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.
Let A be a fixed ring and M be an A-module.
The annihilator of in A is
an ideal of A. Similarly, the annihilator of M in A is
also an ideal of A. If , we say M is a faithful A-module.
Note that if , then M can be regarded as a faithful -module.
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
The support of an A-module M is:
Geometrically, these are points in Spec A at which the module does not vanish.
If M is a finitely generated A-module, then
In particular, the support of M is a closed subspace of .
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 . ♦
If is a short exact sequence of A-modules, then .
If N, P are finitely generated A-modules, then
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.
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
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
Let be a prime ideal. We say is associated to the A-module M if for some .
Let be the set of prime ideals of A associated to M.
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.
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.
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
Let be a short exact sequence of A-modules. Then
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 . ♦
We have .
Follows immediately from proposition 4. ♦
Next, we show that taking the set of associated primes commutes with localization.
Let be a multiplicative subset. Then
(⊇) 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 . ♦