Commutative Algebra/Primary decomposition or Lasker–Noether theory

The following theory was originally developed by world chess champion 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 Emmy Noether to commutative rings satisfying the ascending chain condition (noetherian rings), in her revolutionary 1921 paper "Idealtheorie in Ringbereichen".

Primary ideals

Clearly, every prime ideal is primary.

We have the following characterisations:

Theorem 19.5 (characterisations of primary ideals):

Let $q\le R$, with $r(q)$ denoting the radical ideal of $q$. The following are equivalent:

1. $q$ is primary.
2. If $xy\in q$, then either $x\in q$ or $y\in q$ or $x\in r(q)\wedge y\in r(q)$.
3. Every zerodivisor of $R/q$ is nilpotent.

Proof 1:

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

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

3. $\Rightarrow$ 1.: Let $xy\in q$. Then either $x\in q$ or $y\in q$ or $x+q$ is a zerodivisor within $R/q$, which is why $x^k+q=0$ for a suitable $k$.$◻$

Proof 2:

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

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

2. $\Rightarrow$ 1.: Let $xy\in q$. Assume not $x\in q$ and not $y\in q$. Then in particular $y\in r(q)$, that is, $y^k\in q$ for suitable $k$.$◻$

Theorem 19.6:

If $q\le R$ is any primary ideal, then $r(q)$ is prime.

Proof:

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

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 $I\le R$ 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 $I$ which is not primary. Since $I$ is not primary, there exist $x,y\in R$ such that $xy\in I$, but neither $x\in I$ nor $y^n\in I$ for any $n\in \mathbb{N}$. We form the ascending chain of ideals

$(I:y)\subseteq (I:y^2)\subseteq (I:y^3)\subseteq \cdots$;

this chain is ascending because $r{y}^n\in I\Rightarrow r{y}^{n+1}\in I$. Since we are in a Noetherian ring, this chain eventually stabilizes at some $m\in \mathbb{N}$; that is, for $k\ge m$ we have $(I:y^k)=(I:y^{k+1})$. We now claim that

$I=(I+⟨x⟩)\cap (I+y^{n}R)$.

Indeed, $\subseteq$ is obvious, and for $\supseteq$ we note that if $r\in (I+⟨x⟩)\cap (I+y^{n}R)$, then

$r=i+sx=j+y^{n}t$

for suitable $i,j\in I$ and $s,t\in R$, which is why $sx-y^{n}t\in I$, hence $sxy-y^{n+1}t\in I$, since $xy\in I$ thus $y^{n+1}t\in I$, $t\in (I:y^{n+1})=(I:y^n)$, hence $y^{n}t\in I$ and $sx\in I$. Therefore $r\in I$.

Furthermore, by the choice of $x$ and $y$ both $I+⟨x⟩$ and $I+y^{n}R$ are proper superideals, contradicting the irreducibility of $I$.$◻$

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

$I_1⊊I_2⊊\cdots$,

this chain has an upper bound, since it stabilizes as we are in a Noetherian ring. We may hence choose a maximal element $I$ among all ideals that are not the finite intersection of irreducible ideals. $I$ itself is thus not irreducible. Hence, it can be written as the intersection of strict superideals; that is

$I=J_1\cap \cdots \cap J_n$

for appropriate $J_i⊋I$. Since $I$ is maximal, each $J_i$ is a finite intersection of irreducible ideals, and hence so is $I$, which contradicts the choice of $I$.$◻$

Proof:

Combine lemmas 19.8 and 19.9.$◻$

Minimal decomposition

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 $I_1,\dots ,I_n$ be ideals. Then

$r\left(\bigcap _{j=1}^n{I}_j\right)=\bigcap _{j=1}^{n}r(I_j)$.

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

Proof:

$\Rightarrow$:

$\begin{array}{cc}s\in r\left(\bigcap _{j=1}^n{I}_j\right)& \iff \mathrm{\exists }k\in \mathbb{N}:s^k\in \bigcap _{j=1}^n{I}_j\\ & \iff \mathrm{\exists }k\in \mathbb{N}:\mathrm{\forall }j\in \{1,\dots ,n\}:s^k\in I_j\\ & \Rightarrow \mathrm{\forall }j\in \{1,\dots ,n\}:\mathrm{\exists }k\in \mathbb{N}:s^k\in I_j\\ & \iff s\in \bigcap _{j=1}^{n}r(I_j).\end{array}$

$\Leftarrow$: Let $s\in \bigcap _{j=1}^{n}r(I_j)$. For each $j$, choose $k_j$ such that $s^{k_j}\in I_j$. Set

$k:=\max \{k_1,\dots ,k_n\}$.

Then $s^k\in \bigcap _{j=1}^n{I}_j$, hence $s\in r\left(\bigcap _{j=1}^n{I}_j\right)$.$◻$

Template:DABE Note that for infinite intersections, the lemma need not (!!!) be true.

Proof 1:

First of all, we may exclude all primary ideals $q_t$ for which

$q_t\supseteq \bigcap _{\genfrac{}{}{0ex}{}{s=1}{s\ne t}}^r{q}_s$;

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

$I=\bigcap _{j=1}^n{q}_j$,

and for a fixed prime ideal $p$ set

$q_p:=\bigcap _{r(q_j)=p}{q}_j$;

due to theorem 19.6,

$I=\bigcap _{p\le R\text{ prime}}{q}_p$.

We claim that $q_p$ is primary, and $r(q_p)=p$. For the first claim, note that by the previous lemma

$r\left(\bigcap _{r(q_j)=p}{q}_j\right)=\bigcap _{r(q_j)=p}r(q_j)=p$.

For the second claim, let $xy\in q_p$. If $x\in q_p$ there is nothing to prove. Otherwise let $x\notin q_p$. Then there exists $q_l$ such that $x\notin q_l$, and hence $y^k\in q_l$ for a suitable $k$. Thus $y\in p$, and hence $y^{k_j}\in q_j$ for all $j$ and suitable $k_j$. Pick

$m:=\max \{k_j|r(q_j)=p\}$.

Then $y^m\in q_p$. Hence, $q_p$ is primary.$◻$

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.

Proof:

We begin by deducing an equation. According to theorem 19.2 and lemma 19.12,

$r((I:x))=r\left((\bigcap _{j=1}^n{q}_j:x)\right)=r(\bigcap _{j=1}^n(q_j:x))=\bigcap _{j=1}^{n}r((q_j:x))$.

Now we fix $q_j$ and distinguish a few cases.

1. If $x\in q_j$, then obviously $(q_j:x)=R$.
2. If $x\notin p_j$ (where again $p_j=r(q_j)$), then if $sx\in q_j$ we must have $s\in q_j$ since no power of $x$ is contained within $q_j$.
3. If $x\in p_j$, but $x\notin q_j$, we have $r((q_i:x))=p_i$, since
   $\begin{array}{cc}r\in r((q_i:x))& \iff \mathrm{\exists }k\in \mathbb{N}:r^{k}x\in q_i\\ & \iff \mathrm{\exists }k\in \mathbb{N}:\mathrm{\exists }m\in \mathbb{N}:(r^k{)}^m\in q_i\\ & \iff r\in p_i.\end{array}$

In conclusion, we find

$r((I:x))=\bigcap _{\genfrac{}{}{0ex}{}{j=1}{x\notin q_j}}^n{p}_j$.

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

Let now $p_j$ for $j\in \{1,\dots ,n\}$ be given. Since the given primary decomposition is minimal, we find $x\in R$ such that $x\notin p_j$, but $x\in \bigcap _{\genfrac{}{}{0ex}{}{l=1}{l\ne j}}^n{p}_l$. In this case, $r((I:x))=p_j$ by the above equation.$◻$

This theorem motivates and enables the following definition:

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

Lemma 19.16:

Let $I\le R$ be an ideal which has a primary decomposition

$I=\bigcap _{j=1}^n{q}_j$,

and let again $p_j:=r(q_j)$ for all $j$. If we define

${q_j}^{\prime }:=\{x\in R|(I:x)\subseteq ̸p_j\}$,

then ${q_j}^{\prime }$ is an ideal of $R$ and ${q_j}^{\prime }\subseteq q_j$.

Proof:

Let $a,b\in {q_j}^{\prime }$. There exists $c\notin p_j$ such that $c⟨a⟩\subseteq I$ without $c\in p_j$, and a similar $d$ with an analogous property in regard to $b$. Hence $cd⟨a-b⟩\subseteq I$, but not $cd\in p_j$ since $p_j$ is prime. Also, $cd⟨ab⟩\subseteq I$. Hence, we have an ideal.

Let $x\in {q_j}^{\prime }$. There exists $c\notin p_j$ such that

$c⟨x⟩\subseteq I=\bigcap _{j=1}^n{q}_j$.

In particular, $cx\in q_i$. Since no power of $c$ is in $q_i$, $x\in q_i$.$◻$

Lemma 19.17:

Let $S\subseteq R$ be multiplicatively closed, and let

${\pi }_S:R\to S^{-1}R,r\mapsto r/1$

be the canonical morphism. Let $I$ be a decomposable ideal, that is

$I=\bigcap _{j=1}^n{q}_j$

for primary $q_j$, and number the $q_j$ such that the first $r$ $q_j$ have empty intersection with $S$, and the others nonempty intersection. Then

${\pi }_S^{-1}\circ {\pi }_S(I)=\bigcap _{j=1}^r{q}_r$.

Proof:

We have

${\pi }_S(I)={\pi }_S\left(\bigcap _{j=1}^n{q}_j\right)=\bigcap _{j=1}^n{\pi }_S(q_j)$

by theorem 9.?. If now $S\cap q_j\ne \mathrm{\varnothing }$, lemma 9.? yields ${\pi }_S(q_j)=S^{-1}R$. Hence,

${\pi }_S(I)=\bigcap _{j=1}^n{\pi }_S(q_j)=\bigcap _{j=1}^r{\pi }_S(q_j)$.

Application of ${\pi }_S^{-1}$ on both sides yields

${\pi }_S^{-1}\circ {\pi }_S(I)=\bigcap _{j=1}^r{\pi }_S^{-1}{\pi }_S(q_j)$,

and

${\pi }_S^{-1}{\pi }_S(q_j)=q_j$

since $\supseteq$ holds for general maps, and $x\in {\pi }_S^{-1}{\pi }_S(q_j)$ means ${\pi }_S(x)=r/s$, where $r\in q_j$ and $s\in S$; thus ${\pi }_S(sx)=r/1$, that is $(sx)/1=r/1$. This means that

$\mathrm{\exists }t\in S:tsx=tr$.

Hence $tsx\in q_j$, and since no power of $ts$ is in $q_j$ ($S$ is multiplicatively closed and $S\cap q_j=\mathrm{\varnothing }$), $x\in q_j$.$◻$

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

Proof 1 (using lemma 19.16):

We first reduce the theorem down to the case where $P$ is the set of all prime subideals belonging to $I$ of a prime ideal that belongs to $I$. Let $P$ be any reduced system. For each maximal element of that set $p_r$ (w.r.t. inclusion) define $P_r$ to be the set of all ideals in $P$ contained in $p_r$. Since $P$ is finite,

$P=\bigcup _{p_r\text{ maximal in }P}{P}_r$;

this need not be a disjoint union (note that these are not maximal ideals!). Hence

$\bigcap _{l=1}^k{q}_{i_l}=\bigcap _{p_r\text{ maximal in }P}\bigcap _{p_j\in P_r}{q}_j$.

Hence, let $p$ be an ideal belonging to $I$ and let $P=\{p_{i_1},\dots ,p_{i_k}\}$ be an isolated system of subideals of $p$. Let $p_{j_1},\dots ,p_{j_m}$ be all the primary ideals belonging to $I$ not in $P$. For those ideals, we have $p_{j_l}\subseteq ̸p$, and hence we find $b_{j_l}\in p_{j_l}\setminus p$. For each $l$ take $s(l)\in \mathbb{N}$ large enough so that $b_{j_l}^{s(l)}\in q_{j_l}$. Then

$b:=\prod _{l=1}^m{b}_{j_l}^{s(l)}\in \bigcap _{l=1}^m{q}_{j_l}$,

which is why $b\bigcap _{p_t\in P}{q}_t\subseteq I$. From this follows that

$\bigcap _{p_t\in P}{q}_t\subseteq q^{\prime }$,

where $q$ is the element in the primary decomposition of $I$ to which $p$ is associated, since clearly for each element $x$ of the left hand side, $bx\in I$ and thus $b⟨x⟩\subseteq I$, but also $b\notin p$. But on the other hand, $p_t\subseteq p$ implies $q\subseteq q_t$. Hence for any such $t$ lemma 19.16 implies

$q^{\prime }\subseteq q\subseteq q_t$,

which in turn implies

$\bigcap _{p_t\in P}{q}_t\supseteq q^{\prime }$.$◻$

Proof 2 (using lemma 19.17):

Let $\{p_{i_1},\dots ,p_{i_k}\}$ be an isolated system of prime ideals belonging to $I$. Pick

$S:=R\setminus (p_{i_1}\cup \dots \cup p_{i_k})=(R\setminus p_{i_1})\cap \cdots \cap (R\setminus p_{i_k})$,

which is multiplicatively closed since it's the intersection of multiplicatively closed subsets. The primary ideals of the decomposition of $I$ which correspond to the $p_{i_l}$ are precisely those having empty intersection with $S$, since any other primary ideal $q_j$ in the decomposition of $I$ must contain an element outside all $\{p_{i_1},\dots ,p_{i_k}\}$, since otherwise its radical would be one of them by isolatedness. Hence, lemma 19.17 gives

$\bigcap _{l=1}^k{q}_{i_l}={\pi }_S^{-1}\circ {\pi }_S(I)$

and we have independence of the particular decomposition.$◻$

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.

Proof:

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

Indeed, assume none of the $I_1,\dots ,I_n$ is contained in $p$. Choose $r_j\in I_j\setminus p$ for $j=1,\dots ,n$. Since $p$ is prime, $r_1\cdots r_n\notin p$. But it's in the product, contradiction.$◻$

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

Lemma 19.21:

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

Proof:

If $x^n\in q$, then $x\in p$.$◻$

Proof:

The first assertion follows from proposition 19.20 and lemma 19.21. The second assertion follows since any prime ideal belonging to $I$ contains $I$.$◻$