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

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## 线性代数作业代写linear algebra代考|The Elimination and Extension Theorems

To get a sense of how elimination works, let us look at an example similar to those discussed at the end of Chapter 2 . We will solve the system of equations
\begin{aligned} &x^{2}+y+z=1, \ &x+y^{2}+z=1 \ &x+y+z^{2}=1 . \end{aligned}
If we let $I$ be the ideal
(2) $I=\left\langle x^{2}+y+z-1, x+y^{2}+z-1, x+y+z^{2}-1\right\rangle$,
then a Groebner basis for $I$ with respect to lex order is given by the four polynomials
\begin{aligned} &g_{1}=x+y+z^{2}-1, \ &g_{2}=y^{2}-y-z^{2}+z, \ &g_{3}=2 y z^{2}+z^{4}-z^{2} \ &g_{4}=z^{6}-4 z^{4}+4 z^{3}-z^{2} . \end{aligned}
(3)
It follows that equations (1) and (3) have the same solutions. However, since
$$g_{4}=z^{6}-4 z^{4}+4 z^{3}-z^{2}=z^{2}(z-1)^{2}\left(z^{2}+2 z-1\right)$$
involves only $z$, we see that the possible $z$ ‘s are 0,1 and $-1 \pm \sqrt{2}$. Substituting these values into $g_{2}=y^{2}-y-z^{2}+z=0$ and $g_{3}=2 y z^{2}+z^{4}-z^{2}=0$, we can determine the possible $y$ ‘s, and then finally $g_{1}=x+y+z^{2}-1=0$ gives the corresponding $x$ ‘s. In this way, one can check that equations (1) have exactly five solutions:
\begin{aligned} &(1,0,0),(0,1,0),(0,0,1), \ &(-1+\sqrt{2},-1+\sqrt{2},-1+\sqrt{2}), \ &(-1-\sqrt{2},-1-\sqrt{2},-1-\sqrt{2}) . \end{aligned}
What enabled us to find these solutions? There were two things that made our success possible:

• (Elimination Step) We could find a consequence $g_{4}=z^{6}-4 z^{4}+4 z^{3}-z^{2}=0$ of our original equations which involved only $z$, i.e., we eliminated $x$ and $y$ from the system of equations.
• (Extension Step) Once we solved the simpler equation $g_{4}=0$ to determine the values of $z$, we could extend these solutions to solutions of the original equations.

The basic idea of elimination theory is that both the Elimination Step and the Extension Step can be done in great generality.

## 线性代数作业代写linear algebra代考|The Geometry of Elimination

In this section, we will give a geometric interpretation of the theorems proved in $\S 1$. The main idea is that elimination corresponds to projecting a variety onto a lower dimensional subspace. We will also discuss the Closure Theorem, which describes the relation between partial solutions and elimination ideals. For simplicity, we will work over the field $k=\mathbb{C}$.

Let us start by defining the projection of an affine variety. Suppose that we are given $V=\mathbf{V}\left(f_{1}, \ldots, f_{s}\right) \subset \mathbb{C}^{n}$. To eliminate the first $l$ variables $x_{1}, \ldots, x_{l}$, we will consider the projection map
$$\pi_{i}: \mathbb{C}^{n} \rightarrow \mathbb{C}^{n-l}$$
which sends $\left(a_{1}, \ldots, a_{n}\right)$ to $\left(a_{l+1}, \ldots, a_{n}\right)$. If we apply $\pi_{l}$ to $V \subset \mathbb{C}^{n}$, then we get $\pi_{l}(V) \subset \mathbb{C}^{n-l}$. We can relate $\pi_{l}(V)$ to the $l$-th elimination ideal as follows.

Lemma 1. With the above notation, let $I_{l}=\left\langle f_{1}, \ldots, f_{s}\right\rangle \cap \mathbb{C}\left[x_{l+1}, \ldots, x_{n}\right]$ be the l-th elimination ideal. Then, in $\mathbb{C}^{n-1}$, we have
$$\pi_{l}(V) \subset \mathbf{V}\left(I_{l}\right) .$$
Proof. Fix a polynomial $f \in I_{l}$. If $\left(a_{1}, \ldots, a_{n}\right) \in V$, then $f$ vanishes at $\left(a_{1}, \ldots, a_{n}\right)$ since $f \in\left\langle f_{1}, \ldots, f_{s}\right\rangle$. But $f$ involves only $x_{l+1}, \ldots, x_{n}$, so that we can write
$$f\left(a_{l+1}, \ldots, a_{n}\right)=f\left(\pi_{l}\left(a_{1}, \ldots, a_{n}\right)\right)=0 .$$
This shows that $f$ vanishes at all points of $\pi_{l}(V)$.
As in $\S 1$, points of $\mathbf{V}\left(I_{l}\right)$ will be called partial solutions. Using the lemma, we can write $\pi_{l}(V)$ as follows:
\begin{aligned} \pi_{l}(V)=&\left{\left(a_{l+1}, \ldots, a_{n}\right) \in \mathbf{V}\left(I_{l}\right): \exists a_{1}, \ldots, a_{l} \in \mathbb{C}\right.\ &\text { with } \left.\left.\left(a_{1}, \ldots, a_{l}, a_{l+1}, \ldots, a_{n}\right)\right) \in V\right} . \end{aligned}

## 线性代数作业代写linear algebra代考|The Elimination and Extension Theorems

$$x^{2}+y+z=1, \quad x+y^{2}+z=1 x+y+z^{2}=1 .$$

(2) $I=\left\langle x^{2}+y+z-1, x+y^{2}+z-1, x+y+z^{2}-1\right\rangle$,

$$g_{1}=x+y+z^{2}-1, \quad g_{2}=y^{2}-y-z^{2}+z, g_{3}=2 y z^{2}+z^{4}-z^{2} \quad g_{4}=z^{6}-4 z^{4}+4 z^{3}-z^{2} .$$
(3)

$$g_{4}=z^{6}-4 z^{4}+4 z^{3}-z^{2}=z^{2}(z-1)^{2}\left(z^{2}+2 z-1\right)$$

$$(1,0,0),(0,1,0),(0,0,1), \quad(-1+\sqrt{2},-1+\sqrt{2},-1+\sqrt{2}),(-1-\sqrt{2},-1-\sqrt{2},-1-\sqrt{2}) .$$

• (消除步骤) 我们可以找到一个结果 $g_{4}=z^{6}-4 z^{4}+4 z^{3}-z^{2}=0$ 我们原来的方程 只涉及 $z$ ，即我们消除了 $x$ 和 $y$ 从方程组。
• (扩展步骤) 一旦我们解决了更简单的方程 $g_{4}=0$ 确定的值 $z$ ，我们可以将这些解扩展 到原始方程的解。
消除理论的基本思想是消除步骤和扩展步骤都可以非常普遍地完成。

## 线性代数作业代写linear algebra代考|The Geometry of Elimination

$$\pi_{i}: \mathbb{C}^{n} \rightarrow \mathbb{C}^{n-l}$$

$$\pi_{l}(V) \subset \mathbf{V}\left(I_{l}\right) .$$

$$f\left(a_{l+1}, \ldots, a_{n}\right)=f\left(\pi_{l}\left(a_{1}, \ldots, a_{n}\right)\right)=0 .$$

# 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