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# 数学代写|交换代数代写Commutative Algebra代考|MATH483 Heights and Dimension

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## 数学代写|交换代数代写Commutative Algebra代考|Heights and Dimension

Proposition (9.5.1). – Let A be an integral domain, let $\mathrm{E}$ be its field of fractions and let $\mathrm{E} \rightarrow \mathrm{F}$ be a finite normal extension of fields. Assume that $\mathrm{A}$ is integrally closed in $\mathrm{E}$ and let $\mathrm{B}$ the integral closure of $\mathrm{A}$ in $\mathrm{F}$.

For every prime ideal $\mathrm{P}$ of $\mathrm{A}$, the group $\operatorname{Aut}(\mathrm{F} / \mathrm{E})$ acts transitively on the set of prime ideals $\mathrm{Q}$ of $\mathrm{B}$ such that $\mathrm{Q} \cap \mathrm{A}=\mathrm{P}$.

Proof. – Let $\mathrm{G}=\operatorname{Aut}(\mathrm{F} / \mathrm{E})$. It is a finite group of automorphisms of $\mathrm{F}$; the subfield $\mathrm{F}^{\mathrm{G}}$ of $\mathrm{F}$ is a radicial extension of $\mathrm{E}$ (corollary 4.6.9) and $\mathrm{F}^{\mathrm{G}} \subset \mathrm{F}$ is a Galois extension of group $G$ (Artin’s lemma 4.6.4).

First of all, according to the going up theorem (theorem 9.3.9), the set of prime ideals $Q$ of $B$ such that $Q \cap A=P$ is non-empty.
Let $Q$ and $Q^{\prime}$ be two prime ideals of $B$ such that $Q \cap A=Q^{\prime} \cap A=P$.
Let $x \in \mathrm{Q}^{\prime}$ and let $y=\prod_{\sigma \in \mathrm{G}} \sigma(x)$; since $y \in \mathrm{F}^{\mathrm{G}}$ and since the extension $\mathrm{E} \rightarrow$ $\mathrm{F}^{\mathrm{G}}$ is radicial, there exists an integer $q \geq 1$ such that $y^q \in \mathrm{E}$ (lemma 4.4.16; one has $q=1$ if the characteristic of $\mathrm{E}$ is zero).

By definition of $\mathrm{B}, x$ is integral over $\mathrm{A}$, as well as its images $\sigma(x)$, for $\sigma \in \mathrm{G}$. Consequently, $y$ is integral over $\mathrm{A}$, hence $y^q$ is integral over $\mathrm{A}$. Since $y^q \in \mathrm{E}$ and $\mathrm{A}$ is integrally closed in $\mathrm{E}$, one has $y^q \in \mathrm{A}$. Since $x \in \mathrm{Q}^{\prime}$, it follows that $y^q \in Q^{\prime} \cap A=P$. Using that $P=Q \cap A$, we deduce that $y^q \in Q$. Finally, since $Q$ is a prime ideal, we conclude that $y \in Q$.

Moreover, since $y=\prod_{\sigma \in \mathrm{G}} \sigma(x)$, there exists an element $\sigma \in \mathrm{G}$ such that $\sigma(x) \in Q$. Since $G$ is a group, this gives the inclusion $Q^{\prime} \subset \bigcup_{\sigma \in G} \sigma(Q)$.

## 数学代写|交换代数代写Commutative Algebra代考|Dedekind Rings

Definition (9.6.1). – Let A be an integral domain. One says that $\mathrm{A}$ is a Dedekind ring if every ideal of $\mathrm{A}$ is a projective A-module.
Example (9.6.2). – (i) A field is a Dedekind ring.
(ii) Since non-zero principal ideals of an integral domain A are free Amodules of rank 1 , any principal ideal domain is a Dedekind ring.
(iii) Let $\mathrm{A}$ be a Dedekind ring and let $\mathrm{S}$ be a multiplicative subset of $\mathrm{A}$. Then $\mathrm{S}^{-1} \mathrm{~A}$ is a Dedekind ring.

Indeed, let $\mathrm{J}$ be an ideal of $\mathrm{S}^{-1} \mathrm{~A}$. There exists an ideal $\mathrm{I}$ of $\mathrm{A}$ such that $\mathrm{J}=$ $\mathrm{S}^{-1} \mathrm{I}$. Since $\mathrm{A}$ is a Dedekind ring, the A-module $\mathrm{I}$ is projective. Consequently, the $\mathrm{S}^{-1} \mathrm{~A}$-module $\mathrm{S}^{-1} \mathrm{I}$ is projective (example 7.3.5).
9.6.3. – Let $\mathrm{A}$ be an integral domain and let $\mathrm{K}$ be its field of fractions.
The set $\mathscr{I}(\mathrm{A})$ of non-zero ideals of $\mathrm{A}$ is a commutative monoid with respect to multiplication of ideals. The subset $\mathscr{P}(\mathrm{A})$ of $\mathscr{I}(\mathrm{A})$ consisting of principal ideals is a submonoid. The quotient monoid $\mathscr{C}(\mathrm{A})=\mathscr{I}(\mathrm{A}) / \mathscr{P}(\mathrm{A})$ is called the class monoid of A. This monoid is trivial if and only if every ideal of $A$ is principal. As we will show, Dedekind rings are actually characterized by the property that $\mathscr{C}(\mathrm{A})$ is a group.

## 数学代写|交换代数代写Commutative Algebra代考|Heights and Dimension

$y^q \in Q^{\prime} \cap A=P$. 使用那个 $P=Q \cap A$, 我们推断 $y^q \in Q$. 最后，由于 $Q$ 是素理想，我们得出结 论 $y \in Q$

## 数学代写|交换代数代写Commutative Algebra代考|Dedekind Rings

(ii) 由于积分域 $A$ 的非零主理想是秩为 1 的自由 $A$ 模，因此任何主理想域都是 Dedekind 环。
(iii) 让 $A$ 成为戴德金戒指，让 $S$ 是的乘去子集 $A$. 然后 $S^{-1} A$ 是戴德金戒指。

9.6.3. – 让 $A$ 是一个完整的域，让 $K$ 是它的分数域。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。