Posted on Categories:Commutative Algebra, 交换代数, 数学代写

# 数学代写|交换代数代写Commutative Algebra代考|Fitting Ideals

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|交换代数代写Commutative Algebra代考|Fitting Ideals

In this section we develop the theory of Fitting ideals as a tool for checking locally the freeness of a module. If a finitely generated $A$-module $M$ is given by a finite presentation, then the $k$-th Fitting ideal $F_k(M)$ is the ideal generated by the $(n-k)$-minors of the presentation matrix, $n$ being the number of rows of the matrix. The Fitting ideals do not depend on the presentation of the module. Moreover, they are compatible with base change. We shall prove that $M$ is locally free of constant rank $r$ if and only if $F_r(M)=A$ and $F_{r-1}(M)=0$.

First of all we should like to generalize the notion of rank to arbitrary modules.

Definition 7.2.1. Let $A$ be a ring and $M$ be a finitely generated $A$-module. We say that $M$ has $\operatorname{rank} r$ if $M \otimes_A Q(A)$ is a free $Q(A)$-module of rank $r$, where $Q(A)$ denotes the total quotient ring of $A$.

Remark 7.2.2. If $A$ is an integral domain, then $Q(A)$ is a field and, hence, every finitely generated $A$-module $M$ has a rank, namely
$$\operatorname{rank}(M)=\operatorname{dim}_{Q(A)} M \otimes_A Q(A)$$
For arbitrary rings $A$, not every finitely generated $A$-module needs to have a rank (cf. Exercise 7.2.4).

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

Definition 7.3.1. Let $A$ be a ring. An $A$-module $M$ is called flat if, for every injective homomorphism $i: N \rightarrow L$, the induced map $N \otimes_A M \rightarrow L \otimes_A M$ is again injective. An $A$-algebra $B$ is called flat if it is flat as $A$-module.
Example 7.3.2. Free modules are flat (Exercise 2.7.7).
The following theorem gives a characterization of flatness using Tor.
Theorem 7.3.3. Let $A$ be a ring and $M$ an $A$-module. Then $M$ is flat if and only if $\operatorname{Tor}_1^A(A / I, M)=0$ for all finitely generated ideals $I \subset A$.
Proof. Consider the following exact sequence
$$0 \longrightarrow I \longrightarrow A \longrightarrow A \longrightarrow 0 .$$

We obtain a long exact sequence using Proposition 7.1.2:
$$0=\operatorname{Tor}_1^A(A, M) \longrightarrow \operatorname{Tor}_1^A(A / I, M) \longrightarrow I \otimes_A M \longrightarrow A \otimes_A M=M .$$
If $M$ is flat then $I \otimes_A M \rightarrow M$ is injective and, therefore, $\operatorname{Tor}_1^A(A / I, M)=0$.
Now assume $\operatorname{Tor}_1^A(A / I, M)=0$ for all finitely generated ideals $I \subset A$.
We have to prove that, for any injection $i: N \rightarrow L$, the induced map $N \otimes_A M \rightarrow L \otimes_A M$ is injective.

We consider first the case $N=I, L=A$ and $I \subset A$ an ideal (but not necessarily finitely generated). If $I \otimes_A M \rightarrow M$ is not injective, then there exists $\sum_{\nu=1}^r a_\nu \otimes m_\nu \in I \otimes A M$ different from zero with $\sum_{\nu=1}^r a_\nu m_\nu=0$. Let $I_0:=\left\langle a_1, \ldots, a_r\right\rangle$ then $\sum_{\nu=1}^r a_\nu \otimes m_\nu \in I_0 \otimes A M$ and, therefore, by assumption, it has to be zero. In particular, its image in $I \otimes_A M$ has to be zero, too. This implies that $I \otimes_A M \rightarrow M$ is injective for all ideals $I \subset A$.

## 数学代写|交换代数代写Commutative Algebra代考|Fitting Ideals

7.2.1.定义设$A$为一个环，$M$为一个有限生成的$A$ -模块。如果$M \otimes_A Q(A)$是一个秩为$r$的自由的$Q(A)$ -模块，我们说$M$有$\operatorname{rank} r$，其中$Q(A)$表示$A$的全商环。

$$\operatorname{rank}(M)=\operatorname{dim}_{Q(A)} M \otimes_A Q(A)$$

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

7.3.1.定义让$A$成为一个戒指。如果对于每个内射同态$i: N \rightarrow L$，诱导映射$N \otimes_A M \rightarrow L \otimes_A M$也是内射，则$A$ -模块$M$称为平面。如果一个$A$ -代数$B$像$A$ -模块一样是平坦的，那么它就被称为平坦的。

$$0 \longrightarrow I \longrightarrow A \longrightarrow A \longrightarrow 0 .$$

$$0=\operatorname{Tor}_1^A(A, M) \longrightarrow \operatorname{Tor}_1^A(A / I, M) \longrightarrow I \otimes_A M \longrightarrow A \otimes_A M=M .$$

## 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。