# Prime Composition Series

Throughout this article, *A* is a noetherian ring and all *A*-modules are finitely generated.

Recall (proposition 1 here) that if *M* is a noetherian and artinian module, we can find a sequence of submodules whose consecutive factors are simple modules. Correspondingly we have:

Proposition 1.For finitely generated M, there exists a sequence of submodules

such that each as A-modules for prime ideals .

**Proof**

Assume . Since by proposition 3 here, there is an embedding of *A*-modules . If equality holds, we are done. Otherwise, let be the image of the map and repeat with to obtain a submodule . Repeating this process, this must eventually terminate since we cannot have an infinite ascending chain of submodules . ♦

Corollary 1.Let be as in proposition 1 with for some prime . Then

.

In particular if M is finitely generated then is finite.

**Proof**

Repeatedly applying proposition 4 here, we have

Since each for each *i*, we are done. ♦

# Associated Primes and Support

If , then there is a map ; upon localizing at we get an *A*-linear and so . Hence we have shown:

Lemma 1.For any module M, .

For a partial reverse inclusion, we have:

Proposition 2.If is a minimal element of then .

**Proof**

By proposition 5 here, we have

which is non-empty since . By lemma 1 this set lies in ; but since is minimal in , by exercise B.2 here has exactly one element: so

so by proposition 5 here again. ♦

Definition.If is not a minimal prime of , we call it an

embedded primeof M.

**Example**

Let *k* be a field, and , considered as an *A*-module. Note that with radical so we get (by proposition 1 here)

.

On the other hand, we claim that . Indeed we have:

so and is an embedded prime of *M*.

Conversely, we pick the chain of submodules with generated by . Then ; as shown above, . Also so by corollary 1

.

**Exercise A**

Find an *A*-module *M* and prime ideals with but .

# Existence of Primary Decomposition

In this section we fix an ambient *non-zero* *A*-module *M* (finitely generated of course) and consider its submodules. For each prime , take the set of all submodules such that . Note that

.

Now for each prime , fix a maximal element .

Lemma 2.We have .

**Note**

If then so the above intersection only needs to be taken over , i.e. a finite number of terms.

**Proof**

Let . If it has an associated prime . Since we have , a contradiction. ♦

Lemma 3.Let . Then .

**Proof**

By proposition 4 here, we have . Since we have . On the other hand if , then we have an injection whose image is of the form . But now

which does not contain , contradicting maximality of . Hence no such exists. ♦

Definition.For an associated prime of M, a –

primary submoduleof M is an such that .A

primary decompositionof M is an expressionwhere each is a -primary submodule of M.

The decomposition is

irredundantif for any , . It isminimalif it is irredundant and all are distinct.

By lemmas 2 and 3, a minimal primary decomposition exists for every non-zero module.

It is not true that in every primary decomposition , must be a maximal element of . We will see an example in the next article.

**Exercise B**

Prove that if are -primary submodules, so is .

Hence given any primary decomposition, we can get a minimal one by first removing the redundant terms then taking the intersection of all with the same corresponding .

# Properties of Primary Decomposition

Here, we will discuss properties of a general primary decomposition. Throughout this section we fix:

where is -primary in *M*.

Proposition 3.Every must occur among the .

**Proof**

Since the canonical map is injective. Hence

♦

Proposition 4.If the primary decomposition is irredundant, then every is an associated prime of M.

**Proof**

If then among the injections and , we have since and are disjoint. Thus

is injective

and , contradicting the condition of irredundancy. ♦

**Note**

Thus proposition 4 gives us a way to compute : find a primary decomposition and remove terms until it becomes irredundant. However, note that proposition 4 does not require to be distinct.

Corollary 2.If the primary decomposition is minimal, then

has exactly n elements.

**Proof**

Apply propositions 3 and 4. ♦

Finally, we define the primary decomposition of submodules.

Defintion.Let be a submodule. A

primary decompositionof N in M is an expressionsuch that is a -primary submodule of M.

**Note**

A primary decomposition of *N* in *M* corresponds bijectively to a primary decomposition of *M*/*N*. We say the primary decomposition of *N* in *M* is **irredundant** or **minimal** if the corresponding primary decomposition of *M*/*N* is so.