Atiyah 第四章习题解答

Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 4.

Primary Decomposition

Exercise 1. If an ideal $\mathfrak{a}$ has a primary decomposition, then $\mathrm{Spec}(A/\mathfrak{a})$ has only finitely many irreducible components.

Proof. If $\mathfrak{a}$ has a primary decomposition, then there are finitely many minimal prime ideals containing $\mathfrak{a}$, so there are finitely many minimal prime ideals in $A/\mathfrak{a}$. Thus $\mathrm{Spec}(A/\mathfrak{a})$ has finitely many irreducible components.

$\square$

Exercise 2. If $\mathfrak{a}=r(\mathfrak{a})$, then $\mathfrak{a}$ has no embedded prime ideals.

Proof. If $\mathfrak{a}={\bigcap }_{i=1}^m{\mathfrak{q}}_i$ is a minimal primary decomposition of $\mathfrak{a}$, and ${\mathfrak{q}}_i$ is ${\mathfrak{p}}_i$-primary, then $$\mathfrak{a}=r(\mathfrak{a})=r\text{mleft}(\bigcap _{i=1}^m{\mathfrak{q}}_i\text{mright})=\bigcap _{i=1}^m{\mathfrak{p}}_i.$$

By 1st uniqueness theorem, ${\bigcup }_{i=1}^m{\mathfrak{p}}_i$ is also a minimal primary decomposition of $\mathfrak{a}$, so each ${\mathfrak{p}}_i$ must be minimal along ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$.

$\square$

Exercise 3. If $A$ is absolutely flat, every primary ideal is maximal.

Proof. If $\mathfrak{q}$ is primary, then every zero-divisor of $A/\mathfrak{q}$ is nilpotent. But $A$ is absolute flat, so $A/\mathfrak{q}$ is absolute flat, and every non-unit in it is a zero-divisor, hence nilpotent. This means $A/\mathfrak{q}$ has a unique maximal ideal (the nilradical of it), so it’s a local ring. By its absolute flatness, it’s a field. So $\mathfrak{q}$ is maximal.

$\square$

Exercise 4. In the polynomial ring $\mathbb{Z}[t]$, the ideal $\mathfrak{m}=(2,t)$ is maximal and the ideal $\mathfrak{q}=(4,t)$ is $\mathfrak{m}$-primary, but is not a power of $\mathfrak{m}$.

Proof. $\mathbb{Z}[t]/\mathfrak{m}=\mathbb{Z}/2\mathbb{Z}$ is field, so $\mathfrak{m}$ is maximal. $\mathbb{Z}[t]/\mathfrak{q}=\mathbb{Z}/4\mathbb{Z}$ has no zero-divisor other than nilpotent, so $\mathfrak{q}$ is primary. But ${\mathfrak{m}}^2⊊\mathfrak{q}⊊\mathfrak{m}$, so $\mathfrak{q}$ is not a power of $\mathfrak{m}$.

$\square$

Exercise 5. In the polynomial ring $K[x,y,z]$ where $K$ is a field and $x,y,z$ are independent indeterminates, let ${\mathfrak{p}}_1=(x,y),{\mathfrak{p}}_2=(x,z),\mathfrak{m}=(x,y,z)$; ${\mathfrak{p}}_1$ and ${\mathfrak{p}}_2$ are prime, and $\mathfrak{m}$ is maximal. Let $\mathfrak{a}={\mathfrak{p}}_1{\mathfrak{p}}_2$. Show that $\mathfrak{a}={\mathfrak{p}}_1\cap {\mathfrak{p}}_2\cap {\mathfrak{m}}^2$ is a reduced primary decomposition of $\mathfrak{a}$. Which components are isolated and which are embedded?

Proof. ${\mathfrak{p}}_1,{\mathfrak{p}}_2$ are prime cause $K/{\mathfrak{p}}_1=K[z],K/{\mathfrak{p}}_2=K[y]$ are integral. $\mathfrak{m}$ is maxiaml cause $K/\mathfrak{m}=K$ is a field. We have $$\begin{array}{rl}\mathfrak{a}={\mathfrak{p}}_1{\mathfrak{p}}_2& =(x^2,xy,xz,yz)\\ {\mathfrak{p}}_1\cap {\mathfrak{p}}_2\cap {\mathfrak{m}}^2& =(x,yz)\cap {\mathfrak{m}}^2\\ & =(x^2,xy,xz,yz).\end{array}$$Obviously ${\mathfrak{p}}_{1,2}\notin {\mathfrak{m}}^2,{\mathfrak{m}}^2\notin {\mathfrak{p}}_1$, thus $\mathfrak{a}\subset {\mathfrak{p}}_1\cap {\mathfrak{p}}_2\cap {\mathfrak{m}}^2$ is a reduced primary decomposition of $\mathfrak{a}$. Here ${\mathfrak{p}}_{1,2}\supset \mathfrak{m}$, so $\mathfrak{m}$ is isolated, and ${\mathfrak{p}}_{1,2}$ are embedded.

$\square$

Exercise 6. Let $X$ be an infinite compact Hasusdorff space, $C(X)$ the ring of real-valued continuous function on $X$ (Chapter 1, Exercise 26). Is the zero ideal decomposable in this ring?

Proof. For all $x\in X$, there is a prime ideal $I_x=\{f\in C(x)\mid f(x)=0\}$; and these are all prime ideals. So there are infinite many minimal prime ideals containing zero, hence zero ideal is not decomposable.

$\square$

Exercise 7. Let $A$ be a ring and let $A[x]$ denote the ring of polynomials in one indeterminate over $A$. For each ideal $\mathfrak{a}$ of $A$, let $\mathfrak{a}[x]$ denote the set of all polynomials in $A[x]$ with coefficients in $\mathfrak{a}$.

1.	$\mathfrak{a}[x]$ is the extension of $\mathfrak{a}$ to $A[x]$.
2.	If $\mathfrak{p}$ is a prime ideal in $A$, then $\mathfrak{p}[x]$ is a prime ideal in $A[x]$.
3.	If $\mathfrak{q}$ is a $\mathfrak{p}$-primary ideal in $A$, then $\mathfrak{q}[x]$ is a $\mathfrak{p}[x]$-primary ideal in $A[x]$.
4.	If $\mathfrak{a}={\bigcap }_{i=1}^n{\mathfrak{q}}_i$ is a minimal primary decomposition in $A$, then $\mathfrak{a}[x]={\bigcap }_{i=1}^n{\mathfrak{q}}_i[x]$ is a minimal primary decomposition in $A[x]$.
5.	If $\mathfrak{p}$ is a minimal prime ideal of $\mathfrak{a}$, then $\mathfrak{p}[x]$ is a minimal prime ideal of $\mathfrak{a}[x]$.

Proof.

1.	Trivial.
2.	If $\mathfrak{p}$ is a prime ideal in $A$, then $A/\mathfrak{p}$ is integral. Hence $A[x]/\mathfrak{p}[x]\simeq (A/\mathfrak{p})[x]$ is integral, and $\mathfrak{p}[x]$ is prime in $A[x]$.
3.	If $\mathfrak{q}$ is primary in $A$, then all zero divisors in $A/\mathfrak{q}$ are nilpotents; so if $f\in A[x]/\mathfrak{q}[x]\simeq (A/\mathfrak{q})[x]$ is a zero divisor, there exists $a\in A$ such that $af=0$; thus every coefficients of $f$ are zero divisors, hence nilpotents, and therefore $f$ is nilpotent (Exercise 1.2). This means $\mathfrak{q}[x]$ is primary in $A[x]$. Also, in $A/\mathfrak{q}$ the nilradical is $\mathfrak{p}/\mathfrak{q}$, so by Exercise 1.2 again, the nilradical of $A[x]/\mathfrak{q}[x]$ is $\mathfrak{p}[x]/\mathfrak{q}[x]$. So $\mathfrak{q}[x]$ is $\mathfrak{p}[x]$-primary.
4.	Obviously $\mathfrak{a}[x]={\bigcup }_{i=1}^n\mathfrak{q}[x]$, and $\mathfrak{q}[x]$ are primary by iii). Hence it’s a primary decomposition of $\mathfrak{a}[x]$. Obviously it’s a minimal primary decomposition of $\mathfrak{a}[x]$.
5.	Obvious by iii), iv). $\square$

Exercise 8. Let $k$ be a field. Show that in the polynomial ring $k[x_1,\dots ,x_n]$ the ideals ${\mathfrak{p}}_i=(x_1,\dots ,x_i)$ ($1\le i\le n$) are prime and all their powers are primary.

Proof. In $k[x_1,\dots ,x_i]$ the ideal $(x_1,\dots ,x_i)$ is maximal (the quotient ring is $k$), so it’s prime and all its powers are primary. Hence ${\mathfrak{p}}_i=(x_1,\dots ,x_i)[x_{i+1},\dots ,x_n]$ is prime and their powers are primary, by Exercise 7.

$\square$

Exercise 9. In a ring $A$, let $D(A)$ denote the set of prime ideals $\mathfrak{p}$ which satisfy the following condition: there exists $a\in A$ such that $\mathfrak{p}$ is minimal in the set of prime ideals containing $(0:a)$. Show that $x\in A$ is a zero divisor $⟺x\in \mathfrak{p}$ for some $\mathfrak{p}\in D(A)$.

Proof. If $\mathfrak{p}\in D(A)$ and $x\in \mathfrak{p}$, then there exists $a\in A$, such that $\mathfrak{p}$ is minimal in prime ideals containing $(0:a)$; so $\mathfrak{p}/(0:a)$ is a minimal prime ideal in $A/(0:a)$. By Exercise 3.9 (minimal prime ideals only contains zero divisors), there exists $t\in A$ such that $xt\in (0:a)$; that is, $xta=0$. So $x$ is a zero divisor.

Conversely, if $x$ is zero divisor, then $x$ is contained in some $(0:a)$, so contained in some minimal prime ideal $\mathfrak{p}$ containing $(0:a)$.

$\square$

Let $S$ be a multiplicatively closed subsetof $A$, and identify $\mathrm{Spec}(S^{-1}A)$ with its image in $\mathrm{Spec}(A)$. Show that $$D(S^{-1}A)=D(A)\cap \mathrm{Spec}(S^{-1}A).$$

Proof. $\mathfrak{p}$ is minimal prime ideal containing $(0:a)$, if and only if $S^{-1}\mathfrak{p}$ is minimal prime ideal containing $(0:a/1)$. So $\mathfrak{p}\in D(A)⟺S^{-1}\mathfrak{p}\in D(S^{-1}A)$.

$\square$

If the zero ideal has a primary decomposition, show that $D(A)$ is the set of associated prime ideals of $0$.

Proof. If $$0=\bigcap _{i=1}^n{\mathfrak{q}}_i$$where ${\mathfrak{q}}_i$ is ${\mathfrak{p}}_i$-primary, then every ${\mathfrak{p}}_i$ is $r(0:a)$ for some $a$; hence ${\mathfrak{p}}_i\in D(A)$.

Conversely, $(0:a)={\bigcap }_{i=1}^n({\mathfrak{q}}_i:a)={\bigcap }_{a\notin {\mathfrak{q}}_i}({\mathfrak{q}}_i:a)$ is a minimal primary decomposition of $(0:a)$, and $({\mathfrak{p}}_i:a)$ is ${\mathfrak{p}}_i$-primary. So if $\mathfrak{p}$ is a minimal prime ideal containing $(0:a)$, then $\mathfrak{p}={\mathfrak{p}}_i$ for some $i$. This means $D(A)=\{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n\}$.

$\square$

Exercise 10. For any prime ideal $\mathfrak{p}$ in a ring $A$, let $S_{\mathfrak{p}}(0)$ denote the kernel of the homomorphism $A\to A_{\mathfrak{p}}$. Prove that

1.	$S_{\mathfrak{p}}(0)\subseteq \mathfrak{p}$.
2.	$r(S_{\mathfrak{p}}(0))=\mathfrak{p}⟺\mathfrak{p}$ is a minimal prime ideal of $A$.
3.	If $\mathfrak{p}\supseteq {\mathfrak{p}}^{\prime }$, then $S_{\mathfrak{p}}(0)\subseteq S_{{\mathfrak{p}}^{\prime }}(0)$.
4.	${\bigcap }_{\mathfrak{p}\in D(A)}{S}_{\mathfrak{p}}(0)=0$, where $D(A)$ is defined in Exercise 9.

Proof.

1.	$x\in S_{\mathfrak{p}}(0)⟺\exists t\notin \mathfrak{p},tx=0$. Since $\mathfrak{p}$ is prime, $tx=0\in \mathfrak{p}$ and $t\notin \mathfrak{p}$, there must be $x\in \mathfrak{p}$, so $S_{\mathfrak{p}}(0)\subseteq \mathfrak{p}$.
2.	$r(S_{\mathfrak{p}}(0))=\mathfrak{p}⟺$ the image of $\mathfrak{p}$ in $A_{\mathfrak{p}}$ is the nilradical of $A_{\mathfrak{p}}⟺$ there is no prime ideal contained in ${\mathfrak{p}}^{\mathfrak{e}}$ in $A_{\mathfrak{p}}⟺\mathfrak{p}$ is minimal in $A$.
3.	If $\mathfrak{p}\supseteq {\mathfrak{p}}^{\prime }$, then $A_{{\mathfrak{p}}^{\prime }}$ is localization of $A_{\mathfrak{p}}$ at ${\mathfrak{p}}_{\mathfrak{p}}^{\prime }$, so the map $A\to A_{{\mathfrak{p}}^{\prime }}$ can be factor into $A\to A_{\mathfrak{p}}$ and $A_{\mathfrak{p}}\to A_{{\mathfrak{p}}^{\prime }}$, and $\ker (A\to A_{{\mathfrak{p}}^{\prime }})\supseteq \ker (A\to A_{\mathfrak{p}})$, i.e. $S_{\mathfrak{p}}(0)\subseteq S_{{\mathfrak{p}}^{\prime }}(0)$.
4.	If $x\in S_{\mathfrak{p}}(0)$, then there exists $t\notin \mathfrak{p}$ such that $xt=0$; that is, $(0:x)⊈\mathfrak{p}$. If $x\in {\bigcap }_{\mathfrak{p}\in D(A)}{S}_{\mathfrak{p}}(0)$, then there is no minimal prime ideal containing $(0:x)$, so $(0:x)=A$, and $x=0$. $\square$

Exercise 11. If $\mathfrak{p}$ is a minimal prime ideal of a ring $A$, show that $S_{\mathfrak{p}}(0)$ is the smallest $\mathfrak{p}$-primary ideal.

Proof. Since $\mathfrak{p}$ is a minimal prime ideal, $A_{\mathfrak{p}}$ has unique prime ideal $\mathfrak{p}$; so in $A_{\mathfrak{p}}$ every zero divisor is nilpotent, and so is $A/S_{\mathfrak{p}}(0)\simeq \text{im}(A\to A_{\mathfrak{p}})\subseteq A_{\mathfrak{p}}$. Hence $S_{\mathfrak{p}}(0)$ is primary. By Exercise 10 ii), $S_{\mathfrak{p}}(0)$ is $\mathfrak{p}$-primary.

If $\mathfrak{q}$ is $\mathfrak{p}$-primary, for any $x\in S_{\mathfrak{p}}(0)$, there exists $v\notin \mathfrak{p}$ such that $vx=0\in \mathfrak{q}$. Since $v\notin \mathfrak{p}=r(\mathfrak{q})$, so $x\in \mathfrak{q}$, and $S_{\mathfrak{p}}(0)\subseteq \mathfrak{q}$. This means $S_{\mathfrak{p}}(0)$ is smallest $\mathfrak{p}$-primary ideal.

$\square$

Let $\mathfrak{a}$ be the intersection of the ideals $S_{\mathfrak{p}}(0)$ as $\mathfrak{p}$ runs through the minimal prime ideals of $A$. Show that $\mathfrak{a}$ is contained in the nilradical of $A$.

Proof. If $x\in S_{\mathfrak{p}}(0)$, then $x\in \mathfrak{p}$. So $\mathfrak{a}\subseteq$ intersection of all minimal prime ideal. So $\mathfrak{a}\subseteq$ intersection of all prime ideal $=\mathfrak{R}$ (the nilradical of $A$).

$\square$

Support that the zero ideal is decomposable. Prove that $\mathfrak{a}=0$ if and only if every prime ideal of $0$ is isolated.

Proof. By Exercise 9, $D(A)$ is all minimal prime containing $(0:a)$ for some $a\in A$. but $(0:a)\subseteq \mathfrak{p}⟺a\notin S_{\mathfrak{p}}(0)$; and $D(A)$ is exactly the set of associated prime ideals of $0$.

So $\mathfrak{a}=0⟺$ for every $a\ne 0$, there exists a minimal ideal $\mathfrak{p}$ such that $a\notin S_{\mathfrak{p}}(0)⟺$ for any $a\ne 0$, $(0:a)$ is contained in some minimal prime ideal $⟺D(A)$ only contains minimal prime ideals $⟺$ every prime ideals of $0$ is isolated.

$\square$

Exercise 12. Let $A$ be a ring, $S$ a multiplicatively closed subset of $A$. For any ideal $\mathfrak{a}$, let $S(\mathfrak{a})$ denote the contraction of $S^{-1}\mathfrak{a}$ in $A$. The ideal $S(\mathfrak{a})$ is called the saturation of $\mathfrak{a}$ with respect to $S$. Prove that

1.	$S(\mathfrak{a})\cap S(\mathfrak{b})=S(\mathfrak{a}\cap \mathfrak{b})$.
2.	$S(r(\mathfrak{a}))=r(S(\mathfrak{a}))$.
3.	$S(\mathfrak{a})=(1)⟺\mathfrak{a}$ meets $S$.
4.	$S_1(S_2(\mathfrak{a}))=(S_1{S}_2)(\mathfrak{a})$.

If $\mathfrak{a}$ has a primary decomposition, prove that the set of ideals $S(\mathfrak{a})$ (where $S$ runs through all multiplicatively closed subsets of $A$) is finite.

Proof.

1.	$S^{-1}(\mathfrak{a}\cap \mathfrak{b})=S^{-1}\mathfrak{a}\cap S^{-1}\mathfrak{b}$, so $S(\mathfrak{a}\cap \mathfrak{b})=(S^{-1}\mathfrak{a}\cap S^{-1}\mathfrak{b}{)}^{\mathfrak{c}}=S(\mathfrak{a})\cap S(\mathfrak{b})$.
2.	$S(r(\mathfrak{a}))=(S^{-1}(r(\mathfrak{a})){)}^{\mathfrak{c}}=(r(S^{-1}\mathfrak{a}){)}^{\mathfrak{c}}=r(S(\mathfrak{a}))$.
3.	$S(\mathfrak{a})=(1)$ if and only if $S^{-1}\mathfrak{a}=1$, so it’s equivalent to $\mathfrak{a}$ meets $S$.
4.	By definition, $y\in S(\mathfrak{a})⟺\exists s\in S$ such that $sy\in \mathfrak{a}$. So clearly $y\in S_1(S_2(\mathfrak{a}))⟺\exists s_1\in S_1,s_2\in S_2,s_1{s}_{2}y\in \mathfrak{a}⟺y\in (S_1{S}_2)(\mathfrak{a})$.

If $\mathfrak{a}={\bigcap }_{i=1}^n{\mathfrak{q}}_i$, then $S^{-1}\mathfrak{a}={\bigcap }_{{\mathfrak{q}}_i\cap S=\varnothing }{S}^{-1}{\mathfrak{q}}_i$. By Propsition 4.8, $(S^{-1}{\mathfrak{q}}_i{)}^{\mathfrak{c}}={\mathfrak{q}}_i$, so $S(\mathfrak{a})={\bigcap }_{{\mathfrak{q}}_i\cap S=\varnothing }{\mathfrak{q}}_i$. So the set of ideals $S(\mathfrak{a})$ has at most $2^n$ elements.

$\square$

Exercise 13. Let $A$ be a ring and $\mathfrak{p}$ a prime ideal of $A$. The $n$th symbolic power of $\mathfrak{p}$ is defined to be the ideal (in the notation of Exercise 12) $${\mathfrak{p}}^{(n)}=S_{\mathfrak{p}}({\mathfrak{p}}^n)$$where $S_{\mathfrak{p}}=A-\mathfrak{p}$. Show that

1.	${\mathfrak{p}}^{(n)}$ is a $\mathfrak{p}$-primary ideal;
2.	if ${\mathfrak{p}}^n$ has a primary decomposition, then ${\mathfrak{p}}^{(n)}$ is its $\mathfrak{p}$-primary component;
3.	if ${\mathfrak{p}}^{(m)}{\mathfrak{p}}^{(n)}$ has a primary decomposition, then ${\mathfrak{p}}^{(m+n)}$ is its $\mathfrak{p}$-primary component;
4.	${\mathfrak{p}}^{(n)}={\mathfrak{p}}^n⟺{\mathfrak{p}}^n$ is $\mathfrak{p}$-primary. Misprint in the book: $\cdots ⟺{\mathfrak{p}}^{(n)}$ is $\mathfrak{p}$-primary; but by i) it’s always $\mathfrak{p}$-primary.

Proof.

1.	$r(S_{\mathfrak{p}}^{-1}({\mathfrak{p}}^n))=S_{\mathfrak{p}}^{-1}\mathfrak{p}$ is the maximal ideal in $A_{\mathfrak{p}}$, hence $S_{\mathfrak{p}}^{-1}({\mathfrak{p}}^n)$ is $S_{\mathfrak{p}}^{-1}\mathfrak{p}$-primary. So its contraction ${\mathfrak{p}}^{(n)}$ is $\mathfrak{p}$-primary.
2.	If ${\mathfrak{p}}^n$ has a primary decomposition ${\mathfrak{p}}^n={\bigcap }_{i=1}^m{\mathfrak{q}}_i$, where ${\mathfrak{q}}_i$ is ${\mathfrak{p}}_i$-primary; then $\mathfrak{p}=r({\mathfrak{p}}^n)={\bigcap }_{i=1}^m{\mathfrak{p}}_i$; so there exists some $i$ such that ${\mathfrak{p}}_i=\mathfrak{p}$; and all other ${\mathfrak{p}}_i⊋\mathfrak{p}$. Take $S_{\mathfrak{p}}({\mathfrak{p}}^n)$, we know ${\mathfrak{p}}^{(n)}$ is its $\mathfrak{p}$-primary decomposition. (Theorem 4.11)
3.	Similar to ii), since $r({\mathfrak{p}}^{(m)}{\mathfrak{p}}^{(n)})=\mathfrak{p}$, it’s sufficient to prove $S_{\mathfrak{p}}({\mathfrak{p}}^{(m)}{\mathfrak{p}}^{(n)})={\mathfrak{p}}^{(m+n)}$. For any $x\in {\mathfrak{p}}^{(n)}{\mathfrak{p}}^{(m)}$, by definition there exists $s\in S_{\mathfrak{p}}$ such that $sx\in {\mathfrak{p}}^{n+m}$. Hence ${\mathfrak{p}}^{(n)}{\mathfrak{p}}^{(m)}\subset {\mathfrak{p}}^{(n+m)}$, so $${\mathfrak{p}}^{n+m}\subseteq {\mathfrak{p}}^{(n)}{\mathfrak{p}}^{(m)}\subseteq {\mathfrak{p}}^{(n+m)}.$$Take $S_{\mathfrak{p}}(-)$ we have $S_{\mathfrak{p}}({\mathfrak{p}}^{(n)}{\mathfrak{p}}^{(m)})={\mathfrak{p}}^{(n+m)}$.
4.	If ${\mathfrak{p}}^{(n)}={\mathfrak{p}}^n$, then by i) it’s $\mathfrak{p}$-primary. Conversely if ${\mathfrak{p}}^n$ is $\mathfrak{p}$-primary, then ${\mathfrak{p}}^n=\bigcap \{{\mathfrak{p}}^n\}$ is a primary decomposition of ${\mathfrak{p}}^n$; so by ii) ${\mathfrak{p}}^n={\mathfrak{p}}^{(n)}$. $\square$

Exercise 14. Let $\mathfrak{a}$ be a decomposable ideal in a ring $A$ and let $\mathfrak{p}$ be a maximal element of the set of ideals $(\mathfrak{a}:x)$, where $x\in A$ and $x\notin \mathfrak{a}$. Show that $\mathfrak{p}$ is a prime ideal belong to $\mathfrak{a}$.

Proof. First we show $\mathfrak{p}=(\mathfrak{a}:x_0)$ must be prime. If $y\notin (\mathfrak{a}:x_0)$, then $(\mathfrak{a}:x_0)\subseteq (\mathfrak{a}:x_{0}y)\ne (1)$; so there must be $(\mathfrak{a}:x_0)=(\mathfrak{a}:x_{0}y)$; i.e. if $yz\in (\mathfrak{a}:x_0)$, then $z\in (\mathfrak{a}:x_{0}y)=(\mathfrak{a}:x_0)$. So $\mathfrak{p}=(\mathfrak{a}:x_0)$ is prime.

If $\mathfrak{a}={\bigcap }_{i=1}^n{\mathfrak{q}}_i$, where ${\mathfrak{q}}_i$ is ${\mathfrak{p}}_i$-primary, then $\mathfrak{p}={\bigcap }_{i=1}^n({\mathfrak{q}}_i:x_0)$; so $\mathfrak{p}=({\mathfrak{q}}_i:x_0)$ for some $i$. But $({\mathfrak{q}}_i:x_0)$ must be ${\mathfrak{p}}_i$-primary (if it’s not $(1)$), hence $\mathfrak{p}={\mathfrak{p}}_i$ for some $i$, i.e. $\mathfrak{p}$ is a prime ideal belong to $\mathfrak{a}$.

$\square$

Exercise 15. Let $\mathfrak{a}$ be a decomposable ideal in a ring $A$, let $\Sigma$ be an isolated set of prime ideals belonging to $\mathfrak{a}$, and let ${\mathfrak{q}}_{\Sigma }$ be the intersection of the corresponding primary components. Let $f$ be an element of $A$ such that, for each prime ideal $\mathfrak{p}$ belonging to $\mathfrak{a}$, we have $f\in \mathfrak{p}⟺\mathfrak{p}\notin \Sigma$, and let $S_f$ be the set of all powers of $f$. Show that ${\mathfrak{q}}_{\Sigma }=S_f(\mathfrak{a})=(\mathfrak{a}:f^n)$ for all large $n$.

Proof. If $\mathfrak{a}={\bigcap }_{i=1}^m{\mathfrak{q}}_i$, where ${\mathfrak{q}}_i$ is ${\mathfrak{p}}_i$-primary, and $\Sigma =\{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_p\}$. Since $f\in {\mathfrak{p}}_i⟺\mathfrak{p}\notin \Sigma$, so $S_f\cap {\mathfrak{p}}_i\ne \varnothing ⟺\mathfrak{p}\notin \Sigma$. Thus $S_f(\mathfrak{a})={\bigcap }_{i=1}^m{S}_f({\mathfrak{q}}_i)={\bigcap }_{i=1}^p{\mathfrak{q}}_i={\mathfrak{q}}_{\Sigma }$.

If $f\in {\mathfrak{p}}_i$, then by $r({\mathfrak{q}}_i)={\mathfrak{p}}_i$, we know there exists $N_i$ such that $f^{N_i}\in {\mathfrak{q}}_i$. Let $N=\max (N_{p+1},\dots ,N_m)$, then for all $n\ge N$, $$(\mathfrak{a}:f^n)=\bigcap _{i=1}^m({\mathfrak{q}}_i:f^n)=\bigcap _{i=1}^p{\mathfrak{q}}_i={\mathfrak{q}}_{\Sigma }$$$\square$

Exercise 16. If $A$ is a ring in which every ideal has a primary decomposition, show that every ring of fractions $S^{-1}A$ has the same property.

Proof. For any ideal $\mathfrak{a}\in S^{-1}A$, if ${\mathfrak{a}}^{\mathfrak{c}}={\bigcap }_{i=1}^n{\mathfrak{q}}_i$, then $\mathfrak{a}=S^{-1}({\mathfrak{a}}^{\mathfrak{c}})={\bigcap }_{i=1}^n{S}^{-1}{\mathfrak{q}}_i$, where each $S^{-1}{\mathfrak{q}}_i$ is either $(1)$ or primary ideal. So $\mathfrak{a}$ has a primary decomposition.

$\square$

Exercise 17. Let $A$ be a ring with the following property.

(L1) For every ideal $\mathfrak{a}\ne (1)$ in $A$ and every prime ideal $\mathfrak{p}$, there exists $x\notin \mathfrak{p}$ such that $S_{\mathfrak{p}}(\mathfrak{a})=(\mathfrak{a}:x)$, where $S_{\mathfrak{p}}=A-\mathfrak{p}$.

Then every ideal in $A$ is an intersection of (possibly infinitely many) primary ideals.

Proof. Let $\mathfrak{a}$ be an ideal $\ne (1)$ in $A$, let ${\mathfrak{p}}_1$ be a minimal prime ideal containing $\mathfrak{a}$. By Exercise 11 (apply to $A/\mathfrak{a}$), ${\mathfrak{q}}_1=S_{{\mathfrak{p}}_1}(\mathfrak{a})$ is a ${\mathfrak{p}}_1$-primary ideal. By (L1) property, there exists $x\notin {\mathfrak{p}}_1$ such that ${\mathfrak{q}}_1=(\mathfrak{a}:x)$. Now ${\mathfrak{q}}_1\cap (\mathfrak{a}+(x))\supseteq \mathfrak{a}$, and if $bx\in {\mathfrak{q}}_1$, then by $x\notin {\mathfrak{p}}_1$, $b\in {\mathfrak{q}}_1$; so $bx\in \mathfrak{a}$. This means $\mathfrak{a}={\mathfrak{q}}_1\cap (\mathfrak{a}+(x))$.

Let ${\mathfrak{a}}_1$ be a maximal element of the set of ideals $\mathfrak{b}$ such that: $\mathfrak{b}\supseteq \mathfrak{a}+(x)$ and ${\mathfrak{q}}_1\cap \mathfrak{b}=\mathfrak{a}$. Then ${\mathfrak{a}}_1⊈{\mathfrak{p}}_1$. Repeat the construction starting with ${\mathfrak{a}}_1$, and so on. After $n$ steps we have $\mathfrak{a}={\mathfrak{q}}_1\cap \cdots \cap {\mathfrak{q}}_n\cap {\mathfrak{a}}_n$.

If ${\mathfrak{a}}_n=(1)$ for some $n$, then $$\mathfrak{a}={\mathfrak{q}}_1\cap \cdots \cap {\mathfrak{q}}_n$$is a finite intersection of primary ideals. Otherwise define $${\mathfrak{a}}_{\omega }=\bigcup _{n=1}^{\infty }{\mathfrak{a}}_n,$$then $\mathfrak{a}={\mathfrak{a}}_{\omega }\cap {\bigcap }_{n=1}^{\infty }{\mathfrak{q}}_i$. Similarly, for any ordinal $\lambda$, construct ${\mathfrak{a}}_{\lambda +1}$ from ${\mathfrak{a}}_{\lambda }$ by the construction above (such that ${\mathfrak{a}}_{\lambda }={\mathfrak{q}}_{\lambda }\cap {\mathfrak{a}}_{\lambda +1}$, and ${\mathfrak{a}}_{\lambda }⊊{\mathfrak{a}}_{\lambda +1}$), and for any limit ordinal $\lambda$ define ${\mathfrak{a}}_{\lambda }={\bigcup }_{\beta <\lambda }{\mathfrak{a}}_{\beta }$. Then for any ordinal $\lambda$, we have $$\mathfrak{a}={\mathfrak{a}}_{\lambda }\cap \bigcap _{\beta <\lambda }{\mathfrak{q}}_{\beta }.$$And ${\mathfrak{a}}_{\lambda }⊊{\mathfrak{a}}_{\lambda +1}$. By transfinite induction, it’s easy to show $|{\mathfrak{a}}_{\lambda }|\ge |\lambda |$ (if ${\mathfrak{a}}_{\lambda }\ne (1)$). So for some ordinal larger than $|A|$, there must be ${\mathfrak{a}}_{\lambda }=(1)$, and $$\mathfrak{a}=\bigcap _{\beta <\lambda }{\mathfrak{q}}_{\beta }.$$$\square$

Exercise 18. Consider the following condition on a ring $A$:

(L2) Given an ideal $\mathfrak{a}$ and a descending chain $S_1\supseteq \cdots \supseteq S_n\supseteq \dots$ of multiplicatively closed subsets of $A$, there exists an integer $n$ such that $S_n(\mathfrak{a})=S_{n+1}(\mathfrak{a})=\dots$.

Prove that the following are equivalent:

1.	Every ideal in $A$ has a primary decomposition.
2.	$A$ satisfies (L1) and (L2).

Proof. i)$⟹$ii): For every ideal $\mathfrak{a}\ne (1)$ in $A$ and every prime ideal $\mathfrak{p}$, if $\mathfrak{a}$ has a primary decomposition $\mathfrak{a}={\bigcap }_{i=1}^n{\mathfrak{q}}_i$, where ${\mathfrak{q}}_i$ is ${\mathfrak{p}}_i$-primary, then $$S_{\mathfrak{p}}(\mathfrak{a})=\bigcap _{{\mathfrak{p}}_i\subseteq \mathfrak{p}}{\mathfrak{q}}_i$$Let $\mathfrak{b}={\bigcap }_{{\mathfrak{p}}_i⊈\mathfrak{p}}{\mathfrak{p}}_i$, then $\mathfrak{b}⊈\mathfrak{p}$; so there exists an element $f\in \mathfrak{b}\setminus \mathfrak{p}$, i.e. for any $i=1,\dots ,n,f\in {\mathfrak{p}}_i⟺{\mathfrak{p}}_i⊈\mathfrak{p}$. By Exercise 15), $S_{\mathfrak{p}}(\mathfrak{a})=(\mathfrak{a}:f^n)$ for all large $n$. So (L1) holds.

For every ideal $\mathfrak{a}$, since $\mathfrak{a}$ has a primary decomposition, so by Exercise 12), the set of all $S(\mathfrak{a})$ is finite (where $S$ runs through all multiplicatively closed subsets of $A$). So any descending chain $S_1(\mathfrak{a})\supseteq \cdots \supseteq S_n(\mathfrak{a})\supseteq \dots$ must be stationary. Thus (L2) holds.

ii)$⟹$i): construct ${\mathfrak{a}}_1,{\mathfrak{a}}_2,\dots$ as in Exercise 17), so that $\mathfrak{a}={\mathfrak{q}}_1\cap \cdots \cap {\mathfrak{q}}_n\cap {\mathfrak{a}}_n$, where ${\mathfrak{q}}_k$ is ${\mathfrak{p}}_k$-primary. Let $S_n=S_{{\mathfrak{p}}_1}\cap \cdots \cap S_{{\mathfrak{p}}_n}$. Since ${\mathfrak{a}}_n⊈{\mathfrak{p}}_n$, and ${\mathfrak{a}}_n\supseteq {\mathfrak{a}}_k$ for any $k\le n$, so ${\mathfrak{a}}_n⊈{\mathfrak{p}}_k$ for any $k\le n$, hence ${\mathfrak{a}}_n⊈{\bigcup }_{k=1}^n{\mathfrak{p}}_k$, so $\mathfrak{a}$ meets $S_n=A\setminus {\bigcup }_{k=1}^n{\mathfrak{p}}_k$, therefore $S_n({\mathfrak{a}}_n)=(1)$. This means $S_n(\mathfrak{a})={\mathfrak{q}}_1\cap {\mathfrak{q}}_2\cap \cdots \cap {\mathfrak{q}}_n$. If $\mathfrak{a}$ doesn’t have a primary decomposition, then the construction won’t terminate after a finite number of steps, hence $S_1(\mathfrak{a})⊋S_2(\mathfrak{a})⊋\cdots ⊋S_n(\mathfrak{a})⊋\dots$, contradicts (L2). So every ideal in $A$ has a primary decomposition.

$\square$

Exercise 19. Let $A$ be a ring and $\mathfrak{p}$ a prime ideal of $A$. Show that every $\mathfrak{p}$-primary ideal contains $S_{\mathfrak{p}}(0)$, the kernel of the canonical homomorphism $A\to A_{\mathfrak{p}}$.

Proof. For any $\mathfrak{p}$-primary ideal $\mathfrak{q}$, any $x\in S_{\mathfrak{p}}(0)$, there exists $s\notin \mathfrak{p}$ such that $sx=0\in \mathfrak{q}$. By definition $x\in \mathfrak{q}$, so $S_{\mathfrak{p}}(0)\subseteq \mathfrak{q}$.

$\square$

Suppose that $A$ satisfies the following condition: for every prime ideal $\mathfrak{p}$, the intersection of all $\mathfrak{p}$-primary ideals of $A$ is equal to $S_{\mathfrak{p}}(0)$. (Noetherian rings satisfy this condition: see Chapter 10.) Let ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n$ be distinct prime ideals, none of which is a minimal prime ideal of $A$. Then there exists an ideal $\mathfrak{a}$ in $A$ whose associated prime ideals are ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n$.

Proof. Induction on $n$. The case $n=1$ is trivial (let $\mathfrak{a}={\mathfrak{p}}_1$).

Suppose $n>1$, and ${\mathfrak{p}}_n$ is maximal in $\{{\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n\}$. By inductive hypothesis, there exists an ideal $\mathfrak{b}$ whose associated ideals are ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_{n-1}$. If $\mathfrak{b}\subseteq S_{{\mathfrak{p}}_n}(0)$, choose a minimal ideal $\mathfrak{p}$ contained in ${\mathfrak{p}}_n$, then $\mathfrak{b}\subseteq S_{{\mathfrak{p}}_n}(0)\subseteq S_{\mathfrak{p}}(0)$. Take radicals we have ${\bigcap }_{i=1}^n{\mathfrak{p}}_i\subseteq \mathfrak{p}$, hence ${\mathfrak{p}}_{i_0}\subseteq \mathfrak{p}$ for some $i_0$. Since $\mathfrak{p}$ is minimal, ${\mathfrak{p}}_{i_0}=\mathfrak{p}$, contradiction (cause none of ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_{n-1}$ is minimal). So $\mathfrak{b}⊈S_{{\mathfrak{p}}_n}(0)$, hence $\mathfrak{b}⊈{\mathfrak{q}}_n$ for some ${\mathfrak{p}}_n$-primary ideal ${\mathfrak{q}}_n$.

Let $\mathfrak{a}=\mathfrak{b}\cap {\mathfrak{q}}_n={\bigcap }_{i=1}^n{\mathfrak{q}}_i$ (write $\mathfrak{b}={\bigcap }_{i=1}^{n-1}{\mathfrak{q}}_i$, where ${\mathfrak{q}}_i$ is ${\mathfrak{p}}_i$-primary), then it is sufficient to show: for any $i=1,\dots ,n-1$, $${\mathfrak{q}}_i⊉\bigcap _{j\ne i}{\mathfrak{q}}_j.$$Let $S=S_{{\mathfrak{p}}_1}\cap \cdots \cap S_{{\mathfrak{p}}_{n-1}}$, cause ${\mathfrak{p}}_n⊈{\mathfrak{p}}_k$ for any $k=1,\dots ,n-1$, ${\mathfrak{p}}_n$ meets $S$. So $S({\mathfrak{q}}_i)={\mathfrak{q}}_i,S({\bigcap }_{j\ne i}{\mathfrak{q}}_j)={\bigcap }_{j\ne i,n}{\mathfrak{q}}_j$. So $S({\mathfrak{q}}_i)⊉S({\bigcap }_{j\ne i}{\mathfrak{q}}_j)$, hence ${\mathfrak{q}}_i⊉{\bigcap }_{j\ne i}{\mathfrak{q}}_j$.

$\square$

Primary decomposition of modules

Practically the whole of this chapter can be transposed to the context of modules over a ring $A$. The following exercises indicate how this is done.

Exercise 20. Let $M$ be a fixed $A$-module, $N$ a submodule of $M$. The radical of $N$ in $M$ is defined to be $$r_M(N)=\{x\in A:x^{q}M\subseteq N\text{ for some }q>0\}$$Show that $r_M(N)=r(N:M)=r(\mathrm{Ann}(M/N))$. In particular, $r_M(N)$ is an ideal.

State and prove the formulas for $r_M$ analogous to (1.13).

Proof. $$\begin{array}{rl}{r}_M(N)& =\{x\in A:x^{q}M\subseteq N\text{ for some }q>0\}\\ & =\{x\in A:x^q(M/N)=0\text{ for some }q>0\}\\ & =r(\mathrm{Ann}(M/N))\end{array}$$And clearly:

1.	$r(r_M(N))=r_M(N)$.
2.	$r_M(N_1\cap N_2)=r_M(N_1)\cap r_M(N_2)$.
3.	$r_M(N)=(1)⟺N=M$.
4.	$r_M(N+P)\supseteq r(r_M(N)+r_M(P))$.
5.	if $P\subseteq N\subseteq M$ then $r_M(N)=r_{M/P}(N/P)$. $\square$

Exercise 21. An element $x\in A$ defines an endomorphism ${\varphi }_x$ of $M$, namely $m\mapsto xm$. The element $x$ is said to be a zero-divisor (resp. nilpotent) in $M$ if ${\varphi }_x$ is not injective (resp. is nilpotent). A submodule $Q$ of $M$ is primary in $M$ if $Q\ne M$ and every zero-divisor in $M/Q$ is nilpotent.

Show that if $Q$ is primary in $M$, then $(Q:M)$ is a primary ideal and hence $r_M(Q)$ is a prime ideal $\mathfrak{p}$. We say that $Q$ is $\mathfrak{p}$-primary (in $M$).

Proof. Since $Q\ne M$, we have $(Q:M)\ne (1)$. If $xy\in (Q:M)=\mathrm{Ann}(M/Q)$ and $y\notin \mathrm{Ann}(M/Q)$, then $\ker {\varphi }_{xy}=M/Q\ne \ker {\varphi }_y$, so $x$ is a zero-divisor in $M/Q$. Since $Q$ is primary in $M$, $x$ is also a nilpotent in $M/Q$, i.e. ${\varphi }_{x^n}=0$ for some $n$. Thus $x^n\in (Q:M)$; so $(Q:M)$ is a primary ideal in $A$.

$\square$

Prove the analogues of (4.3) and (4.4).

Proof. Analogues of (4.3): If $Q_i(1\le i\le n)$ are $\mathfrak{p}$-primary, then $Q={\bigcap }_{i=1}^n{Q}_i$ is $\mathfrak{p}$-primary.

First we have $r_M(Q)={\bigcap }_{i=1}^n{r}_M(Q_i)=\mathfrak{p}$. If $x$ is a zero divisor in $M/Q$, then there exists $v\notin Q$ such that $xv\in Q$. Since $Q={\bigcap }_{i=1}^n{Q}_i$, there exists $j$ such that $v\notin Q_j$, but $xv\in Q\subseteq Q_j$, thus $x\in \mathfrak{p}=r_M(Q)$. Hence $Q$ is $\mathfrak{p}$-primary.

Analogues of (4.4): If $Q$ is $\mathfrak{p}$-primary submodule of $M$, then

1.	if $v\in Q$ then $(Q:v)=(1)$;
2.	if $v\notin Q$ then $(Q:v)$ is a $\mathfrak{p}$-primary ideal.
3.	if $x\notin \mathfrak{p}$ then $(Q:x)(:=\{v\in M:xv\in Q\})=Q$.

i) and iii): by definition.

ii): if $xv\in Q$, then $x\in \mathfrak{p}$. So $(Q:M)\subseteq (Q:v)\subseteq \mathfrak{p}$. Taking radicals, we have $r(Q:v)=\mathfrak{p}$. If $yz\in (Q:v)$ but $y\notin \mathfrak{p}$, then $yzv\in Q$, hence $zv\in Q$, and $z\in (Q:v)$. So $(Q:v)$ is $\mathfrak{p}$-primary.

$\square$

Exercise 22. A primary decomposition of $N$ in $M$ is a representation of $N$ as an intersection $$N=Q_1\cap \cdots \cap Q_n$$of primary submodules of $M$; it is a minimal primary decomposition if the ideals ${\mathfrak{p}}_i=r_M(Q_i)$ are all distinct and if none of the components $Q_i$ can be omitted from the intersection, that is if $Q_i⊉{\cap }_{j\ne i}{Q}_j(1\le i\le n)$.

Prove the analogue of (4.5), that the prime ideals ${\mathfrak{p}}_i$ depend only on $N$ (and $M$). They are called the prime ideals belonging to $N$ in $M$. Show that they are also the prime ideals belonging to $0$ in $M/N$.

Proof. We will prove that the prime ideals ${\mathfrak{p}}_i$ are exactly all prime ideals which occur in the set of ideals $r(N:v)(v\in M)$; so that they only depend on $N$ (and $M$).

Suppose $N=Q_1\cap \cdots \cap Q_n$ is a minimal primary decomposition of $N$, where $Q_i$ is ${\mathfrak{p}}_i$-primary. For any $v\in M$, we have $$r(N:v)=\bigcap _{i=1}^{n}r(Q_i:v)$$If it’s a prime ideal $\mathfrak{p}$, then $\mathfrak{p}=r(Q_i:v)$ for some $1\le i\le n$. But $r(Q_i:v)={\mathfrak{p}}_i$ (if not $(1)$), hence $\mathfrak{p}={\mathfrak{p}}_i$. Conversely, for each $1\le i\le n$, there exists $v_i\notin Q_i$, $v_i\in {\bigcap }_{j\ne i}{Q}_j$. In this case we have $r(N:v_i)=r(Q_i:v_i)={\mathfrak{p}}_i$.

Now if the image of $v$ in $M/N$ is $\bar{v}$, then $(N:v)=(0:\bar{v})$, so particularly $r(N:v)=r(0:\bar{v})$. Hence the prime ideals belonging to $N$ in $M$ are precisely the prime ideals belonging to $0$ in $M/N$. (Equivalently, if $N=Q_1\cap \cdots \cap Q_n$ is a minimal primary decomposition of $N$ in $M$, then $0=(Q_1/N)\cap \cdots \cap (Q_n/N)$ is also a minimal primary decomposition of $0$ in $M/N$; and clearly $r_{M/N}(Q_i/N)={\mathfrak{p}}_i$.)

$\square$

Exercise 23. State and prove the analogues of (4.6) - (4.11) inclusive. (There is no loss of generality in taking $N=0$.)

(4.6)	Let $N$ be a decomposable submodule of $M$. Then any prime ideal $\mathfrak{p}\supseteq r_M(N)$ contains a minimal prime ideal belonging to $N$ in $M$, and thus the minimal prime ideals of $N$ in $M$ are precisely the minimal elements in the set of all prime ideals containing $r_M(N)$. Proof. If $\mathfrak{p}\supseteq r_M(N)={\bigcap }_{i=1}^n{r}_M(Q_i)={\bigcap }_{i=1}^n{\mathfrak{p}}_i$, then $\mathfrak{p}\supseteq {\mathfrak{p}}_i$ for some $i$. Hence $\mathfrak{p}$ contains a minimal prime ideal of $N$ in $M$. $\square$
(4.7)	Let $N$ be a decomposable submodule of $M$, let $N={\bigcap }_{i=1}^n{Q}_i$ be a minimal primary decomposition, and let $r_M(Q_i)={\mathfrak{p}}_i$. Then $$\bigcup _{i=1}^n{\mathfrak{p}}_i=\{x\in A:(N:x)\ne N\}$$where $(N:x)\stackrel{def}{=}\{v\in M:xv\in N\}$. In particular, if $N=0$, then the set $D_M\subseteq A$ of zero-divisors in $M$ of $A$ is the union of the prime ideals belonging to $0$. Proof. Without loss of generality, we can assume $N=0$. Clearly $D_M={\bigcup }_{v\ne 0\in M}r(0:v)$. For each $v\ne 0\in M$, $$r(0:v)=\bigcap _{i=1}^{n}r(Q_i:v)=\bigcap _{v\notin Q_i}{\mathfrak{p}}_i.$$Hence $r(0:v)\subseteq {\mathfrak{p}}_i$ for some $1\le i\le n$, and thus $D_M\subseteq {\bigcup }_{i=1}^n{\mathfrak{p}}_i$. Conversely, for each $1\le i\le n$, there exists $v_i\notin Q_i,v_i\in {\bigcap }_{j\ne i}{Q}_j$, so ${\mathfrak{p}}_i=r(0:v_i)\subseteq D_M$. $\square$
(4.8)	Let $S$ be a multiplicatively closed subset of $A$, and let $Q\subseteq M$ be a $\mathfrak{p}$-primary submodule. 1. if $S\cap \mathfrak{p}\ne \varnothing$, then $S^{-1}Q=S^{-1}M$. 2. if $S\cap \mathfrak{p}=\varnothing$, then $S^{-1}Q$ is a $S^{-1}\mathfrak{p}$-primary submodule of $S^{-1}M$ (as a $S^{-1}A$-module); and its preimage (contraction) under the canonical map $M\to S^{-1}M$ is $Q$. Hence primary $S^{-1}A$-submodules of $S^{-1}M$ correspond to primary $A$-submodules of $M$. Proof. If $S\cap \mathfrak{p}\ne \varnothing$, then there exists $x\in S$ satisfies $x\in \mathfrak{p}=r_M(Q)$, so there is $n\ge 1$ such that $x^{n}M\subseteq Q$. Hence for each $y\in S^{-1}M$, $y=(x^{n}y)/x^n\in S^{-1}Q$. If $S\cap \mathfrak{p}=\varnothing$. First we show $r_{S^{-1}M}(S^{-1}Q)=S^{-1}\mathfrak{p}$. Suppose $x/s\in r_{S^{-1}M}(S^{-1}Q)$, then also $x=x/1\in r_{S^{-1}M}(S^{-1}Q)$, so there is $n\ge 1$ such that $x^n(S^{-1}M)\subseteq (S^{-1}Q)$. In particular, for each $m\in M$, $x^{n}m/1\in S^{-1}Q$, hence exists $t\in S$ that $x^{n}tm\in Q$. So $x^{n}t\in r_M(Q)=\mathfrak{p}$. Since $t\notin \mathfrak{p}$, we have $x\in \mathfrak{p}$. Therefore $x/s\in S^{-1}\mathfrak{p}$ and $r_{S^{-1}M}(S^{-1}Q)\subseteq \mathfrak{p}$. The other direction follows immediately from definition. Suppose $x/t\in S^{-1}A,m/s\in S^{-1}M-S^{-1}Q$ but $xm/st\in S^{-1}Q$. Then there exist $u\in S$ such that $uxm\in Q$, but $m\notin Q$. hence $ux\in \mathfrak{p}$ and so $x\in \mathfrak{p}$. Thus $x/t\in S^{-1}\mathfrak{p}=r_{S^{-1}M}(S^{-1}Q)$, and $S^{-1}Q$ is a $S^{-1}\mathfrak{p}$-primary submodule. Finally, if $\mathfrak{m}/1\in S^{-1}Q$, then there exists $s\in S$ such that $sm\in Q$. But $s\notin \mathfrak{p}$, hence $m\in Q$. Therefore the preimage of $S^{-1}Q$ in $M$ is $Q$. $\square$
(4.9)	Let $S$ be a multiplicatively closed subset of $A$ and let $N$ be a decomposable submodule of $M$. Let $N={\bigcap }_{i=1}^n{Q}_i$ be a minimal primary decomposition of $N$. Let ${\mathfrak{p}}_i=r_M(Q_i)$ and suppose the $Q_i$ numbered so that $S$ meets ${\mathfrak{p}}_{m+1},\dots ,{\mathfrak{p}}_n$ but not ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$. Then $$S^{-1}N=\bigcap _{i=1}^m{S}^{-1}{Q}_i,S(N)=\bigcap _{i=1}^M{Q}_i$$ Proof. Clearly by (4.8). $\square$
(4.10)	Let $N$ be a decomposable submodule of $M$, let $N={\bigcap }_{i=1}^n{Q}_i$ be a minimal primary decomposition of $N$, and let ${\mathfrak{p}}_{i_1},\dots ,{\mathfrak{p}}_{i_m}$ be an isolated set of prime ideals of $N$ in $M$. Then $Q_{i_1}\cap \cdots \cap Q_{i_m}$ is independent of the decomposition. Proof. Take $S=A-{\bigcup }_{j=1}^m{\mathfrak{p}}_{i_j}$ in (4.9), then $Q_{i_1}\cap \cdots \cap Q_{i_m}=S(N)$, hence independent of the decomposition. $\square$ In particular,
(4.11)	: The isolated primary components ($m=1$ in (4.10), the corresponding prime ideal is minimal) are uniquely determined by $N$ (and $M$).