (Jordan):

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module which has a composition series

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$.

We say that this composition series has length ${\displaystyle n}$, and then it follows that

1. each normal series in ${\displaystyle M}$ without repetitions has length ${\displaystyle \leq n}$,
2. every other composition series of ${\displaystyle M}$ also has length ${\displaystyle n}$, and
3. every normal series in ${\displaystyle M}$ can be refined to a composition series of ${\displaystyle M}$.

Proof:

First, we note that 1. implies 3., since whenever a normal series has a refinement that has no repetitions, we may apply that refinement, and due to 1., we must eventually reach a composition series.

Then we prove 1. and 2. by induction on ${\displaystyle n}$. Indeed, for ${\displaystyle n=1}$, this theorem follows since then ${\displaystyle M}$ is simple, and therefore any normal series of length ${\displaystyle >n=1}$ must have repetitions, which is why the trivial normal series is the only one without repetitions, and there is only one composition series.

Assume now the case ${\displaystyle n-1}$ to be valid. Let there be a composition series

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$

of length ${\displaystyle n}$, and assume that there is any other normal series

${\displaystyle M\geq L_{1}\geq L_{2}\geq \cdots \geq L_{m-1}\geq \langle 0\rangle }$

without repetition of length ${\displaystyle m}$. Now ${\displaystyle N_{1}}$ hence has a composition series of length ${\displaystyle n-1}$. By induction, we have:

1. If ${\displaystyle N_{1}=L_{1}}$, then ${\displaystyle L_{1}\geq \cdots \geq L_{m-1}\geq \langle 0\rangle }$ is a normal series in ${\displaystyle L_{1}}$ and hence has length at most ${\displaystyle n-1}$, whence the complete normal series ${\displaystyle M\geq L_{1}\geq L_{2}\geq \cdots \geq L_{m-1}\geq \langle 0\rangle }$ has length at most ${\displaystyle n}$.
2. If ${\displaystyle L_{1}, then ${\displaystyle N_{1}}$ has a normal series without repetitions of length ${\displaystyle n}$, which is a contradiction.
3. If not ${\displaystyle L_{1}\leq N_{1}}$, we have ${\displaystyle N_{1}+L_{1}=M}$, for otherwise the composition series ${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$ would have a proper refinement. Then we have two normal series
${\displaystyle M=N_{1}+L_{1}\geq N_{1}\geq N_{1}\cap L_{1}\geq \cdots \geq N_{n-1}\cap L_{1}\geq \langle 0\rangle }$
and
${\displaystyle M=N_{1}+L_{1}\geq L_{1}\geq N_{1}\cap L_{1}\geq \cdots \geq N_{n-1}\cap L_{1}\geq \langle 0\rangle }$.
Now ${\displaystyle N_{1}}$ has a composition series of length ${\displaystyle n-1}$, whence ${\displaystyle N_{1}\cap L_{1}}$ has a composition series of length ${\displaystyle \leq n-2}$. Furthermore, ${\displaystyle L_{1}/(N_{1}\cap L_{1})\cong (L_{1}+N_{1})/N_{1}=M/N_{1}}$, which is why any such composition series then extends to a composition series of ${\displaystyle L_{1}}$ of length ${\displaystyle n-1}$. Therefore, the partial series
${\displaystyle L_{1}\geq \cdots \geq L_{m-1}\geq \langle 0\rangle }$
has length at most ${\displaystyle n-1}$.

This proves 1. by induction. Furthermore, by induction, ${\displaystyle M}$ can not have a composition series of length ${\displaystyle , since then also the composition series above would have length ${\displaystyle n-1}$, whence 2. is proven by 1. and induction.${\displaystyle \Box }$

Definition 12.7:

Assume ${\displaystyle M}$ has a composition series. Then the length of the module ${\displaystyle M}$ is defined to be the length of such a composition series.

If ${\displaystyle M}$ doesn't have a composition series, we set the length of ${\displaystyle M}$ to be ${\displaystyle \infty }$.

Furthermore, composition series are essentially unique, as given by the following theorem:

Theorem 12.8 (Hölder):

If

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$

and

${\displaystyle M\geq L_{1}\geq L_{2}\geq \cdots \geq L_{n-1}\geq \langle 0\rangle }$

are two composition series, then there exists a permutation ${\displaystyle \sigma :\{0,1,\ldots ,n-1\}\to \{0,1,\ldots ,n-1\}}$ such that for all ${\displaystyle k\in \{0,1,\ldots ,n-1\}}$

${\displaystyle N_{\sigma (k)}/N_{\sigma (k)+1}\cong L_{k}/L_{k+1}}$

(again ${\displaystyle N_{0}:=M}$ and ${\displaystyle N_{n}:=\langle 0\rangle }$, and analogous for ${\displaystyle L}$).

We say that the two series are equivalent.

Proof:

We proceed by induction on ${\displaystyle n}$. For ${\displaystyle n=1}$, we have only the trivial composition series as composition series. Now assume the theorem for ${\displaystyle n-1}$. Let two composition series

${\displaystyle M\geq N_{1}\geq N_{2}\geq \cdots \geq N_{n-1}\geq \langle 0\rangle }$

and

${\displaystyle M\geq L_{1}\geq L_{2}\geq \cdots \geq L_{n-1}\geq \langle 0\rangle }$

be given. If ${\displaystyle L_{1}=N_{1}}$, we have equivalence by induction. If not, we have once again ${\displaystyle L_{1}+N_{1}=M}$ (since neither can be properly contained in the other, for else we would obtain a contradiction to the previous theorem of Jordan). Now ${\displaystyle L_{1}\cap N_{1}}$ must have a composition series, since by the previous theorem we may refine the series

${\displaystyle M\geq L_{1}\cap N_{1}\geq \langle 0\rangle }$

to a composition series of ${\displaystyle M}$. Further, we again have

${\displaystyle N_{1}/(L_{1}\cap N_{1})\cong M/L_{1}}$ and ${\displaystyle L_{1}/(N_{1}\cap L_{1})\cong M/N_{1}}$;

both modules on the right of the isomorphisms are simple, whence we get two composition series of ${\displaystyle M}$ given by

${\displaystyle M\geq N_{1}\geq L_{1}\cap N_{1}\geq \cdots \geq \langle 0\rangle }$

and

${\displaystyle M\geq L_{1}\geq L_{1}\cap N_{1}\geq \cdots \geq \langle 0\rangle }$.

Now the two above isomorphisms also imply that these two are equivalent, and by induction, the first one is equivalent to the first composition series, and the second one equivalent to the second composition series.${\displaystyle \Box }$

Proposition 12.9:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module, let ${\displaystyle N\leq M}$ be a submodule.

${\displaystyle M}$ has a composition series if and only if both ${\displaystyle N}$ and ${\displaystyle M/N}$ have composition series.

Proof:

If ${\displaystyle M}$ has a composition series, then intersecting this series or projecting this series gives normal series of ${\displaystyle N}$ or ${\displaystyle M/N}$ respectively. When the repetitions are crossed out, no refinements are possible (else they induce a refinement of the original composition series, in the latter case by the correspondence theorem).

If ${\displaystyle N}$ and ${\displaystyle M/N}$ both have composition series, we take a composition series

${\displaystyle N>N_{1}>N_{2}>\cdots >N_{k-1}>\langle 0\rangle }$

of ${\displaystyle N}$ and another one of ${\displaystyle M/N}$ given by

${\displaystyle M/N>L_{1}>\cdots >L_{m-1}>\langle 0\rangle }$.

By the correspondence theorem, we write ${\displaystyle L_{j}=(K_{j}+N)/N}$ for suitable ${\displaystyle K_{j}}$. Then

${\displaystyle M>K_{1}+N>\cdots >K_{m-1}+N>N>N_{1}>\cdots >N_{k-1}>\langle 0\rangle }$

is a composition series of ${\displaystyle M}$.${\displaystyle \Box }$

## Normal series between modules

Definition 12.10:

Let ${\displaystyle M}$ be a module and ${\displaystyle N\leq M}$ a submodule. A series

${\displaystyle N\leq L_{1}\leq L_{2}\leq \cdots \leq L_{n-1}\leq M}$

is called a normal series between ${\displaystyle M}$ and ${\displaystyle N}$, and if each inclusion is strict and there does not exist a refinement which leaves each inclusion strict, it is called a composition series between ${\displaystyle M}$ and ${\displaystyle N}$.

By the correspondence theorem, we get a bijection between normal (or composition) series

${\displaystyle N\leq L_{1}\leq L_{2}\leq \cdots \leq L_{n-1}\leq M}$

between ${\displaystyle M}$ and ${\displaystyle N}$ on the one hand, and of normal (or composition) series

${\displaystyle N/N\leq L_{1}/N\leq L_{2}/N\leq \cdots \leq L_{n-1}/N\leq M/N}$

of ${\displaystyle M/N}$. Then by the above and the third isomorphism theorem, composition series between ${\displaystyle M}$ and ${\displaystyle N}$ are essentially unique. Further, if there is a composition series, normal series can be refined to composition series of same length.