线性代数网课代修|交换代数代写Commutative Algebra代考|MAT4200

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• 数值分析
• 高等线性代数
• 矩阵论
• 优化理论
• 线性规划
• 逼近论

线性代数作业代写linear algebra代考|Unique Factorization and Resultants

This definition says that if a nonconstant polynomial $f$ is irreducible over $k$, then up to a constant multiple, its only nonconstant factor is $f$ itself. Also note that the concept of irreducibility depends on the field. For example, $x^{2}+1$ is irreducible over $\mathbb{Q}$ and $\mathbb{R}$, but, over $\mathbb{C}$, we have $x^{2}+1=(x-i)(x+i)$.
Every polynomial is a product of irreducible polynomials as follows.
Proposition 2. Every nonconstant polynomial $f \in k\left[x_{1}, \ldots, x_{n}\right]$ can be written as a product of polynomials which are irreducible over $k$.

Proof. If $f$ is irreducible over $k$, then we are done. Otherwise, we can write $f=g h$, where $g, h \in k\left[x_{1}, \ldots, x_{n}\right]$ are nonconstant. Note that the total degrees of $g$ and $h$ are less than the total degree of $f$. Now apply this process to $g$ and $h$ : if either fails to be irreducible over $k$, we factor it into nonconstant factors. Since the total degree drops each time we factor, this process can be repeated at most finitely many times. Thus, $f$ must be a product of irreducibles.

In Theorem 5 we will show that the factorization of Proposition 2 is essentially unique. But first, we have to prove the following crucial property of irreducible polynomials.

Theorem 3. Let $f \in k\left[x_{1}, \ldots, x_{n}\right]$ be irreducible over $k$ and suppose that $f$ divides the product $g h$, where $g, h \in k\left[x_{1}, \ldots, x_{n}\right]$. Then $f$ divides $g$ or $h$.

Proof. We will use induction on the number of variables. When $n=1$, we can use the GCD theory developed in $\S 5$ of Chapter 1. If $f$ divides $g h$, then consider $p=$ $\operatorname{GCD}(f, g)$. If $p$ is nonconstant, then $f$ must be a constant multiple of $p$ since $f$ is irreducible, and it follows that $f$ divides $g$. On the other hand, if $p$ is constant, we can assume $p=1$, and then we can find $A, B \in k\left[x_{1}\right]$ such that $A f+B g=1$ (see Proposition 6 of Chapter $1, \S 5$ ). If we multiply this by $h$, we get
$$h=h(A f+B g)=A h f+B g h .$$
Since $f$ divides $g h, f$ is a factor of $A h f+B g h$, and, thus, $f$ divides $h$. This proves the case $n=1$.

线性代数作业代写linear algebra代考|Resultants and the Extension Theorem

In this section we will prove the Extension Theorem using the results of $\S$. Our first task will be to adapt the theory of resultants to the case of polynomials in $n$ variables. Thus, suppose we are given $f, g \in k\left[x_{1}, \ldots, x_{n}\right]$ of positive degree in $x_{1}$. As in $\S 5$, we write
\begin{aligned} &f=a_{0} x_{1}^{\prime}+\cdots+a_{l}, \quad a_{0} \neq 0 \ &g=b_{0} x_{1}^{m}+\cdots+b_{m}, \quad b_{0} \neq 0, \end{aligned}
where $a_{i}, b_{i} \in k\left[x_{2}, \ldots, x_{n}\right]$, and we define the resultant of $f$ and $g$ with respect to $x_{1}$

where the empty spaces are filled by zeros.
For resultants of polynomials in several variables, the results of $\$ 5$can be stated as follows. Proposition 1. Let$f, g \in k\left[x_{1}, \ldots, x_{n}\right]$have positive degree in$x_{1}$. Then: (i)$\operatorname{Res}\left(f, g, x_{1}\right)$is in the first elimination ideal$\langle f, g\rangle \cap k\left[x_{2}, \ldots, x_{n}\right]$. (ii)$\operatorname{Res}\left(f, g, x_{1}\right)=0$if and only if$f$and$g$have a common factor in$k\left[x_{1}, \ldots, x_{n}\right]$which has positive degree in$x_{1}$. Proof. When we write$f, g$in terms of$x_{1}$, the coefficients$a_{i}, b_{i}$, lie in$k\left[x_{2}, \ldots, x_{n}\right]$. Since the resultant is an integer polynomial in$a_{i}, b_{i}$, (Proposition 9 of$\S 5$), it follows that$\operatorname{Res}\left(f, g, x_{1}\right) \in k\left[x_{2}, \ldots, x_{n}\right]$. We also know that $$A f+B g=\operatorname{Res}\left(f, g, x_{1}\right),$$ where$A$and$B$are polynomials in$x_{1}$whose coefficients are again integer polynomials in$a_{i}, b_{i}$(Proposition 9 of$\S$5). Thus,$A, B \in k\left[x_{2}, \ldots, x_{n}\right]\left[x_{1}\right]=k\left[x_{1}, \ldots, x_{n}\right]$, and then the above equation implies$\operatorname{Res}\left(f, g, x_{1}\right) \in\langle f, g\rangle$. This proves part (i) of the proposition. 线性代数作业代写linear algebra代考|Unique Factorization and Resultants 这个定义说，如果一个非常数多项式$f$是不可约的$k$，那么直到一个常数倍数，它唯一的非 常数因子是$f$本身。另请注意，不可约性的概念取决于场。例如，$x^{2}+1$是不可约的$Q$和$\mathbb{R}$, 但是, 结束$\mathbb{C}$， 我们有$x^{2}+1=(x-i)(x+i)$. 每个多项式都是如下不可约多项式的乘积。 命题 2. 每个非常数多项式$f \in k\left[x_{1}, \ldots, x_{n}\right]$可以写成多项式的乘积，这些多项式是不可约 的$k$. 证明。如果$f$是不可约的$k$，那么我们就完成了。否则，我们可以写$f=g h$，在哪里$g, h \in k\left[x_{1}, \ldots, x_{n}\right]$是非恒定的。注意总度数$g$和$h$小于总度数$f$. 现在将此过程应用于$g$和$h$: 如果其中任何一个都不能不可约$k$，我们将其分解为非常数因子。由于每次我们考虑总度 数都会下降，所以这个过程最多可以重复多次。因此，f一定是不可约数的乘积。 在定理 5 中，我们将证明命题 2 的因式分解本质上是唯一的。但首先，我们必须证明不可 约多项式的以下关键性质。 定理 3. 让$f \in k\left[x_{1}, \ldots, x_{n}\right]$不可约$k$并假设$f$划分产品$g h$， 在哪里$g, h \in k\left[x_{1}, \ldots, x_{n}\right]$. 然后$f$划分$g$或者$h$. 证明。我们将对变量的数量使用归纳法。什么时候$n=1$, 我们可以使用 GCD 理论$\S 5$第 1 章。如果$f$划分$g h$，然后考虑$p=\operatorname{GCD}(f, g)$. 如果$p$是非常数，那么$f$必须是的常数倍$p$自 从$f$是不可约的，因此$f$划分$g$. 另一方面，如果$p$是常数，我们可以假设$p=1$，然后我们可 以找到$A, B \in k\left[x_{1}\right]$这样$A f+B g=1$（见第 6 章命题$1, \S 5$)。如果我们将其乘以$h$，我 们得到 $$h=h(A f+B g)=A h f+B g h .$$ 自从$f$划分$g h, f$是一个因素$A h f+B g h ，$因此，$f$划分$h$. 这证明了情况$n=1$. 线性代数作业代写linear algebra代考|Resultants and the Extension Theorem 在本节中，我们将使用以下结果证明扩展定理$\S$. 我们的首要任务是使结果式理论适应多项 式的情况$n$变量。因此，假设我们有$f, g \in k\left[x_{1}, \ldots, x_{n}\right]$在积极程度$x_{1}$. 如在$\S 5$，我们写 $$f=a_{0} x_{1}^{\prime}+\cdots+a_{l}, \quad a_{0} \neq 0 \quad g=b_{0} x_{1}^{m}+\cdots+b_{m}, \quad b_{0} \neq 0,$$ 在哪里$a_{i}, b_{i} \in k\left[x_{2}, \ldots, x_{n}\right]$，我们定义的结果$f$和$g$关于$x_{1}$其中空白处用零填充。 对于多个变量的多项式的结果，结果$\$5$ 可以表述如下。

(二) $\operatorname{Res}\left(f, g, x_{1}\right)=0$ 当且仅当 $f$ 和 $g$ 有一个共同的因素 $k\left[x_{1}, \ldots, x_{n}\right]$ 度数为正的 $x_{1}$.

$$A f+B g=\operatorname{Res}\left(f, g, x_{1}\right),$$

计量经济学代写

在这种情况下，如何学好线性代数？如何保证线性代数能获得高分呢？

1.1 mark on book

【重点的误解】划重点不是书上粗体，更不是每个定义，线代概念这么多，很多朋友强迫症似的把每个定义整整齐齐用荧光笔标出来，然后整本书都是重点，那期末怎么复习呀。我认为需要标出的重点为

A. 不懂，或是生涩，或是不熟悉的部分。这点很重要，有的定义浅显，但证明方法很奇怪。我会将晦涩的定义，证明方法标出。在看书时，所有例题将答案遮住，自己做，卡住了就说明不熟悉这个例题的方法，也标出。

B. 老师课上总结或强调的部分。这个没啥好讲的，跟着老师走就对了

C. 你自己做题过程中，发现模糊的知识点

1.2 take note

1.3 understand the relation between definitions