Composition Series

Positive integers can be uniquely factored as a product of primes. Here, we would like to prove a counterpart for modules. Now there are two ways to “factor” a module M; a more liberal way takes a submodule N which gives us composition factors (NM/N). A stricter way is to insist upon a direct sum MN ⊕ N’. There are uniqueness theorems in both cases, but we’ll focus on the former here.

Here, all modules are over a fixed ring R, possibly non-commutative.

Definition. A composition series of M is a sequence of submodules

0 = M_0 \subset M_1 \subset M_2 \subset \ldots \subset M_n = M

such that M_{i+1}/M_i is simple for each i=0,…,n-1. The length of the composition series is then n. The modules (M_{i+1}/M_i)_{i=0}^{n-1} are called the composition factors of the series.

Example

Let R be the ring of upper triangular matrices \begin{pmatrix} * & * & * \\ 0 & * & * \\ 0 & 0 & * \end{pmatrix} with real entries. Let R act on MR3 by left multiplication, so we get an R-module. We get a sequence of submodules:

e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, e_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, e_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \implies 0 \subset \mathbf{R} e_1 \subset \mathbf{R} e_1 \oplus \mathbf{R} e_2 \subset M

which is a composition series since the consecutive factors are of dimension 1 over R so they must be simple.

Theorem. M has a composition series if and only if it is noetherian and artinian.

Proof

⇒ : a simple module is both noetherian and artinian, so M1 is noetherian and artinian. Since the same holds for M2/M1 we see that so is M2. Iterating, we see that Mn = M is artinian and noetherian.

⇐ : suppose M is noetherian and artinian. Assume M≠0. Taking the collection of non-zero submodules of M, this collection has a minimal element M1, which is necessarily simple. If M = M1 we’re done. Otherwise M/M1 is also noetherian, artinian and non-zero so it has a simple submodule, which must be of the form M2/M1 for some M1 ⊂ M2 ⊆ M. Keep repeating this procedure and we get an increasing sequence of M_0 \subset M_1 \subset \ldots of submodules of M. Since M is noetherian this must terminate after finitely many terms. ♦

 Corollary. If M has a composition series, so do any submodule N and its quotient M/N.

Furthermore, we can concatenate the composition series of N and M/N together, since a composition series for M/N corresponds to an increasing sequence N = M_0 \subset M_1 \subset \ldots \subset M_k = M such that the consecutive quotients are simple.

Now the main theorem we would like to prove is:

Theorem. If

\begin{aligned} 0 &=M_0 \subset M_1 \subset \ldots \subset M_m = M\\ 0 &=N_0 \subset N_1 \subset \ldots \subset N_n = M\end{aligned}

are two composition series for M, then the composition factors are identical up to isomorphism and permutation. In particular, they are of the same length.

Proof

For notation convenience, we shall denote the composition factors of (Mi) by CF(Mi).

The proof is by contradiction: suppose not. We let ∑ be the set of all submodules N of M such that there exist two composition series for N with distinct composition factors. Since M is artinian, ∑ has a minimal element. Replacing M with this, we may assume:

  • 0 = M_0 \subset M_1 \subset \ldots \subset M_m = M and 0 =N_0 \subset N_1 \subset \ldots \subset N_n = M are composition series of M with distinct composition factors;
  • for any proper submodule N of M, any two composition series have the same composition factors.

Now suppose M_{m-1} = N_{n-1} are the same submodule of M, then M_0 \subset M_1 \subset \ldots \subset M_{m-1} and N_0 \subset N_1 \subset \ldots \subset N_{n-1} are composition series for this submodule with distinct composition factors, which contradicts the minimality of M.

Otherwise M_{m-1} + N_{n-1} = M so we have

M_m/M_{m-1} = (M_{m-1} + N_{n-1})/M_{m-1} \cong N_{n-1}/(M_{m-1} \cap N_{n-1}) (#)

as well as N_n/N_{n-1} \cong M_{m-1}/(N_{n-1} \cap M_{m-1}). Now pick a composition series (Pi) for N_{n-1} \cap M_{m-1}. We have:

\begin{aligned}\text{CF}(M_i) &= (M/M_{m-1}) + \text{CF}(M_i)_{i=0}^{m-1}\\ &\cong (M/M_{m-1}, M_{m-1}/(M_{m-1} \cap N_{n-1})) + \text{CF}(P_i) \\ &\cong (N_{n-1}/(M_{m-1} \cap N_{n-1}), M/N_{n-1}) + \text{CF}(P_i)\\ &\cong (M/N_{n-1}) + \text{CF}(N_i)_{i=0}^{n-1} = \text{CF}(N_i)\end{aligned}

where the second and third equivalences follow from that M_{m-1}, N_{n-1} \subset M are proper submodules, so any two composition series must have identical factors. Pictorially, we have:

composition_uniq

where the first and third arrows follow from minimality of M and the second follows from the isomorphism (#). ♦

Hence, we define the length and composition factors of M to be those of any composition series of M. If l(M) denotes its length, we have l(M) = l(N) + l(M/N) for any submodule N of M.

If R is an artinian and noetherian ring, every simple module must occur among the composition factors of R, since a simple module must be a quotient of R. Let us consider the above example of R = set of 3 × 3 upper triangular real matrices. Then

R = \overbrace{\left \{ \begin{pmatrix} * & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\right \}}^I \oplus \overbrace{\left \{ \begin{pmatrix} 0 & * & 0 \\ 0 & * & 0 \\ 0 & 0 & 0 \end{pmatrix} \right\}}^J \oplus \overbrace{ \left\{ \begin{pmatrix} 0 & 0 & * \\ 0& 0 & * \\ 0 & 0 & *\end{pmatrix} \right\}}^K

is a direct sum of three left ideals. I is clearly simple, while J has a submodule J' =\left\{ \begin{pmatrix} 0 & * & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\right\} which is isomorphic to I. Similarly, K has a submodule which is isomorphic to J. Hence, as composition factors, we can write:

IA,   JA+B,   KA+B+C

so the composition series for R has 3 copies of A, 2 copies of B and 1 copy of C; thus R has exactly 3 simple modules up to isomorphism. Note that the composition series for J and K do not decompose them into direct sums.

Exercise

Prove that there is no submodule of M which is isomorphic to B or C. [ Hint: an element of B can be represented by (0, *, 0). Explicitly write down its R-module structure. Suppose fB → R takes (0, 1, 0) to the matrix T; write down equations for the 6 entries of T.  ]

Prove that AB and C are pairwise non-isomorphic.

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