Commutative Algebra 59

Prime Composition Series

Throughout this article, A is a noetherian ring and all A-modules are finitely generated.

Recall (proposition 1 here) that if M is a noetherian and artinian module, we can find a sequence of submodules whose consecutive factors are simple modules. Correspondingly we have:

Proposition 1.

For finitely generated M, there exists a sequence of submodules

0 = M_0 \subsetneq M_1 \subsetneq \ldots \subsetneq M_n = M

such that each M_{i} / M_{i-1} \cong A/\mathfrak p_i as A-modules for prime ideals \mathfrak p_i\subseteq A.

Proof

Assume M\ne 0. Since \mathrm{Ass}_A M \ne \emptyset by proposition 3 here, there is an embedding of A-modules A/\mathfrak p \hookrightarrow M. If equality holds, we are done. Otherwise, let M_1 be the image of the map and repeat with M/M_1 to obtain a submodule M_2/M_1 \cong A/\mathfrak p'. Repeating this process, this must eventually terminate since we cannot have an infinite ascending chain of submodules M_0 \subsetneq M_1 \subsetneq M_2 \subsetneq \ldots. ♦

Corollary 1.

Let 0 = M_0 \subset M_1 \subset \dots \subset M_n = M be as in proposition 1 with M_i / M_{i-1} \cong A/\mathfrak p_i for some prime \mathfrak p_i \subset A. Then

\mathrm{Ass}_A M \subseteq \{\mathfrak p_1, \ldots, \mathfrak p_n\}.

In particular if M is finitely generated then \mathrm{Ass}_A M is finite.

Proof

Repeatedly applying proposition 4 here, we have

\mathrm{Ass}_A M \subseteq \mathrm{Ass}_A (M_1/M_0) \cup \ldots \cup \mathrm{Ass}_A (M_n / M_{n-1}).

Since each \mathrm{Ass}_A M_i/M_{i-1} = \mathrm{Ass}_A (A/\mathfrak p_i) = \{\mathfrak p_i\} for each i, we are done. ♦

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Associated Primes and Support

If \mathfrak p\in \mathrm{Ass}_A M, then there is a map A/\mathfrak p \hookrightarrow M; upon localizing at \mathfrak p we get an A-linear k(\mathfrak p)\hookrightarrow M_{\mathfrak p} and so M_{\mathfrak p} \ne 0. Hence we have shown:

Lemma 1.

For any module M, \mathrm{Ass}_A M \subseteq \mathrm{Supp}_A M.

For a partial reverse inclusion, we have:

Proposition 2.

If \mathfrak p is a minimal element of \mathrm{Supp}_A M then \mathfrak p \in \mathrm{Ass}_A M.

Proof

By proposition 5 here, we have

\mathrm{Ass}_{A_{\mathfrak p}} M_{\mathfrak p} = \{\mathfrak q A_{\mathfrak p} : \mathfrak q \in \mathrm{Ass}_A M \text{ contained in } \mathfrak p\},

which is non-empty since M_{\mathfrak p} \ne 0. By lemma 1 this set lies in \mathrm{Supp}_{A_{\mathfrak p}} M_{\mathfrak p}; but since \mathfrak p is minimal in \mathrm{Supp}_A M, by exercise B.2 here \mathrm{Supp}_{A_{\mathfrak p}} M_{\mathfrak p} has exactly one element: \mathfrak p A_{\mathfrak p} so

\mathrm{Ass}_{A_{\mathfrak p}} M_{\mathfrak p} = \{\mathfrak p A_{\mathfrak p}\}

so \mathfrak p \in \mathrm{Ass}_A M by proposition 5 here again. ♦

Definition.

If \mathfrak p \in \mathrm{Ass}_A M is not a minimal prime of \mathrm{Supp}_A M, we call it an embedded prime of M.

Example

Let k be a field, A = k[X, Y] and M = k[X, Y]/(X^2, XY), considered as an A-module. Note that \mathrm{Ann}_A M = (X^2, XY) with radical (X) so we get (by proposition 1 here)

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

On the other hand, we claim that \mathrm{Ass}_A M = \{(X), (X, Y)\}. Indeed we have:

\begin{aligned} f = Y \in M &\implies \mathrm{Ann}_A f = (X), \\ g = X \in M &\implies \mathrm{Ann}_A g = (X, Y),\end{aligned}

so (X), (X, Y) \in \mathrm{Ass}_A M and (X, Y) is an embedded prime of M.

Conversely, we pick the chain of submodules 0 \subset M_1 \subset M with M_1 generated by g = X\in M. Then M_1 = (X\cdot A)/(X^2\cdot A + XY\cdot A); as shown above, M_1 \cong A/(X,Y). Also M/M_1 \cong A/(X) so by corollary 1

\mathrm{Ass}_A M \subseteq \{(X), (X,Y)\}.

Exercise A

Find an A-module M and prime ideals \mathfrak p_1 \subsetneq \mathfrak p_2 \subsetneq \mathfrak p_3 with \mathfrak p_1, \mathfrak p_3 \in \mathrm{Ass}_A M but \mathfrak p_2 \not\in \mathrm{Ass}_A M.

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Existence of Primary Decomposition

In this section we fix an ambient non-zero A-module M (finitely generated of course) and consider its submodules. For each prime \mathfrak p \subset A, take the set \Sigma(\mathfrak p) of all submodules N\subseteq M such that \mathfrak p \not\in \mathrm{Ass}_A N. Note that

0\in \Sigma(\mathfrak p) \implies \Sigma(\mathfrak p) \ne \emptyset.

Now for each prime \mathfrak p, fix a maximal element E(\mathfrak p) \in \Sigma(\mathfrak p).

Lemma 2.

We have \cap_{\mathfrak p} E(\mathfrak p) = 0.

Note

If \mathfrak p\not\in \mathrm{Ass}_A M then E(\mathfrak p) = M so the above intersection only needs to be taken over \mathfrak p\in \mathrm{Ass}_A M, i.e. a finite number of terms.

Proof

Let N := \cap_{\mathfrak p} E(\mathfrak p). If N\ne 0 it has an associated prime \mathfrak p. Since N\subseteq E(\mathfrak p) we have \mathfrak p \in \mathrm{Ass}_A E(\mathfrak p), a contradiction. ♦

Lemma 3.

Let \mathfrak p \in \mathrm{Ass}_A M. Then \mathrm{Ass}_A (M/E(\mathfrak p)) = \{\mathfrak p \}.

Proof

By proposition 4 here, we have \mathrm{Ass}_A M \subseteq \mathrm{Ass}_A E(\mathfrak p) \cup \mathrm{Ass}_A (M/ E(\mathfrak p)). Since \mathfrak p \not\in \mathrm{Ass}_A E(\mathfrak p) we have \mathfrak p \in \mathrm{Ass}_A M/E(\mathfrak p). On the other hand if \mathfrak q \in \mathrm{Ass}_A (M/E(\mathfrak p)) - \{\mathfrak p\}, then we have an injection A/\mathfrak q \hookrightarrow M/E(\mathfrak p) whose image is of the form Q/E(\mathfrak p). But now

\mathrm{Ass}_A Q \subseteq \mathrm{Ass}_A E(\mathfrak p) \cup \mathrm{Ass}_A Q/E(\mathfrak p) = \mathrm{Ass}_A E(\mathfrak p) \cup \{\mathfrak q\}

which does not contain \mathfrak p, contradicting maximality of E(\mathfrak p). Hence no such \mathfrak q exists. ♦

Definition.

For an associated prime \mathfrak p of M, a \mathfrak pprimary submodule of M is an N\subseteq M such that \mathrm{Ass}_A M/N = \{\mathfrak p\}.

A primary decomposition of M is an expression

0 = M_1 \cap M_2 \cap \ldots \cap M_n

where each M_i is a \mathfrak p_i-primary submodule of M.

The decomposition is irredundant if for any 1\le i\le n, \cap_{j\ne i} M_j \ne 0. It is minimal if it is irredundant and all \mathfrak p_i are distinct.

By lemmas 2 and 3, a minimal primary decomposition exists for every non-zero module.

warningIt is not true that in every primary decomposition 0 =\cap_{\mathfrak p} M(\mathfrak p), M(\mathfrak p) must be a maximal element of \Sigma(\mathfrak p). We will see an example in the next article.

Exercise B

Prove that if N_1, N_2 \subseteq M are \mathfrak p-primary submodules, so is N_1\cap N_2 \subseteq M.

Hence given any primary decomposition, we can get a minimal one by first removing the redundant terms then taking the intersection of all M_i with the same corresponding \mathfrak p_i.

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Properties of Primary Decomposition

Here, we will discuss properties of a general primary decomposition. Throughout this section we fix:

0 = M_1 \cap M_2 \cap \ldots \cap M_n where M_i is \mathfrak p_i-primary in M.

Proposition 3.

Every \mathfrak p \in \mathrm{Ass}_A M must occur among the \mathfrak p_i.

Proof

Since \cap_i M_i = 0 the canonical map M\hookrightarrow M/M_1 \oplus \ldots \oplus M/M_n is injective. Hence

\mathrm{Ass}_A M \subseteq \cup_{i=1}^n \mathrm{Ass}_A (M/M_i) = \{\mathfrak p_1, \ldots, \mathfrak p_n\}.

Proposition 4.

If the primary decomposition is irredundant, then every \mathfrak p_i is an associated prime of M.

Proof

If \mathfrak p_i \not\in \mathrm{Ass}_A M then among the injections M \hookrightarrow \oplus_j M/M_j and M/M_i \hookrightarrow \oplus_j M/M_j, we have M \cap (M/M_i) = 0 since \mathrm{Ass}_A M and \mathrm{Ass}_A (M/M_i) = \{\mathfrak p_i\} are disjoint. Thus

M\longrightarrow \oplus_{j\ne i} M/M_j is injective

and \cap_{j\ne i} M_j = 0, contradicting the condition of irredundancy. ♦

Note

Thus proposition 4 gives us a way to compute \mathrm{Ass}_A M: find a primary decomposition and remove terms until it becomes irredundant. However, note that proposition 4 does not require \mathfrak p_1, \ldots, \mathfrak p_n to be distinct.

Corollary 2.

If the primary decomposition is minimal, then

\mathrm{Ass}_A M = \{\mathfrak p_1, \ldots, \mathfrak p_n\}

has exactly n elements.

Proof

Apply propositions 3 and 4. ♦

Finally, we define the primary decomposition of submodules.

Defintion.

Let N\subseteq M be a submodule. A primary decomposition of N in M is an expression

N = M_1 \cap M_2 \cap \ldots \cap M_n

such that M_i\subseteq M is a \mathfrak p_i-primary submodule of M.

Note

A primary decomposition of N in M corresponds bijectively to a primary decomposition of M/N. We say the primary decomposition of N in M is irredundant or minimal if the corresponding primary decomposition of M/N is so.

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