View source for Commutative Algebra/Primary decomposition or Lasker–Noether theory

The following theory was originally developed by world chess champion [[w:Emmanuel Lasker|Emmanuel Lasker]] in his 1905 paper "Zur theorie der Moduln und Ideale" ("On the theory of modules and ideals") on polynomial rings, and then generalised by [[w:Emmy Noether|Emmy Noether]] to commutative rings satisfying the ascending chain condition (noetherian rings), in her revolutionary 1921 paper "Idealtheorie in Ringbereichen".

== Primary ideals ==

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 19.4''':

An ideal <math>q \le R</math> is called '''primary ideal''' if and only if the following holds:
:<math>xy \in q \Rightarrow x \in q \vee \exists n \in \mathbb N: y^n \in q</math>.}}

Clearly, every prime ideal is primary.

We have the following characterisations:

'''Theorem 19.5 (characterisations of primary ideals)''':

Let <math>q \le R</math>, with <math>r(q)</math> denoting the radical ideal of <math>q</math>. The following are equivalent:
#<math>q</math> is primary.
#If <math>xy \in q</math>, then either <math>x \in q</math> or <math>y \in q</math> or <math>x \in r(q) \wedge y \in r(q)</math>.
#Every zerodivisor of <math>R/q</math> is nilpotent.

'''Proof 1''':

1. <math>\Rightarrow</math> 2.: Let <math>q</math> be primary. Assume <math>xy \in q</math> and neither <math>x \in q</math> nor <math>y \in q</math>. Since <math>x \notin q</math>, <math>y^k \in q</math> for a suitable <math>k \ge 2</math>. Since <math>yx \in q</math> and <math>y \notin q</math>, <math>x^m \in q</math> for a suitable <math>m \ge 2</math>.

2. <math>\Rightarrow</math> 3.: Let <math>d + q</math> be a zerodivisor of <math>R/q</math>, that is, <math>cd \in q</math> for a certain <math>c \in R</math> such that <math>c \notin q</math>. Hence <math>d \in r(q)</math>, that is, <math>d^k \in q</math> for a suitable <math>k</math>.

3. <math>\Rightarrow</math> 1.: Let <math>xy \in q</math>. Then either <math>x \in q</math> or <math>y \in q</math> or <math>x + q</math> is a zerodivisor within <math>R/q</math>, which is why <math>x^k + q = 0</math> for a suitable <math>k</math>.{{QED}}

'''Proof 2''':

1. <math>\Rightarrow</math> 3.: Let <math>q</math> be primary, and let <math>x + q</math> be a zerodivisor within <math>R/q</math>. Then <math>xy \in q</math> for a <math>y \notin q</math> and hence <math>x^k \in q</math> for a suitable <math>k</math>.

3. <math>\Rightarrow</math> 2.: Let <math>xy \in q</math>. Assume neither <math>x \in q</math> nor <math>y \in q</math>. Then both <math>x + q</math> and <math>y + q</math> are zerodivisors in <math>R/q</math>, and hence are nilpotent, which is why <math>x^k, y^m \in q</math> for suitable <math>k, m</math> and hence <math>x, y \in r(q)</math>.

2. <math>\Rightarrow</math> 1.: Let <math>xy \in q</math>. Assume not <math>x \in q</math> and not <math>y \in q</math>. Then in particular <math>y \in r(q)</math>, that is, <math>y^k \in q</math> for suitable <math>k</math>.{{QED}}

'''Theorem 19.6''':

If <math>q \le R</math> is any primary ideal, then <math>r(q)</math> is prime.

'''Proof''':

Let <math>xy \in r(q)</math>. Then <math>(xy)^n \in q</math> for a suitable <math>n \in \mathbb N</math>. Hence either <math>x^n \in q</math> and thus <math>x \in r(q)</math> or <math>(y^n)^m \in q</math> for a suitable <math>m \in \mathbb N</math> and hence <math>y \in r(q)</math>.{{QED}}

== Existence ==

=== Existence in the Noetherian case ===

Following the exposition of Zariski, Samuel and Cohen, we deduce the classical Noetherian existence theorem from two lemmas and a definition.

'''Definition 19.7''':

An ideal <math>I \le R</math> is called '''irreducible''' if and only if it can not be written as the intersection of finitely many proper superideals.

'''Lemma 19.8''':

In a Noetherian ring, every irreducible ideal is primary.

'''Proof''':

Assume there exists an irreducible ideal <math>I</math> which is not primary. Since <math>I</math> is not primary, there exist <math>x, y \in R</math> such that <math>xy \in I</math>, but neither <math>x \in I</math> nor <math>y^n \in I</math> for any <math>n \in \mathbb N</math>. We form the ascending chain of ideals
:<math>(I:y) \subseteq (I:y^2) \subseteq (I:y^3) \subseteq \cdots</math>;
this chain is ascending because <math>ry^n \in I \Rightarrow ry^{n+1} \in I</math>. Since we are in a Noetherian ring, this chain eventually stabilizes at some <math>m \in \mathbb N</math>; that is, for <math>k \ge m</math> we have <math>(I:y^k) = (I:y^{k+1})</math>. We now claim that
:<math>I = (I + \langle x \rangle) \cap (I + y^n R)</math>.
Indeed, <math>\subseteq</math> is obvious, and for <math>\supseteq</math> we note that if <math>r \in (I + \langle x \rangle) \cap (I + y^n R)</math>, then
:<math>r = i + sx = j + y^n t</math>
for suitable <math>i, j \in I</math> and <math>s, t \in R</math>, which is why <math>sx - y^n t \in I</math>, hence <math>sxy - y^{n+1} t \in I</math>, since <math>xy \in I</math> thus <math>y^{n+1} t \in I</math>, <math>t \in (I:y^{n+1}) = (I:y^n)</math>, hence <math>y^n t \in I</math> and <math>sx \in I</math>. Therefore <math>r \in I</math>.

Furthermore, by the choice of <math>x</math> and <math>y</math> both <math>I + \langle x \rangle</math> and <math>I + y^n R</math> are proper superideals, contradicting the irreducibility of <math>I</math>.{{QED}}

'''Lemma 19.9''':

In a Noetherian ring, every ideal can be written as the finite intersection of irreducible ideals.

'''Proof''':

Assume otherwise. Consider the set of all ideals that are not the finite intersection of irreducible ideals. If we are given an ascending chain within that set
:<math>I_1 \subsetneq I_2 \subsetneq \cdots</math>,
this chain has an upper bound, since it stabilizes as we are in a Noetherian ring. We may hence choose a maximal element <math>I</math> among all ideals that are not the finite intersection of irreducible ideals. <math>I</math> itself is thus not irreducible. Hence, it can be written as the intersection of strict superideals; that is
:<math>I = J_1 \cap \cdots \cap J_n</math>
for appropriate <math>J_i \supsetneq I</math>. Since <math>I</math> is maximal, each <math>J_i</math> is a finite intersection of irreducible ideals, and hence so is <math>I</math>, which contradicts the choice of <math>I</math>.{{QED}}

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Corollary 19.10''':

In a Noetherian ring, every ideal can be written as the finite intersection of primary ideals.}}

'''Proof''':

Combine lemmas 19.8 and 19.9.{{QED}}

=== Minimal decomposition ===

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 19.11''':

Let <math>I \le R</math> be an ideal in a ring, and let
:<math>I = \bigcap_{s = 1}^r q_s</math>
be a primary decomposition of <math>I</math>. This decomposition is called '''minimal''' if and only if
#there does not exist <math>t \in \{1, \ldots, r\}</math> with <math>q_t \supseteq \bigcap_{s=1 \atop s \neq t}^r q_s</math>, and
#for all <math>i \neq j</math>, <math>r(q_i) \neq r(q_j)</math> (that is, the radicals of the prime ideals are pairwise distinct).}}

In fact, once we have a primary composition for a given ideal, we can find a minimal primary decomposition of that ideal. But before we prove that, we need a general fact about radicals first.

'''Lemma 19.12''':

Let <math>I_1, \ldots, I_n</math> be ideals. Then
:<math>r\left( \bigcap_{j=1}^n I_j \right) = \bigcap_{j=1}^n r(I_j)</math>.

One could phrase this lemma as "radical interchanges with finite intersections".

'''Proof''':

<math>\Rightarrow</math>:
:<math>\begin{align}
s \in r\left( \bigcap_{j=1}^n I_j \right) & \Leftrightarrow \exists k \in \mathbb N: s^k \in \bigcap_{j=1}^n I_j \\
& \Leftrightarrow \exists k \in \mathbb N: \forall j \in \{1, \ldots, n\}: s^k \in I_j \\
& \Rightarrow \forall j \in \{1, \ldots, n\}: \exists k \in \mathbb N: s^k \in I_j \\
& \Leftrightarrow s \in \bigcap_{j=1}^n r(I_j).
\end{align}</math>

<math>\Leftarrow</math>:
Let <math>s \in \bigcap_{j=1}^n r(I_j)</math>. For each <math>j</math>, choose <math>k_j</math> such that <math>s^{k_j} \in I_j</math>. Set
:<math>k := \max\{k_1, \ldots, k_n\}</math>.
Then <math>s^k \in \bigcap_{j=1}^n I_j</math>, hence <math>s \in r\left( \bigcap_{j=1}^n I_j \right)</math>.{{QED}}

{{DABE}} Note that for infinite intersections, the lemma need ''not'' (!!!) be true.

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Theorem 19.13''':

Let <math>I \le R</math> be an ideal in a ring that has a primary decomposition. Then <math>I</math> also has a minimal primary decomposition.}}

'''Proof 1''':

First of all, we may exclude all primary ideals <math>q_t</math> for which
:<math>q_t \supseteq \bigcap_{s=1 \atop s \neq t}^r q_s</math>;
the intersection won't change if we do that, for intersecting with a superset changes nothing in general.

Then assume we are given a decomposition
:<math>I = \bigcap_{j=1}^n q_j</math>,
and for a fixed prime ideal <math>p</math> set
:<math>q_p := \bigcap_{r(q_j) = p} q_j</math>;
due to theorem 19.6,
:<math>I = \bigcap_{p \le R \text{ prime}} q_p</math>.
We claim that <math>q_p</math> is primary, and <math>r(q_p) = p</math>. For the first claim, note that by the previous lemma
:<math>r \left( \bigcap_{r(q_j) = p} q_j \right) = \bigcap_{r(q_j) = p} r(q_j) = p</math>.
For the second claim, let <math>xy \in q_p</math>. If <math>x \in q_p</math> there is nothing to prove. Otherwise let <math>x \notin q_p</math>. Then there exists <math>q_l</math> such that <math>x \notin q_l</math>, and hence <math>y^k \in q_l</math> for a suitable <math>k</math>. Thus <math>y \in p</math>, and hence <math>y^{k_j} \in q_j</math> for all <math>j</math> and suitable <math>k_j</math>. Pick
:<math>m:=\max\{k_j | r(q_j) = p\}</math>.
Then <math>y^m \in q_p</math>. Hence, <math>q_p</math> is primary.{{QED}}

== Uniqueness properties ==

In general, we don't have uniqueness for primary decompositions, but still, any two primary decompositions of the same ideal in a ring look somewhat similar. The classical first and second uniqueness theorems uncover some of these similarities.

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Theorem 19.14 (first uniqueness theorem)''':

Let <math>I \le R</math> be an ideal within a ring <math>r</math>, and assume we are given a ''minimal'' primary decomposition
:<math>I = \bigcap_{j=1}^n q_j</math>.
Then the prime (theorem 19.6) ideals <math>p_j := r(q_j)</math> are exactly the prime ideals among the ideals <math>r((I: x)), x \in R</math> and hence are independent of the choice of the particular decomposition. That is, the ideals <math>p_j</math> are uniquely determined by <math>I</math>.}}

'''Proof''':

We begin by deducing an equation. According to theorem 19.2 and lemma 19.12,
:<math>r((I: x)) = r \left( \left( \bigcap_{j=1}^n q_j : x \right) \right) = r \left( \bigcap_{j=1}^n (q_j : x) \right) = \bigcap_{j=1}^n r((q_j : x))</math>.
Now we fix <math>q_j</math> and distinguish a few cases.
#If <math>x \in q_j</math>, then obviously <math>(q_j:x) = R</math>.
#If <math>x \notin p_j</math> (where again <math>p_j = r(q_j)</math>), then if <math>sx \in q_j</math> we must have <math>s \in q_j</math> since no power of <math>x</math> is contained within <math>q_j</math>.
#If <math>x \in p_j</math>, but <math>x \notin q_j</math>, we have <math>r((q_i:x)) = p_i</math>, since<br /><math>\begin{align}
r \in r((q_i:x)) & \Leftrightarrow \exists k \in \mathbb N: r^k x \in q_i \\
& \Leftrightarrow \exists k \in \mathbb N: \exists m \in \mathbb N: (r^k)^m \in q_i \\
& \Leftrightarrow r \in p_i.
\end{align}</math>
In conclusion, we find
:<math>r((I: x)) = \bigcap_{j=1 \atop x \notin q_j}^n p_j</math>.

Assume first that <math>r((I:x))</math> is prime. Then the prime avoidance lemma implies that <math>r((I:x))</math> is contained within one of the <math>p_j</math>, <math>x \notin q_j</math>, and since <math>p_j = r((q_j,x)) \subseteq r((I:x))</math>, <math>r((I:x)) = p_j</math>.

Let now <math>p_j</math> for <math>j \in \{1, \ldots, n\}</math> be given. Since the given primary decomposition is minimal, we find <math>x \in R</math> such that <math>x \notin p_j</math>, but <math>x \in \bigcap_{l=1 \atop l \neq j}^n p_l</math>. In this case, <math>r((I:x)) = p_j</math> by the above equation.{{QED}}

This theorem motivates and enables the following definition:

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 19.15''':

Let <math>I</math> be any ideal that has a minimal primary decomposition
:<math>I = \bigcap_{j=1}^n q_j</math>.
Then the ideals <math>p_j := r(q_j)</math> are called the '''prime ideals belonging''' to <math>I</math>.}}

We now prove two lemmas, each of which will below yield a proof of the second uniqueness theorem (see below).

'''Lemma 19.16''':

Let <math>I \le R</math> be an ideal which has a primary decomposition
:<math>I = \bigcap_{j=1}^n q_j</math>,
and let again <math>p_j := r(q_j)</math> for all <math>j</math>. If we define
:<math>q_j' := \{x \in R | (I:x) \not\subseteq p_j \}</math>,
then <math>q_j'</math> is an ideal of <math>R</math> and <math>q_j' \subseteq q_j</math>.

'''Proof''':

Let <math>a, b \in  q_j'</math>. There exists <math>c \notin p_j</math> such that <math>c \langle a \rangle \subseteq I</math> without <math>c \in p_j</math>, and a similar <math>d</math> with an analogous property in regard to <math>b</math>. Hence <math>cd \langle a-b \rangle \subseteq I</math>, but not <math>cd \in p_j</math> since <math>p_j</math> is prime. Also, <math>cd \langle ab \rangle \subseteq I</math>. Hence, we have an ideal.

Let <math>x \in q_j'</math>. There exists <math>c \notin p_j</math> such that
:<math>c \langle x \rangle \subseteq I = \bigcap_{j=1}^n q_j</math>.
In particular, <math>c x \in q_i</math>. Since no power of <math>c</math> is in <math>q_i</math>, <math>x \in q_i</math>.{{QED}}

'''Lemma 19.17''':

Let <math>S \subseteq R</math> be multiplicatively closed, and let
:<math>\pi_S: R \to S^{-1}R, r \mapsto r/1</math>
be the canonical morphism. Let <math>I</math> be a decomposable ideal, that is
:<math>I = \bigcap_{j=1}^n q_j</math>
for primary <math>q_j</math>, and number the <math>q_j</math> such that the first <math>r</math> <math>q_j</math> have empty intersection with <math>S</math>, and the others nonempty intersection. Then
:<math>\pi_S^{-1} \circ \pi_S (I) = \bigcap_{j=1}^r q_r</math>.

'''Proof''':

We have
:<math>\pi_S(I) = \pi_S \left( \bigcap_{j=1}^n q_j \right) = \bigcap_{j=1}^n \pi_S (q_j)</math>
by theorem 9.?. If now <math>S \cap q_j \neq \emptyset</math>, lemma 9.? yields <math>\pi_S(q_j) = S^{-1} R</math>. Hence,
:<math>\pi_S(I) = \bigcap_{j=1}^n \pi_S (q_j) = \bigcap_{j=1}^r \pi_S (q_j)</math>.
Application of <math>\pi_S^{-1}</math> on both sides yields
:<math>\pi_S^{-1} \circ \pi_S (I) = \bigcap_{j=1}^r \pi_S^{-1} \pi_S (q_j)</math>,
and
:<math>\pi_S^{-1} \pi_S (q_j) = q_j</math>
since <math>\supseteq</math> holds for general maps, and <math>x \in \pi_S^{-1} \pi_S (q_j)</math> means <math>\pi_S(x) = r/s</math>, where <math>r \in q_j</math> and <math>s \in S</math>; thus <math>\pi_S(sx) = r/1</math>, that is <math>(sx)/1 = r/1</math>. This means that
:<math>\exists t \in S: tsx = tr</math>.
Hence <math>tsx \in q_j</math>, and since no power of <math>ts</math> is in <math>q_j</math> (<math>S</math> is multiplicatively closed and <math>S \cap q_j = \emptyset</math>), <math>x \in q_j</math>.{{QED}}

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 19.18''':

Let <math>I</math> be an ideal which admits a primary decomposition, and let <math>P \subseteq \operatorname{Spec} R</math> be a set of prime ideals of <math>R</math> that all belong to <math>I</math>. <math>P</math> is called '''isolated''' if and only if for every prime ideal <math>p \in P</math>, if <math>p'</math> is a prime ideal belonging to <math>I</math> such that <math>p' \subseteq p</math>, then <math>p' \in P</math> as well.}}

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Theorem 19.19 (second uniqueness theorem)''':

Let <math>I</math> be an ideal that has a minimal primary decomposition. If <math>P = \{p_{i_1}, \ldots, p_{i_k}\}</math> is a subset of the set of the prime ideals belonging to <math>I</math> which is isolated, then
:<math>\bigcap_{l=1}^k q_{i_l}</math>
is independent of the particular minimal primary decomposition from which the <math>q_{i_l}</math> are coming.}}

Note that applied to reduced sets consisting of only one prime ideal, this means that if all prime subideals of a prime ideal <math>p_i</math> belonging to <math>I</math> also belong to <math>I</math>, then the corresponding <math>q_i</math> is predetermined.

'''Proof 1 (using lemma 19.16)''':

We first reduce the theorem down to the case where <math>P</math> is the set of all prime subideals belonging to <math>I</math> of a prime ideal that belongs to <math>I</math>. Let <math>P</math> be any reduced system. For each maximal element of that set <math>p_r</math> (w.r.t. inclusion) define <math>P_r</math> to be the set of all ideals in <math>P</math> contained in <math>p_r</math>. Since <math>P</math> is finite,
:<math>P = \bigcup_{p_r \text{ maximal in } P} P_r</math>;
this need not be a disjoint union (note that these are not maximal ''ideals''!). Hence
:<math>\bigcap_{l=1}^k q_{i_l} = \bigcap_{p_r \text{ maximal in } P} \bigcap_{p_j \in P_r} q_j</math>.

Hence, let <math>p</math> be an ideal belonging to <math>I</math> and let <math>P = \{p_{i_1}, \ldots, p_{i_k}\}</math> be an isolated system of subideals of <math>p</math>. Let <math>p_{j_1}, \ldots, p_{j_m}</math> be all the primary ideals belonging to <math>I</math> not in <math>P</math>. For those ideals, we have <math>p_{j_l} \not\subseteq p</math>, and hence we find <math>b_{j_l} \in p_{j_l} \setminus p</math>. For each <math>l</math> take <math>s(l) \in \mathbb N</math> large enough so that <math>b_{j_l}^{s(l)} \in q_{j_l}</math>. Then
:<math>b := \prod_{l=1}^m b_{j_l}^{s(l)} \in \bigcap_{l=1}^m q_{j_l}</math>,
which is why
<math>b \bigcap_{p_t \in P} q_t \subseteq I</math>.
From this follows that
:<math>\bigcap_{p_t \in P} q_t \subseteq q'</math>,
where <math>q</math> is the element in the primary decomposition of <math>I</math> to which <math>p</math> is associated, since clearly for each element <math>x</math> of the left hand side, <math>bx \in I</math> and thus <math>b \langle x \rangle \subseteq I</math>, but also <math>b \notin p</math>. But on the other hand, <math>p_t \subseteq p</math> implies <math>q \subseteq q_t</math>. Hence for any such <math>t</math> lemma 19.16 implies
:<math>q' \subseteq q \subseteq q_t</math>,
which in turn implies
:<math>\bigcap_{p_t \in P} q_t \supseteq q'</math>.{{QED}}

'''Proof 2 (using lemma 19.17)''':

Let <math>\{p_{i_1}, \ldots, p_{i_k}\}</math> be an isolated system of prime ideals belonging to <math>I</math>. Pick
:<math>S := R \setminus (p_{i_1} \cup \ldots \cup p_{i_k}) = (R \setminus p_{i_1}) \cap \cdots \cap (R \setminus p_{i_k})</math>,
which is multiplicatively closed since it's the intersection of multiplicatively closed subsets. The primary ideals of the decomposition of <math>I</math> which correspond to the <math>p_{i_l}</math> are precisely those having empty intersection with <math>S</math>, since any other primary ideal <math>q_j</math> in the decomposition of <math>I</math> must contain an element outside all <math>\{p_{i_1}, \ldots, p_{i_k}\}</math>, since otherwise its radical would be one of them by isolatedness. Hence, lemma 19.17 gives
:<math>\bigcap_{l=1}^k q_{i_l} = \pi_S^{-1} \circ \pi_S (I)</math>
and we have independence of the particular decomposition.{{QED}}

== Characterisation of prime ideals belonging to an ideal ==

The following are useful further theorems on primary decomposition.

First of all, we give a proposition on general prime ideals.

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Proposition 19.20''':

Let <math>R</math> be a (commutative) ring, and let <math>p \le R</math> be a prime ideal. If <math>p</math> contains
:either the intersection <math>\bigcap_{j=1}^n I_j</math> or the product <math>I_1 \cdot I_2 \cdots I_n</math>
of certain arbitrary ideals, then it contains one of the <math>I_j</math> ''completely''.}}

'''Proof''':

Since the product is contained in the intersection, it suffices to prove the theorem under the assumption that <math>I_1 \cdots I_n \subseteq p</math>.

Indeed, assume none of the <math>I_1, \ldots, I_n</math> is contained in <math>p</math>. Choose <math>r_j \in I_j \setminus p</math> for <math>j = 1, \ldots, n</math>. Since <math>p</math> is prime, <math>r_1 \cdots r_n \notin p</math>. But it's in the product, contradiction.{{QED}}

This proposition has far-reaching consequences for primary decomposition, given in Corollary 19.22. But first, we need a lemma.

'''Lemma 19.21''':

Let <math>q \le R</math> be a primary ideal, and assume <math>p \le R</math> is prime such that <math>q \subseteq p</math>. Then <math>r(q) \subseteq p</math>.

'''Proof''':

If <math>x^n \in q</math>, then <math>x \in p</math>.{{QED}}

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Corollary 19.22''':

Let <math>I</math> be an ideal admitting a prime decomposition
:<math>I = \bigcap_{j=1}^n q_j</math>.
If <math>p \le R</math> is any prime ideal that contains <math>I</math>, then it also contains a prime ideal belonging to <math>I</math>. Further, the prime ideals belonging to <math>I</math> are exactly those that are minimal with respect to the partial order induced by inclusion on <math>V(I) \subseteq \operatorname{Spec}(R)</math>.}}

'''Proof''':

The first assertion follows from proposition 19.20 and lemma 19.21. The second assertion follows since any prime ideal belonging to <math>I</math> contains <math>I</math>.{{QED}}

{{BookCat}}