4 Primary Decomposition

4.8 不讲，10.21 用 4.8，但不讲10.2！
4.9-4.11 不讲。 包含4.1-4.7 + ch7 part2.

Definition

We call $q⊊A$ as a primary ideal, if all zero-divisors of $A/q$ are nilpotent.

TFAE:

• $q⊊A$ primary
• for any $x,y\in q$, $x\in q$ or $y^n\in q$ for some $n>0$
• for any $xy\in q$, $y\in q$ or $x^n\in q$ for some $n>0$
• for any $xy\in q$, one of $x,y\in q$, or, both $x^n,y^n\in q$ for some $n>0$.

\begin{proof} i)←>iv) is easy. Note that $xy\in q$ iff $\overline{xy}=0$ in $A/q$.

The rest is trivial. \end{proof}

Examples.

• Any prime ideal is primary.
• For a ring homomorphism $f:A\to B$ with $q\subseteq B$ primary, then $q^c:f^{-1}(q)\subseteq A$ is primary. (Consider $A/q^c\hookrightarrow B/q$. )

Proposition

Let $q\subseteq A$ be a primary ideal. Then $r(q)=\text{smallest prime containing }q$. In this case, if $p=r(q)$, then we say $q$ is $p$-primary.

\begin{proof} Recall that $r(q)={\cap }_{p\supseteq q}p$, so it suffices to show $r(q)$ is a prime. If $xy\in r(q)$, then $x^n{y}^n\in q$. By definition, $x^{n{n}^{\prime }}\in q$ or $y^{n{n}^{\prime }}\in q$. Then one of $x$, $y\in r(q)$ and so $r(q)$ is prime. \end{proof}

Examples.

• $(p^n)\subseteq \mathbb{Z}$ is $(p)$-primary.
• A primary ideal is not necessary powers of prime ideal.
  • $q=(x,y^2)\subseteq A=k[x,y]$. Then $A/q=k[y]/(y^2)$ and $q$ is primary. Note that $r(q)=(x,y)$ is a maximal ideal of $A$ and $p^2⊊q⊊p$.
• A prime power is not necessary primary.
  • Consider $p=(\overline{x},\overline{z})\subseteq A=k[x,y,z]/(xy-z^2)$. Notice that $A/p\simeq k[xy,z]/(x,z,xy-z^2)\simeq k[y]$ and so $p$ is prime.
  • $p^2$ is not primary. Indeed, $\overline{x}\cdot \overline{y}={\overline{z}}^2\in p^2$. Since $y^n\notin p^2$ and $\overline{x},\overline{y}\notin p^2$, we know $p^2$ is not primary.

Proposition

Let $\alpha \subseteq A$ be an ideal. If $r(\alpha )$ is a maximal ideal, then $\alpha$ is a primary. In particular, if $m\subseteq A$ is a maximal ideal, then $m^n$ is primary.

\begin{proof} Note that $\alpha \subseteq r(\alpha )={\cap }_{p\supseteq \alpha }p=m$ is maximal. Then $m$ is the unique prime ideal containing $\alpha$, and so $m$ is the unique maximal ideal in $A/\alpha$. Hence $A/\alpha$ is a local ring. Suppose $xy\in \alpha$, if $x\notin m=r(\alpha )$, i.e., $x^n\notin \alpha$, then we aim to show $y\in \alpha$. Since $A/\alpha$ is local, $\overline{x}\in (A/\alpha {)}^{\times }$ and so $\overline{y}=0\in A/\alpha$. Thus, $y\in \alpha$.

For “in particular” part, because $r(m^n)\supseteq m$ and $m$ is maximal, one have $r(m^n)=m$ and so $m^n$ is primary. \end{proof}

Lemma

Let $q_i$ with $1⩽i⩽n$ be $p$-primary. Then $q={\cap }_{i=1}^n{q}_i$ is $p$-primary.

\begin{proof} First, $r(q)=\cap r(q^i)=p$. Now, let $xy\in q$. Suppose $x\notin q$, then $x\notin q_i$ for some $i$ and so $y^n\in q_i$ for some $n$. Then $y\in r(q_i)=p=r(q)$ and so $y^n\in q$. \end{proof}

Lemma

Suppose $q$ is $p$-primary. For $x\in A$, we have

• if $x\in q$, then $(q:x)=(1)=A$
• if $x\notin q$, then $(q:x)$ is $p$-primary
• if $x\notin p$, then $(q:x)=q$.

\begin{proof} i) is trivial.

iii) If $y\in (q:x)$, then $xy\in q$. Since $x\notin r(q)$, $y\in q$.

ii) Let $y\in (q:x)$, then $xy\in q$. Since $x\notin q$, there is $y\in r(q)=p$. Then $q\subseteq (q:x)\subseteq p$ and so $p=r(q)\subseteq r(q:x)\subseteq r(p)=p$. It remains to show $(q:x)$ is primary. For $yz\in (q:x)$ and $y\notin p$, we aim to show $z\in (q:x)$. Since $yz\in (q:x)$, we have $xyz\in q$. By $y\notin p$, $xy\in q$ and so $z\in (q:x)$. Now we finish the proof. \end{proof}

Definition

A primary decomposition of ideal $\alpha \subseteq A$ is an expression $\alpha ={\cap }_{i=1}^n{q}_i$ for finitely many primary ideal.

• If it exists, then we can apply ^188cbb and assume $r(q_i)=p_i$ are all distinct. $(\ast )$
• We can also “throw away” the big deals, that is, we can assume for all $i$, $q_i⊉{\cap }_{j\ne i}{q}_j$. $(\ast \ast )$

We call a decomposition satisfying $(\ast )$ and $(\ast \ast )$ a minimal decomposition.

Example. In $\mathbb{Z}$, $(n)=\cap (p_i^{n_i})$ is the unique minimal decomposition.

Remarks.

• possibly $\alpha$ has no primary decomposition
• could have more than one primary decomposition
• in theorem 7.13, $\alpha \subseteq A$ with Noetherian $A$, $\alpha$ has primary decomposition (not necessary unique)
• if $\alpha$ has a primary decomposition, then $n$ is unique. in theorem 4.5
• will see: primary decomposition is related with irreducible component of $\mathrm{Spec}(A/\alpha )$.

First uniqueness theorem

Suppose $\alpha \in A$ is decomposable, and suppose $\alpha ={\cap }_{i=1}^n{q}_i$ is a minimal decomposition. Let $p_i=r(q_i)$. Then

$$\{p_1,\cdots ,p_n\}=\{\text{prime ideals of form }r(\alpha :x)\text{ with some }x\in A\}.\text{\hspace{1em}(*)}$$

In particular, LHS is independent of the primary decomposition.

\begin{proof} Step 1. Show LHS $(\ast )\supseteq$ RHS $(\ast )$.

For $x\in A$, one have $(\alpha :x)={\cap }_{i=1}^n(q_i:x)$. It deduces $r(\alpha :x)={\cap }_{i=1}^{n}r(q_i:x)$. By ^2df6cb,

$${\cap }_{i=1}^{n}r(q_i:x)={\cap }_{x\notin q_j}{p}_j.$$

Now suppose $r(\alpha :x)\in \text{RHS}(\ast )$, that is, ${\cap }_{x\notin q_j}{p}_j=p$ prime, by ^tepoo1 $p=p_j$ for some $j$. Then $r(\alpha :x)=p=p_j\in \text{LHS}(\ast )$.

Step 2. Show LHS $(\ast )\subseteq$ RHS $(\ast )$.

Take $p_i\in \text{LHS}$. Suppose $\alpha ={\cap }_{i=1}^n{q}_i$ is a minimal decomposition. Then there exists $x\notin q_i$ such that $x={\cap }_{j\ne i}{q}_j$. By ^2df6cb, $r(\alpha :x)={\cap }_{i=1}^{n}r(q_i:x)=r(q_i:x)=p_i$ and so $p_i\in \text{RHS}$. \end{proof}

Definition

In ^3049a9, we call $\{p_1,\cdots ,p_n\}$ the associated primes of $\alpha$; or say $p_1,\cdots ,p_n$ belong to $\alpha$.

Example. $\alpha =(x^2,xy)=(x)\cap (x,y{)}^2$ is a minimal decomposition;

$A=k[x,y]=(x)\cap (x^2,y)$ is a minimal decomposition.

Associated primes are $\{(x),(xy)\}$.

Definition

In the associated primes $\{p_1,\cdots ,p_n\}$, a minimal element is called a minimal (or isolated) associated prime ideal. Others are called embedded associated prime ideals.

Proposition

Suppose $\alpha$ is decomposable. Then minimal associated prime ideals

\{p_1,\cdots,p_n\}\leftrightarrow\{\text{minimal elements in }\mathrm{Spec}(A/\alpha)\}\,,p_i\stackrel{\theta}{\mapsto} p_i/\alpha. $$ ^6e2c2b

\begin{proof} If $p$ is prime, then $p\supseteq \alpha =\cap q_i$ yields $p\supseteq r(\alpha )=\cap r(q_i)=\cap p_i$. By ^tepoo1, $p\supseteq p_i$ for some $i$. So given a minimal element $p/\alpha$ on RHS, $p\supseteq p_i$. By minimality, $p=p_i$ and $p_i$ has to be minimal in LHS. So $\theta$ is surjective.

In remains to show, if $p_i$ is minimal element in $\{p_1,\cdots ,p_n\}$, then $p_i/\alpha$ is a minimal element in $\mathrm{Spec}(A/\alpha )$. Otherwise, by Zorn’s lemma, there is a minimal element $p_i⊋p$ for some $p$ such that $p/\alpha$ is minimal in $\mathrm{Spec}(A/\alpha )$. But by Step 1, $p=p_j$ for some $j$ but $p_i⊋p_j$ leading to a contradiction. \end{proof}

Interlude: relation between primary decomposition and irreducible components

Recall that

$\mathrm{Spec}A$ is irreducible iff $\mathrm{Nil}(A)$ is a prime ideal.

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and an irreducible component is defined as a maximal irreducible subspace. In general, irreducible component of $\mathrm{Spec}A$ are $V(p)$‘s where $p$ is a minimal prime ideal.

If $\alpha$ is decomposable, then by ^6e2c2b, $\mathrm{Spec}(A/\alpha )$ has finitely many irreducible components.

Example. $k[x,y]\supseteq \alpha =(x^2,xy)=(x)\cap (x,y{)}^2$, then $(x)$ is the minimal associated prime ideal.

Remark. So above discussion, together with ^6e2c2b tells us, to compute irreducible components of $\mathrm{Spec}(A/\alpha )$, it suffices to find a primary decomposition of $\alpha$.

Remark. Consider the closed subspace $V(\alpha )\subseteq \mathrm{Spec}(A)$ with induced topology, also $\mathrm{Spec}(A/\alpha )$ with its own topology. Then the map

$$V(\alpha )\to \mathrm{Spec}(A/\alpha ),p\mapsto p/\alpha$$

is a homeomorphism. The proof is easy.

Proposition

Let $\alpha ={\cap }_{i=1}^n{q}_i$ be a minimal decomposition with $r(q_i)=p_i$. Then ${\cup }_{i=1}^n{p}_i=\{x\in A:(\alpha :x)\ne \alpha \}$. In particular, if $0$ is decomposable, then the set of zero divisors of $A$, denoted by $D$, is the union of associate prime s of $0$.

\begin{proof} $\alpha ={\cap }_{i=1}^n{q}_i$ iff $\overline{0}={\cap }_{i=1}^n\overline{q_i}$ in $A/\alpha$.

$(\alpha :x)\ne \alpha$ iff $(\overline{0}:\overline{x})\ne \overline{0}$ in $A/\alpha$.

So WLOG we can assume $\alpha =0$.

So $D={\cup }_{x\ne 0}(0:x)={\cup }_{x\ne 0}r(0:x)$ because $r(D)=D$. Since $0={\cap }_{i=1}^n{q}_i$, in the proof of ^3049a9, $r(0:x)={\cap }_{x\notin q_j}{p}_j\subseteq p_j$ for some $j$. Then $D\subseteq {\cup }_{i=1}^n{p}_i$.

Also by ^3049a9, each $p_i=r(0:x)$ with $x\notin q_i$, $x\notin {\cap }_{j\ne i}{p}_j$. Then ${\cup }_{i=1}^n{p}_i\subseteq D$ \end{proof}

Part 2 of Chapter 7: primary decomposition in Noetherian rings

Primary Decomposition

Definition

Say an ideal $\alpha$ is irreducible, if for any $\alpha =\beta \cap \gamma$, then $\alpha =\beta$ or $\alpha =\gamma$. Say $\alpha$ is reducible, if $\alpha$ can be written as $\alpha =\beta \cap \gamma$ with $\beta ,\gamma ⊋\alpha$.

Lemma

Let $A$ be a Noetherian ring. Then any ideal is a finite intersection of irreducible ideals, that is, for any ideal $\alpha$, $\alpha$ can be written as $\alpha ={\cap }_{i=1}^n{\beta }_i$.

\begin{proof} Let $S=\{\alpha \subseteq A:\alpha \text{ is not a finite intersection of irreducible ideals}\}$. If $S\ne \varnothing$, then by $A$ Noetherian $S$ has a maximal element $\alpha$. Since $\alpha$ is reducible, $\alpha$ can be written as $\alpha =\beta \cap \gamma$ and both $\beta ,\gamma$ are finite intersection of irreducible ideals, leading to contradiction. \end{proof}

Lemma

For a Noetherian ring $A$, an irreducible ideal is primary.

\begin{proof} Note that $\alpha \subseteq A$ is irreducible iff $\overline{0}\subseteq A/\alpha$ is irreducible, and $\alpha \subseteq A$ is primary iff $\overline{0}\subseteq A/\alpha$ is primary. So WLOG just consider the case where $0\subseteq A$ is an irreducible ideal. We aim to show $0$ irreducible yields $0$ primary. Suppose $xy=0$ and $y\ne 0$, it is enough to show $x^n=0$ for some $n$. Consider the chain

$$\mathrm{Ann}(x)\subseteq \mathrm{Ann}(x^2)\subseteq \cdots .$$

As $A$ Noetherian, there exists $n$ such that $\mathrm{Ann}(x^n)=\mathrm{Ann}(x^{n+1})=\cdots$.

We claim that $(x^n)\cap (y)=(0)$. For any $a\in (x^n)\cap (y)$, we have $ax=0$ by $xy=0$ and $a=x^{n}b$ for some $b\in A$. It deduces that $b{x}^{n+1}=0$ and $b\in \mathrm{Ann}(x^{n+1})=\mathrm{Ann}(x^n)$ and so $a=x^{n}b=0$. Now we prove the claim.

Since $(0)$ is irreducible and $(y)\ne 0$, we have $(x^n)=(0)$ and so we finish the proof. \end{proof}

Theorem

A Noetherian proper ideals have primary decomposition.

\begin{proof} By ^98d425 and ^3e8c56. \end{proof}

Proposition

Let $A$ be a Noetherian ring, and let $\alpha$ be an ideal of $A$. Then $\alpha \supseteq (r(\alpha ){)}^n$ for some $n$.

\begin{proof} Suppose $r(\alpha )$ is finitely generated by $x_1,\cdots ,x_k$ with $x_i^n\in \alpha$ for big enough $n$. Then $\alpha \supseteq (r(\alpha ){)}^{nk+1}$ and we have done. \end{proof}

Corollary

For a Noetherian ring $A$, then there exists $n$ such that $(\mathrm{Nil}(A){)}^n=0$.

\begin{proof} Take $\alpha =(0)$ in ^384015. \end{proof}

Corollary

Let $A$ be a Noetherian ring, and let $m\subseteq A$ be a maximal ideal. For an ideal $q\subseteq A$, TFAE:

• $q$ is $m$-primary
• $r(q)=m$
• $m^n\subseteq q\subseteq m$ for some $n$

\begin{proof} i)→ii) by definition.

ii)→i) By ^vsnd02.

ii)→iii) by ^384015.

iii)→ii) obvious. \end{proof}

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