HW8

Text-Ex. 4.12. Let $A$ be a ring, $S$ a multiplicatively closed subset of $A$. For any ideal $a$, let $S(a)$ denote the contraction of $S^{-1}a$ in $A$. The ideal $S(a)$ is called the saturation of $a$ with respect to $S$. Prove that

• i) $S(a)\cap S(b)=S(a\cap b)$
• ii) $S(r(a))=r(S(a))$
• iii) $S(a)=(1)\iff a$ meets $S$
• iv) $S_1\left(S_2(a)\right)=\left(S_1{S}_2\right)(a)$.

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

\begin{proof} For any $x\in S^{-1}a$, we know $x/1=a^{\prime }/s^{\prime }$ for some $a^{\prime }\in A$ and $s^{\prime }\in S$. Then there exists $s\in S$ such that $xs\in a$. Conversely, if $x\in A$ such that $xs\in a$ for some $s$, then $xs/s\sim x/1$ in $S^{-1}a$ and so $x\in S(a)$. Therefore, $S(a)=\{x\in A:xs\in a\text{ for some }s\in S\}$.

i) For any $x\in S(a)\cap S(b)$, there exists $s_1,s_2\in S$ such that $x{s}_1\in a$ and $x_2\in b$. Then $x(s_1{s}_2)\in a\cap b$ and so $x\in S(a\cap b)$. On the other hand, if $x\in S(a\cap b)$, then there exists $s\in S$ such that $sx\in a\cap b\subseteq a$ and $sx\in a\cap b\subseteq b$. Hence $x\in S(a)\cap S(b)$.

ii) For any $x\in S(r(a))$, there exists $s\in S$ such that $sx\in r(a)$ and so $(sx{)}^n\in a$ for some $n\in {\mathbb{N}}_{+}$. It deduces that $s^n{x}^n\in a$ and so $x^n\in S(a)$. Thus $x\in r(S(a))$. On the other hand, if $x\in r(S(a))$, there exists $m\in {\mathbb{N}}_{+}$ such that $x^m\in S(a)$ and there exists $s^{\prime }\in S$ such that $s^{\prime }{x}^m\in a$. Then $(s^{\prime }x{)}^m=(s^{\prime }{)}^{m-1}(s^{\prime }{x}^m)\in a$ and $s^{\prime }x\in r(a)$. It yields $x\in S(r(a))$.

iii) If $S(a)=(1)$, then $1\in \{x\in A:xs\in a\text{ for some }s\in S\}$ and $1\cdot s\in a$ for some $s\in S$. Thus $s\in S\cap a$. Conversely, if $a$ meets $S$, then there is $s\in S\cap a$ and $r\cdot s\in a$ for all $r\in A$. So $(1)=S(a)$.

iv) For any $x\in S_1(S_2(a))$, there exists $s_1\in S_1$ such that $s_{1}x\in S_2(a)$, and then there exists $s_2\in S_2$ such that $s_2(s_{1}x)\in a$. Then $x\in (S_1{S}_2)(a)$. On the other hand, for any $x\in (S_1{S}_2)(a)$, there exists $s_1{s}_2\in S_1{S}_2$ such that $s_1{s}_{2}x\in a$. Then $s_2(s_{1}x)\in a$ yields $s_{1}x\in S_2(a)$ and so $x\in S_1(S_2(a))$. Now we proved iv).

If $a$ has a primary decomposition, $a$ can be written as $a={\cap }_{i=1}^n{q}_i$ for finitely many primary ideals $q_i$. Then $S(a)=S({\cap }_{i=1}^n{q}_i)={\cap }_{i=1}^{n}S(q_i)$. For each $q_i$ with $r(q_i)=p_i$, we have

p_i,&\text{if }p_i\cap S=\emptyset, \\ (1),&\text{otherwise.} \end{cases}$$ For the former case, for any $x\in S(q_i)$, there exists $s\in S$ such that $sx\in q_i$. If $x\notin q_i$, then $s^n\in q_i\subseteq p_i$ and $s\in p_i$, which is a contradiction. Hence $x\in q_i$ and $S(q_i)\subseteq q_i$. Since $q_i\subseteq S(q_i)$ is trivial, we have $S(q_i)=q_i$. For the later case, since $p_i\cap S\neq \emptyset$, there exists $s\in p_i\cap S$ and so $s^n\in q_i$. Thus $q_i\cap S\neq\emptyset$ and $S(q_i)=(1)$. Now we have proved that $$S(q_i)=\begin{cases} q_i,&\text{if }p_i\cap S=\emptyset, \\ (1),&\text{otherwise.} \end{cases}$$ Therefore, $S(a)=\cap_{i=1}^n S(q_i)$ has finitely many possibilities when $S$ runs through all multiplicatively closed subsets of $A$ as each $S(q_i)$ has only two possibilities. `\end{proof}` **Text-Ex. 4.13. Let $A$ be a ring and $\mathfrak{p}$ a prime ideal of $A$. The nth symbolic power of $\mathfrak{p}$ is defined to be the ideal (in the notation of Exercise 12)**

\mathfrak{p} ^{(n)}=S_\mathfrak{p} \left(\mathfrak{p} ^n\right)

**where $S_\mathfrak{p}=A-\mathfrak{p}$. Show that** - **i) $\mathfrak{p}^{(n)}$ is a $\mathfrak{p}$-primary ideal.** `\begin{proof}` i) We firstly show $\mathfrak{p}^{(n)}$ is a primary ideal. For any $xy\in \mathfrak{p}^{(n)}$, there exists $s_0\notin \mathfrak{p}$ such that $s_0xy\in \mathfrak{p}^n$. If $x\notin \mathfrak{p}^{(n)}$, then for any $s\notin \mathfrak{p}$, we have $sx\notin \mathfrak{p}^n$. Thus $s_0y\in \mathfrak{p}$, otherwise $s_0y\notin\mathfrak{p}$ and $(s_0y)x\in \mathfrak{p}^n$, which is impossible. Since $s_0\notin\mathfrak{p}$ and $\mathfrak{p}$ is a prime ideal, we have $y\in\mathfrak{p}$. Because $y\in \mathfrak{p}$, we know $y^n\in\mathfrak{p}^n\subseteq \mathfrak{p}^{(n)}$ and so $\mathfrak{p}^{(n)}$ is primary. Now we prove $r(\mathfrak{p}^{(n)})=\mathfrak{p}$. For any $x\in \mathfrak{p}^{(n)}$, there exists $s\notin p$ such that $sx\in \mathfrak{p}$ and so $x\in \mathfrak{p}$. Hence $\mathfrak{p}^{(n)}\subseteq \mathfrak{p}$ and so $r(\mathfrak{p} ^{(n)})\subseteq r(\mathfrak{p})=\mathfrak{p}$. On the other hand, for any $x\in \mathfrak{p}$, there is $x^n\in \mathfrak{p}^n$ and so $sx^n\in\mathfrak{p}^n$ for all $s\notin \mathfrak{p}$, which deduces that $x^n\in S_\mathfrak{p}(\mathfrak{p}^n)$ and $x\in r(S_\mathfrak{p}(\mathfrak{p}^n))$. Therefore, $r(\mathfrak{p}^{(n)})=\mathfrak{p}$ and so $\mathfrak{p}^{(n)}$ is a $\mathfrak{p}$-primary ideal. `\end{proof}` **Problem C.（往年考试题)** **Let $A$ be a ring and $M$ a Noetherian $A$-module. Let $\mathfrak{a} \subset A$ be the annihilator of $M$, i.e., $\mathfrak{a}=\{x \in A \mid x M=0\}$. Prove that the ring $A / \mathfrak{a}$ is Noetherian.** `\begin{proof}` Since $M$ is finitely generated, we assume $M=\left\langle m_1,\cdots,m_k\right\rangle_A$. Define

\varphi:A\to M^k,a\mapsto (am_1,\cdots,am_k)

and then $\ker\varphi=\mathfrak{a}$. Hence $\varphi$ induces a map $\overline \varphi:A/\mathfrak{a}\hookrightarrow M^k$ and so $A/\frak a$ can be identified as a $M^k$-submodule. Then it is a Noetherian $A$-module. Since $\mathfrak{a}$ is trivial on $A/\mathfrak{a}$, $A/\mathfrak{a}$ is a Noetherian $A/\mathfrak{a}$-module and so the ring $A / \mathfrak{a}$ is Noetherian. `\end{proof}`