# 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. ♦

# 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 $\mathfrak p_2 \not\in \mathrm{Ass}_A M$.

# 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 $\mathfrak p$primary 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.

It 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 $\mathfrak p_i$.

# 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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