Commutative Algebra 58

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 \mathbb Z, every non-zero ideal is an intersection of p^m \mathbb Z \subset \mathbb Z, ideals generated by prime powers.

Annihilators

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

Definition.

The annihilator of m\in M in A is

\mathrm{Ann}_A m := \{ a \in A : am = 0\},

an ideal of A. Similarly, the annihilator of M in A is

\mathrm{Ann}_A M := \{a \in A : \forall m\in M, am = 0\},

also an ideal of A. If \mathrm{Ann}_A M = 0, we say M is a faithful A-module.

Note that if \mathfrak a = \mathrm{Ann}_A M, then M can be regarded as a faithful A/\mathfrak a-module.

Exercise A

1. Given A-submodules N, P \subseteq M, let

(N : P) = \{ a \in A: aP\subseteq N\}.

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

\mathrm{Ann}_A m = (0 : Am), \quad \mathrm{Ann}_A M = (0 : M).

Conversely, express (N: P) in terms of annihilators.

2. Prove that (S^{-1}N : S^{-1}P) = S^{-1}(N : P) if P is a finitely generated submodule of M.

In particular if M is finitely generated then

\mathrm{Ann}_{S^{-1}A} S^{-1}M = S^{-1} (\mathrm{Ann}_A M).

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Support of a Module

Definition.

The support of an A-module M is:

\mathrm{Supp}_A M := \{ \mathfrak p \in \mathrm{Spec} A : M_{\mathfrak p} \ne 0\}.

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

\mathrm{Supp}_A M = V(\mathrm{Ann}_A M).

In particular, the support of M is a closed subspace of \mathrm{Spec} A.

Proof

Let m\in M. For \mathfrak p \in \mathrm{Spec} A we have \frac m 1 \in M_{\mathfrak p} is zero if and only if there exists a \in A - \mathfrak p such that am = 0, equivalently if there exists a \in (\mathrm{Ann}_A m) - \mathfrak p. Hence \frac m 1 \ne 0 if and only if \mathrm{Ann}_A m \subseteq \mathfrak p.

In the general case, let m_1, \ldots, m_n generate M as an A-module. Then

\begin{aligned} M_{\mathfrak p} \ne 0 &\iff \text{for some } i,\ \tfrac {m_i} 1 \in M_{\mathfrak p} \text{ is not zero } \\ &\iff \text{for some } i,\ \mathfrak p \supseteq \mathrm{Ann}_A m_i \\ &\iff \mathfrak p \supseteq \cap_i \mathrm{Ann}_A m_i = \mathrm{Ann}_A M.\end{aligned}

For the last equivalence, recall that \mathfrak p contains \mathfrak a \cap \mathfrak b if and only if it contains either \mathfrak a or \mathfrak b. ♦

Proposition 2.

If 0 \to N \to M \to P \to 0 is a short exact sequence of A-modules, then \mathrm{Supp}_A M = (\mathrm{Supp}_A N) \cup (\mathrm{Supp}_A P).

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

\mathrm{Supp}_A (N\otimes_A P) = (\mathrm{Supp}_A N) \cap (\mathrm{Supp}_A P).

Note

Philosophically, if we imagine the first case as NP  and the second case as N × P, then the result says N_{\mathfrak p} + P_{\mathfrak p} = 0 if and only if both terms are zero whereas N_{\mathfrak p} \times P_{\mathfrak p} = 0 if and only if at least one term is zero.

Proof

Let \mathfrak p \subset A be prime.

For the first claim, we have a short exact sequence 0 \to N_{\mathfrak p} \to M_{\mathfrak p} \to P_{\mathfrak p} \to 0 and it follows that M_{\mathfrak p} = 0 if and only if N_{\mathfrak p} = P_{\mathfrak p} = 0.

For the second claim, we have

(N \otimes_A P)_{\mathfrak p} = N_{\mathfrak p} \otimes_{A_{\mathfrak p}} P_{\mathfrak p}

If N_{\mathfrak p} = 0 or P_{\mathfrak p} = 0, clearly the RHS is zero. Conversely if both are non-zero, since they are finitely generated A_{\mathfrak p}-modules, Nakayama’s lemma gives k(\mathfrak p) \otimes_A N = N_{\mathfrak p}/\mathfrak p N_{\mathfrak p} \ne 0 and k(\mathfrak p) \otimes_A P \ne 0. But these are vector spaces over k(\mathfrak p) so we have

k(\mathfrak p) \otimes_A N \otimes_A P = [k(\mathfrak p) \otimes_A N] \otimes_{k(\mathfrak p)} [k(\mathfrak p) \otimes_A P] \ne 0 \implies (N\otimes_A P)_{\mathfrak p} \ne 0.

Exercise B

1. Let f:M\to N be a homomorphism of finitely generated A-modules; prove that \mathrm{Supp} f = \{\mathfrak p \in \mathrm{Spec} A : f_{\mathfrak p} \ne 0\} is a closed subset of Spec A.

2. Prove: if S\subseteq A is multiplicative then

\begin{aligned} \mathrm{Supp}_{S^{-1}A} S^{-1}M &= (\mathrm{Supp}_A M) \cap (\mathrm{Spec} S^{-1}A) \\ &= \{ \mathfrak p (S^{-1}A) : \mathfrak p \in \mathrm{Supp}_A M, \mathfrak p \cap S = \emptyset\}. \end{aligned}

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Associated Primes

Definition.

Let \mathfrak p \subset A be a prime ideal. We say \mathfrak p is associated to the A-module M if \mathfrak p = \mathrm{Ann}_A m for some m\in M.

Let \mathrm{Ass}_A M be the set of prime ideals of A associated to M.

Note

We have: \mathfrak p \in \mathrm{Ass}_A M if and only if there is an injective A-linear map A/\mathfrak p \hookrightarrow M. Observe that if the annihilator of m \in M is \mathfrak p, 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 A/\mathfrak a-module; we can compute both \mathrm{Ass}_A M and \mathrm{Ass}_{A/\mathfrak a} M. There is a bijection

\mathrm{Ass}_A M \cong \mathrm{Ass}_{A/\mathfrak a} M, \quad \mathfrak p \mapsto \mathfrak p/\mathfrak a \subset A/\mathfrak a.

In particular, we can compute \mathrm{Ass} (A/\mathfrak a) by considering A/\mathfrak a 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

\{a \in A : am = 0 \text{ for some } m \in M, m\ne 0\}.

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

Proof

Fix an a\in A and m' \in M - \{0\} such that am' = 0; we need to show a is contained in an associated prime of M.

So let \Sigma be the set of ideals of A containing a of the form \mathrm{Ann}_A m for m \in M-\{0\}. Since A is noetherian and \mathrm{Ann}_A m' \in \Sigma, there is a maximal \mathfrak p \in \Sigma. We will show \mathfrak p is prime; first write \mathfrak p = \mathrm{Ann}_A m_0 for some m_0 \in M-\{0\}.

Pick b,c\in A such that bc\in \mathfrak p and b \not\in \mathfrak p. Since bm_0 \ne 0, we have \mathfrak a := \mathrm{Ann}_A (bm_0) \in \Sigma. And since \mathfrak a \supseteq \mathfrak p, by maximality of \mathfrak p we have \mathfrak a = \mathfrak p so since bc m_0 = 0 we have c\in \mathfrak a = \mathfrak p. ♦

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

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Properties of Associated Primes

Proposition 4.

Let 0\to N \to M \to P \to 0 be a short exact sequence of A-modules. Then

\mathrm{Ass}_A N \subseteq \mathrm{Ass}_A M \subseteq \mathrm{Ass}_A N \cup \mathrm{Ass}_A P.

Proof

We may assume N\subseteq M is a submodule and P = M/N. The first containment in the claim is obvious. For the second, suppose we have an A-linear map f : A/\mathfrak p \hookrightarrow M.

If (\mathrm{im } f )\cap N \ne 0, then any non-zero m\in (\mathrm{im }f) \cap N has annihilator \mathfrak p so \mathfrak p \in \mathrm{Ass}_A N. If (\mathrm{im } f)\cap N = 0, then composing A/\mathfrak p \stackrel f\to M \to M/N is still injective, so \mathfrak p \in \mathrm{Ass}_A (M/N). ♦

Corollary 1.

We have \mathrm{Ass}_A (M \oplus N) = \mathrm{Ass}_A M \cup \mathrm{Ass}_A N.

Proof

Follows immediately from proposition 4. ♦

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

Proposition 5.

Let S\subseteq A be a multiplicative subset. Then

\begin{aligned} \mathrm{Ass}_{S^{-1}A} (S^{-1}M) &= \mathrm{Ass}_A M \cap \mathrm{Spec} S^{-1}A \\ &= \{\mathfrak p S^{-1}A : \mathfrak p \in \mathrm{Ass}_A M, \ \mathfrak p \cap S = \emptyset\}.\end{aligned}

Proof

(⊇) If A/\mathfrak p \hookrightarrow M and \mathfrak p \cap S = \emptyset, then localization at S gives an A-linear map S^{-1}A/(\mathfrak p S^{-1}A) \hookrightarrow S^{-1}M.

(⊆) Suppose we have an S^{-1}A-linear map (S^{-1}A)/\mathfrak q \hookrightarrow S^{-1}M where \mathfrak q \subset S^{-1}A is prime. By theorem 1 here, we can write \mathfrak q = \mathfrak p S^{-1}A for some prime \mathfrak p \subset A such that \mathfrak p \cap S = \emptyset. Now in the injection (S^{-1}A)/\mathfrak p(S^{-1}A) \hookrightarrow S^{-1}M we let \frac m s be the image of 1. It remains to show: there exists t\in S such that \mathrm{Ann}_A (tm) = \mathfrak p.

For each a\in \mathfrak p, we have \frac{am}s = 0 in S^{-1}M so that s'am = 0 in M for some s'\in S, i.e. a \in \mathrm{Ann}_A (s'm). Pick a generating set a_1, \ldots, a_n of \mathfrak p; then for each i there exists s_i' \in S with a_i \in \mathrm{Ann}_A (s_i' m). Then t := s_1' \ldots s_n' satisfies \mathfrak p \subseteq \mathrm{Ann}_A (tm).

Conversely if a\in \mathrm{Ann}_A (tm) then atm = 0 so \frac{at}1 \in S^{-1}A lies in the annihilator of \frac m s, i.e. in \mathfrak q. Then at \in \mathfrak q \cap A = \mathfrak p; since \mathfrak p \cap S = \emptyset we have a\in \mathfrak p. ♦

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