# 线性代数网课代修|同调代数代写homological algebra代考|MATH813

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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|Quasi-Frobenius Rings

We are now going to assume that a ring $R$ is self-injective; that is, $R$ is injective as a left $R$-module. (There is no need to consider self-projective or self-flat, for the left $R$-module $R$ is always projective, and hence it is always flat.) Self-injectivity is most interesting when it is coupled with chain conditions.

It can be shown that the apparent asymmetry of the definition is only virtual: if $R$ is quasi-Frobenius, then $R$ is an injective right $R$-module (see Jans, Rings and Homology, p. 78, or Lam, Lectures on Modules and Rings, p. 409).

Clearly, semisimple rings are quasi-Frobenius (for they are both left and right noetherian, and every module is injective); in particular, $k G$ is quasiFrobenius when $G$ is a finite group and $k$ is a field whose characteristic does not divide $|G|$. Although there are other examples, as we shall see, the most important examples of quasi-Frobenius rings are group rings $k G$ for $G$ finite and $k$ a field of any characteristic (see Theorem 4.46). Such rings arise naturally in the theory of modular group representations. For example, if $G$ is a finite solvable group, then a minimal normal subgroup $V$ of $G$ is a vector space over $\mathbb{F}{p}$ for some prime $p$ (see Rotman, An Introduction to the Theory of Groups, p. 105). Since $V \triangleleft G$, the group $G$ acts on $V$ by conjugation, and so $V$ is an $\mathbb{F}{p} G$-module.

## 线性代数作业代写linear algebra代考|Semiperfect Rings

There is a notion dual to that of injective envelope, called projective cover. In contrast to injective envelopes, which exist for modules over any ring, projective covers exist only for certain rings, called perfect. A semiperfect ring is one for which every finitely generated module has a projective cover. We shall see that local rings and artinian rings are semiperfect. We begin with some basic ring theory. Definition. If $R$ is a ring, then its Jacobson radical $J(R)$ is defined to be the intersection of all the maximal left ideals in $R$.

Clearly, we can define another Jacobson radical: the intersection of all the maximal right ideals. It turns out, however, that both of these coincide (see Rotman, Advanced Modern Algebra, p. 547), so that $J(R)$ is a two-sided ideal. Consequently, $R / J(R)$ is a ring.

Proposition 4.50. If $x$ is an element in a ring $R$, then $x \in J(R)$ if and only if, for each $a \in R$, the element $1-a x$ has a left inverse; that is, there is $u \in R$ with $u(1-a x)=1$.

Proof. If $R(1-a x)$ is a proper left ideal, then Zorn’s Lemma shows that there is some maximal left ideal containing it; say, $R(1-a x) \subseteq M$. By definition, $a x \in J \subseteq M$, so that $1=(1-a x)+a x \in M$, contradicting $M$ being a proper ideal. Therefore, $R(1-a x)=R$, and so there is $u \in R$ with $u(1-a x)=1$.

Conversely, if $x \notin J$, then there is a maximal left ideal $M$ with $x \notin M$. Since $M \subsetneq M+R x$, we have $M+R x=R$, so that there are $m \in M$ and $a \in R$ with $m+a x=1$. If $m=1-a x$ has a left inverse $u$, then $1=u m \in M$, contradicting $M$ being a proper left ideal.

## 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