Commutative Algebra/Normal and composition series

Normal series

Note that a normal series of a module is a normal series of the underlying group $(M,+)$; 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 $R$.

Refinements and composition series

Note that this implies $m\ge n$. Refinements arise from a normal series

$M\ge N_1\ge N_2\ge \cdots \ge N_{n-1}\ge ⟨0⟩$

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

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:

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 $n$. Indeed, for $n=1$, this theorem follows since then $M$ is simple, and therefore any normal series of length $>n=1$ 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 $n-1$ to be valid. Let there be a composition series

$M\ge N_1\ge N_2\ge \cdots \ge N_{n-1}\ge ⟨0⟩$

of length $n$, and assume that there is any other normal series

$M\ge L_1\ge L_2\ge \cdots \ge L_{m-1}\ge ⟨0⟩$

without repetition of length $m$. Now $N_1$ hence has a composition series of length $n-1$. By induction, we have:

1. If $N_1=L_1$, then $L_1\ge \cdots \ge L_{m-1}\ge ⟨0⟩$ is a normal series in $L_1$ and hence has length at most $n-1$, whence the complete normal series $M\ge L_1\ge L_2\ge \cdots \ge L_{m-1}\ge ⟨0⟩$ has length at most $n$.
2. If $L_1<N_1$, then $N_1$ has a normal series without repetitions of length $n$, which is a contradiction.
3. If not $L_1\le N_1$, we have $N_1+L_1=M$, for otherwise the composition series $M\ge N_1\ge N_2\ge \cdots \ge N_{n-1}\ge ⟨0⟩$ would have a proper refinement. Then we have two normal series

$M=N_1+L_1\ge N_1\ge N_1\cap L_1\ge \cdots \ge N_{n-1}\cap L_1\ge ⟨0⟩$

and

$M=N_1+L_1\ge L_1\ge N_1\cap L_1\ge \cdots \ge N_{n-1}\cap L_1\ge ⟨0⟩$.

Now $N_1$ has a composition series of length $n-1$, whence $N_1\cap L_1$ has a composition series of length $\le n-2$. Furthermore, $L_1/(N_1\cap L_1)\cong (L_1+N_1)/N_1=M/N_1$, which is why any such composition series then extends to a composition series of $L_1$ of length $n-1$. Therefore, the partial series

$L_1\ge \cdots \ge L_{m-1}\ge ⟨0⟩$

has length at most $n-1$.

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

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

We say that the two series are equivalent.

Proof:

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

$M\ge N_1\ge N_2\ge \cdots \ge N_{n-1}\ge ⟨0⟩$

and

$M\ge L_1\ge L_2\ge \cdots \ge L_{n-1}\ge ⟨0⟩$

be given. If $L_1=N_1$, we have equivalence by induction. If not, we have once again $L_1+N_1=M$ (since neither can be properly contained in the other, for else we would obtain a contradiction to the previous theorem of Jordan). Now $L_1\cap N_1$ must have a composition series, since by the previous theorem we may refine the series

$M\ge L_1\cap N_1\ge ⟨0⟩$

to a composition series of $M$. Further, we again have

$N_1/(L_1\cap N_1)\cong M/L_1$ and $L_1/(N_1\cap L_1)\cong M/N_1$;

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

$M\ge N_1\ge L_1\cap N_1\ge \cdots \ge ⟨0⟩$

and

$M\ge L_1\ge L_1\cap N_1\ge \cdots \ge ⟨0⟩$.

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.$◻$

Proof:

If $M$ has a composition series, then intersecting this series or projecting this series gives normal series of $N$ or $M/N$ 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 $N$ and $M/N$ both have composition series, we take a composition series

$N>N_1>N_2>\cdots >N_{k-1}>⟨0⟩$

of $N$ and another one of $M/N$ given by

$M/N>L_1>\cdots >L_{m-1}>⟨0⟩$.

By the correspondence theorem, we write $L_j=(K_j+N)/N$ for suitable $K_j$. Then

$M>K_1+N>\cdots >K_{m-1}+N>N>N_1>\cdots >N_{k-1}>⟨0⟩$

is a composition series of $M$.$◻$

Normal series between modules

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

$N\le L_1\le L_2\le \cdots \le L_{n-1}\le M$

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

$N/N\le L_1/N\le L_2/N\le \cdots \le L_{n-1}/N\le M/N$

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