Primary Decomposition of Ideals
Let be a proper ideal. A primary decomposition of 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.
A proper ideal is -primary for some if and only if:
in which case .
Recall that is the radical of the ideal .
(⇒) 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.
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.
There is a bijection between:
- primary ideals such that ;
- primary ideals
Furthermore, if is in the first set and , then
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.
Suppose we have minimal primary decompositions
with for each i. If is minimal in then .
We localize at . But since is minimal in , we have
Hence . By proposition 2 this gives which is uniquely determined by . ♦
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 .
Throughout this section, k denotes a field.
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.
Let with . Then
We have , an intersection of primary ideals with and . This translates to
with and is already prime.
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 .