## 数学代写|交换代数代写commutative algebra代考|MATH0021

statistics-lab™ 为您的留学生涯保驾护航 在代写交换代数commutative algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写交换代数commutative algebra代写方面经验极为丰富，各种代写交换代数commutative algebra相关的作业也就用不着说。

## 数学代写|交换代数代写commutative algebra代考|Canonical forms for square matrices

Definitions and Remarks. Let $M$ be a module over the commutative ring $R$. An $R$-endomorphism of $M$, or simply an endomorphism of $M$, is just an $R$-homomorphism from $M$ to itself. We denote by $\operatorname{End}_R(M)$ the set of all $R$-endomorphisms of $M$. It is routine to check that $\operatorname{End}_R(M)$ is a ring under the addition defined in 6.27 and ‘multiplication’ given by composition of mappings: the identity element of this ring is $\operatorname{Id}_M$, the identity mapping of $M$ onto itself, while the zero element of $\operatorname{End}_R(M)$ is the zero homomorphism $0: M \rightarrow M$ defined in 6.27.

The reader should be able to construct easy examples from vector space theory which show that, in general, the ring $\operatorname{End}_R(M)$ need not be commutative.

For each $\psi \in \operatorname{End}_R(M)$ and each $r \in R$, we define $r \psi: M \rightarrow M$ by the rule $(r \psi)(m)=r \psi(m)$ for all $m \in M$. It is routine to check that $r \psi$ is again an endomorphism of $M$. Observe that the effect of $r \operatorname{Id}_M$ (for $r \in R$ ) on an element $m \in M$ is just to multiply $m$ by $r$. Note also that each $\psi \in \operatorname{End}_R(M)$ commutes with $r \operatorname{Id}_M$ for all $r \in R$ : in fact, an Abelian group homomorphism $\theta: M \rightarrow M$ belongs to $\operatorname{End}_R(M)$ if and only if it commutes with $r \operatorname{Id}_M$ for all $r \in R$.

For each $\psi \in \operatorname{End}_R(M)$ and each $r \in R$, we define $r \psi: M \rightarrow M$ by the rule $(r \psi)(m)=r \psi(m)$ for all $m \in M$. It is routine to check that $r \psi$ is again an endomorphism of $M$. Observe that the effect of $r \operatorname{Id}_M$ (for $r \in R$ ) on an element $m \in M$ is just to multiply $m$ by $r$. Note also that each $\psi \in \operatorname{End}_R(M)$ commutes with $r \operatorname{Id}_M$ for all $r \in R$ : in fact, an Abelian group homomorphism $\theta: M \rightarrow M$ belongs to $\operatorname{End}_R(M)$ if and only if it commutes with $r \operatorname{Id}_M$ for all $r \in R$.

## 数学代写|交换代数代写commutative algebra代考|Some applications to field theory

12.1 Definitions. A subset $F$ of a field $K$ is said to be a subfield of $K$ precisely when $F$ is itself a field with respect to the operations in $K$. We shall also describe this situation by saying that ‘ $K$ is an extension field of $F$ ‘, or ‘ $F \subseteq K$ is an extension of fields’.

When this is the case, $1_F=1_K$, so that $F$ is a subring of $K$ in the sense of 1.4 , because $1_F^2=1_F=1_F 1_K$ in $K$.

We say that $K$ is an intermediate field between $F$ and $L$ precisely when $F \subseteq K$ and $K \subseteq L$ are extensions of fields.

A mapping $f: K_1 \rightarrow K_2$, where $K_1, K_2$ are fields, is a homomorphism, or a field homomorphism, precisely when it is a ring homomorphism. When this is the case, $\operatorname{Ker} f=\left{0_{K_1}\right}$ (because it must be a proper ideal of $K_1$ ), and so $f$ is injective by 2.2 .
12.2 ExAmples. Let $K$ be a field and let $X$ be an indeterminate.
(i) Denote by $K(X)$ the field of fractions of the integral domain $K[X]$. The composition $K \rightarrow K[X] \rightarrow K(X)$ of the natural injective ring homomorphisms enables us to consider $K(X)$ as a field extension of $K$. We refer to $K(X)$ as the field of rational functions in $X$ with coefficients in $K$. A typical element of $K(X)$ can be written in the form $f / g$, where $f, g$ are polynomials in $X$ with coefficients in $K$ and $g \neq 0$.
(ii) Let $m \in K[X]$ be a monic irreducible polynomial in $X$ with coefficients in $K$. By 3.34, the ring $L:=K[X] / m K[X]$ is a field, and the composition
$$K \rightarrow K[X] \rightarrow K[X] / m K[X]=L$$
of the natural ring homomorphisms must be injective (by 12.1) even though the second ring homomorphism is not; this composition enables us to regard $L$ as an extension field of $K$. Observe also that, if we denote by $\alpha$ the natural image $X+m K[X]$ of $X$ in $L$, then $m(\alpha)=0$ : to see this, let $m=\sum_{i=0}^n a_i X^i$ (where $a_n=1$ ), and note that
$$m(\alpha)=\sum_{i=0}^n a_i(X+m K[X])^i=m+m K[X]=0_L$$
(because an $a \in K$ is identified with its natural image $a+m K[X] \in L$ ).

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Some applications to field theory

12.1定义。就$K$中的操作而言，当$F$本身就是一个字段时，将字段$K$的子集$F$称为$K$的子字段。我们还可以这样描述这种情况:“$K$是$F$的扩展字段”，或者“$F \subseteq K$是字段的扩展字段”。

12.2示例。设$K$为字段，$X$为不确定值。
(i)用$K(X)$表示积分域$K[X]$的分数域。天然注入环同态的组成$K \rightarrow K[X] \rightarrow K(X)$使我们可以把$K(X)$看作$K$的域扩展。我们称$K(X)$为$X$中有理函数的域，其系数在$K$中。$K(X)$的一个典型元素可以写成$f / g$的形式，其中$f, g$是$X$中的多项式，系数在$K$和$g \neq 0$中。
(ii)设$m \in K[X]$为$X$中的一元不可约多项式，其系数在$K$中。通过3.34，圆环$L:=K[X] / m K[X]$是一个场，并组成
$$K \rightarrow K[X] \rightarrow K[X] / m K[X]=L$$

$$m(\alpha)=\sum_{i=0}^n a_i(X+m K[X])^i=m+m K[X]=0_L$$
(因为$a \in K$与它的自然图像$a+m K[X] \in L$是一致的)。

## 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。

## 数学代写|交换代数代写commutative algebra代考|MATH5020

statistics-lab™ 为您的留学生涯保驾护航 在代写交换代数commutative algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写交换代数commutative algebra代写方面经验极为丰富，各种代写交换代数commutative algebra相关的作业也就用不着说。

## 数学代写|交换代数代写commutative algebra代考|Rings of fractions

9.1 Remarks. Let $M$ be a module over the commutative ring $R$. By a minimal generating set for $M$ we shall mean a subset, say $\Delta$, of $M$ such that $\Delta$ generates $M$ but no proper subset of $\Delta$ generates $M$.

Observe that if $M$ is finitely generated, by $g_1, \ldots, g_n$ say, then a minimal generating set $\Delta$ for $M$ must be finite: this is because each $g_i(1 \leq i \leq n)$ can be expressed as
$$g_i=\sum_{\delta \in \Delta} r_{i \delta} \delta$$
with $r_{i \delta} \in R$ for all $\delta \in \Delta$ and almost all the $r_{i \delta}$ zero, so that the finite subset $\Delta^{\prime}$ of $\Delta$ given by
$$\Delta^{\prime}=\bigcup_{i=1}^n\left{\delta \in \Delta: r_{i \delta} \neq 0\right}$$
also generates $M$. (Recall that ‘almost all’ is an abbreviation for ‘all except possibly finitely many’.)

However, even a finitely generated $R$-module $M$ may have two minimal generating sets which have different cardinalities; that is, the number of elements in one minimal generating set for $M$ need not be the same as the number in another. To give an example of this phenomenon, consider the $\mathbb{Z}$-module $\mathbb{Z} / 6 \mathbb{Z}$ : it is easy to check that ${1+6 \mathbb{Z}}$ and ${2+6 \mathbb{Z}, 3+6 \mathbb{Z}}$ are both minimal generating sets for $\mathbb{Z} / 6 \mathbb{Z}$.

Of course, this unpleasant situation does not occur in vector space theory: a minimal generating set for a finitely generated vector space over a field forms a basis. We show now that it cannot occur for finitely generated modules over quasi-local rings. An interesting aspect of the discussion is that we use Nakayama’s Lemma to reduce to a situation where we can use vector space theory.

## 数学代写|交换代数代写commutative algebra代考|Modules over principal ideal domains

10.1 $\sharp$ EXERCISE. Let $R$ be a commutative ring, let $n \in \mathbb{N}$ and let $F$ be a free $R$-module with a base $\left(e_i\right)_{i=1}^n$ of $n$ elements. Let $c_1, \ldots, c_n \in R$. By

$$f: F \longrightarrow \bigoplus_{i=1}^n R / R c_i$$
such that $f\left(\sum_{i=1}^n r_i e_i\right)=\left(r_1+R c_1, \ldots, r_n+R c_n\right)$ for all $r_1, \ldots, r_n \in R$, and clearly $f$ is an epimorphism.

Show that $\operatorname{Ker} f$ is generated by $c_1 e_1, \ldots, c_n e_n$. Show further that, if $R$ is an integral domain, then $\operatorname{Ker} f$ is free, and determine $\operatorname{rank}(\operatorname{Ker} f)$ in this case.

Now let us return to our finitely generated module, $G$ say, over our PID $R$. Suppose that $G$ has a generating set with $n$ elements, where $n \in \mathbb{N}$. By $6.57, G$ can be expressed as a homomorphic image of a free $R$-module $F$ with a base $\left(e_i\right){i=1}^n$ of $n$ elements. Thus there is a submodule $H$ of $F$ such that $F / H \cong G$. If we could find $c_1, \ldots, c_n \in R$ such that $H$ is generated by $c_1 e_1, \ldots, c_n e_n$, then it would follow from 10.1 above that $$G \cong F / H \cong R / R c_1 \oplus \cdots \oplus R / R c_n,$$ a direct sum of cyclic $R$-modules. In general, it is too much to hope that, for a specified base $\left(e_i\right){i=1}^n$ for $F$, it will always be possible to find such $c_1, \ldots, c_n \in R$ with the property that $H$ is generated by $c_1 e_1, \ldots, c_n e_n$ : just consider the $\mathbb{Z}$-submodule of $F^{\prime}:=\mathbb{Z} \oplus \mathbb{Z}$ generated by $(1,3)$ and the base $\left(\tilde{e}i\right){i=1}^2$ for $F^{\prime}$ given by $\tilde{e}1=(1,0), \tilde{e}_2=(0,1)$. However, in this example, $(1,3)$ and $(0,1)$ form another base for $F^{\prime}$, and this is symptomatic of the general situation: we shall see that, given the submodule $H$ of the above finitely generated free module $F$ over the PID $R$, then it is always possible to find a base $\left(e_i^{\prime}\right){i=1}^n$ for $F$ and $c_1^{\prime}, \ldots, c_n^{\prime} \in R$ such that $H$ is generated by $c_1^{\prime} e_1^{\prime}, \ldots, c_n^{\prime} e_n^{\prime}$. This result provides the key to some of the main results of the chapter; our proof of it makes significant use of the fact that $R$ is a PID.

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Rings of fractions

9.1备注设$M$为可交换环$R$上的一个模。所谓$M$的最小生成集，我们指的是$M$的一个子集，比如$\Delta$，使得$\Delta$生成$M$，而$\Delta$的适当子集不能生成$M$。

$$g_i=\sum_{\delta \in \Delta} r_{i \delta} \delta$$

$$\Delta^{\prime}=\bigcup_{i=1}^n\left{\delta \in \Delta: r_{i \delta} \neq 0\right}$$

## 数学代写|交换代数代写commutative algebra代考|Modules over principal ideal domains

10.1 $\sharp$ 锻炼。让 $R$ 是交换环，设 $n \in \mathbb{N}$ 让 $F$ 做一个自由的人 $R$-带底座的模块 $\left(e_i\right)_{i=1}^n$ 的 $n$ 元素。让 $c_1, \ldots, c_n \in R$． By

$$f: F \longrightarrow \bigoplus_{i=1}^n R / R c_i$$

## 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。

## 数学代写|交换代数代写commutative algebra代考|MATH662

statistics-lab™ 为您的留学生涯保驾护航 在代写交换代数commutative algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写交换代数commutative algebra代写方面经验极为丰富，各种代写交换代数commutative algebra相关的作业也就用不着说。

## 数学代写|交换代数代写commutative algebra代考|Rings of fractions

Recall the construction: if $R$ is an integral domain, then $S:=R \backslash{0}$ is a multiplicatively closed subset of $R$ in the sense of 3.43 (that is $1 \in S$ and $S$ is closed under multiplication); an equivalence relation $\sim$ on $R \times S$ given by, for $(a, s),(b, t) \in R \times S$,
$$(a, s) \sim(b, t) \quad \Longleftrightarrow \quad a t-b s=0$$
is considered; the equivalence class which contains $(a, s)$ (where $(a, s) \in$ $R \times S$ ) is denoted by $a / s$; and the set of all the equivalence classes of $\sim$ can be given the structure of a field in such a way that the rules for addition and multiplication resemble exactly the familiar high school rules for addition and multiplication of fractions.

The generalization which concerns us in this chapter applies to any multiplicatively closed subset $S$ of an arbitrary commutative ring $R$ : once again, we consider an equivalence relation on the set $R \times S$, but in this case the definition of the relation is more complicated in order to overcome problems created by the possible presence of zerodivisors. Apart from this added complication, the construction is remarkably similar to that of the field of fractions of an integral domain, although the end product does not have quite such good properties: we do not often get a field, and, in fact, the general construction yields what is known as the ring of fractions $S^{-1} R$ of $R$ with respect to the multiplicatively closed subset $S$; this ring of fractions may have non-zero zerodivisors; and, although there is a natural ring homomorphism $f: R \rightarrow S^{-1} R$, this map is not automatically injective.
However, on the credit side, we should point out right at the beginning that one of the absolutely fundamental examples of this construction arises when we take for the multiplicatively closed subset $S$ of $R$ the complement $R \backslash P$ of a prime ideal $P$ of $R$ : in this case, the new ring of fractions $S^{-1} R$ turns out to be a quasi-local ring, denoted by $R_P$; furthermore, the passage from $R$ to $R_P$ for appropriate $P$, referred to as ‘localization at $P$ ‘, is often a powerful tool in commutative algebra.

## 数学代写|交换代数代写commutative algebra代考|Modules

Definition. Let $R$ be a commutative ring. A module over $R$, or an $R$-module, is an additively written Abelian group $M$ furnished with a ‘scalar multiplication’ of its elements by elements of $R$, that is, a mapping
$$\text { . }: R \times M \rightarrow M \text {, }$$

such that
(i) $r .\left(m+m^{\prime}\right)=r . m+r . m^{\prime}$ for all $r \in R, m, m^{\prime} \in M$,
(ii) $\left(r+r^{\prime}\right) \cdot m=r . m+r^{\prime} \cdot m$ for all $r, r^{\prime} \in R, m \in M$,
(iii) $\left(r r^{\prime}\right) \cdot m=r \cdot\left(r^{\prime} \cdot m\right)$ for all $r, r^{\prime} \in R, m \in M$, and
(iv) $1_R \cdot m=m$ for all $m \in M$.
6.2 REMARKs. (i) In practice, the ‘ ‘ denoting scalar multiplication of a module element by a ring element is usually omitted.
(ii) The axioms in 6.1 should be familiar to the reader from his undergraduate studies of vector spaces. Indeed, a module over a field $K$ is just a vector space over $K$. In our study of module theory, certain fundamental facts about vector spaces will play a crucial rôle: it will be convenient for us to introduce the abbreviation $K$-space for the more cumbersome ‘vector space over $K$ ‘.
(iii) The axioms in 6.1 have various easy consequences regarding the manipulation of expressions involving addition, subtraction and scalar multiplication, such as, for example, the fact that
$$\left(r-r^{\prime}\right) m=r m-r^{\prime} m \quad \text { for all } r, r^{\prime} \in R \text { and } m \in M \text {. }$$
We shall not dwell on such points.

Examples. Let $R$ be a commutative ring, and let $I$ be an ideal of $R$.
(i) A very important example of an $R$-module is $R$ itself: $R$ is, of course, an Abelian group, the multiplication in $R$ gives us a mapping
$$\text { . : } R \times R \rightarrow R,$$
and the ring axioms ensure that this ‘scalar multiplication’ turns $R$ into an $R$-module.
(ii) Since $I$ is closed under addition and under multiplication by arbitrary elements of $R$, it follows that $I$ too is an $R$-module under the addition and multiplication of $R$.
(iii) We show next that the residue class ring $R / I$ can be viewed as an $R$-module. Of course, $R / I$ has a natural Abelian group structure; we need to provide it with a scalar multiplication by elements of $R$. To this end, let $s, s^{\prime} \in R$ be such that $s+I=s^{\prime}+I$ in $R / I$, and let $r \in R$. Thus $s-s^{\prime} \in I$, and so $r s-r s^{\prime}=r\left(s-s^{\prime}\right) \in I$; hence $r s+I=r s^{\prime}+I$. It follows that we can unambiguously define a mapping
$$\begin{array}{ccc} R \times R / I & \longrightarrow & R / I \ (r, s+I) & \longmapsto & r s+I, \end{array}$$
and it is routine to check that $R / I$ becomes an $R$-module with respect to this ‘scalar multiplication’.

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Modules

$$\text { . }: R \times M \rightarrow M \text {, }$$

(i) $r .\left(m+m^{\prime}\right)=r . m+r . m^{\prime}$适用于所有$r \in R, m, m^{\prime} \in M$;
(ii) $\left(r+r^{\prime}\right) \cdot m=r . m+r^{\prime} \cdot m$适用于所有$r, r^{\prime} \in R, m \in M$;
(iii) $\left(r r^{\prime}\right) \cdot m=r \cdot\left(r^{\prime} \cdot m\right)$适用于所有$r, r^{\prime} \in R, m \in M$，以及
(iv) $1_R \cdot m=m$适用于所有$m \in M$。
6.2备注(i)在实践中，表示模块元素与环元素的标量乘法的“”通常被省略。
(ii)在本科学习过向量空间的读者应该熟悉6.1中的公理。事实上，一个模在一个场$K$上就是一个向量空间在$K$上。在我们对模块理论的研究中，关于向量空间的一些基本事实将起到至关重要的rôle作用:为了方便起见，我们引入了缩写$K$ -space来代替更繁琐的“$K$上的向量空间”。
(iii) 6.1中的公理对于涉及加法、减法和标量乘法的表达式的操作有各种简单的结果，例如
$$\left(r-r^{\prime}\right) m=r m-r^{\prime} m \quad \text { for all } r, r^{\prime} \in R \text { and } m \in M \text {. }$$

(i) $R$ -模块的一个非常重要的例子是$R$本身:$R$当然是一个阿贝尔群，在$R$中的乘法给我们一个映射
$$\text { . : } R \times R \rightarrow R,$$

(ii)由于$I$对$R$的任意元素的加法和乘法是封闭的，因此$I$在$R$的加法和乘法下也是一个$R$ -模块。
(iii)接下来我们将证明剩余类环$R / I$可以看作是一个$R$ -模块。当然，$R / I$有一个天然的阿贝尔群结构;我们需要为它提供一个标量乘以$R$的元素。为此目的，让$s, s^{\prime} \in R$变成$s+I=s^{\prime}+I$在$R / I$中，让$r \in R$。于是$s-s^{\prime} \in I$，于是$r s-r s^{\prime}=r\left(s-s^{\prime}\right) \in I$;因此，$r s+I=r s^{\prime}+I$。因此，我们可以明确地定义映射
$$\begin{array}{ccc} R \times R / I & \longrightarrow & R / I \ (r, s+I) & \longmapsto & r s+I, \end{array}$$

## 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。

## 数学代写|交换代数代写commutative algebra代考|Math6170

statistics-lab™ 为您的留学生涯保驾护航 在代写交换代数commutative algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写交换代数commutative algebra代写方面经验极为丰富，各种代写交换代数commutative algebra相关的作业也就用不着说。

## 数学代写|交换代数代写commutative algebra代考|Big modules

Lemma 3.66. (Kaplansky) Let $R$ be a ring, and let $F$ be an $R$-module which is a direct sum of countably generated submodules: say $F=\bigoplus_{\lambda \in \Lambda} E_\lambda$. Then every direct summand of $F$ is again a direct sum of countably generated submodules.
Proof. We ClaIM that there is an ordinal filtration $\left{F_i\right}_{i \leq \alpha}$ on $F$ satisfying all of the following properties. (i) For all $i<\alpha, F_{i+1} / F_i$ is countably generated. (ii) If $M_i=F_i \cap M, N_i=F_i \cap N$, then $F_i=M_i \oplus N_i$.
(iii) For each $i$ there is a subset $\Lambda_i$ of $\Lambda$ such that $F_i=\bigoplus_{\lambda \in \Lambda_i} \Lambda_i$.
SUFFICIENCY OF CLAIM: If so, $\left{M_i\right}_{i \leq \alpha}$ is an ordinal filtration on $M$. Moreover, since $M_i \subset M_{i+1}$ are both direct summands of $F, M_i$ is a direct summand of $M_{i+1}$. The Transfinite Dévissage Lemma (Lemma 3.51) applies to give
$$M \cong \operatorname{Gr}(M)=\bigoplus_{i<\alpha} M_{i+1} / M_i .$$
Moreover, for all $i<\alpha$ we have
$$F_{i+1} / F_i=\left(M_{i+1} \oplus N_{i+1}\right) /\left(M_i \oplus N_i\right) \cong M_{i+1} / M_i \oplus N_{i+1} / N_i,$$
which shows that each successive quotient $M_{i+1} / M_i$ is countably generated. Therefore $M$ is a direct sum of countably generated submodules.
PROOF OF CLAIM: We will construct the filtration by transfinite induction. The base case and the limit ordinal induction step are forced upon us by the definition of ordinal filtration: we must have $F_0={0}$, and for any limit ordinal $\beta \leq \alpha$, assuming we have defined $F_i$ for all $i<\beta$ we must have $F_\beta=\bigcup_{i<\beta} F_i$.

So consider the case of a successor ordinal $\beta=\beta^{\prime}+1$. Let $Q_1$ be any $E_\lambda$ which is not contained in $F_{\beta^{\prime}}$. (Otherwise we have $F_{\beta^{\prime}}=F$ and we may just define $F_i=F$ for all $\beta \leq i \leq \alpha$.) Let $x_{11}, x_{12}, \ldots$ be a sequence of generators of $Q_1$, and decompose $x_{11}$ into its $M$ – and $N$-components. Let $Q_2$ be the direct sum of the finitely many $E_\lambda$ which are necessary to write both of these components, and let $x_{21}, x_{22}, \ldots$ be a sequence of generators for $Q_2$. Similarly decompose $x_{12}$ into $M$ and $N$ components, and let $Q_3$ be the direct sum of the finitely many $E_\lambda$ needed to write out these components, and let $x_{31}, x_{32}, \ldots$ be a sequence of generators of $Q_3$. We continue to carry out this procedure for all $x_{i j}$, proceeding according to a diagonal enumeration of $\mathbb{Z}^{+} \times \mathbb{Z}^{+}$: i.e., $x_{11}, x_{12}, x_{21}, x_{13}, x_{22}, x_{31}, \ldots$ Put $F_\beta=\left\langle F_{\beta^{\prime}},\left{x_{i j}\right}_{i, j \in \mathbb{Z}^{+}}\right\rangle_R$. This works!

## 数学代写|交换代数代写commutative algebra代考|Co/chain complexes

Let $R$ be a ring. A chain complex $C$. of $R$-modules is a family $\left{C_n\right}_{n \in \mathbb{Z}}$ of $R$-modules together with for all $n \in \mathbb{Z}$, an $R$-module map $d_n: C_n \rightarrow C_{n-1}$ such that for all $n, d_{n-1} \circ d_n=0$. (It is often the case that $C_n=0$ for all $n<0$, but this is not a required part of the definition.)

An example of a chain complex of $R$-modules is any long exact sequence. However, from the perspective of homology theory this is a trivial example in the following precise sense: for any chain complex we may define its homology modules: for all $n \in \mathbb{Z}$, we put
$$H_n(C)=\operatorname{Ker}\left(d_n\right) / \operatorname{Im}\left(d_{n+1}\right) .$$
Example: Let $X$ be any topological space. For any ring $R$, we have the singular chain complex $S(X)$. $S(X)n=0$ for $n<0$, and for $n \geq 0, S(X)_n$ is the free $R$-module with basis the set of all continuous maps $\Delta_n \rightarrow X$, where $\Delta_n$ is the standard $n$-dimensional simplex. A certain carefully defined alternating sum of restrictions to faces of $\Delta_n$ gives rise to a boundary map $d_n: S(X)_n \rightarrow S(X){n-1}$, and the indeed the homology groups of this complex are nothing else than the singular homology groups $H_n(X, R)$ with coefficients in $R$.

If $C_{\boldsymbol{\bullet}}$ and $D_{\boldsymbol{\bullet}}$ are two chain complexes of $R$-modules, a homomorphism $\eta: C_{\boldsymbol{\bullet}} \rightarrow$ $D_{\bullet}$ is given by maps $\eta_n: C_n \rightarrow D_n$ for all $n$ rendering the following infinite ladder commutative:
INSERT ME!.
In this way one has evident notions of a monomorphism and epimorphisms of chain complexes. In fact the chain complexes of $R$-modules form an abelian category and thus these notions have a general categorical meaning, but it turns out they are equivalent to the much more concrete naive conditions: $\eta$ is a monomorphism iff each $\eta_n$ is injective and is an epiomorphism iff each $\eta_n$ is surjective.
In particular it makes sense to consider a short exact sequence of chain complexes:
$$0 \longrightarrow A_{\bullet} \longrightarrow B_{\bullet} \longrightarrow C_{\bullet}$$

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Big modules

(iii)对于每个$i$, $\Lambda$有一个子集$\Lambda_i$，使得$F_i=\bigoplus_{\lambda \in \Lambda_i} \Lambda_i$。

$$M \cong \operatorname{Gr}(M)=\bigoplus_{i<\alpha} M_{i+1} / M_i .$$

$$F_{i+1} / F_i=\left(M_{i+1} \oplus N_{i+1}\right) /\left(M_i \oplus N_i\right) \cong M_{i+1} / M_i \oplus N_{i+1} / N_i,$$

## 数学代写|交换代数代写commutative algebra代考|Co/chain complexes

$$H_n(C)=\operatorname{Ker}\left(d_n\right) / \operatorname{Im}\left(d_{n+1}\right) .$$

9.8事实设$M$为一个有限生成的a模块。

## 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。

## 数学代写|交换代数代写commutative algebra代考|MATH662

statistics-lab™ 为您的留学生涯保驾护航 在代写交换代数commutative algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写交换代数commutative algebra代写方面经验极为丰富，各种代写交换代数commutative algebra相关的作业也就用不着说。

## 数学代写|交换代数代写commutative algebra代考|Quasi-Integral Rings

6.1 Definition A ring is said to be integral if every element is null or regular. ${ }^4$ A ring $\mathbf{A}$ is said to be quasi-integral when every element admits as its annihilator an (ideal generated by an) idempotent. In the literature, a quasi-integral ring is sometimes called a pp-ring (principal ideals are projective, cf. Sect. V-7).

As usual, the “or” in the previous definition must be read as an explicit or. An integral ring is therefore a discrete set if and only if furthermore it is trivial or nontrivial. So, our nontrivial integral rings are precisely the “discrete domains” of [MRR].

In this work, sometimes we speak of a “nonzero element” in an integral ring, but we should actually say “regular element” in order not to exclude the trivial ring case.
6.2 Fact A pp-ring is reduced.
$D$ If $e$ is the idempotent annihilator of $x$ and if $x^2=0$, then $x \in\langle e\rangle$, therefore $x=e x=0$.

A discrete field is an integral ring. A ring $\mathbf{A}$ is integral if and only if its total ring of fractions Frac $\mathbf{A}$ is a discrete field. A finite product of pp-rings is a pp-ring. A ring is integral if and only if it is a connected pp-ring.

## 数学代写|交换代数代写commutative algebra代考|Equational Definition of pp-rings

In a pp-ring, for $a \in \mathbf{A}$, let $e_a$ be the unique idempotent such that $\operatorname{Ann}(a)=\left\langle 1-e_a\right\rangle$. We have $\mathbf{A} \simeq \mathbf{A}\left[1 / e_a\right] \times \mathbf{A} /\left\langle e_a\right\rangle$.
In the ring $\mathbf{A}\left[1 / e_a\right]$, the element $a$ is regular, and in $\mathbf{A} /\left\langle e_a\right\rangle, a$ is null.
We then have $e_{a b}=e_a e_b, e_a a=a$ and $e_0=0$.
Conversely, suppose that a commutative ring is equipped with a unary law $a \mapsto a^{\circ}$ which satisfies the following three axioms
$$a^{\circ} a=a, \quad(a b)^{\circ}=a^{\circ} b^{\circ}, \quad 0^{\circ}=0 .$$
Then, for all $a \in \mathbf{A}$, we have $\operatorname{Ann}(a)=\left\langle 1-a^{\circ}\right\rangle$, and $a^{\circ}$ is idempotent, such that the ring is a pp-ring.
Indeed, first of all $\left(1-a^{\circ}\right) a=0$, and if $a x=0$, then
$$a^{\circ} x=a^{\circ} x^{\circ} x=(a x)^{\circ} x=0^{\circ} x=0,$$
so $x=\left(1-a^{\circ}\right) x$. Hence $\operatorname{Ann}(a)=\left\langle 1-a^{\circ}\right\rangle$. Next let us show that $a^{\circ}$ is idempotent. Apply the previous result to $x=1-a^{\circ}$ which satisfies $a x=0$ (by the first axiom); the equality $x=\left(1-a^{\circ}\right) x$ gives $x=x^2$, i.e. the element $1-a^{\circ}$ is idempotent.
The following splitting lemma is almost immediate.

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Quasi-Integral Rings

6.1定义如果环的每个元素都为空或正则，则称环为整环。${ }^4$当每个元素都承认一个(由一个产生的理想)幂等子为它的湮灭子时，一个环$\mathbf{A}$被称为准积分。在文献中，拟整环有时被称为pp环(主理想是射影的，参见第V-7节)。

6.2事实一个pp环被简化了。
$D$如果$e$是$x$的幂等湮灭子，如果$x^2=0$，那么$x \in\langle e\rangle$，因此$x=e x=0$。

## 数学代写|交换代数代写commutative algebra代考|Equational Definition of pp-rings

$$a^{\circ} a=a, \quad(a b)^{\circ}=a^{\circ} b^{\circ}, \quad 0^{\circ}=0 .$$

$$a^{\circ} x=a^{\circ} x^{\circ} x=(a x)^{\circ} x=0^{\circ} x=0,$$

## 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。

## 数学代写|交换代数代写commutative algebra代考|Newton’s Method in Algebra

statistics-lab™ 为您的留学生涯保驾护航 在代写交换代数commutative algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写交换代数commutative algebra代写方面经验极为丰富，各种代写交换代数commutative algebra相关的作业也就用不着说。

## 数学代写|交换代数代写commutative algebra代考|Newton’s Method in Algebra

Let $\mathbf{k}$ be a ring and $f_1, \ldots, f_s \in \mathbf{k}[X]=\mathbf{k}\left[X_1, \ldots, X_n\right]$. The Jacobian matrix of the system is the matrix
$$\mathrm{JAC}{X_1, \ldots, X_n}\left(f_1, \ldots, f_s\right)=\left(\frac{\partial f_i}{\partial X_j}\right){i \in \llbracket 1 . . s \rrbracket, j \in \llbracket 1 . . n \rrbracket} \in \mathbf{k}[X]^{s \times n} .$$
It is also denoted by $\mathrm{JAC}_{X}(f)$ or $\mathrm{JAC}(f)$. It is visualized as follows

If $s=n$, we denote by $\operatorname{Jac}{X}(f)$ or $\operatorname{Jac}{X_1, \ldots, X_n}\left(f_1, \ldots, f_n\right)$ or $\operatorname{Jac}(f)$ the Jacobian of the system $(f)$, i.e. the determinant of the Jacobian matrix.

In analysis Newton’s method to approximate a root of a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ is the following. Starting from a point $x_0$ “near a root,” at which the derivative is “far from 0 “, we construct a series $\left(x_m\right){m \in \mathbb{N}}$ by induction by letting $$x{m+1}=x_m-\frac{f\left(x_m\right)}{f^{\prime}\left(x_m\right)} .$$
The method can be generalized for a system of $p$ equations with $p$ unknowns. A solution of such a system is a zero of a function $f: \mathbb{R}^p \rightarrow \mathbb{R}^p$. We apply “the same formula” as above
$$x_{m+1}=x_m-f^{\prime}\left(x_m\right)^{-1} \cdot f\left(x_m\right),$$
where $f^{\prime}(x)$ is the differential (the Jacobian matrix) of $f$ at the point $x \in \mathbb{R}^p$, which must be invertible in a neighborhood of $x_0$.

## 数学代写|交换代数代写commutative algebra代考|Definition, Changing Generator Set

A finitely presented module is an A-module $M$ given by a finite number of generators and relations. Therefore it is a module with a finite generator set having a finitely generated syzygy module. Equivalently, it is a module $M$ isomorphic to the cokernel of a linear map
$$\gamma: \mathbf{A}^m \longrightarrow \mathbf{A}^q$$
The matrix $G \in \mathbf{A}^{q \times m}$ of $\gamma$ has as its columns a generator set of the syzygy module between the generators $g_i$ which are the images of the canonical base of $\mathbf{A}^q$ by the surjection $\pi: \mathbf{A}^q \rightarrow M$. Such a matrix is called a presentation matrix of the module $M$ for the generator set $\left(g_1, \ldots, g_q\right)$. This translates into

• $\left[g_1 \cdots g_q\right] G=0$, and
• every syzygy between the $g_i$ ‘s is a linear combination of the columns of $G$, i.e.: if $\left[g_1 \cdots g_q\right] C=0$ with $C \in \mathbf{A}^{q \times 1}$, there exists a $C^{\prime} \in \mathbf{A}^{m \times 1}$ such that $C=G C^{\prime}$.

Examples
1) A free module of rank $k$ is a finitely presented module presented by a matrix column formed of $k$ zeros. ${ }^1$ More generally every simple matrix is the presentation matrix of a free module of finite rank.
2) Recall that a finitely generated projective module is a module $P$ isomorphic to the image of a projection matrix $F \in \mathbb{M}_n(\mathbf{A})$ for a specific integer $n$. Since $\mathbf{A}^n=\operatorname{Im}(F) \oplus \operatorname{Im}\left(\mathrm{I}_n-F\right)$, we obtain $P \simeq \operatorname{Coker}\left(\mathrm{I}_n-F\right)$. This shows that every finitely generated projective module is finitely presented.
3) Let $\varphi: V \rightarrow V$ be an endomorphism of a finite-dimensional vector space over a discrete field $\mathbf{K}$. Consider $V$ as a $\mathbf{K}[X]$-module with the following external law
\left{\begin{aligned} \mathbf{K}[X] \times V & \rightarrow V \ (P, u) & \mapsto P \cdot u:=P(\varphi)(u) . \end{aligned}\right.
Let $\left(u_1, \ldots, u_n\right)$ be a basis of $V$ as a $\mathbf{K}$-vector space and $A$ be the matrix of $\varphi$ with respect to this basis. Then we can show that a presentation matrix of $V$ as a $\mathbf{K}[X]$-module for the generator set $\left(u_1, \ldots, u_n\right)$ is the matrix $X \mathrm{I}_n-A$ (see Exercise 3).

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Newton’s Method in Algebra

$$\mathrm{JAC}{X_1, \ldots, X_n}\left(f_1, \ldots, f_s\right)=\left(\frac{\partial f_i}{\partial X_j}\right){i \in \llbracket 1 . . s \rrbracket, j \in \llbracket 1 . . n \rrbracket} \in \mathbf{k}[X]^{s \times n} .$$

$$x_{m+1}=x_m-f^{\prime}\left(x_m\right)^{-1} \cdot f\left(x_m\right),$$

## 数学代写|交换代数代写commutative algebra代考|Definition, Changing Generator Set

$$\gamma: \mathbf{A}^m \longrightarrow \mathbf{A}^q$$
$\gamma$的矩阵$G \in \mathbf{A}^{q \times m}$的列是生成器$g_i$之间的syzygy模块的生成集，这些生成器是通过射$\pi: \mathbf{A}^q \rightarrow M$得到的$\mathbf{A}^q$的规范基的图像。这样的矩阵称为发电机组$\left(g_1, \ldots, g_q\right)$模块$M$的表示矩阵。这转化为

$\left[g_1 \cdots g_q\right] G=0$，和

$g_i$之间的每一个协同都是$G$的列的线性组合，即:如果$\left[g_1 \cdots g_q\right] C=0$与$C \in \mathbf{A}^{q \times 1}$，则存在一个$C^{\prime} \in \mathbf{A}^{m \times 1}$，使得$C=G C^{\prime}$。

1)秩为$k$的自由模块是由$k$个零组成的矩阵列表示的有限呈现模块。${ }^1$更一般地说，每个简单矩阵都是有限秩自由模的表示矩阵。
2)回想一下，一个有限生成的投影模块是一个模块$P$同构于一个特定整数$n$的投影矩阵$F \in \mathbb{M}_n(\mathbf{A})$的像。由于$\mathbf{A}^n=\operatorname{Im}(F) \oplus \operatorname{Im}\left(\mathrm{I}_n-F\right)$，我们得到$P \simeq \operatorname{Coker}\left(\mathrm{I}_n-F\right)$。这说明每一个有限生成的投影模都是有限表示的。
3)设$\varphi: V \rightarrow V$为离散场$\mathbf{K}$上有限维向量空间的自同态。将$V$视为具有以下外部律的$\mathbf{K}[X]$ -模块
\left{\begin{aligned} \mathbf{K}[X] \times V & \rightarrow V \ (P, u) & \mapsto P \cdot u:=P(\varphi)(u) . \end{aligned}\right.