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## 数学代写|交换代数代写commutative algebra代考|Some Solutions, or Sketches of Solutions

1. Assume without loss of generality $a_0=b_0=1$. When you write $f g=1$, you get $0=a_n b_m, 0=a_n b_{m-1}+a_{n-1} b_m, 0=a_n b_{m-2}+a_{n-1} b_{m-1}+a_{n-2} b_m$,
and so on up to degree 1.
Then prove by induction over $j$ that $\operatorname{deg}\left(a_n^j g\right) \leqslant m-j$.
In particular, for $j=m+1$, we get $\operatorname{deg}\left(a_n^{m+1} g\right) \leqslant-1$, i.e. $a_n^{m+1} g=0$. Whence $a_n^{m+1}=0$. Finally, by reasoning modulo $\mathrm{D}{\mathbf{B}}(0)$, we obtain successively nilpotent $a_j$ ‘s for $j=n-1, \ldots, 1$. $2 a$. Consider the polynomials over the commutative ring $\mathbf{B}[A]$ : $$f(T)=\operatorname{det}\left(\mathrm{I}_n-T A\right) \text { and } g(T)=\operatorname{det}\left(\mathrm{I}_n+T A+T^2 A^2+\cdots+T^{e-1} A^{e-1}\right) .$$ We have $f(T) g(T)=\operatorname{det}\left(\mathrm{I}_n-T^e A^e\right)=1$. The coefficient of degree $n-i$ of $f$ is $\pm a_i$. Apply 1 . 2b. It suffices to prove that $\operatorname{Tr}(A)^{(\ell-1) n+1}=0$, because $a_i=\pm \operatorname{Tr}\left(\bigwedge^{n-i}(A)\right)$. Consider the determinant defined with respect to a fixed basis $\mathcal{B}$ of $\mathbf{A}^n$. If we take the canonical basis formed by the $e_i$ ‘s, we have an obvious equality $$\operatorname{Tr}(f)=\operatorname{det}{\mathcal{B}}\left(f\left(e_1\right), e_2, \ldots, e_n\right)+\cdots+\operatorname{det}_{\mathcal{B}}\left(e_1, e_2, \ldots, f\left(e_n\right)\right)$$

It can be written in the following form:
$$\operatorname{Tr}(f) \operatorname{det}{\mathcal{B}}\left(e_1, \ldots, e_n\right)=\operatorname{det}{\mathcal{B}}\left(f\left(e_1\right), e_2, \ldots, e_n\right)+\cdots+\operatorname{det}_{\mathcal{B}}\left(e_1, e_2, \ldots, f\left(e_n\right)\right) .$$
In this form we can replace the $e_i$ ‘s by any system of $n$ vectors of $\mathbf{A}^n$ : both sides are $n$-multilinear alternating forms (at the $e_i$ ‘s) over $\mathbf{A}^n$, therefore are equal because they coincide on a basis.

Thus, multiplying a determinant by $\operatorname{Tr}(f)$ reduces to replacing it by a sum of determinants in which $f$ acts on each vector.

One deduces that the expression $\operatorname{Tr}(f)^{n(e-1)+1} \operatorname{det}{\mathcal{B}}\left(e_1, \ldots, e_n\right)$ is equal to a sum of which each term is a determinant of the form $$\operatorname{det}{\mathcal{B}}\left(f^{m_1}\left(e_1\right), f^{m_2}\left(e_2\right), \ldots, f^{m_n}\left(e_n\right)\right),$$
with $\sum_i m_i=n(e-1)+1$, therefore at least one of the exponents $m_i$ is $\geqslant e$.
Remark This solution for the bound $n(e-1)+1$ is due to Gert Almkvist. See on this matter: ZEILBERGER D. Gert Almkvist’s generalization of a mistake of Bourbaki. Contemporary Mathematics $\mathbf{1 4 3}$ (1993), pp. 609-612.

## 数学代写|交换代数代写commutative algebra代考|Partial Factorization Algorithm

We assume the reader to be familiar with the extended Euclid algorithm which computes the monic gcd of two monic polynomials in $\mathbf{K}[X]$ when $\mathbf{K}$ is a discrete field (see for example Problem 2 ).
1.1 Lemma If $\mathbf{K}$ is a discrete field, we have a partial factorization algorithm for the finite families of monic polynomials in $\mathbf{K}[X]$ : a partial factorization for a finite family $\left(g_1, \ldots, g_r\right)$ is given by a finite pairwise comaximal family $\left(f_1, \ldots, f_s\right)$ of monic polynomials and by the expression of each $g_i$ in the form
$$g_i=\prod_{k=1}^s f_k^{m_{k, i}}\left(m_{k, i} \in \mathbb{N}\right) .$$
The family $\left(f_1, \ldots, f_s\right)$ is called a partial factorization basis for the family $\left(g_1, \ldots, g_r\right)$

D If the $g_i$ ‘s are pairwise comaximal, there is nothing left to prove. Otherwise, assume for example that $\operatorname{gcd}\left(g_1, g_2\right)=h_0, g_1=h_0 h_1$ and $g_2=h_0 h_2$ with $\operatorname{deg}\left(h_0\right) \geqslant 1$. We replace the family $\left(g_1, \ldots, g_r\right)$ with the family $\left(h_0, h_1, h_2, g_3, \ldots, g_r\right)$. We note that the sum of the degrees has decreased. We also note that we can delete from the list the polynomials equal to 1 , or any repeats of a polynomial. We finish by induction on the sum of the degrees. The details are left to the reader.

# 交换代数代考

## 数学代写|交换代数代写交换代数代考|一些解决方案，或解决方案的草图

.

1. 假设不丧失一般性$a_0=b_0=1$。当你写$f g=1$时，你会得到$0=a_n b_m, 0=a_n b_{m-1}+a_{n-1} b_m, 0=a_n b_{m-2}+a_{n-1} b_{m-1}+a_{n-2} b_m$，
等等，直到1阶。
然后对$j$用归纳法证明$\operatorname{deg}\left(a_n^j g\right) \leqslant m-j$ .
特别地，对于$j=m+1$，我们得到$\operatorname{deg}\left(a_n^{m+1} g\right) \leqslant-1$，即$a_n^{m+1} g=0$。从那里$a_n^{m+1}=0$。最后，对$\mathrm{D}{\mathbf{B}}(0)$进行模推理，得到了$j=n-1, \ldots, 1$的幂零$a_j$。$2 a$。考虑交换环上的多项式$\mathbf{B}[A]$: $$f(T)=\operatorname{det}\left(\mathrm{I}_n-T A\right) \text { and } g(T)=\operatorname{det}\left(\mathrm{I}_n+T A+T^2 A^2+\cdots+T^{e-1} A^{e-1}\right) .$$我们有$f(T) g(T)=\operatorname{det}\left(\mathrm{I}_n-T^e A^e\right)=1$。$f$的度$n-i$的系数是$\pm a_i$。应用1。2b。它足以证明$\operatorname{Tr}(A)^{(\ell-1) n+1}=0$，因为$a_i=\pm \operatorname{Tr}\left(\bigwedge^{n-i}(A)\right)$。考虑关于固定基底$\mathcal{B}$或$\mathbf{A}^n$定义的行列式。如果我们取由$e_i$ ‘s组成的正则基，我们有一个明显的等式$$\operatorname{Tr}(f)=\operatorname{det}{\mathcal{B}}\left(f\left(e_1\right), e_2, \ldots, e_n\right)+\cdots+\operatorname{det}_{\mathcal{B}}\left(e_1, e_2, \ldots, f\left(e_n\right)\right)$$

$$\operatorname{Tr}(f) \operatorname{det}{\mathcal{B}}\left(e_1, \ldots, e_n\right)=\operatorname{det}{\mathcal{B}}\left(f\left(e_1\right), e_2, \ldots, e_n\right)+\cdots+\operatorname{det}_{\mathcal{B}}\left(e_1, e_2, \ldots, f\left(e_n\right)\right) .$$

## 数学代写|交换代数代写交换代数代考|部分分解算法

$$g_i=\prod_{k=1}^s f_k^{m_{k, i}}\left(m_{k, i} \in \mathbb{N}\right) .$$

D如果$g_i$是成对的同极大值，就没有什么需要证明的了。否则，假设$\operatorname{gcd}\left(g_1, g_2\right)=h_0, g_1=h_0 h_1$和$g_2=h_0 h_2$带有$\operatorname{deg}\left(h_0\right) \geqslant 1$。我们用家庭$\left(h_0, h_1, h_2, g_3, \ldots, g_r\right)$替换家庭$\left(g_1, \ldots, g_r\right)$。我们注意到度数的总和下降了。我们还注意到，我们可以从列表中删除多项式等于1，或一个多项式的任何重复。最后，我们用归纳法把度数相加。细节留给读者。

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## 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。

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