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

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线性代数作业代写linear algebra代考|Properties of Groebner Bases

As shown in $\S 5$, every nonzero ideal $I \subset k\left[x_{1}, \ldots, x_{n}\right]$ has a Groebner basis. In this section, we will study the properties of Groebner bases and learn how to detect when a given basis is a Groebner basis. We begin by showing that the undesirable behavior of the division algorithm in $k\left[x_{1}, \ldots, x_{n}\right]$ noted in $\S 3$ does not occur when we divide by the elements of a Groebner basis.

Let us first prove that the remainder is uniquely determined when we divide by a Groehner basis.

Proposition 1. Let $G=\left{g_{1}, \ldots, g_{t}\right}$ be a Groebner basis for an ideal $I \subset$ $k\left[x_{1}, \ldots, x_{n}\right]$ and let $f \in k\left[x_{1}, \ldots, x_{n}\right]$. Then there is a unique $r \in k\left[x_{1}, \ldots, x_{n}\right]$ with the following two properties:
(i) No term of $r$ is divisible by any of $\operatorname{LT}\left(g_{1}\right), \ldots, \operatorname{LT}\left(g_{t}\right)$.
(ii) There is $g \in I$ such that $f=g+r$.
In particular, $r$ is the remainder on division of $f$ by $G$ no matter how the elements of $G$ are listed when using the division algorithm.

Proof. The division algorithm gives $f=a_{1} g_{1}+\cdots+a_{t} g_{t}+r$, where $r$ satisfies (i). We can also satisfy (ii) by setting $g=a_{1} g_{1}+\cdots+a_{t} g_{t} \in I$. This proves the existence of $r$.

To prove uniqueness, suppose that $f=g+r=g^{\prime}+r^{\prime}$ satisfy (i) and (ii). Then $r-$ $r^{\prime}=g^{\prime}-g \in I$, so that if $r \neq r^{\prime}$, then $\operatorname{LT}\left(r-r^{\prime}\right) \in\langle\operatorname{LT}(I)\rangle=\left\langle\operatorname{LT}\left(g_{1}\right), \ldots, \operatorname{LT}\left(g_{t}\right)\right\rangle$. By Lemma 2 of $\S 4$, it follows that $\operatorname{LT}\left(r-r^{\prime}\right)$ is divisible by some $\operatorname{LT}\left(g_{i}\right)$. This is impossible since no term of $r, r^{\prime}$ is divisible by one of $\operatorname{LT}\left(g_{1}\right), \ldots, \operatorname{LT}\left(g_{t}\right)$. Thus $r-r^{\prime}$ must be zero, and uniqueness is proved.
The final part of the proposition follows from the uniqueness of $r$.

线性代数作业代写linear algebra代考|Buchberger’s Algorithm

In Corollary 6 of $\S 5$, we saw that every ideal in $k\left[x_{1}, \ldots, x_{n}\right]$ other than 0 has a Groebner basis. Unfortunately, the proof given was nonconstructive in the sense that it did not tell us how to produce the Groebner basis. So we now turn to the question: given an ideal $I \subset k\left[x_{1}, \ldots, x_{n}\right]$, how can we actually construct a Groebner basis for $I$ ? To see the main ideas behind the method we will use, we return to the ideal of Example 2 from $\S 5$ and proceed as follows.

Example 1. Consider the ring $k[x, y]$ with grlex order, and let $I=\left\langle f_{1}, f_{2}\right\rangle=$ $\left(x^{3}-2 x y, x^{2} y-2 y^{2}+x\right)$. Recall that $\left{f_{1}, f_{2}\right}$ is not a Groebner basis for I since $\operatorname{LT}\left(S\left(f_{1}, f_{2}\right)\right)=-x^{2} \notin\left\langle\operatorname{LT}\left(f_{1}\right), \operatorname{LT}\left(f_{2}\right)\right\rangle$.

To produce a Groebner basis, one natural idea is to try first to extend the original generating set to a Groebner basis by adding more polynomials in $I$. In one sense, this adds nothing new, and even introduces an element of redundancy. However, the extra information we get from a Groebner basis more than makes up for this.

What new generators should we add? By what we have said about the S-polynomials in $\S$ 6, the following should come as no surprise. We have $S\left(f_{1}, f_{2}\right)=-x^{2} \in I$, and its remainder on division by $F=\left(f_{1}, f_{2}\right)$ is $-x^{2}$, which is nonzero. Hence, we should include that remainder in our generating set, as a new generator $f_{3}=-x^{2}$. If we set $F=\left(f_{1}, f_{2}, f_{3}\right)$, we can use Theorem 6 of $\S 6$ to test if this new set is a Groebner basis for $I$. We compute
\begin{aligned} &S\left(f_{1}, f_{2}\right)=f_{3}, \text { so } \ &\frac{S\left(f_{1}, f_{2}\right)}{F}=0 \ &S\left(f_{1}, f_{3}\right)=\left(x^{3}-2 x y\right)-(-x)\left(-x^{2}\right)=-2 x y, \text { but } \ &\frac{S\left(f_{1}, f_{3}\right)}{F}=-2 x y \neq 0 \end{aligned}

线性代数作业代写linear algebra代考|Properties of Groebner Bases

(i) 没有 $r$ 可以被任何一个整除 $\mathrm{LT}\left(g_{1}\right), \ldots, \mathrm{LT}\left(g_{t}\right)$.
(ii) 有 $g \in I$ 这样 $f=g+r$.

线性代数作业代写linear algebra代考|Buchberger’s Algorithm

$$S\left(f_{1}, f_{2}\right)=f_{3}, \text { so } \quad \frac{S\left(f_{1}, f_{2}\right)}{F}=0 S\left(f_{1}, f_{3}\right)=\left(x^{3}-2 x y\right)-(-x)\left(-x^{2}\right)=-2 x y, \text { but }$$

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