# Artinian Modules

Instead of the ascending chain condition, we can take its reverse.

Definition.Let M be an A-module. Consider the set of submodules of M, ordered by inclusion, i.e. if and only if . We say M is

artinianif is noetherian.The ring A is said to be

artinianif it is artinian as a module over itself.

Again *M* is artinian if either of the following equivalent conditions holds.

- Every non-empty collection of submodules of
*M*has a minimal element. - If is a sequence of submodules of
*M*, then for some .

### Examples

Let . This is a noetherian ring as we saw; it is not artinian because we have a decreasing sequence of ideals .

Let as a -module where . Then *M* is artinian but not noetherian (exercise). We thus see that an artinian module is not finitely generated in general.

**Exercise A**

1. Prove that in an exact sequence of A-modules:

,

M is artinian if and only if N and P are. In particular, the direct sum of two artinian modules is artinian.

2. Prove that if *M* is a noetherian *A*-module, any surjective linear is also injective. State and prove the dual statement for an artinian module.

Decide if each of the following statements is true.

3. The product of any two artinian rings is artinian.

4. Any quotient of an artinian ring is artinian.

5. Any localization of an artinian ring is artinian.

# Simple Modules

Let *A* be a fixed ring throughout this article and *M* be an *A*-module.

Definition.M is said to be

simpleif it is non-zero and has no submodules except 0 and itself.

We immediately have the following.

Lemma 1.M is simple if and only if for some maximal ideal .

**Proof**

(⇒) Pick a non-zero and consider the *A*-linear map , . Its image is a non-zero submodule of *M* so it is the whole *M*; hence *f* is surjective. We have and submodules of *M* correspond to ideals of *A* containing ker *f*. Thus ker *f* is maximal.

(⇐) Submodules of correspond to ideals of *A* containing . Thus *M* is simple ⟹ is maximal. ♦

**Example**

Thus we see that simple modules over a coordinate ring *k*[*V*] (*k* algebraically closed) correspond to points on *V*. Even though we have as *k*-algebras, different points correspond to different *A*-modules!

**Easy Exercise**

Prove that for any ideals of *A*, as *A*-modules if and only if .

# Composition Series

Next we would like to “factor” a module into its constituents comprising of simple modules.

Definition.A

composition seriesfor M is a sequence of submodulessuch that is simple for all . The

lengthof the composition series is n; thecomposition factorsof the series are the isomorphism classes of , treated as a multi-set.

**Example**

Let and be varieties cut out by the equations and respectively.

Recall that their scheme intersection (exercise D here) has coordinate ring

by Chinese Remainder Theorem. In fact, the above are all isomorphisms as algebras over where we have identified with . Thus we have the following composition series

This has length 3; the composition factors are two copies of and one copy of .

Proposition 1.M has a composition series if and only if it is a noetherian and artinian module.

**Proof**

(⇒) Suppose is a composition series. Each is simple hence both artinian and noetherian. By induction one shows that all are noetherian and artinian.

(⇐) Conversely suppose *M* is both noetherian and artinian; we may assume . If *M* has no simple submodule, we can find an infinitely decreasing sequence of submodules , contradicting the fact that *M* is artinian. Hence we pick any simple submodule . If we are done; otherwise, we repeat the process with and obtain a simple submodule . This gives the next term for a composition series , etc. Since *M* is noetherian, the process must eventually terminate. ♦

Corollary 1.If M has a composition series, so do any submodule and quotient module of M.

# Uniqueness of Composition Series

Theorem 1.Let M be a noetherian and artinian A-module. The composition factors of any composition series are unique up to isomorphism and reordering.

**Note**

For a composition series we write for the multi-set of its composition factors.

**Proof**

The proof is by contradiction; say a module is *bad* if it has two composition series with different sets of composition factors.

**Step 1: Assume M is minimal.**

Take the collection Σ of all bad submodules of *M*. Since *M* is artinian, we can replace *M* by a minimal element of Σ. Thus *M* has two composition series

such that ; and no other submodule of *M* has this property.

**Step 2: Consider the easy case.**

Clearly so . If then by minimality this is a good module so and have the same composition factors, a contradiction.

**Step 3: Do the “diamond argument”.**

Otherwise we have which gives us

and similarly .

**Step 4: Apply induction.**

Pick any composition series for . Using (*P*) to denote the isomorphism class of a module *P* we have:

The second and fourth equalities follow from the fact that and are not bad, since *M* is a minimal bad module. The third equality follows from step 3. In diagram form we have the following.

This completes our proof. ♦

Corollary 2.If M is a noetherian and artinian module, write (

lengthof M) for the length of any composition series of M. This is a well-defined value.Similarly, we can define the

composition factorsof M and denote it by .

Note that if is an exact sequence of *A*-modules and *M* has finite length, then

This gives another application of short exact sequences.

**Exercise B**

1. Compute the length and composition series of

as an *B*-module.

2. Prove that for a ring quotient , a *B*-module *M* is noetherian (resp. artinian) as a *B*-module if and only if it is so as an *A*-module.

In the next article, we will look at *artinian rings*. It turns out this is a small subclass of the collection of noetherian rings.

**Optional Note**

All the results in this section can be generalized to left modules over a non-commutative ring. This has applications in representation theory, since a simple module over the group algebra (*k* = any field) corresponds to an irreducible *k*-representation of *G*.