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

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

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}

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.

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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4 Responses to Commutative Algebra 58

1. Vanya says:

How to conclude that $(N \otimes_A P)_{\mathfrak{\p}) \neq 0$ in the last line in proposition 2?

• limsup says:

Recall that $(N\otimes_A P)_{\mathfrak p} = N\otimes_A P \otimes_A A_{\mathfrak p}$ thus the module $k(\mathfrak p) \otimes_A N \otimes_A P$ is obtained by quotienting the former module (i.e. by tensoring with $A_{\mathfrak p} / \mathfrak p A_{\mathfrak p}$). So if the latter is non-zero so is the former.

2. Vanya says:

In statement of proposition 3 , it is mentioned in the end that a y nonzero module has an associated prime. It may happen that M is a faithful A module and A need not be a integral domain. Am i missing something obvious?

• limsup says:

In your example, the associated prime would be 0. Thus we get an injection $A/0 = A \hookrightarrow M$.