Primary Decomposition

edit_note

This post is being revised. The text below is the version as of 2026-06-05, so some statements may be out of date and references to this post from other posts may not line up.

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In this post we assume that \(A\) is Noetherian and that \(M\) is a finitely generated \(A\)-module.

Primary Submodules

Definition 1 A submodule \(N\) of \(M\) is primary if \(\Ass(M/N)\) consists of a single prime ideal. In this case, if \(\Ass(M/N)=\{\mathfrak{p}\}\), we call \(N\) a \(\mathfrak{p}\)-primary submodule. If \(\Ass(M)\) consists of a single prime ideal, we call \(M\) coprimary.

That is, if \(M/N\) is coprimary then \(N\) is primary. Moreover, by §Associated Primes of Ideals, ⁋Lemma 5, any finite intersection of \(\mathfrak{p}\)-primary submodules is \(\mathfrak{p}\)-primary.

Now the following holds.

Proposition 2 For a ring \(A\) and a prime ideal \(\mathfrak{p}\), the following are equivalent.

1. The \(A\)-module \(M\) is \(\mathfrak{p}\)-coprimary.
2. \(\mathfrak{p}\) is minimal among prime ideals containing \(\ann(M)\), and elements not in \(\mathfrak{p}\) are not zero divisors on \(M\).
3. For some \(k\), \(\mathfrak{p}^k\) annihilates \(M\), and elements not in \(\mathfrak{p}\) are not zero divisors on \(M\).

Proof

First, suppose the first condition holds. Then by definition \(\mathfrak{p}\) is the unique associated prime of \(M\). Now by condition 1 of §Associated Primes of Ideals, ⁋Theorem 7, \(\mathfrak{p}\) must be minimal among prime ideals containing \(\ann(M)\), and by condition 2, elements outside \(\mathfrak{p}\) are not zero divisors on \(M\).

Now assume the second condition holds. Since elements of \(A\setminus \mathfrak{p}\) are not zero divisors on \(M\), it suffices to prove the claim after localizing at \(\mathfrak{p}\). That is, we may assume \((A, \mathfrak{p})\) is a local ring, and the desired result now follows from the assumption that \(\mathfrak{p}\) is minimal over \(\ann(M)\) together with §Properties of Localization, ⁋Corollary 8.

Finally, suppose the third condition holds. Then it is immediate that \(\mathfrak{p}\) is minimal among prime ideals containing \(\ann M\), and therefore by the first condition of §Associated Primes of Ideals, ⁋Theorem 7, \(\mathfrak{p}\) is an associated prime of \(M\). Moreover, since elements outside \(\mathfrak{p}\) are all non-zero-divisors, by the second condition of §Associated Primes of Ideals, ⁋Theorem 7 we know that every associated prime is contained in \(\mathfrak{p}\). Hence \(\mathfrak{p}\) is the unique associated prime of \(M\).

Primary Decomposition

Our goal in this post is to prove the following theorem.

Theorem 3 (Primary decomposition) Any submodule \(M'\) of \(M\) is an intersection of primary submodules. That is, for prime ideals \(\mathfrak{p}_1,\ldots, \mathfrak{p}_n\) and \(\mathfrak{p}_k\)-primary submodules \(M_k\), we can write \(M'=\bigcap_{k=1}^n M_k\). We call this a primary decomposition, and then the following hold.

1. The associated primes of \(M/M'\) are among the \(\mathfrak{p}_k\).
2. If there are no redundant \(M_k\) in the expression for \(M'\), then the \(\mathfrak{p}_i\) are exactly the associated primes of \(M/M'\).
3. If there is no way to express \(M'\) using fewer \(M_k\), then the associated primes of \(M/M'\) correspond exactly to one \(\mathfrak{p}_k\) per index. If in addition \(\mathfrak{p}_i\) is minimal among prime ideals containing the annihilator ideal of \(M/M'\), then \(M_i\) is the \(\mathfrak{p}_i\)-primary component of \(M'\).

4. For a given minimal primary decomposition and any multiplicative subset \(S\) of \(A\), let \(\mathfrak{p}_1,\ldots, \mathfrak{p}_m\) be the prime ideals not meeting \(S\). Then

   \[S^{-1}M'=\bigcap_{i=1}^m S^{-1}M_i\]

   is a minimal primary decomposition of \(S^{-1}M\) over \(S^{-1}A\).

To prove this, we first define the irreducible decomposition of a module.

Definition 4 A submodule \(N\) of an \(A\)-module \(M\) is irreducible if there do not exist \(N_1,N_2\supsetneq N\) with \(N=N_1\cap N_2\).

Then the following holds.

Lemma 5 (Noether) Any submodule of \(M\) can be expressed as an intersection of irreducible submodules.

Proof

We argue by contradiction. Since \(M\) is Noetherian, we may choose a maximal element among the submodules that cannot be expressed as an intersection of irreducible submodules. Call it \(N\). Then \(N\) is not irreducible, so there exist \(N_1,N_2\supsetneq N\) with \(N=N_1\cap N_2\). But by maximality of \(N\), both \(N_1\) and \(N_2\) are intersections of irreducible submodules, and hence so is \(N\), a contradiction.

From this we know that for any submodule \(M'\) of \(M\), an irreducible decomposition

\[M'=\bigcap_{k=1}^n M_k,\qquad \text{$M_k$ irreducible}\]

always exists.

Lemma 6 The above irreducible decomposition is a primary decomposition.

Proof

For this it suffices to show that any irreducible submodule \(P\) is primary, which is equivalent to showing that \(M/P\) is coprimary. Suppose for contradiction that \(M/P\) has two associated primes \(\mathfrak{p},\mathfrak{q}\). Then \(M/P\) has submodules isomorphic to \(A/\mathfrak{p}\) and \(A/\mathfrak{q}\), respectively. By definition, the annihilator of any nonzero element of \(A/\mathfrak{p}\) is \(\mathfrak{p}\), and the annihilator of any nonzero element of \(A/\mathfrak{q}\) is \(\mathfrak{q}\), so these submodules intersect trivially. That is, the zero submodule of \(M/P\) is reducible. Hence \(P\) is reducible in \(M\), a contradiction.

Therefore any submodule of \(M\) admits a primary decomposition. We now prove the remaining assertions of Theorem 3 (Primary decomposition). As in the proof of the preceding lemma, it suffices to work with \(M/M'\), so without loss of generality we may assume \(M'=0\).

Proof of Theorem 3

First, to prove the first assertion, suppose a primary decomposition of the zero submodule of \(M\) is given:

\[0=\bigcap_{k=1}^n M_k\]

Then, by a generalization of the exact sequence from [Multilinear Algebra] §Exact Sequences, ⁋Proposition 7,

\[M\subseteq \bigoplus_{k=1}^n M/M_k\]

so by §Associated Primes of Ideals, ⁋Lemma 5 we know that any prime in \(\Ass M\) occurs among the \(\mathfrak{p}_k\).

Now we prove the second assertion. In particular, for each \(j\) we have

\[\bigcap_{k\neq j} M_k\neq 0\]

Since \(M_j\cap \bigcap_{k\neq j}M_k=0\), we obtain

\[\bigcap_{k\neq j} M_k=\left(\bigcap_{k\neq j} M_k\right)\bigg/\left(M_k\cap \bigcap_{k\neq j}M_k\right)\cong \left(\bigcap_{k\neq j} M_k + M_j\right)\bigg/M_j\subseteq M/M_j\]

so \(\bigcap_{k\neq j} M_k\) is \(\mathfrak{p}_j\)-coprimary, which yields the desired result.

Now we prove the third assertion. In general, since the intersection of \(\mathfrak{p}\)-primary submodules is again \(\mathfrak{p}\)-primary, the given condition forces the \(\mathfrak{p}_k\) to be distinct prime ideals. Now assume that the \(\mathfrak{p}_k\) are minimal among those containing the annihilator ideal, and let us show that \(\Ass(M/M_k)=\{\mathfrak{p}_k\}\). For this we must show that for any nonzero \(x+M_k\in M/M_k\), we have \(\ann(x)=\mathfrak{p}_k\), and by §Localization, ⁋Proposition 5 it suffices to show that the kernel of \(\varepsilon: M \rightarrow M_{\mathfrak{p}_k}\) is \(M_k\).

Now consider the following commutative diagram:

Since the kernel of \(M \rightarrow M/M_k\) is \(M_k\), to prove the desired claim it suffices to show that both \(M_{\mathfrak{p}_k}\rightarrow (M/M_k)_{\mathfrak{p}_k}\) and \(M/M_k \rightarrow (M/M_k)_{\mathfrak{p}_k}\) are injective. First, the injectivity of \(M/M_k \rightarrow (M/M_k)_{\mathfrak{p}_k}\) is immediate from the fact that \(M_k\) is \(\mathfrak{p}_k\)-primary. Then, as we saw at the outset,

\[M \rightarrow \bigoplus_{k=1}^n M/M_k\]

is injective, and hence its localization

\[M_{\mathfrak{p}_k} \rightarrow \left(\bigoplus_{k=1}^n M/M_k\right)_{\mathfrak{p}_k}\]

is also injective. On the other hand, for each \(j\neq k\), the module \(M/M_j\) is \(\mathfrak{p}_j\)-coprimary, and by minimality \(\mathfrak{p}_j\) is not contained in \(\mathfrak{p}_k\), so \((M/M_j)_{\mathfrak{p}_k}=0\). The resulting map is exactly \(M_{\mathfrak{p}_k}\rightarrow (M/M_k)_{\mathfrak{p}_k}\), and we obtain the desired result.

The last assertion is almost immediate.

Primary Decomposition and Factorization

Meanwhile, the following theorem shows that primary decomposition generalizes the familiar notion of factorization.

Theorem 7 For a Noetherian domain \(A\), the following hold.

1. Suppose \(f\in A\) factors as \(f=u p_1^{e_1}\cdots p_n^{e_n}\), where \(u\) is a unit and the \(p_i\) are elements such that the \((p_i)\) are pairwise distinct prime ideals. Then \((f)=\bigcap(p_i^{e_i})\) is a minimal primary decomposition of \((f)\).
2. \(A\) is a UFD if and only if all minimal prime ideals over principal ideals are principal.

Proof

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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