Primary Decomposition

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

Primary Submodules

Definition 1 In \(M\), a submodule \(N\) is a primary submodule if \(\Ass(M/N)\) consists of only 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 only a single prime ideal, we call \(M\) a coprimary submodule.

That is, if \(M/N\) is a coprimary module, then \(N\) is a primary submodule. 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 \(A\), a prime ideal \(\mathfrak{p}\), and a non-\(0\) \(A\)-module \(M\), the following are all equivalent.

1. The \(A\)-module \(M\) is a \(\mathfrak{p}\)-coprimary module.
2. \(\mathfrak{p}\) is minimal among the 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 ideal of \(M\). Now, by the first condition of §Associated Primes, ⁋Theorem 7, \(\mathfrak{p}\) must be minimal among the prime ideals containing \(\ann(M)\), and by the second condition, elements outside \(\mathfrak{p}\) are not zero divisors on \(M\).

Now suppose the second condition holds. Since elements of \(A\setminus \mathfrak{p}\) are not zero divisors on \(M\), it suffices to prove the given claim in the localization \(M_\mathfrak{p}\). That is, we may assume that \((A, \mathfrak{p})\) is a local ring, and now from the assumption that \(\mathfrak{p}\) is minimal over \(\ann(M)\) and §Properties of Localization, ⁋Corollary 8, we obtain \(\sqrt{\ann(M)}=\mathfrak{p}\). But since \(A\) is Noetherian, \(\mathfrak{p}\) is finitely generated, and since a suitable power of each generator belongs to \(\ann(M)\), choosing \(k\) larger than the sum of all these exponents yields \(\mathfrak{p}^k\subseteq \ann(M)\).

Finally, suppose the third condition holds. Then it is clear that \(\mathfrak{p}\) is minimal among the prime ideals containing \(\ann M\), and therefore, by the first condition of §Associated Primes, ⁋Theorem 7, \(\mathfrak{p}\) is an associated prime ideal of \(M\). Moreover, since elements outside \(\mathfrak{p}\) are all not zero divisors, we know again by the second condition of §Associated Primes, ⁋Theorem 7 that any associated prime is always contained in \(\mathfrak{p}\). That is, \(\mathfrak{p}\) is the unique associated prime ideal of \(M\).

Primary Decomposition

In this post, our goal is to show the following theorem.

Theorem 3 (Primary decomposition) In \(M\), every proper submodule \(M'\) 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. Every associated prime of \(M/M'\) is one of the \(\mathfrak{p}_k\).
2. If, in expressing \(M'\), no \(M_k\) is redundant, then the \(\mathfrak{p}_i\) are precisely the associated primes of \(M/M'\).
3. If there is no way to express \(M'\) using fewer \(M_k\), then the associated primes of \(M/M'\) are precisely the \(\mathfrak{p}_k\), one for each index. If moreover \(\mathfrak{p}_i\) is minimal among the prime ideals containing the annihilator ideal of \(M/M'\), then \(M_i/M'\) equals the kernel of \(M/M' \rightarrow (M/M')_{\mathfrak{p}_i}\), and hence \(M_i\) is determined by \(M'\) and \(\mathfrak{p}_i\) alone.

4. Given a minimal primary decomposition, for any of \(A\)’s multiplicative subsets \(S\), let \(\mathfrak{p}_1,\ldots, \mathfrak{p}_m\) be the prime ideals that do not meet \(S\). Then

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

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

In particular, when \(M=A\) and \(M'=\mathfrak{a}\) is an ideal of \(A\), in the third result of Theorem 3 (Primary decomposition), those prime ideals containing \(\mathfrak{a}\) that are minimal with respect to inclusion are called the minimal prime ideals of \(\mathfrak{a}\); when \(\mathfrak{a}=(0)\), we simply call them the minimal prime ideals of \(A\).

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

Definition 4 For an \(A\)-module \(M\), a proper submodule \(N\) is irreducible if \(N=N_1\cap N_2\) holds for no \(N_1,N_2\supsetneq N\).

Then the following holds.

Lemma 5 (Noether) Every proper submodule of \(M\) is an intersection of irreducible submodules.

Proof

We argue by contradiction. Since \(M\) is Noetherian, we can choose a maximal element among the proper submodules that are not intersections of irreducible submodules; call it \(N\). Since \(N\) is not an irreducible submodule, \(N=N_1\cap N_2\) holds for some \(N_1,N_2\supsetneq N\). Here, if \(N_1=M\), then \(N=N_2\), contradicting \(N_2\supsetneq N\), and the case \(N_2=M\) is similar; hence \(N_1,N_2\) are both proper submodules. But by the maximality of \(N\), \(N_1,N_2\) are both intersections of irreducible submodules, and therefore so is \(N\), which is a contradiction.

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

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

always exists.

Lemma 6 The irreducible decomposition above is a primary decomposition.

Proof

For this, it suffices to show that any irreducible submodule \(P\) is a primary submodule, which is equivalent to showing that \(M/P\) is a coprimary submodule. First, since \(P\) is a proper submodule, \(M/P\neq 0\), and by the first result of §Associated Primes, ⁋Theorem 7, \(\Ass(M/P)\) is nonempty. Therefore, contrary to the conclusion, suppose that \(M/P\) has two distinct 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 element of \(A/\mathfrak{p}\) other than \(0\) is \(\mathfrak{p}\), and the annihilator of any element of \(A/\mathfrak{q}\) other than \(0\) is \(\mathfrak{q}\), so they have only \(0\) as a common element. That is, in \(M/P\), the zero submodule \(0\) is a reducible submodule. From this, in \(M\), \(P\) is a reducible submodule, which yields a contradiction.

Therefore, every proper submodule of \(M\) always has a primary decomposition. Now the remaining parts of Theorem 3 (Primary decomposition) must be proved. As in the proof of the preceding lemma, in proving them it suffices to prove them for \(M/M'\), so without loss of generality we may assume \(M'=0\).

Proof of Theorem 3

First, to show the first result, suppose that for \(M\), a primary decomposition of the zero submodule \(0\)

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

is given. Then, since from generalizing the exact sequence of [Multilinear Algebra] §Exact Sequences, ⁋Proposition 7 we have

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

we know from §Associated Primes, ⁋Lemma 5 that any primes in \(\Ass M\) are obtained from among the \(\mathfrak{p}_k\).

Now let us show the second result. Then, in particular, for each \(j\),

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

holds. Then, since \(M_j\cap \bigcap_{k\neq j}M_k=0\),

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

so \(\bigcap_{k\neq j} M_k\) is \(\mathfrak{p}_j\)-coprimary. From this, we obtain the desired result.

Now let us show the third result. In general, since an 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 suppose that the \(\mathfrak{p}_k\) are minimal among those containing the annihilator ideal. What we must show here is that \(M_k\) is determined by \(M\) and \(\mathfrak{p}_k\) alone; since by §Localization, ⁋Proposition 5 the kernel of \(\varepsilon: M \rightarrow M_{\mathfrak{p}_k}\) is the collection of elements for which \(sx=0\) holds for some \(s\in A\setminus \mathfrak{p}_k\), that is, of such \(x\), it is determined by \(M\) and \(\mathfrak{p}_k\) alone, and therefore it suffices to show that the kernel of \(\varepsilon\) is \(M_k\).

Now consider the following commutative diagram:

Then, since the kernel of \(M \rightarrow M/M_k\) is \(M_k\), to show 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, that \(M/M_k \rightarrow (M/M_k)_{\mathfrak{p}_k}\) is injective is immediate from the fact that \(M_k\) is \(\mathfrak{p}_k\)-primary. Then, as we observed at the very beginning,

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

is injective, and therefore the localization of this map

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

is also injective. Meanwhile, for each \(j\neq k\), \(M/M_j\) is \(\mathfrak{p}_j\)-coprimary, and by minimality \(\mathfrak{p}_j\) cannot be contained in \(\mathfrak{p}_k\), so \((M/M_j)_{\mathfrak{p}_k}=0\) holds; since the map thus obtained is precisely \(M_{\mathfrak{p}_k}\rightarrow (M/M_k)_{\mathfrak{p}_k}\), we obtain the desired result.

Finally, let us show the fourth result. First, since localization is an exact functor (§Properties of Localization, ⁋Proposition 2), it commutes with finite intersections, and therefore we obtain \(0=\bigcap_{k=1}^n S^{-1}M_k\). Now, following the notation of the statement, let \(\mathfrak{p}_1,\ldots, \mathfrak{p}_m\) be the only prime ideals that do not meet \(S\). If \(\mathfrak{p}_j\cap S\neq\emptyset\), then since \(M/M_j\) is \(\mathfrak{p}_j\)-coprimary, by the third condition of Proposition 2, for some \(t\), \(\mathfrak{p}_j^t\) annihilates \(M/M_j\), and choosing \(s\in \mathfrak{p}_j\cap S\), the element \(s^t\in \mathfrak{p}_j^t\) also annihilates \(M/M_j\), so again by §Localization, ⁋Proposition 5 we know that \(S^{-1}(M/M_j)=0\), that is, \(S^{-1}M_j=S^{-1}M\). In other words, these components play no role in the intersection above, so

\[0=\bigcap_{i=1}^m S^{-1}M_i\]

holds. Now, for each \(i\) appearing here, \(\Ass(M/M_i)=\{\mathfrak{p}_i\}\) and \(\mathfrak{p}_i\cap S=\emptyset\), so by the third result of §Associated Primes, ⁋Theorem 7, \(\Ass_{S^{-1}A}S^{-1}(M/M_i)=\{\mathfrak{p}_iS^{-1}A\}\), and thus \(S^{-1}M_i\) is a \(\mathfrak{p}_iS^{-1}A\)-primary submodule.

Now let us show that this decomposition is minimal. By §Localization, ⁋Proposition 8, since prime ideals not meeting \(S\) in \(A\) are in one-to-one correspondence with the prime ideals of \(S^{-1}A\), the ideals \(\mathfrak{p}_1S^{-1}A,\ldots, \mathfrak{p}_mS^{-1}A\) are \(m\) distinct prime ideals, and by the first result of the same proposition, in \(S^{-1}A\), any ideal \(\mathfrak{b}\) is generated by the image of the preimage of \(\mathfrak{b}\) in \(A\). Now, from the fact that \(A\) is Noetherian, \(S^{-1}A\) is also Noetherian, and since the images of the generators of \(M\) generate \(S^{-1}M\), the module \(S^{-1}M\) is a finitely generated \(S^{-1}A\)-module. Then, from the third result of §Associated Primes, ⁋Theorem 7 and the second result already proven, \(\Ass_{S^{-1}A}S^{-1}M\) consists of precisely these \(m\) prime ideals, and applying the first result over \(S^{-1}A\), any primary decomposition of the zero submodule of \(S^{-1}M\) must have all of them among its prime ideals, so it has at least \(m\) components. That is, the decomposition above is minimal.

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, \(e_i\geq 1\), and the \(p_i\) are prime elements such that the \((p_i)\) are 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 every prime ideal minimal over a principal ideal is principal.

Proof

We first prove the first claim. To begin with, we check that \((p_i^{e_i})\) is \((p_i)\)-primary for each \(i\). Since it is clear that \((p_i)^{e_i}=(p_i^{e_i})\) annihilates \(A/(p_i^{e_i})\), by the third condition of Proposition 2 it suffices to show that the elements outside \((p_i)\) are not zero divisors on \(A/(p_i^{e_i})\). For this, we show by induction on \(e_i\) that whenever \(x\not\in (p_i)\) and \(a\in A\) satisfy \(xa\in (p_i^{e_i})\), then \(a\in(p_i^{e_i})\). The case \(e_i=0\) is trivial. If \(e_i\geq 1\), then \(xa\in(p_i^{e_i})\subseteq (p_i)\), and since \((p_i)\) is prime and \(x\not\in(p_i)\), we have \(a\in (p_i)\). Thus we can write \(a=p_ib\), and setting \(xa=p_i^{e_i}c\), we have \(p_i(xb-p_i^{e_i-1}c)=0\); since \(A\) is a domain and \(p_i\neq 0\), this gives \(xb=p_i^{e_i-1}c\in (p_i^{e_i-1})\). By the induction hypothesis \(b\in (p_i^{e_i-1})\), and therefore \(a=p_ib\in(p_i^{e_i})\).

We now show that \((f)=\bigcap_{i=1}^n (p_i^{e_i})\). Since \(f\in (p_i^{e_i})\) for each \(i\), one inclusion is trivial. Conversely, let \(g\in\bigcap_{i=1}^n (p_i^{e_i})\); we show \(g\in (p_1^{e_1}\cdots p_j^{e_j})\) by induction on \(j\). The case \(j=0\) is trivial. Now assume we can write \(g=p_1^{e_1}\cdots p_j^{e_j}h\). First, we check that \(p_i\not\in (p_k)\) for any two distinct indices \(i\neq k\): if \(p_i=cp_k\), then since the prime element \(p_i\) is irreducible ([Ring Theory] §Integral Domains, ⁋Proposition 12) and \(p_k\) is not a unit, \(c\) must be a unit, giving \((p_i)=(p_k)\) and contradicting the assumption that the \((p_i)\) are distinct. Therefore, since \((p_{j+1})\) is prime, \(p_1^{e_1}\cdots p_j^{e_j}\not\in (p_{j+1})\), and since \(p_1^{e_1}\cdots p_j^{e_j}h=g\in (p_{j+1}^{e_{j+1}})\), what we showed in the previous paragraph gives \(h\in (p_{j+1}^{e_{j+1}})\). That is, \(g\in (p_1^{e_1}\cdots p_{j+1}^{e_{j+1}})\), and by induction we obtain \(g\in (p_1^{e_1}\cdots p_n^{e_n})=(f)\). The last equality holds because \(u\) is a unit.

To show that this decomposition is minimal, we first check that no component can be omitted. For each \(i\), the product \(\prod_{j\neq i} p_j^{e_j}\) lies in \(\bigcap_{j\neq i}(p_j^{e_j})\) but not in \((p_i^{e_i})\): if it did, then in particular \(\prod_{j\neq i}p_j^{e_j}\in (p_i)\), and since \((p_i)\) is prime, we would have \(p_j\in (p_i)\) for some \(j\neq i\), contradicting what we checked in the previous paragraph. Therefore, by the second part of Theorem 3 (Primary decomposition), \(\Ass(A/(f))=\{(p_1),\ldots,(p_n)\}\), and in particular these are \(n\) distinct prime ideals. Meanwhile, by the first part of Theorem 3 (Primary decomposition), every primary decomposition of \((f)\) must have all the elements of \(\Ass(A/(f))\) among its primes, so it must have at least \(n\) components. That is, the decomposition above is minimal.

We now prove the second claim. First, suppose \(A\) is a UFD, and let a principal ideal \((f)\) and a prime ideal \(\mathfrak{p}\) minimal among those containing \((f)\) be given. If \(f=0\), then since \(A\) is a domain, \((0)\) is prime, and hence \(\mathfrak{p}=(0)\) is principal. If \(f\) is a unit, there is no prime ideal containing \((f)=A\), so there is nothing to prove. Now suppose \(f\) is a nonzero non-unit, and factor it as \(f=up_1^{e_1}\cdots p_n^{e_n}\). Then \(f\in\mathfrak{p}\), and since \(\mathfrak{p}\) is prime, \(p_i\in \mathfrak{p}\) for some \(i\). That is, \((f)\subseteq (p_i)\subseteq \mathfrak{p}\), and since \((p_i)\) is prime, the minimality of \(\mathfrak{p}\) forces \(\mathfrak{p}=(p_i)\), which is principal.

Conversely, suppose that every prime ideal minimal over a principal ideal is principal. First, since \(A\) is Noetherian, every nonzero non-unit element is expressed as a product of irreducible elements. Supposing otherwise, the collection of principal ideals \((a)\) formed by nonzero non-unit elements \(a\) that cannot be expressed as products of irreducible elements is nonempty, so by the Noetherian condition we can choose a maximal element \((a)\) of this collection. Then \(a\) is not irreducible, so we can write \(a=bc\) for non-units \(b,c\); if \((a)=(b)\), then \(b=ad\) for some \(d\) and \(a=adc\), and since \(A\) is a domain this makes \(c\) a unit, a contradiction; hence \((a)\subsetneq (b)\), and for the same reason \((a)\subsetneq(c)\). Then by the maximality of \((a)\), \(b,c\) are both expressed as products of irreducible elements, and therefore so is \(a=bc\), a contradiction.

Next, we show that every irreducible element \(p\) is prime. Since \(A\) is Noetherian, there exists a prime ideal minimal among those containing \((p)\); letting it be \(\mathfrak{p}\), by assumption \(\mathfrak{p}=(q)\) is principal. From \(p\in (q)\) we can write \(p=qc\), and since \(p\) is irreducible and \(q\) is a non-unit, \(c\) must be a unit; therefore \((p)=(q)=\mathfrak{p}\) is a prime ideal. That is, \(p\) is a prime element.

Finally, we show the uniqueness of factorization. Suppose \(up_1\cdots p_m=vq_1\cdots q_k\) are products of irreducible elements and \(u,v\) are units; we use induction on \(m\). If \(m=0\), then the left-hand side is a unit, so we must have \(k=0\). If \(m\geq 1\), then since \(p_m\) is prime, \(p_m\mid q_j\) for some \(j\), and since \(q_j\) is irreducible and \(p_m\) is a non-unit, there exists a unit \(w\) such that \(q_j=wp_m\). Since \(A\) is a domain, canceling \(p_m\) from both sides and applying the induction hypothesis yields that, after a suitable rearrangement, each \(p_i\) and \(q_i\) are associates. Therefore \(A\) is a UFD. ([Ring Theory] §Integral Domains, ⁋Definition 16)

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References

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

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