# 线性代数网课代修|交换代数代写Commutative Algebra代考|MATH4312

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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|A Division Algorithm in k

In $\S 1$, we saw how the division algorithm could be used to solve the ideal membership problem for polynomials of one variable. To study this problem when there are more variables, we will formulate a division algorithm for polynomials in $k\left[x_{1}, \ldots, x_{n}\right]$ that extends the algorithm for $k[x]$. In the general case, the goal is to divide $f \in$ $k\left[x_{1}, \ldots, x_{n}\right]$ by $f_{1}, \ldots, f_{s} \in k\left[x_{1}, \ldots, x_{n}\right]$. As we will see, this means expressing $f$ in the form
$$f=a_{1} f_{1}+\cdots+a_{s} f_{s}+r,$$
where the “quotients” $a_{1}, \ldots, a_{s}$ and remainder $r$ lie in $k\left[x_{1}, \ldots, x_{n}\right]$. Some care will be needed in deciding how to characterize the remainder. This is where we will use the monomial orderings introduced in $\S$. We will then see how the division algorithm applies to the ideal membership problem.

The basic idea of the algorithm is the same as in the one-variable case: we want to cancel the leading term of $f$ (with respect to a fixed monomial order) by multiplying some $f_{i}$ by an appropriate monomial and subtracting. Then this monomial becomes a term in the corresponding $a_{i}$. Rather than state the algorithm in general, let us first work through some examples to see what is involved.

## 线性代数作业代写linear algebra代考|The Hilbert Basis Theorem and Groebner Bases

In this section, we will give a complete solution of the ideal description problem from $\S$ 1. Our treatment will also lead to ideal bases with “good” properties relative to the division algorithm introduced in $\S \dot{3}$. The key idea we will use is that once we choose a monomial ordering, each $f \in k\left[x_{1}, \ldots, x_{n}\right]$ has a unique leading term $\operatorname{LT}(f)$. Then, for any ideal $I$, we can define its ideal of leading terms as follows.
Definition 1. Let $I \subset k\left[x_{1}, \ldots, x_{n}\right]$ be an ideal other than ${0}$.
(i) We denote by $\mathrm{LT}(I)$ the set of leading terms of elements of $I$. Thus,
$$\operatorname{LT}(I)=\left{c x^{\alpha}: \text { there exists } f \in I \text { with } \operatorname{LT}(f)=c x^{\alpha}\right} .$$
(ii) We denote by $\langle\operatorname{LT}(I)\rangle$ the ideal generated by the elements of $\operatorname{LT}(I)$.
We have already seen that leading terms play an important role in the division algorithm. This brings up a subtle but important point concerning $\langle\mathrm{LT}(I)\rangle$. Namely, if we are given a finite generating set for $I$, say $I=\left\langle f_{1}, \ldots, f_{s}\right\rangle$, then $\left\langle\operatorname{LT}\left(f_{1}\right), \ldots, \operatorname{LT}\left(f_{s}\right)\right\rangle$ and $\langle\mathrm{LT}(I)\rangle$ may be different ideals. It is true that $\mathrm{LT}\left(f_{i}\right) \in \operatorname{LT}(I) \subset\langle\mathrm{LT}(I)\rangle$ by definition, which implies $\left\langle\operatorname{LT}\left(f_{1}\right), \ldots, \operatorname{LT}\left(f_{s}\right)\right\rangle \subset\langle\operatorname{LT}(I)\rangle$. However, $\langle\operatorname{LT}(I)\rangle$ can be strictly larger. To see this, consider the following example.

Example 2. Let $I=\left\langle f_{1}, f_{2}\right\rangle$, where $f_{1}=x^{3}-2 x y$ and $f_{2}=x^{2} y-2 y^{2}+x$, and use the grlex ordering on monomials in $k[x, y]$. Then
$$x \cdot\left(x^{2} y-2 y^{2}+x\right)-y \cdot\left(x^{3}-2 x y\right)=x^{2},$$ so that $x^{2} \in I$. Thus $x^{2}=\operatorname{LT}\left(x^{2}\right) \in\langle\operatorname{LT}(I)\rangle$. However $x^{2}$ is not divisible by LT $\left(f_{1}\right)=x^{3}$, or $\operatorname{LT}\left(f_{2}\right)=x^{2} y$, so that $x^{2} \notin\left\langle\operatorname{LT}\left(f_{1}\right), \operatorname{LT}\left(f_{2}\right)\right\rangle$ by Lemma 2 of $\S 4$.

## 线性代数作业代写linear algebra代考|A Division Algorithm in k

$$f=a_{1} f_{1}+\cdots+a_{s} f_{s}+r$$
“商”在哪里 $a_{1}, \ldots, a_{s}$ 和剩余的 $r$ 位于 $k\left[x_{1}, \ldots, x_{n}\right]$. 在决定如何表征其余部分时需要小心谨 慎。这是我们将使用引入的单项式排序的地方 $\S$. 然后我们将看到除法算法如何应用于理想 的隶属问题。

## 线性代数作业代写linear algebra代考|The Hilbert Basis Theorem and Groebner Bases

(i) 我们用 $\mathrm{LT}(I)$ 元素的前导项的集合 $I$. 因此，
(ii) 我们表示 $\langle\mathrm{LT}(I)\rangle$ 由元素产生的理想LT $(I)$.

$\langle\mathrm{LT}(I)\rangle$. 也就是说，如果给定一个有限生成集 $I$ ，说 $I=\left\langle f_{1}, \ldots, f_{s}\right\rangle$ ，然后
$\left\langle\operatorname{LT}\left(f_{1}\right), \ldots, \operatorname{LT}\left(f_{s}\right)\right\rangle$ 和 $\langle\mathrm{LT}(I)\rangle$ 可能是不同的理想。的确如此
$\operatorname{LT}\left(f_{i}\right) \in \operatorname{LT}(I) \subset\langle\mathrm{LT}(I)\rangle$ 根据定义，这意味着 $\left\langle\mathrm{LT}\left(f_{1}\right), \ldots, \mathrm{LT}\left(f_{s}\right)\right\rangle \subset\langle\mathrm{LT}(I)\rangle$. 然

$$x \cdot\left(x^{2} y-2 y^{2}+x\right)-y \cdot\left(x^{3}-2 x y\right)=x^{2},$$

# 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