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

Definition 19.4:

An ideal $q\leq R$ is called primary ideal if and only if the following holds:

$xy\in q\Rightarrow x\in q\vee \exists n\in \mathbb {N} :y^{n}\in q$.

Clearly, every prime ideal is primary.

We have the following characterisations:

Theorem 19.5 (characterisations of primary ideals):

Let $q\leq R$, 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 $k\geq 2$. Since $yx\in q$ and $y\notin q$, $x^{m}\in q$ for a suitable $m\geq 2$.

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 $k$.$\Box$

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 $k$.$\Box$

Theorem 19.6:

If $q\leq R$ 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 $y\in r(q)$.$\Box$

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

$I=(I+\langle x\rangle )\cap (I+y^{n}R)$.

Indeed, $\subseteq$ is obvious, and for $\supseteq$ we note that if $r\in (I+\langle x\rangle )\cap (I+y^{n}R)$, 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 $I+\langle x\rangle$ and $I+y^{n}R$ are proper superideals, contradicting the irreducibility of $I$.$\Box$

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

$I_{1}\subsetneq I_{2}\subsetneq \cdots$,

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 $J_{i}\supsetneq I$. Since $I$ is maximal, each $J_{i}$ is a finite intersection of irreducible ideals, and hence so is $I$, which contradicts the choice of $I$.$\Box$

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.$\Box$

Minimal decomposition

Definition 19.11:

Let $I\leq R$ be an ideal in a ring, and let

$I=\bigcap _{s=1}^{r}q_{s}$

be a primary decomposition of $I$. This decomposition is called minimal if and only if

1. there does not exist $t\in \{1,\ldots ,r\}$ with $q_{t}\supseteq \bigcap _{s=1 \atop s\neq t}^{r}q_{s}$, and
2. for all $i\neq j$, $r(q_{i})\neq r(q_{j})$ (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 $I_{1},\ldots ,I_{n}$ 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:

$\Rightarrow$:

${\begin{aligned}s\in r\left(\bigcap _{j=1}^{n}I_{j}\right)&\Leftrightarrow \exists k\in \mathbb {N} :s^{k}\in \bigcap _{j=1}^{n}I_{j}\\&\Leftrightarrow \exists k\in \mathbb {N} :\forall j\in \{1,\ldots ,n\}:s^{k}\in I_{j}\\&\Rightarrow \forall j\in \{1,\ldots ,n\}:\exists k\in \mathbb {N} :s^{k}\in I_{j}\\&\Leftrightarrow s\in \bigcap _{j=1}^{n}r(I_{j}).\end{aligned}}$

$\Leftarrow$: 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

$k:=\max\{k_{1},\ldots ,k_{n}\}$.

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

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

Theorem 19.13:

Let $I\leq R$ be an ideal in a ring that has a primary decomposition. Then $I$ also has a minimal primary decomposition.

Proof 1:

First of all, we may exclude all primary ideals $q_{t}$ for which

$q_{t}\supseteq \bigcap _{s=1 \atop s\neq t}^{r}q_{s}$;

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,

$I=\bigcap _{p\leq R{\text{ prime}}}q_{p}$.

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.$\Box$

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.

Theorem 19.14 (first uniqueness theorem):

Let $I\leq R$ be an ideal within a ring $r$, and assume we are given a minimal primary decomposition

$I=\bigcap _{j=1}^{n}q_{j}$.

Then the prime (theorem 19.6) ideals $p_{j}:=r(q_{j})$ are exactly the prime ideals among the ideals $r((I:x)),x\in R$ and hence are independent of the choice of the particular decomposition. That is, the ideals $p_{j}$ are uniquely determined by $I$.

Proof:

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

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

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
   ${\begin{aligned}r\in r((q_{i}:x))&\Leftrightarrow \exists k\in \mathbb {N} :r^{k}x\in q_{i}\\&\Leftrightarrow \exists k\in \mathbb {N} :\exists m\in \mathbb {N} :(r^{k})^{m}\in q_{i}\\&\Leftrightarrow r\in p_{i}.\end{aligned}}$

In conclusion, we find

$r((I:x))=\bigcap _{j=1 \atop x\notin q_{j}}^{n}p_{j}$.

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 $j\in \{1,\ldots ,n\}$ be given. Since the given primary decomposition is minimal, we find $x\in R$ such that $x\notin p_{j}$, but $x\in \bigcap _{l=1 \atop l\neq j}^{n}p_{l}$. In this case, $r((I:x))=p_{j}$ by the above equation.$\Box$

This theorem motivates and enables the following definition:

Definition 19.15:

Let $I$ be any ideal that has a minimal primary decomposition

$I=\bigcap _{j=1}^{n}q_{j}$.

Then the ideals $p_{j}:=r(q_{j})$ are called the prime ideals belonging to $I$.

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

Lemma 19.16:

Let $I\leq R$ 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

$q_{j}':=\{x\in R|(I:x)\not \subseteq p_{j}\}$,

then $q_{j}'$ is an ideal of $R$ and $q_{j}'\subseteq q_{j}$.

Proof:

Let $a,b\in q_{j}'$. There exists $c\notin p_{j}$ such that $c\langle a\rangle \subseteq I$ without $c\in p_{j}$, and a similar $d$ with an analogous property in regard to $b$. Hence $cd\langle a-b\rangle \subseteq I$, but not $cd\in p_{j}$ since $p_{j}$ is prime. Also, $cd\langle ab\rangle \subseteq I$. Hence, we have an ideal.

Let $x\in q_{j}'$. There exists $c\notin p_{j}$ such that

$c\langle x\rangle \subseteq I=\bigcap _{j=1}^{n}q_{j}$.

In particular, $cx\in q_{i}$. Since no power of $c$ is in $q_{i}$, $x\in q_{i}$.$\Box$

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 $S\cap q_{j}\neq \emptyset$, 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

$\exists t\in S:tsx=tr$.

Hence $tsx\in q_{j}$, and since no power of $ts$ is in $q_{j}$ ($S$ is multiplicatively closed and $S\cap q_{j}=\emptyset$), $x\in q_{j}$.$\Box$

Definition 19.18:

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

Theorem 19.19 (second uniqueness theorem):

Let $I$ be an ideal that has a minimal primary decomposition. If $P=\{p_{i_{1}},\ldots ,p_{i_{k}}\}$ is a subset of the set of the prime ideals belonging to $I$ which is isolated, then

$\bigcap _{l=1}^{k}q_{i_{l}}$

is independent of the particular minimal primary decomposition from which the $q_{i_{l}}$ 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 $p_{i}$ 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 $P=\{p_{i_{1}},\ldots ,p_{i_{k}}\}$ be an isolated system of subideals of $p$. Let $p_{j_{1}},\ldots ,p_{j_{m}}$ be all the primary ideals belonging to $I$ not in $P$. For those ideals, we have $p_{j_{l}}\not \subseteq p$, 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

$\bigcap _{p_{t}\in P}q_{t}\subseteq q'$,

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 $b\langle x\rangle \subseteq I$, 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

$q'\subseteq q\subseteq q_{t}$,

which in turn implies

$\bigcap _{p_{t}\in P}q_{t}\supseteq q'$.$\Box$

Proof 2 (using lemma 19.17):

Let $\{p_{i_{1}},\ldots ,p_{i_{k}}\}$ be an isolated system of prime ideals belonging to $I$. Pick

$S:=R\setminus (p_{i_{1}}\cup \ldots \cup p_{i_{k}})=(R\setminus p_{i_{1}})\cap \cdots \cap (R\setminus p_{i_{k}})$,

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 $\{p_{i_{1}},\ldots ,p_{i_{k}}\}$, 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.$\Box$

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.

Proposition 19.20:

Let $R$ be a (commutative) ring, and let $p\leq R$ be a prime ideal. If $p$ contains

either the intersection $\bigcap _{j=1}^{n}I_{j}$ or the product $I_{1}\cdot I_{2}\cdots I_{n}$

of certain arbitrary ideals, then it contains one of the $I_{j}$ completely.

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 $I_{1},\ldots ,I_{n}$ is contained in $p$. Choose $r_{j}\in I_{j}\setminus p$ for $j=1,\ldots ,n$. Since $p$ is prime, $r_{1}\cdots r_{n}\notin p$. But it's in the product, contradiction.$\Box$

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

Lemma 19.21:

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

Proof:

If $x^{n}\in q$, then $x\in p$.$\Box$

Corollary 19.22:

Let $I$ be an ideal admitting a prime decomposition

$I=\bigcap _{j=1}^{n}q_{j}$.

If $p\leq R$ is any prime ideal that contains $I$, then it also contains a prime ideal belonging to $I$. Further, the prime ideals belonging to $I$ are exactly those that are minimal with respect to the partial order induced by inclusion on $V(I)\subseteq \operatorname {Spec} (R)$.

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 $I$.$\Box$