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 (*N*, *M*/*N*). A stricter way is to insist upon a direct sum *M* = *N* ⊕ *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. Acomposition seriesof M is a sequence of submodulessuch that is simple for each i=0,…,n-1. The

lengthof the composition series is then n. The modules are called thecomposition factorsof the series.

**Example**

Let *R* be the ring of upper triangular matrices with real entries. Let *R* act on *M* = **R**^{3} by left multiplication, so we get an *R*-module. We get a sequence of submodules:

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 *M*_{1} is noetherian and artinian. Since the same holds for *M*_{2}/*M*_{1} we see that so is *M*_{2}. Iterating, we see that *M*_{n} = *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 *M*_{1}, which is necessarily simple. If *M* = *M*_{1} we’re done. Otherwise *M*/*M*_{1} is also noetherian, artinian and non-zero so it has a simple submodule, which must be of the form *M*_{2}/*M*_{1} for some *M*_{1} ⊂ *M*_{2} ⊆ *M*. Keep repeating this procedure and we get an increasing sequence of 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 such that the consecutive quotients are simple.

Now the main theorem we would like to prove is:

Theorem. Ifare 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 (*M _{i}*) by CF(

*M*).

_{i}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:

- and 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 are the *same* submodule of *M*, then and are composition series for this submodule with distinct composition factors, which contradicts the minimality of *M*.

Otherwise so we have

(#)

as well as . Now pick a composition series (*P _{i}*) for We have:

where the second and third equivalences follow from that are proper submodules, so any two composition series must have identical factors. Pictorially, we have:

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

is a direct sum of three left ideals. *I* is clearly simple, while *J* has a submodule which is isomorphic to *I*. Similarly, *K* has a submodule which is isomorphic to *J*. Hence, as composition factors, we can write:

*I* = *A*, *J* = *A*+*B*, *K* = *A*+*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 *f* : *B* → *R* takes (0, 1, 0) to the matrix *T*; write down equations for the 6 entries of *T*. ]

Prove that *A*, *B* and *C* are pairwise non-isomorphic.