This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In this post we assume that \(A\) is Noetherian and \(M\) 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. Also, from §Associated Primes, ⁋Lemma 5 we know that 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.
- The \(A\)-module \(M\) is \(\mathfrak{p}\)-coprimary.
- \(\mathfrak{p}\) is minimal among prime ideals containing \(\ann(M)\), and elements not in \(\mathfrak{p}\) are not zero divisors on \(M\).
- 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, ⁋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. Then 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 now the desired result follows from the assumption that \(\mathfrak{p}\) is minimal over \(\ann(M)\) and §Properties of Localization, ⁋Corollary 8.
Finally, suppose the third condition holds. Then it is obvious that \(\mathfrak{p}\) is minimal among prime ideals containing \(\ann M\), and therefore by the first condition of §Associated Primes, ⁋Theorem 7, \(\mathfrak{p}\) is an associated prime of \(M\). Also, since elements outside \(\mathfrak{p}\) are all not zero divisors, by the second condition of §Associated Primes, ⁋Theorem 7 we know that any associated prime is always contained in \(\mathfrak{p}\). That is, \(\mathfrak{p}\) 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.
- The associated primes of \(M/M'\) are among the \(\mathfrak{p}_k\).
- If there are no redundant \(M_k\) in the expression for \(M'\), then the \(\mathfrak{p}_i\) are exactly the associated primes of \(M/M'\).
- 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 \(M'\).
-
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 use proof by contradiction. Since \(M\) is Noetherian, we can choose a maximal submodule among those that cannot be expressed as an intersection of irreducible submodules. Call this \(N\). 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 therefore so is \(N\), which is 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 the same as 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. Then 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 they have only \(0\) in common. That is, the zero submodule of \(M/P\) is reducible. From this we obtain that \(P\) is reducible in \(M\), a contradiction.
Therefore any submodule of \(M\) always has a primary decomposition. We now need to prove the remaining parts of Theorem 3. As in the proof of the preceding lemma, when proving these 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 result, suppose a primary decomposition of the zero submodule of \(M\) is given:
\[0=\bigcap_{k=1}^n M_k\]Then from a generalization of the exact sequence of [Multilinear Algebra] §Exact Sequences, ⁋Proposition 7,
\[M\subseteq \bigoplus_{k=1}^n M/M_k\]so by §Associated Primes, ⁋Lemma 5 we know that any prime of \(\Ass M\) is obtained among the \(\mathfrak{p}_k\).
Now we prove the second result. Then in particular, for each \(j\),
\[\bigcap_{k\neq j} M_k\neq 0\]holds. Then since \(M_j\cap \bigcap_{k\neq j}M_k=0\),
\[\bigcap_{k\neq j} M_k=\left(\bigcap_{k\neq j} M_k\right)\bigg/\left(M_k\cap \bigcap_{k\neq j}M_k\right)\cong \left(\bigcap_{k\neq j} M_k + M_j\right)\bigg/M_j\subseteq M/M_j\]so \(\bigcap_{k\neq j} M_k\) is \(\mathfrak{p}_j\)-coprimary. From this we obtain the desired result.
Now we prove the third result. In general, since the intersection of \(\mathfrak{p}\)-primary submodules is also \(\mathfrak{p}\)-primary, to satisfy the given condition the \(\mathfrak{p}_k\) must all be distinct prime ideals. Now assume 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 need to show that for any nonzero \(x+M_k\in M/M_k\), \(\ann(x)=\mathfrak{p}_k\) holds, so 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
Then 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 obvious from the fact that \(M_k\) is \(\mathfrak{p}_k\)-primary. Then as we saw at the beginning,
\[M \rightarrow \bigoplus_{k=1}^n M/M_k\]is injective, and therefore 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 \(j\neq k\), \(M/M_j\) is \(\mathfrak{p}_j\)-coprimary, and by minimality \(\mathfrak{p}_j\) must not be contained in \(\mathfrak{p}_i\), so \((M/M_j)_{\mathfrak{p}_k}=0\) holds, and the resulting map is exactly \(M_{\mathfrak{p}_k}\rightarrow (M/M_k)_{\mathfrak{p}_k}\), so we obtain the desired result.
The last claim is almost obvious.
Primary Decomposition and Factorization
Meanwhile, the following theorem shows that primary decomposition generalizes the notion of factorization we already know.
Theorem 7 For a Noetherian domain \(A\), the following hold.
- Suppose \(f\in A\) factors as \(f=u p_1^{e_1}\cdots p_n^{e_n}\). Here \(u\) 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 \((f)\).
- \(A\) is a UFD if and only if all minimal prime ideals over principal ideals are principal.
Proof
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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