# Primary Decomposition of Ideals

Definition.Let be a proper ideal. A

primary decompositionof is its primary decomposition as an A-submodule of A:where each is -primary for some prime , i.e. .

Here is a quick way to determine if an ideal is primary.

Proposition 1.A proper ideal is -primary for some if and only if:

,

in which case .

Recall that is the *radical* of the ideal .

**Proof**

(⇒) Suppose . By proposition 3 here, the set of zero-divisors (in *A*) of is . Now if and , then is a zero-divisor of so .

To prove that , suppose ; then is a zero-divisor for so and we have . Conversely, if , then since every minimal prime in is an associated prime of (by proposition 2 here), *x* is contained in every minimal prime of . By proposition 5 here, .

(⇐) Suppose satisfies the given condition; we first show that is prime. Suppose and . For some *n* > 0, . But so by the given condition and .

Let ; it remains to show . First, is the unique minimal element of so .

Conversely, it remains to show any zero-divisor of as an *A*-module lies in . But if and are such that , then by the given condition . ♦

We thus say a proper ideal is **primary** if:

.

Note that *prime ideals are primary, and by proposition 1, the radical of a primary ideal is prime.*

**Exercise A**

1. Prove that if is a ring homomorphism and is a primary ideal, then is a primary ideal of A. Also .

2. Prove that for an ideal , there is a bijection between primary ideals of *A* containing and primary ideals of .

# Primary Decomposition and Localization

Throughout this section is a fixed multiplicative subset.

Proposition 2.There is a bijection between:

- primary ideals such that ;
- primary ideals
Furthermore, if is in the first set and , then

.

**Proof**

By exercise A.2 above and proposition 3 here, it suffices to show: if is primary and then

- is primary, and
- .

For the first claim, note that is a proper ideal since . Suppose satisfy ; then for some we have . Since no power of *s’* is contained in so

.

For the second claim, clearly . Conversely let satisfy . For some we have . As above so . ♦

Finally, we will see later that unlike factorization of ideals in a Dedekind domain, primary decompositions are not unique. However we can still salvage the following.

Proposition 3.Suppose we have minimal primary decompositions

with for each i. If is minimal in then .

**Proof**

We localize at . But since is minimal in , we have

.

Hence . By proposition 2 this gives which is uniquely determined by . ♦

**Exercise B**

Prove the following.

1. If is a reduced ideal, then it (i.e. ) has no embedded primes.

2. If is maximal, then is -primary if and only if for some *n* > 0.

A power of a prime ideal is not primary in general. E.g. let *k* be a field and with , which is prime since

.

Then is not primary because but and .

# Worked Examples

Throughout this section, *k* denotes a field.

### Example 1

In an earlier example, we saw that for and , the *A*-module has two associated primes: (*X*) and (*X*, *Y*). The following are primary decompositions, both of which are clearly minimal:

.

To check that these ideals are primary:

- is prime and hence primary;
- is a power of the maximal ideal ; by exercise B.2 it is primary;
- so is primary by exercise B.2.

Note that . Geometrically, the *k*-scheme with coordinate ring looks like the following.

### Example 2

Let with . Then

.

We have , an intersection of primary ideals with and . This translates to

with and is already prime.

### Example 3

Let with . Then each of , and is primary (the first ideal is prime; the remaining two ideals are powers of a maximal ideal). Hence this gives a primary decomposition of so its associated primes are , and , with the latter two embedded.

Geometrically, the *k*-scheme with coordinate ring looks like:

Computing an explicit set of generators for is not trivial, but it can be done with Buchberger’s algorithm.

**Exercise C (from Atiyah & MacDonald, Exercise 4.5)**

Let and , , be ideals of *A*. Set . Prove that

is a minimal primary decomposition of .