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# 数学代写|交换代数代写Commutative Algebra代考|MATH483 Quotient Rings

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

Given a ring and an adequate equivalence relation on that ring, the goal of this section is to endow the set of equivalence classes with the structure of a ring. This will allow us to formally make all elements of an ideal equal to zero without modifying the other rules for computation.
1.5.1. Construction – Let $\mathscr{R}$ be binary relation on a set $\mathrm{X}$. Recall that one says that $\mathscr{R}$ is an equivalence relation if it is reflexive (for any $x, x \mathscr{R} x$ ), symmetric (if $x \mathscr{R} y$, then $y \mathscr{R} x$ ) and transitive (if $x \mathscr{R} y$ and $y \mathscr{R} z$, then $x \mathscr{R} z$ ). The set of all equivalence classes of $\mathrm{X}$ for the relation $\mathscr{R}$ is denoted by $\mathrm{X} / \mathscr{R}$; the map $\mathrm{cl}{\mathscr{R}}: \mathrm{X} \rightarrow \mathrm{X} / \mathscr{R}$ (such that, for every $x \in \mathrm{X}, \mathrm{cl}{\mathscr{R}}(x)$ is the equivalence class of $x$ ) is a surjection; its important property is that any two elements have the same image if and only if they are in relation by $\mathscr{R}$.

Let $\mathrm{A}$ be a ring. Let us search for all equivalence relations which are compatible with the ring structure, namely
if $x \mathscr{R} y$ and $x^{\prime} \mathscr{R} y^{\prime}$, then $x+x^{\prime} \mathscr{R} y+y^{\prime}$ and $x x^{\prime} \mathscr{R} y y^{\prime}$.
Let I be the equivalence class of 0 . If $x \mathscr{R} y$, since $(-y) \mathscr{R}(-y)$, one gets $x-y \mathscr{R} 0$, hence $x-y \in \mathrm{I}$, and conversely. Therefore, the relation $\mathscr{R}$ can be recovered from I by the property: $x \mathscr{R} y$ if and only if $x-y \in \mathrm{I}$.

On the other hand, let us show that I is a two-sided ideal of $A$. Of course, one has $0 \in \mathrm{I}$. Moreover, for any $x \in \mathrm{I}$ and $y \in \mathrm{I}$, one has $x \mathscr{R} 0$ and $y \mathscr{R} 0$, so that $(x+y) \mathscr{R} 0$, which shows that $x+y \in \mathrm{I}$. Finally, for $x \in \mathrm{I}$ and $a \in \mathrm{A}$, one has $x \mathscr{R} 0$, hence $a x \mathscr{R} a 0$ and $x a \mathscr{R} 0 a$; since $a 0=0 a=0$, we get that $a x \in \mathrm{I}$ and $x a \in \mathrm{I}$.

In the opposite direction, the preceding computations show that the following theorem holds.

## 数学代写|交换代数代写Commutative Algebra代考|Chinese remainder theorem

1.5.7. Chinese remainder theorem – We say that two two-sided ideals I and $\mathrm{J}$ of a ring A are comaximal if $\mathrm{I}+\mathrm{J}=\mathrm{A}$. This notion gives rise to the general formulation of the Chinese remainder theorem.

Theorem (1.5.8). – Let $\mathrm{A}$ be a ring and let $\mathrm{I}$ and $\mathrm{J}$ be two-sided ideals of $\mathrm{A}$. Let us assume that $\mathrm{I}$ and $\mathrm{J}$ are comaximal. Then the canonical morphism $\mathrm{A} \rightarrow(\mathrm{A} / \mathrm{I}) \times(\mathrm{A} / \mathrm{J})$ given by $a \mapsto\left(\mathrm{cl}{\mathrm{I}}(a), \mathrm{cl}{\mathrm{J}}(a)\right)$ is surjective; its kernel is the two-sided ideal $\mathrm{I} \cap \mathrm{J}$ of $\mathrm{A}$. Passing to the quotient, we thus have an isomorphism
$$\mathrm{A} /(\mathrm{I} \cap \mathrm{J}) \simeq \mathrm{A} / \mathrm{I} \times \mathrm{A} / \mathrm{J} .$$
Corollary (1.5.9). – Let I and J be comaximal two-sided ideals of a ring A. For any pair $(x, y)$ of elements of $\mathrm{A}$, there exists an element $a \in \mathrm{A}$ such that $a \in x+\mathrm{I}$ and $a \in y+J$
Proof. – In the diagram of rings
we have to show the existence of a unique morphism $\varphi$, drawn as a dashed arrow, so that this diagram is commutative, and that $\varphi$ is an isomorphism. But the morphism A $\rightarrow \mathrm{A} / \mathrm{I} \times \mathrm{A} / \mathrm{J}$ maps $a \in \mathrm{A}$ to $\left(\mathrm{cl}{\mathrm{I}}(a), \mathrm{cl}{\mathrm{J}}(a)\right)$. Its kernel is thus $I \cap J$. By the factorization theorem, there exists a unique morphism $\varphi$ that makes the diagram commutative, and this morphism is injective. For any $a \in \mathrm{A}$, one has $\varphi\left(\mathrm{cl}{\mathrm{I} \cap \mathrm{J}}(a)\right)=\left(\mathrm{cl}{\mathrm{I}}(a), \mathrm{cl}_{\mathrm{J}}(a)\right)$.

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

1.5.1。建设一一让 $\mathscr{R}$ 是集合上的二元关系 $\mathrm{X}$. 回想一下，有人说过 $\mathscr{R}$ 是一个等价关系，如果它是自反的（对于任何 $x, x \mathscr{R} x$ )，对称 的 (如果 $x \mathscr{R} y$ ，然后 $y \mathscr{R} x$ ) 和传递（如果 $x \mathscr{R} y$ 和 $y \mathscr{R} z$ ，然后 $x \mathscr{R} z$ ) 。的所有等价类的集合 $\mathrm{X}$ 对于关系 $\mathscr{R}$ 表示为 $\mathrm{X} / \mathscr{R}$; 地图 $\operatorname{cl} \mathscr{R}: \mathrm{X} \rightarrow \mathrm{X} / \mathscr{R}$ (这样，对于每个 $x \in \mathrm{X}, \operatorname{cl} \mathscr{R}(x)$ 是等价美 $x$ )是一个名词；它的重要特性是任何两个元駇具有相同的图像当 且仅当它们通过 $\mathscr{R}$.

## 数学代写|交换代数代写Commutative Algebra代考|Chinese remainder theorem

1.5.7。中国剩余定理一一我们说两个两侧的理想 I 和 $\mathrm{J}$ 个个环 $\mathrm{A}$ 是共极大的，如果 $\mathrm{I}+\mathrm{J}=\mathrm{A}$. 这个概念产生了中国剩余定理的一 般表述。

$$\mathrm{A} /(\mathrm{I} \cap \mathrm{J}) \simeq \mathrm{A} / \mathrm{I} \times \mathrm{A} / \mathrm{J}$$

## 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。