# 线性代数网课代修|同调代数代写homological algebra代考|MATH6350

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• 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|Localization

The ring $\mathbb{Z}$ has infinitely many prime ideals, but the ring $\mathbb{Z}{(2)}={a / b \in \mathbb{Q}$ : $b$ is odd} has only one prime ideal, namely, (2) (all other primes in $\mathbb{Z}$ are invertible in $\mathbb{Z}{(2)}$ ). Now $\mathbb{Z}{(2)}$-modules are much simpler than $\mathbb{Z}$-modules. For example, there are only two $\mathbb{Z}{(2)}$-submodules of $\mathbb{Q}$ (to isomorphism): $\mathbb{Z}{(2)}$ and $\mathbb{Q}$. On the other hand, there are uncountably many nonisomorphic subgroups of $\mathbb{Q}$. Similar observations lead to a localization-globalization strategy to attack algebraic and number-theoretic problems. The fundamental assumption underlying this strategy is that the local case is simpler than the global. Evidence for this can be seen in the structure of projective $R$-modules: for arbitrary commutative rings $R$, projectives can be quite complicated, but Theorem $4.58$ says that projective modules over local rings are always free. Given a prime ideal $\mathfrak{p}$ in a commutative ring $R$, we will construct local rings $R{\mathfrak{p}}$. Localization looks at problems involving the rings $R_{p}$, while globalization uses all such local information to answer questions about $R$.

Definition. A subset $S \subseteq R$ of a commutative ring $R$ is multiplicative if $S$ is a monoid not containing 0 ; that is, $0 \notin S, 1 \in S$, and $S$ is closed under multiplication: if $s, s^{\prime} \in S$, then $s s^{\prime} \in S$.
Example 4.67.
(i) If $\mathfrak{p}$ is a prime ideal in $R$, then its set-theoretic complement $S=R-\mathfrak{p}$ is multiplicative.
(ii) If $R$ is a domain, then the set $S=R^{\times}$of all its nonzero elements is multiplicative [this is a special case of part (i), for ${0}$ is a prime ideal in a domain].
(iii) If $a \in R$ is not nilpotent, then the set of its powers $S=\left{a^{n}: n \geq 0\right}$ is multiplicative. More generally, any submonoid of $R$ not containing 0 is multiplicative.

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

In the mid-1950s, Serre proved that if $R=k\left[x_{1}, \ldots, x_{m}\right]$, where $k$ is a field, then every finitely generated projective $R$-module $P$ has a finitely generated free complement $F$; that is, $P \oplus F$ is free. Serre wondered ${ }^{5}$ whether every projective $k\left[x_{1}, \ldots, x_{m}\right]$-module is free (for projective and free modules over $R$ have natural interpetations in algebraic geometry). This problem was the subject of much investigation until 1976, when it was solved in the affirmative by Quillen and Suslin, independently. We refer the reader to T. Y. Lam, Serre’s Problem on Projective Modules, for a more thorough account. The last chapter of Lam’s book describes recent work, after 1976, inspired by and flowing out of the work of Serre, Quillen, and Suslin.

Definition. Let $R$ be a commutative ring and let $R^{n}$ denote the free $R$ module of rank $n$ (recall that every commutative ring has IBN). A unimodular column is an element $\alpha=\left(a_{1}, \ldots, a_{n}\right) \in R^{n}$ for which there exist $b_{i} \in R$ with $a_{1} b_{1}+\cdots+a_{n} b_{n}=1$.

A commutative ring $R$ has the unimodular column property if, for every $n$, every unimodular column is the first column of some $n \times n$ invertible matrix uver $R$.

If $\varepsilon_{1}$ denotes the column vector having first coordinate 1 and all other entries 0 , then $\alpha \in R^{n}$ is the first column of a matrix $M$ over $R$ if and only if
$$\alpha=M \varepsilon_{1}$$
The first column $\alpha=\left[a_{i 1}\right]$ of an invertible matrix $M=\left[a_{i j}\right]$ is always unimodular. Since $M$ is invertible, $\operatorname{det}(M)=u$, where $u$ is a unit in $R$, and Laplace expansion down the first column gives $\operatorname{det}(M)=u=\sum_{i} a_{i 1} d_{i}$. Hence, $\sum_{i} a_{i 1}\left(u^{-1} d_{i}\right)=1$, and $\alpha=M \varepsilon_{1}$ is a unimodular column. The unimodular column property for $R$ asserts the converse.

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

$\mathbb{Z}(2)=a / b \in \mathbb{Q} \$: \$b$ isoddhasonlyoneprimeideal, namely, (2)(allotherprimesin Imathbb ${Z}$ areinvertiblein $\backslash$ mathbb ${Z}{(2)})$. Now mathbb ${Z}{(2)}$

• modulesaremuchsimplerthan $\backslash$ mathbb ${Z}$
• modules. Forexample, thereareonlytwolmathbb ${\mathrm{Z}}{(2)}-$ submodulesof
Imathbb ${Q}$ (toisomorphism) :Imathbb ${Z}{(2)} a n d$ mathbb ${Q}$
. Ontheotherhand, thereareuncountablymanynonisomorphicsubgroupsof
Imathbb ${\mathrm{Q}}$
$\mathrm{R}$-modules : forarbitrarycommutativerings $\mathrm{R}$
, projectivescanbequitecomplicated, butTheorem $4.58$
saysthatprojectivemodulesoverlocalringsarealways free. Givenaprimeideal
Imathfrak{p}inacommutativering $\mathrm{R}$, wewillconstructlocalrings $\mathrm{R}{\mathrm{Imathfrak { \textrm {p } } }}$
. LocalizationlooksatproblemsinvolvingtheringsR_{p}
定义。一个子集 $S \subseteq R$ 交换环的 $R$ 如果是乘法 $S$ 是一个不包含 0 的么半群；那是，
$0 \notin S, 1 \in S$ ，和 $S$ 在乘法下闭合: 如果 $s, s^{\prime} \in S$ ，然后 $s s^{\prime} \in S$.
例 4.67。
(一) 如果 $p$ 是一个主要理想 $R$, 那么它的集合论补集 $S=R-\mathfrak{p}$ 是乘法的。
(ii) 如果 $R$ 是一个域，那么集合 $S=R^{\times}$它的所有非零元素都是乘法的[这是第 (i) 部分的一个 特例，因为 0 是域中的素理想]。
(iii) 如果 $a \in R$ 不是幂零的，那么它的幂集| $\mathrm{S}=\mid$ left ${\mathrm{a} \wedge{\mathrm{n}}: \mathrm{n} \mid \mathrm{geq}$ O|right $}$ 是乘法的。更一般 地，任何亚么半群 $R$ 不包含 0 是乘法。

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

$$\alpha=M \varepsilon_{1}$$

# 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