# 线性代数网课代修|同调代数代写homological algebra代考|MA3204

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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|Injective Modules

There is another type of module that turns out to be interesting.
Definition. A left $R$-module $E$ is injective if, whenever $i$ is an injection, a dashed arrow exists making the following diagram commute.

In words, every homomorphism from a submodule into $E$ can always be extended to a homomorphism from the big module into $E$.

Proposition 3.25. A left $R$-module $E$ is injective if and only if $\operatorname{Hom}{R}(\square, E$ ) is an exact functor. Proof. If $$0 \rightarrow A \stackrel{i}{\longrightarrow} B \stackrel{p}{\longrightarrow} C \rightarrow 0$$ is a short exact sequence, we must prove exactness of $$0 \rightarrow \operatorname{Hom}{R}(C, E) \stackrel{p^{}}{\rightarrow} \operatorname{Hom}{R}(B, E) \stackrel{i^{}}{\longrightarrow} \operatorname{Hom}{R}(A, E) \rightarrow 0 .$$
Since $\operatorname{Hom}{R}(\square, E)$ is a left exact contravariant functor, the thrust of the proposition is that the induced map $i^{}$ is surjective whenever $i$ is injective. If $f \in \operatorname{Hom}{R}(A, E)$, there exists $g \in \operatorname{Hom}{R}(B, E)$ with $f=i^{}(g)=g i$; that is, the appropriate diagram commutes, showing that $E$ is an injective module.
For the converse, if $E$ is injective, then given $f: A \rightarrow E$, there exists $g: B \rightarrow E$ with $g i=f$. Thus, if $f \in \operatorname{Hom}{R}(A, E)$, then $f=g i=i^{}(g) \in$ $\operatorname{im} i^{}$, and so the induced map $i^{*}$ is surjective. Therefore, $\operatorname{Hom}(\square, E)$ is an exact functor.
Compare the next result to Proposition 3.3.

## 线性代数作业代写linear algebra代考|Flat Modules

The next type of module arises from tensor products in the same way that projective and injective modules arise from Hom.

Definition. If $R$ is a ring, then a right $R$-module $A$ is flat $^{3}$ if $A \otimes_{R} \square$ is an exact functor; that is, whenever
$$0 \rightarrow B^{\prime} \stackrel{i}{\rightarrow} B \stackrel{p}{\rightarrow} B^{\prime \prime} \rightarrow 0$$
is an exact sequence of left $R$-modules, then
$$0 \rightarrow A \otimes_{K} B^{\prime} \stackrel{1_{A} \otimes i}{\longrightarrow} A \otimes_{K} B \stackrel{1_{A} \otimes p}{\longrightarrow} A \otimes_{K} B^{\prime \prime} \rightarrow 0$$
is an exact sequence of abelian groups. Flatness of left $R$-modules is defined similarly.

Because the functors $A \otimes_{R} \square:{ }{R} \operatorname{Mod} \rightarrow \mathbf{A b}$ are right exact, we see that a right $R$-module $A$ is flat if and only if, whenever $i: B^{\prime} \rightarrow B$ is an injection, then $1{A} \otimes i: A \otimes_{R} B^{\prime} \rightarrow A \otimes_{R} B$ is also an injection.

## 线性代数作业代写linear algebra代考|Injective Modules

$$0 \rightarrow A \stackrel{i}{\longrightarrow} B \stackrel{p}{\longrightarrow} C \rightarrow 0$$

$$0 \rightarrow \operatorname{Hom} R(C, E) \stackrel{p}{\rightarrow} \operatorname{Hom} R(B, E) \stackrel{i}{\longrightarrow} \operatorname{Hom} R(A, E) \rightarrow 0 .$$

$\operatorname{Hom}(\square, E)$ 是一个精确函子。

## 线性代数作业代写linear algebra代考|Flat Modules

$$0 \rightarrow B^{\prime} \stackrel{i}{\rightarrow} B \stackrel{p}{\rightarrow} B^{\prime \prime} \rightarrow 0$$

$$0 \rightarrow A \otimes_{K} B^{\prime} \stackrel{1_{A} \otimes i}{\longrightarrow} A \otimes_{K} B \stackrel{1_{A} \otimes p}{\longrightarrow} A \otimes_{K} B^{\prime \prime} \rightarrow 0$$

# 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