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|Saturation

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|Saturation

Let $I_1, I_2 \subset K[x]{>}$be as in Section 1.8.7. We consider the quotient of $I_1$ by powers of $I_2$ $$I_1=I_1: I_2^0 \subset I_1: I_2^1 \subset I_1: I_2^2 \subset I_1: I_2^3 \subset \ldots \subset K[x]{>} .$$
Since $K[x]{>}$is Noetherian, there exists an $s$ such that $I_1: I_2^s=I_1: I_2^{s+i}$ for all $i \geq 0$. Such an $s$ satisfies $$I_1: I_2^{\infty}:=\bigcup{i \geq 0} I_1: I_2^i=I_1: I_2^s$$
and $I_1: I_2^s$ is called the saturation of $I_1$ with respect to $I_2$.
The minimal such $s$ is called the saturation exponent. If $I_1$ is radical, then the saturation exponent is 1 .

Problem: Given ideals $I_1, I_2 \subset K[x]_{>}$, we want to compute generators for $I_1: I_2^{\infty}$ and the saturation exponent.
Solution: Set $I^{(0)}=I_1$ and compute successively $I^{(j+1)}=I^{(j)}: I_2, j \geq 0$, by any of the methods of Section 1.8.8. In each step check whether $I^{(j+1)} \subset I^{(j)}$, by using Section 1.8.1. If $s$ is the first $j$ when this happens, then $I^{(s)}=I_1: I_2^{\infty}$ and $s$ is the saturation exponent.

Correctness follows from $I^{(j)}=I_1: I_2^j$, which is a consequence of Lemma 1.8.14 (1). The above method is usually much faster than computing $I_1: I_2^j$, since $I_2^j$ can become quite large.

## 数学代写|交换代数代写Commutative Algebra代考|Algebraic Dependence and Subalgebra Membership

Recall that a sequence of polynomials $f_1, \ldots, f_k \in K\left[x_1, \ldots, x_n\right]$ is called algebraically dependent if there exists a polynomial $g \in K\left[y_1, \ldots, y_k\right] \backslash{0}$ satisfying $g\left(f_1, \ldots, f_k\right)=0$. This is equivalent to $\operatorname{Ker}(\varphi) \neq 0$, where $\varphi: K\left[y_1, \ldots, y_k\right] \rightarrow K\left[x_1, \ldots, x_n\right]$ is defined by $\varphi\left(y_i\right)=f_i . \operatorname{Ker}(\varphi)$ can be computed according to Section 1.8.10, and any $g \in \operatorname{Ker}(\varphi) \backslash{0}$ defines an algebraic relation between the $f_1, \ldots, f_k$. In particular, $f_1, \ldots, f_k$ are algebraically independent if and only if $\operatorname{Ker}(\varphi)=0$ and this problem was solved in Section 1.8.10.
Related, but slightly different is the subalgebra-membership problem.
Problem: Given $f \in K\left[x_1, \ldots, x_n\right]$, we may ask whether $f$ is an element of the subalgebra $K\left[f_1, \ldots, f_k\right] \subset K\left[x_1, \ldots, x_n\right]=K[x]$.
Solution 1: Define $\psi: K\left[y_0, \ldots, y_k\right] \rightarrow K[x], y_0 \mapsto f, y_i \mapsto f_i$, compute $\operatorname{Ker}(\psi)$ according to Section 1.8 .10 and check whether $\operatorname{Ker}(\psi)$ contains an element of the form $y_0-g\left(y_1, \ldots, y_k\right)$. That is, we define an elimination ordering for $x_1, \ldots, x_n$ on $\operatorname{Mon}\left(x_1, \ldots, x_n, y_0, \ldots, y_k\right)$ with $y_0$ greater than $y_1, \ldots, y_k$ (for example, $(\mathrm{dp}(\mathrm{n}), \mathrm{dp}(1), \operatorname{dp}(\mathrm{k}))$ ) and compute a standard basis $G$ of $\left\langle y_0-f, y_1-f_1, \ldots y_k-f_k\right\rangle$. Then $G$ contains an element with leading monomial $y_0$ if and only if $f \in K\left[f_1, \ldots, f_k\right]$.
Solution 2: Compute a standard basis of $\left\langle y_1-f_1, \ldots, y_k-f_k\right\rangle$ for an elimination ordering for $x_1, \ldots, x_n$ on $\operatorname{Mon}\left(x_1, \ldots, x_n, y_1, \ldots, y_k\right)$ and check whether the normal form of $f$ with respect to this standard basis does not involve any $x_i$. This is the case if and only if $f \in K\left[f_1, \ldots, f_k\right]$ and the normal form expresses $f$ as a polynomial in $f_1, \ldots, f_k$.
We omit the proofs for these statements (cf. Exercise 1.8.10).
Note that $f \in K\left[f_1, \ldots, f_k\right]$ implies a relation $h\left(f, f_1, \ldots, f_k\right)=0$ with $h\left(y_0, y_1, \ldots, y_k\right)=y_0-g\left(y_1, \ldots, y_n\right)$, hence $f, f_1, \ldots, f_k$ are algebraically dependent (the converse does not need to be true).

Note further that the $\operatorname{map} \varphi: K\left[y_1, \ldots, y_k\right] \rightarrow K\left[x_1, \ldots, x_n\right], y_i \rightarrow f_i(x)$ is surjective if and only if $x_i \in K\left[f_1, \ldots, f_k\right]$ for all $i$. Hence, Solution 1 or Solution 2 can be used to check whether a given ring map is surjective.

## 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。