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

2023年4月3日

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
• (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
couryes™为您提供可以保分的包课服务

The reader will certainly find our will to give to the trivial ring every property under the sun a little arbitrary, especially through our use of a weakened version of negation (cf. footnote 1 p. 478). We hope to convince them of the practical use of such a convention by way of the examples. On the proper use of the trivial ring, see [157, Richman].

The “proof by Azumaya” of the local freeness Lemma 2.2 is extracted from the proof of the Azumaya theorem III.6.2 in [MRR], in the case that concerns us here. In other words, we have given the “matrix” content of the proof of the local freeness lemma in [MRR].

Monomial curves (example on p. 498) are treated in [Kunz], Chap. V, Example 3.13.f.

Decomposed rings play an important role in the classical theory of Henselian local rings for example in the works [Raynaud] or [Lafon \& Marot].

A local-global ring is sometimes called a “ring with many units” in the literature. Local-global rings have been particularly studied in [78, Estes \& Guralnick]. Other “rings with many units” have appeared long beforehand, under the terminology “unit-irreducible rings” (see for example [112]). Those are the rings A for which the following property is satisfied: if two polynomials of $\mathbf{A}[X]$ represent an inverse, then their product also represents an inverse. Also introduced were the “primitive” or “strongly U-irreducible” rings which are the rings for which the following property is satisfied: every primitive polynomial represents an inverse. They are special localglobal rings. In the proof of Fact 6.7 we have shown that a Nagata ring is always “primitive.”

Concerning the Nagata ring $\mathbf{A}(X)$, given Fact 6.7 and the good properties of localglobal rings, it is not surprising that this ring plays a crucial role for the uniform solution of systems of linear equations with parameters over a discrete field and more generally for the uniform computations “in a reasonable amount of time” over arbitrary commutative rings (see [56, 57, Díaz-Toca et al.]).

## 数学代写|交换代数代写commutative algebra代考|Complements on Exterior Powers of a Finitely Generated

The following lemma is immediate.
1.1 Lemma Let $P$ be a free $\mathbf{A}$-module of rank $h$ and $\varphi \in \operatorname{End}(P)$ be a diagonalizable endomorphism, with a matrix similar to $\operatorname{Diag}\left(\lambda_1, \ldots, \lambda_h\right)$, then for the fundamental polynomial of $\varphi$ we get
$$\mathrm{F}{\varphi}(X) \stackrel{\text { def }}{=} \operatorname{det}\left(\operatorname{Id}{P[X]}+X \varphi\right)=\left(1+\lambda_1 X\right) \cdots\left(1+\lambda_h X\right) .$$
We now establish the crucial result.
1.2 Proposition (Exterior powers) Let $P$ be a finitely generated projective module.

1. The $k^{\text {th }}$ exterior power of $P$, denoted by $\wedge^k P$, is also a finitely generated projective module. If $P=\operatorname{Im}(F)$ for $F \in \mathbb{A} \mathbb{G}(\mathbf{A})$, the module $\wedge^k P$ is (isomorphic to) the image of the projection matrix $\wedge^k F$.
2. If $\varphi$ is an endomorphism of $P$, the fundamental polynomial $\mathrm{F}{\wedge^k \varphi}(X)$ only depends on $k$ and on the polynomial $\mathrm{F}{\varphi}(X)$. In particular, the rank polynomial of $\Lambda^k P$ only depends on $k$ and on the rank polynomial of $P$.
3. a. If $P$ is of constant rank $h<k$, the module $\bigwedge^k P$ is null.
b. If $P$ is of constant rank $h \geqslant k$, the module $\bigwedge^k P$ is of constant rank $\left(\begin{array}{l}h \ k\end{array}\right)$.
c. In this case, if $\varphi$ is an endomorphism whose fundamental polynomial is $\mathrm{F}{\varphi}=\left(1+\lambda_1 X\right) \cdots\left(1+\lambda_h X\right)$, we have $$\mathrm{F}{\wedge^k \varphi}(X)=\prod_{1 \leqslant i_1<\cdots<i_k \leqslant h}\left(1+\lambda_{i_1} \cdots \lambda_{i_k} X\right) .$$
4. If a projection matrix $F$ has as its image a projective module of constant rank $k$, then $\mathcal{D}_{k+1}(F)=0$.

D 1. Let $M$ and $N$ be two A-modules and consider the first exterior powers of their direct sum $M \oplus N$. By examining the universal problem that the $k^{\text {th }}$ exterior power of a module solves, we obtain the canonical isomorphisms
\begin{aligned} & \bigwedge^2(M \oplus N) \simeq \Lambda^2 M \oplus(M \otimes N) \oplus \Lambda^2 N \ & \Lambda^3(M \oplus N) \simeq \Lambda^3 M \oplus\left(\left(\Lambda^2 M\right) \otimes N\right) \oplus\left(M \otimes\left(\Lambda^2 N\right)\right) \oplus \Lambda^3 N, \end{aligned} and more generally
$$\bigwedge^m(M \oplus N) \simeq \bigoplus_{k=0}^m\left(\left(\bigwedge^k M\right) \otimes\left(\bigwedge^{m-k} N\right)\right)$$
(with $\bigwedge^0 M=\mathbf{A}$ and $\bigwedge^1 M=M$ ). In particular, if $P \oplus Q \simeq \mathbf{A}^m, \bigwedge^k P$ is a direct matrix $F, \Lambda^k P$ is (isomorphic to) the image of the projection matrix $\wedge^k F$, because this matrix represents the identity over $\Lambda^k P$ and 0 over all the other summands of the direct sum.

# 交换代数代考

## 数学代写|交换代数代写commutative algebra代考|Complements on Exterior Powers of a Finitely Generated

1.1 引理 设 $\$ P \$$是秩为 \ h \$$ 的自由 \$\mathbf${A} \$$-模， \lvarphi \in loperatorname{End}(P)\ 是可对角化的自 同态，其矩阵类似于 \ 1 operatorname{Diag} } eft(lambda_1, \dots, Vambda_h right)\，然后对于 \ivarphi\ 的基本多项 式，我们得到 P 成为一个自由的 \backslash m a t h b f{A} \mathbf{A}-排名模块 h 和 \varphi \in \operatorname{End}(P) 是可对角化的自同态，其矩阵类似于 \operatorname{Diag}\left(\lambda_1, \ldots, \lambda_h\right), 然后对于基本多项式 \varphi 我们得到$$
\mathrm{F} \varphi(X) \stackrel{\text { def }}{=} \operatorname{det}(\operatorname{Id} P[X]+X \varphi)=\left(1+\lambda_1 X\right)
$$P 是一个有限生成的投影模块。 1. \ P \$$ 的 $\$ k^{\wedge}{1$text${$th$}} \$$外幂，记为 \lwedge \wedge k P \ ， 也是一个有限生成的射影模。如果 \ P= loperatorname {\operatorname{Im}}(F) \ 对于 \ F \backslash i n \backslash Imathbb {A} \backslash mathbb {G}(\backslash \operatorname{mathbf}{\mathrm{A}}) \，模块 \lwedge \wedge k P \ 是 (同构于) 投影矩阵 \lwedge^k F\ 的图像。 k^{\text {th }} P \wedge^k P P=\operatorname{Im}(F) F \in \mathbb{A} \mathbb{G}(\mathbf{A}) \wedge^k P \wedge^k F 2. 如果 \lvarphi\ 是 \P\ 的自同态，则基本多项式 \Imathrm{F}{lwedge^k Ivarphi}(X)\ 仅取决于 \k\ 和多项式 \\mathrm{F} {lvarphi}(X)\。特别 地，\Lambda^k P \ 的秩多项式仅取决于 \k 和 \P\ 的秩多项式。 \varphi P \mathrm{~F} \wedge^k \varphi(X) k \mathrm{~F} \varphi(X) \Lambda^k P k P3. A。如果 \ P \$$ 的秩为 $\$ h<k \$$，则模 \ \backslash b i g w e d g e^{\wedge} k P \$$ 为空。 $P h<k \bigwedge^k P$
b. 如果 $\$ P \$$的秩为 \h lgeqslant k \$$ ，则模块 \$\bigwedge^$\mathrm{k} P \$$的秩为 \ \backslash left( \backslash begin { array } {\mid} h \backslash k \backslash kend array}\ight ) \ \circ P h \geqslant k \bigwedge^k P (h k) C。在这种情况下，如果 \lvarphi\ 是一个自同 态，其基本多项式是 \\mathrm {\mathrm{F}} { lvarphi }=\backslash left(1+Vlambda_1 XIright ) Icdots Veft(1+\lambda_h XIright )\, 我们有 \\ Imathrm{F}{lwedge^k Ivarphi }(X)= Iprod_ {1 Veqslant i_1<lcdots<i_k \eqslant h}\left(1+\lambda_{i_1} \cdots \lambda_{i_k} XIright) \Leftarrow \mathrm{F} \varphi=\left(1+\lambda_1 X\right) \cdots\left(1+\lambda_h X\right) \mathrm{F} \wedge^k \varphi(X)=\prod_{1 \leqslant i_1<\cdots<i_k \leqslant h}\left(1+\lambda_{i_1} \cdots \lambda_{i_k} X\right). 3. 如果投影矩阵 \ F \$$ 的图像是常秩 $\$ \mathrm{k}$的投影模， 则$\$\backslash$ mathcal ${D} _{k+1}(F)=0 \$ 。 F k\mathcal{D}{k+1}(F)=0$$\mathrm{D} 1 1.让 M 和 N 是两个 A 模并考虑它们的直和的第一个外 部幂 M \oplus N. 通过检查普遍存在的问题 k^{\text {th }} 求解一个模 块的外幂，我们得到规范同构$$ \bigwedge^2(M \oplus N) \simeq \Lambda^2 M \oplus(M \otimes N) \oplus \Lambda^2 N $$更一般地说$$ \bigwedge^m(M \oplus N) \simeq \bigoplus{k=0}^m\left(\left(\bigwedge^k M\right) \otimes\left(\bigwedge^{m-k} N\right)\right)

(和 $\wedge^0 M=\mathbf{A}$ 和 $\bigwedge^1 M=M$ ). 特别是，如果 $P \oplus Q \simeq \mathbf{A}^m, \Lambda^k P$ 是直接矩阵 $F, \Lambda^k P$ 是 (同构) 投影矩阵的图像 $\wedge^k F$ ，因为这个矩阵代表身份 $\Lambda^k P$ 和 0 在直和的所有其他被加数上。

## 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。