## 数学代写|交换代数代写commutative algebra代考|MATH3033

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。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|交换代数代写commutative algebra代考|Injectivity and Surjectivity Criteria

Two famous propositions are contained in the following theorem.
5.22 Theorem Let $\varphi: \mathbf{A}^n \rightarrow \mathbf{A}^m$ be a linear map with matrix $A$.

1. The map $\varphi$ is surjective if and only if $\varphi$ is of rank $m$, i.e. here $\mathcal{D}_m(\varphi)=\langle 1\rangle$ (we then say that $A$ is unimodular).
2. (McCoy’s theorem) The map $\varphi$ is injective if and only if $\mathcal{D}_n(\varphi)$ is faithful, i.e. if the annihilator of $\mathcal{D}_n(\varphi)$ is reduced to ${0}$.

D 1. If $\varphi$ is surjective, it admits a right inverse $\psi_*$ and Fact $5.6$ gives $\langle 1\rangle=\mathcal{D}_m\left(\mathrm{I}_m\right) \subseteq$ $\mathcal{D}_m(\varphi) \mathcal{D}_m(\psi)$, so $\mathcal{D}_m(\varphi)=\langle 1\rangle$. Conversely, if $A$ is of rank $m$, Eq. (18) shows that $A$ admits a right inverse, and $\varphi$ is surjective.

1. Assume that $\mathcal{D}n(A)$ is faithful. By equality (16), if $A V=0$, then $\mu{\alpha, 1 . . n} V=0$ for all the generators $\mu_{\alpha, 1 . . n}$ of $\mathcal{D}n(A)$, and so $V=0$. For the converse, we will prove by induction on $k$ the following property: if $k$ column vectors $x_1, \ldots, x_k$ are linearly independent, then the annihilator of the vector $x_1 \wedge \cdots \wedge x_k$ is reduced to 0 . For $k=1$ it is trivial. To pass from $k$ to $k+1$ we proceed as follows. Let $z$ be a scalar that annihilates $x_1 \wedge \cdots \wedge x{k+1}$. For $\alpha \in \mathcal{P}{k, m}$, we denote by $d\alpha\left(y_1, \ldots, y_k\right)$ the minor extracted on the index rows of $\alpha$ for the column vectors $y_1, \ldots, y_k$ of $\mathbf{A}^m$. Since $z\left(x_1 \wedge \cdots \wedge x_{k+1}\right)=0$, and by the Cramer formulas, we have the equality
$$z\left(d_\alpha\left(x_1, \ldots, x_k\right) x_{k+1}-d_\alpha\left(x_1, \ldots, x_{k-1}, x_{k+1}\right) x_k+\cdots\right)=0,$$
so $z d_\alpha\left(x_1, \ldots, x_k\right)=0$.
As this is true for any $\alpha$, this gives $z\left(x_1 \wedge \cdots \wedge x_k\right)=0$, and by the induction hypothesis, $z=0$.
Remark Theorem $5.22$ can also be read in the following way.
2. The linear map $\varphi: \mathbf{A}^n \rightarrow \mathbf{A}^m$ is surjective if and only if the map $\bigwedge^m \varphi: \mathbf{A}^{\left(\begin{array}{c}n \ m\end{array}\right)} \rightarrow$ $\mathbf{A}$ is surjective.
3. The linear map $\varphi: \mathbf{A}^n \rightarrow \mathbf{A}^m$ is injective if and only if the map $\wedge^n \varphi: \mathbf{A} \rightarrow \mathbf{A}^{\left(\begin{array}{c}m \ n\end{array}\right)}$ is injective.

## 数学代写|交换代数代写commutative algebra代考|Gram Determinants and Discriminants

5.32 Definition Let $M$ be an A-module, $\varphi: M \times M \rightarrow \mathbf{A}$ be a symmetric bilinear form and $(x)=\left(x_1, \ldots, x_k\right)$ be a list of elements of $M$. We call the matrix
$$\operatorname{Gram}{\mathbf{A}}(\varphi, x) \stackrel{\text { def }}{=}\left(\varphi\left(x_i, x_j\right)\right){i, j \in[1 . . k]}$$
the Gram matrix of $\left(x_1, \ldots, x_k\right)$ for $\varphi$. Its determinant is called the Gram determinant of $\left(x_1, \ldots, x_k\right)$ for $\varphi$ and is denoted by $\operatorname{gram}_{\mathrm{A}}(\varphi, x)$.
If $\mathbf{A} y_1+\cdots+\mathbf{A} y_k \subseteq \mathbf{A} x_1+\cdots+\mathbf{A} x_k$ we have an equality
$$\operatorname{gram}\left(\varphi, y_1, \ldots, y_k\right)=\operatorname{det}(A)^2 \operatorname{gram}\left(\varphi, x_1, \ldots, x_k\right),$$
where $A$ is a $k \times k$ matrix which expresses the $y_j$ ‘s in terms of the $x_i$ ‘s.
We now introduce an important case of a Gram determinant, the discriminant. Recall that two elements $a, b$ of a ring $\mathbf{A}$ are said to be associated if there exists a $u \in \mathbf{A}^{\times}$such that $a=u b$. In the literature such elements are also referred to as associates.
5.33 Proposition and definition Let $\mathbf{C} \supseteq \mathbf{A}$ be an $\mathbf{A}$-algebra which is a free $\mathbf{A}$-module of finite rank and $x_1, \ldots, x_k, y_1, \ldots, y_k \in \mathbf{C}$.

1. We call the determinant of the matrix
$$\left(\operatorname{Tr}{\mathbf{C} / \mathbf{A}}\left(x_i x_j\right)\right){i, j \in \llbracket 1 . . k \rrbracket}$$
the discriminant of $\left(x_1, \ldots, x_k\right)$. We denote it by $\operatorname{disc} \mathbf{C}_{\mathbf{C} / \mathrm{A}}\left(x_1, \ldots, x_k\right)$ or $\operatorname{disc}\left(x_1, \ldots, x_k\right)$.
2. If $\mathbf{A} y_1+\cdots+\mathbf{A} y_k \subseteq \mathbf{A} x_1+\cdots+\mathbf{A} x_k$ we have
$$\operatorname{disc}\left(y_1, \ldots, y_k\right)=\operatorname{det}(A)^2 \operatorname{disc}\left(x_1, \ldots, x_k\right),$$
where A is a $k \times k$ matrix which expresses the $y_j$ ‘s in terms of the $x_i$ ‘s.

## 数学代写|交换代数代写交换代数代考|注入和满射标准

.

5.22定理设$\varphi: \mathbf{A}^n \rightarrow \mathbf{A}^m$是矩阵$A$的线性映射

1. 当且仅当$\varphi$的秩为$m$，即这里的$\mathcal{D}_m(\varphi)=\langle 1\rangle$(我们然后说$A$是单模的)，地图$\varphi$是满射的。
2. (McCoy’s定理)当且仅当$\mathcal{D}_n(\varphi)$是忠实的，即当$\mathcal{D}_n(\varphi)$的湮灭子减少到${0}$时，映射$\varphi$是单射的D 1。如果$\varphi$是满射，它承认一个右逆$\psi_*$，事实$5.6$给出$\langle 1\rangle=\mathcal{D}_m\left(\mathrm{I}_m\right) \subseteq$$\mathcal{D}_m(\varphi) \mathcal{D}_m(\psi)，因此是\mathcal{D}_m(\varphi)=\langle 1\rangle。相反，如果A的秩为m，则Eq.(18)表明A允许一个右逆，而\varphi是满射 1. 假设\mathcal{D}n(A)是忠实的。根据等式(16)，如果A V=0，那么对于\mathcal{D}n(A)的所有生成器\mu_{\alpha, 1 . . n}，则\mu{\alpha, 1 . . n} V=0，因此V=0。反之，我们将在k上通过归纳法证明以下性质:如果k列向量x_1, \ldots, x_k是线性无关的，则向量x_1 \wedge \cdots \wedge x_k的湮灭子化简为0。对于k=1来说，这是微不足道的。要从k传递到k+1，我们按照以下步骤进行。设z是一个灭掉x_1 \wedge \cdots \wedge x{k+1}的标量。对于\alpha \in \mathcal{P}{k, m}，我们用d\alpha\left(y_1, \ldots, y_k\right)表示在\alpha的索引行上提取的子项，用于\mathbf{A}^m的列向量y_1, \ldots, y_k。从z\left(x_1 \wedge \cdots \wedge x_{k+1}\right)=0开始，根据克莱默公式，我们得到等式$$ z\left(d_\alpha\left(x_1, \ldots, x_k\right) x_{k+1}-d_\alpha\left(x_1, \ldots, x_{k-1}, x_{k+1}\right) x_k+\cdots\right)=0, $$所以z d_\alpha\left(x_1, \ldots, x_k\right)=0 . 因为这对任何\alpha都成立，所以得到z\left(x_1 \wedge \cdots \wedge x_k\right)=0，根据归纳假设，得到z=0 备注定理5.22也可以这样理解。 2. 线性映射\varphi: \mathbf{A}^n \rightarrow \mathbf{A}^m是满射的当且仅当映射\bigwedge^m \varphi: \mathbf{A}^{\left(\begin{array}{c}n \ m\end{array}\right)} \rightarrow$$\mathbf{A}$是满射的。
3. 线性映射$\varphi: \mathbf{A}^n \rightarrow \mathbf{A}^m$是单射的，当且仅当映射$\wedge^n \varphi: \mathbf{A} \rightarrow \mathbf{A}^{\left(\begin{array}{c}m \ n\end{array}\right)}$是单射的 数学代写|交换代数代写交换代数代考|克行列式和鉴别 5.32 $M$ 做一个a模块， $\varphi: M \times M \rightarrow \mathbf{A}$ 是对称的双线性形式 $(x)=\left(x_1, \ldots, x_k\right)$ 的元素列表 $M$。我们称矩阵
$$\operatorname{Gram}{\mathbf{A}}(\varphi, x) \stackrel{\text { def }}{=}\left(\varphi\left(x_i, x_j\right)\right){i, j \in[1 . . k]}$$
的克矩阵 $\left(x_1, \ldots, x_k\right)$ 为 $\varphi$。它的行列式叫做 $\left(x_1, \ldots, x_k\right)$ 为 $\varphi$ 表示为 $\operatorname{gram}_{\mathrm{A}}(\varphi, x)$.
如果 $\mathbf{A} y_1+\cdots+\mathbf{A} y_k \subseteq \mathbf{A} x_1+\cdots+\mathbf{A} x_k$ 我们有一个等式
$$\operatorname{gram}\left(\varphi, y_1, \ldots, y_k\right)=\operatorname{det}(A)^2 \operatorname{gram}\left(\varphi, x_1, \ldots, x_k\right),$$
where $A$ 是 $k \times k$ 矩阵表示 $y_j$ 的表达式 $x_i$
我们现在介绍克行列式的一个重要情况，即判别式。回想一下这两个元素 $a, b$ 一枚戒指 $\mathbf{A}$ 如果存在 $u \in \mathbf{A}^{\times}$如此这般 $a=u b$。在文献中，这类元素也被称为关联 $\mathbf{C} \supseteq \mathbf{A}$ 做一个 $\mathbf{A}$-代数是免费的 $\mathbf{A}$-有限秩和的模 $x_1, \ldots, x_k, y_1, \ldots, y_k \in \mathbf{C}$. 我们称矩阵的行列式
$$\left(\operatorname{Tr}{\mathbf{C} / \mathbf{A}}\left(x_i x_j\right)\right){i, j \in \llbracket 1 . . k \rrbracket}$$
的鉴别 $\left(x_1, \ldots, x_k\right)$。我们用 $\operatorname{disc} \mathbf{C}_{\mathbf{C} / \mathrm{A}}\left(x_1, \ldots, x_k\right)$ 或 $\operatorname{disc}\left(x_1, \ldots, x_k\right)$.
4. $\mathbf{A} y_1+\cdots+\mathbf{A} y_k \subseteq \mathbf{A} x_1+\cdots+\mathbf{A} x_k$ 我们有
$$\operatorname{disc}\left(y_1, \ldots, y_k\right)=\operatorname{det}(A)^2 \operatorname{disc}\left(x_1, \ldots, x_k\right),$$
其中A是A $k \times k$ 矩阵表示 $y_j$ 的表达式 $x_i$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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|MATH2301

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。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|交换代数代写commutative algebra代考|Generalized Cramer Formula

We study in this subsection some generalizations of the usual Cramer formulas. We will exploit these in the following paragraphs.

For a matrix $A \in \mathbf{A}^{m \times n}$ we denote by $A_{\alpha, \beta}$ the matrix extracted on the rows $\alpha=\left{\alpha_1, \ldots, \alpha_r\right} \subseteq \llbracket 1 . . m \rrbracket$ and the columns $\beta=\left{\beta_1, \ldots, \beta_s\right} \subseteq \llbracket 1 . . n \rrbracket$.

Suppose that the matrix $A$ is of rank $\leqslant k$. Let $V \in \mathbf{A}^{m \times 1}$ be a column vector such that the bordered matrix $[A \mid V]$ is also of rank $\leqslant k$. Let us call $A_j$ the $j$-th column of $A$. Let $\mu_{\alpha, \beta}=\operatorname{det}\left(A_{\alpha, \beta}\right)$ be the minor of order $k$ of the matrix $A$ extracted on the rows $\alpha=\left{\alpha_1, \ldots, \alpha_k\right}$ and the columns $\beta=\left{\beta_1, \ldots, \beta_k\right}$. For $j \in \llbracket 1 . . k \rrbracket$ let $\nu_{\alpha, \beta, j}$ be the determinant of the same extracted matrix, except that the column $j$ has been replaced with the extracted column of $V$ on the rows $\alpha$. Then, we obtain for each pair $(\alpha, \beta)$ of multi-indices a Cramer identity:
$$\mu_{\alpha, \beta} V=\sum_{j=1}^k \nu_{\alpha, \beta, j} A_{\beta_j}$$
due to the fact that the rank of the bordered matrix $\left[A_{1 . . m, \beta} \mid V\right]$ is $\leqslant k$. This can be read as follows:
\begin{aligned} \mu_{\alpha, \beta} V &=\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot\left[\begin{array}{c} \nu_{\alpha, \beta, 1} \ \vdots \ \nu_{\alpha, \beta, k} \end{array}\right] \ &=\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left[\begin{array}{c} v_{\alpha_1} \ \vdots \ v_{\alpha_k} \end{array}\right] \ &=A \cdot\left(\mathrm{I}n\right){1 \ldots n, \beta} \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left(\mathrm{I}m\right){\alpha, 1 . m} \cdot V \end{aligned}
This leads us to introduce the following notation.
5.12 Notation We denote by $\mathcal{P}{\ell}$ the set of parts of $\llbracket 1 . . \ell \rrbracket$ and $\mathcal{P}{k, \ell}$ the set of parts of $\llbracket 1 . . \ell \rrbracket$ with $k$ elements. For $A \in \mathbf{A}^{m \times n}$ and $\alpha \in \mathcal{P}{k, m}, \beta \in \mathcal{P}{k, n}$
$$\operatorname{Adj}{\alpha, \beta}(A):=\left(\mathrm{I}_n\right){1 \ldots n, \beta} \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left(\mathrm{I}m\right){\alpha, 1 \ldots m} .$$

## 数学代写|交换代数代写commutative algebra代考|Generalized Inverses and Locally Simple Maps

Let $E$ and $F$ be two $\mathbf{A}$-modules, and $\varphi: E \rightarrow F$ be a linear map. We can see this as some sort of generalized system of linear equations (a usual system of linear equations corresponds to the free modules of finite rank case). Informally such a system of linear equations is considered to be “well-conditioned” if there is a systematic way to solve the equation $\varphi(x)=y$ for $x$ from a given $y$, when such a solution exists. More precisely, we ask if there exists a linear map $\psi: F \rightarrow E$ satisfying $\varphi(\psi(y))=y$ each time there exists a solution $x$. This amounts to asking $\varphi(\psi(\varphi(x)))=\varphi(x)$ for all $x \in E$.

This clarifies the importance of the Eq. (17) and leads to the notion of a generalized inverse.

The terminology regarding generalized inverses does not seem fully fixed. We adopt that of [Lancaster \& Tismenetsky].
In the book [Bhaskara Rao], the author uses the term “reflexive g-inverse.”
5.16 Definition Let $E$ and $F$ be two A-modules, and $\varphi: E \rightarrow F$ be a linear map. A linear map $\psi: F \rightarrow E$ is called a generalized inverse of $\varphi$ if we have
$$\varphi \circ \psi \circ \varphi=\varphi \text { and } \psi \circ \varphi \circ \psi=\psi$$
A linear map is said to be locally simple when it has a generalized inverse. The following fact is immediate.

5.17 Fact When $\psi$ is a generalized inverse of $\varphi$, we have:

• $\varphi \psi$ and $\psi \varphi$ are projections,
$-\operatorname{Im} \varphi=\operatorname{Im} \varphi \psi, \operatorname{Im} \psi=\operatorname{Im} \psi \varphi, \operatorname{Ker} \varphi=\operatorname{Ker} \psi \varphi, \operatorname{Ker} \psi=\operatorname{Ker} \varphi \psi$,
$-E=\operatorname{Ker} \varphi \oplus \operatorname{Im} \psi$ and $F=\operatorname{Ker} \psi \oplus \operatorname{Im} \varphi$,
$-\operatorname{Ker} \varphi \simeq \operatorname{Coker} \psi$ and $\operatorname{Ker} \psi \simeq \operatorname{Coker} \varphi$.
Moreover $\varphi$ and $\psi$ provide by restriction reciprocal isomorphisms $\varphi_1$ and $\psi_1$ between $\operatorname{Im} \psi$ and $\operatorname{Im} \varphi$. In matrix form we obtain:
Remarks
1) If we have a linear map $\psi_0$ satisfying as in Theorem $5.14$ the equality $\varphi \psi_0 \varphi=\varphi$, we obtain a generalized inverse of $\varphi$ by stating $\psi=\psi_0 \varphi \psi_0$. In other words, a linear map $\varphi$ is locally simple if and only if there exists a $\psi$ satisfying $\varphi \psi \varphi=\varphi$.
2) A simple linear map between free modules of finite rank is locally simple (immediate verification).
3) Theorem $5.14$ informs us that a linear map which has rank $k$ in the sense of Definition $5.7$ is locally simple.

## 数学代写|交换代数代写交换代数代考|广义Cramer公式

$$\mu_{\alpha, \beta} V=\sum_{j=1}^k \nu_{\alpha, \beta, j} A_{\beta_j}$$
，这是因为有边界矩阵$\left[A_{1 . . m, \beta} \mid V\right]$的秩为$\leqslant k$。
\begin{aligned} \mu_{\alpha, \beta} V &=\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot\left[\begin{array}{c} \nu_{\alpha, \beta, 1} \ \vdots \ \nu_{\alpha, \beta, k} \end{array}\right] \ &=\left[A_{\beta_1} \ldots A_{\beta_k}\right] \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left[\begin{array}{c} v_{\alpha_1} \ \vdots \ v_{\alpha_k} \end{array}\right] \ &=A \cdot\left(\mathrm{I}n\right){1 \ldots n, \beta} \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left(\mathrm{I}m\right){\alpha, 1 . m} \cdot V \end{aligned}

5.12表示法我们用$\mathcal{P}{\ell}$表示$\llbracket 1 . . \ell \rrbracket$的部分的集合，用$\mathcal{P}{k, \ell}$表示$\llbracket 1 . . \ell \rrbracket$的部分的集合，其中包含$k$元素。对于$A \in \mathbf{A}^{m \times n}$和$\alpha \in \mathcal{P}{k, m}, \beta \in \mathcal{P}{k, n}$
$$\operatorname{Adj}{\alpha, \beta}(A):=\left(\mathrm{I}_n\right){1 \ldots n, \beta} \cdot \operatorname{Adj}\left(A_{\alpha, \beta}\right) \cdot\left(\mathrm{I}m\right){\alpha, 1 \ldots m} .$$

## 数学代写|交换代数代写交换代数代考|广义逆和局部简单映射

5.16定义设$E$和$F$是两个a模，$\varphi: E \rightarrow F$是一个线性映射。如果我们有
$$\varphi \circ \psi \circ \varphi=\varphi \text { and } \psi \circ \varphi \circ \psi=\psi$$

• $\varphi \psi$ 和 $\psi \varphi$ 是投影，
$-\operatorname{Im} \varphi=\operatorname{Im} \varphi \psi, \operatorname{Im} \psi=\operatorname{Im} \psi \varphi, \operatorname{Ker} \varphi=\operatorname{Ker} \psi \varphi, \operatorname{Ker} \psi=\operatorname{Ker} \varphi \psi$，
$-E=\operatorname{Ker} \varphi \oplus \operatorname{Im} \psi$ 和 $F=\operatorname{Ker} \psi \oplus \operatorname{Im} \varphi$，
$-\operatorname{Ker} \varphi \simeq \operatorname{Coker} \psi$ 和 $\operatorname{Ker} \psi \simeq \operatorname{Coker} \varphi$
此外 $\varphi$ 和 $\psi$ 通过限制提供互同构 $\varphi_1$ 和 $\psi_1$ 之间 $\operatorname{Im} \psi$ 和 $\operatorname{Im} \varphi$。在矩阵形式中，我们得到:
备注
1)如果我们有一个线性映射 $\psi_0$ 在定理中满足 $5.14$ 平等 $\varphi \psi_0 \varphi=\varphi$的广义逆 $\varphi$ 通过说明 $\psi=\psi_0 \varphi \psi_0$。换句话说，一个线性映射 $\varphi$ 当且仅当存在 $\psi$ 令人满意的 $\varphi \psi \varphi=\varphi$.
2)有限秩自由模之间的简单线性映射是局部简单的(立即验证) $5.14$ 告诉我们有秩的线性映射 $k$ 在定义的意义上 $5.7$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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|MATH3303

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。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|交换代数代写commutative algebra代考|A Little Exterior Algebra

Let $k \in \mathbb{N}$. A free module of rank $k$ is by definition an $\mathbf{A}$-module isomorphic to $\mathbf{A}^k$. If $k$ is not specified, we will say free module of finite rank.

When $\mathbf{A}$ is a discrete field we speak of a finite dimensional vector space or a finite rank vector space interchangeably.

The modules whose structure is the simplest are the free modules of finite rank. We are thus interested in the possibility of constructing an arbitrary module $M$ in the form $L \oplus N$ where $L$ is a free module of finite rank. A (partial) answer to this question is given by the exterior algebra.
5.1 Proposition (Splitting OIf) Let $a_1, \ldots, a_k$ be elements of an A-module $M$, then the following properties are equivalent.

1. The submodule $L=\left\langle a_1, \ldots, a_k\right\rangle$ of $M$ is free with basis $\left(a_1, \ldots, a_k\right)$ and is $a$ direct summand of $M$.
2. There exists a k-multilinear alternating form $\varphi: M^k \rightarrow \mathbf{A}$ which satisfies the equality $\varphi\left(a_1, \ldots, a_k\right)=1$.

D $1 \Rightarrow 2$. If $L \oplus N=M$, if $\pi: M \rightarrow L$ is the projection parallel to $N$, and if $\theta_j: L \rightarrow \mathbf{A}$ is the $j$-th coordinate form for the basis $\left(a_1, \ldots, a_k\right)$, we define
$$\varphi\left(x_1, \ldots, x_k\right)=\operatorname{det}\left(\left(\theta_j\left(\pi\left(x_i\right)\right)\right){i, j \in \llbracket 1 . . k \rrbracket}\right)$$ $2 \Rightarrow 1$. We define the linear map $\pi: M \rightarrow M$ as $$\pi(x)=\sum{j=1}^k \varphi(\underbrace{a_1, \ldots, x, \ldots, a_k}_{(x \text { is in position } j)}) a_j .$$

We immediately have $\pi\left(a_i\right)=a_i$ and $\operatorname{Im} \pi \subseteq L:=\left\langle a_1, \ldots, a_k\right\rangle$, thus $\pi^2=\pi$ and $\operatorname{Im} \pi=L$. Finally, if $x=\sum_j \lambda_j a_j=0$, then $\varphi\left(a_1, \ldots, x, \ldots, a_k\right)=\lambda_j=0$ (with $x$ in position $j$ ).

Special case: for $k=1$ we say that the element $a_1$ of $M$ is unimodular when there exists a linear form $\varphi: M \rightarrow \mathbf{A}$ such that $\varphi\left(a_1\right)=1$. The vector $b=\left(b_1, \ldots, b_n\right) \in$ $\mathbf{A}^n$ is unimodular if and only if the $b_i$ ‘s are comaximal. In this case we also say that the sequence $\left(b_1, \ldots, b_n\right)$ is unimodular.

## 数学代写|交换代数代写commutative algebra代考|The Rank of a Free Module

As we will see, the rank of a free module is a well-determined integer if the ring is nontrivial. In other words, two $\mathbf{A}$-modules $M \simeq \mathbf{A}^m$ and $P \simeq \mathbf{A}^p$ with $m \neq p$ can only be isomorphic if $1=\mathrm{A} 0$.

We will use the notation $\operatorname{rk}_{\mathbf{A}}(M)=k$ (or $\operatorname{rk}(M)=k$ if $\mathbf{A}$ is clear from the context) to indicate that a (supposedly free) module has rank $k$.

A scholarly proof consists to say that, if $m>p$, the $m$-th exterior power of $P$ is ${0}$ whereas that of $M$ is isomorphic to $\mathbf{A}$ (this is essentially the proof for Corollary $5.23$ ).
The same proof can be presented in a more elementary way as follows. First recall the basic Cramer formula. If $B$ is a square matrix of order $n$, we denote by $\widetilde{B}$ or Adj $B$ the cotransposed matrix (sometimes called adjoint). The elementary form of Cramer’s identities is then expressed as:
$$A \operatorname{Adj}(A)=\operatorname{Adj}(A) A=\operatorname{det}(A) \mathrm{I}_n .$$
This formula, in combination with the product formula
$$\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)$$
has a couple of implications regarding square matrices. First, that a square matrix $A$ is invertible on one side if and only if $A$ is invertible if and only if its determinant is invertible. Second, that the inverse of $A$ is equal to (det $A)^{-1} \operatorname{Adj} A$.

We now consider two $\mathbf{A}$-modules $M \simeq \mathbf{A}^m$ and $P \simeq \mathbf{A}^p$ with $m \geqslant p$ and a surjective linear map $\varphi: P \rightarrow M$. Therefore there exists a linear map $\psi: M \rightarrow P$ such that $\varphi \circ \psi=\operatorname{Id}_M$. This corresponds to two matrices $A \in \mathbf{A}^{m \times p}$ and $B \in \mathbf{A}^{p \times m}$ with $A B=\mathrm{I}_m$. If $m=p$, the matrix $A$ is invertible with inverse $B$ and $\varphi$ and $\psi$ are reciprocal isomorphisms. If $m>p$, we have $A B=A_1 B_1$ with square $A_1$ and $B_1$ respectively obtained from $A$ and $B$ by filling in with zeros ( $m-p$ columns for $A_1$, $m-p$ rows for $\left.B_1\right)$

## 数学代写|交换代数代写交换代数代考|一个小的外部代数

5.1命题(拆分OIf)设$a_1, \ldots, a_k$为A模块$M$的元素，则下列属性等价

. . .
5.1命题(拆分OIf
$M$的子模块$L=\left\langle a_1, \ldots, a_k\right\rangle$是基$\left(a_1, \ldots, a_k\right)$的自由子模块，是$M$的$a$直接求和存在一个k-多线性交替形式$\varphi: M^k \rightarrow \mathbf{A}$，它满足等式$\varphi\left(a_1, \ldots, a_k\right)=1$ .

D $1 \Rightarrow 2$。如果$L \oplus N=M$，如果$\pi: M \rightarrow L$是平行于$N$的投影，如果$\theta_j: L \rightarrow \mathbf{A}$是基$\left(a_1, \ldots, a_k\right)$的$j$ -th坐标形式，我们定义
$$\varphi\left(x_1, \ldots, x_k\right)=\operatorname{det}\left(\left(\theta_j\left(\pi\left(x_i\right)\right)\right){i, j \in \llbracket 1 . . k \rrbracket}\right)$$$2 \Rightarrow 1$。我们将线性映射$\pi: M \rightarrow M$定义为$$\pi(x)=\sum{j=1}^k \varphi(\underbrace{a_1, \ldots, x, \ldots, a_k}_{(x \text { is in position } j)}) a_j .$$

## 数学代写|交换代数代写commutative algebra代考|Integral closure of ideals

X米+一个米−1X米−1+⋯+一个1X+一个0

(i)一个∈我~.
(ii) 存在一个有限生成的R-模块米这样一个∈我米:R米并且这样0:R米⊂0:R一个

(二)⇒(i) 应用与证明 2.2.1 相同的方法，(iii)⇒(i), 到有限的一组生成器米, 使用假设一个米⊂我米，得到某个方阵一个. 然后使用 Remark 中的想法2.2.2至派生⁡这⁡(一个)米=0然后应用剩余的假设得到形式的积分依赖方程(一个这⁡(一个))l=0对于一些l.

## 数学代写|交换代数代写commutative algebra代考|Krull dimension and Noether normalization

H T 我:=分钟H吨磷∣磷⊃我 一个首要理想 .

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。