We have already seen two forms of unique factorization.

- In a UFD, every non-zero element is a unique product of irreducible (also prime) elements.
- In a Dedekind domain, every non-zero ideal is a unique product of maximal ideals.

Here, we will introduce yet another type of factorization, called *primary decomposition*. The main idea is that in a noetherian ring, every ideal (even the zero ideal) is an intersection of *primary ideals*. E.g. in , every non-zero ideal is an intersection of , ideals generated by prime powers.

# Annihilators

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

Definition.The

annihilatorof in A is,

an ideal of A. Similarly, the

annihilatorof M in A is,

also an ideal of A. If , we say M is a

faithfulA-module.

Note that if , then *M* can be regarded as a faithful -module.

**Exercise A**

1. Given *A*-submodules , let

.

State and prove the analogue of proposition 2 here for submodules of *M*. Observe that we can write the annihilators as

Conversely, express in terms of annihilators.

2. Prove that if *P *is a finitely generated submodule of *M*.

In particular if *M* is finitely generated then

.

# Support of a Module

Definition.The

supportof an A-module M is:

Geometrically, these are points in Spec *A* at which the module does not vanish.

Proposition 1.If M is a finitely generated A-module, then

In particular, the support of M is a closed subspace of .

**Proof**

Let . For we have is zero if and only if there exists such that , equivalently if there exists . Hence if and only if .

In the general case, let generate *M* as an *A*-module. Then

For the last equivalence, recall that contains if and only if it contains either or . ♦

Proposition 2.If is a short exact sequence of A-modules, then .

If N, P are finitely generated A-modules, then

.

**Note**

Philosophically, if we imagine the first case as *N* + *P* and the second case as *N* × *P*, then the result says if and only if both terms are zero whereas if and only if at least one term is zero.

**Proof**

Let be prime.

For the first claim, we have a short exact sequence and it follows that if and only if .

For the second claim, we have

If or , clearly the RHS is zero. Conversely if both are non-zero, since they are finitely generated -modules, Nakayama’s lemma gives and . But these are vector spaces over so we have

♦

**Exercise B**

1. Let be a homomorphism of finitely generated *A*-modules; prove that is a closed subset of Spec A.

2. Prove: if is multiplicative then

# Associated Primes

Definition.Let be a prime ideal. We say is

associatedto the A-module M if for some .Let be the set of prime ideals of A associated to M.

**Note**

We have: if and only if there is an injective *A*-linear map . Observe that if the annihilator of is , then so is that of any non-zero multiple of *m*; this follows immediately from the definition of prime ideals.

Next suppose *M* is an -module; we can compute both and . There is a bijection

.

In particular, we can compute by considering as an *A*-module or a module over itself. The above shows that there is effectively no difference.

Proposition 3.Suppose A is noetherian. The union of all associated primes of M is its set of zero-divisors

.

In particular, any non-zero A-module M has an associated prime.

**Proof**

Fix an and such that ; we need to show *a* is contained in an associated prime of *M*.

So let be the set of ideals of *A* containing *a* of the form for . Since *A* is noetherian and , there is a maximal . We will show is prime; first write for some .

Pick such that and . Since , we have . And since , by maximality of we have so since we have . ♦

Henceforth, *A* denotes a noetherian ring. *However, we do not assume all modules are finitely generated at first.*

# Properties of Associated Primes

Proposition 4.Let be a short exact sequence of A-modules. Then

**Proof**

We may assume is a submodule and . The first containment in the claim is obvious. For the second, suppose we have an *A*-linear map .

If , then any non-zero has annihilator so . If , then composing is still injective, so . ♦

Corollary 1.We have .

**Proof**

Follows immediately from proposition 4. ♦

Next, we show that taking the set of associated primes commutes with localization.

Proposition 5.Let be a multiplicative subset. Then

**Proof**

(⊇) If and , then localization at *S* gives an *A*-linear map .

(⊆) Suppose we have an -linear map where is prime. By theorem 1 here, we can write for some prime such that . Now in the injection we let be the image of 1. It remains to show: there exists such that .

For each , we have in so that in *M* for some , i.e. . Pick a generating set of ; then for each *i* there exists with . Then satisfies .

Conversely if then so lies in the annihilator of , i.e. in . Then ; since we have . ♦