### 数学代写|交换代数代写commutative algebra代考|MATH4312

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• Statistical Inference 统计推断
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• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|交换代数代写commutative algebra代考|Local Rings and Localization of Rings

Proposition 2. Let $R$ be a ring and $\mathfrak{m} \subsetneq R$ a proper ideal. The following conditions on $\mathrm{m}$ are equivalent:
(i) $R$ is a local ring with maximal ideal $\mathfrak{m}$.
(ii) Every element of $R-\mathfrak{m}$ is a unit in $R$.
(iii) $\mathfrak{m}$ is a maximal ideal and every element of type $1+m$ with $m \in \mathfrak{m}$ is a unit in $R$.

Proof. We start by showing that conditions (i) and (ii) are equivalent. If (i) holds and $a \in R$ is not a unit, we can use $1.1 / 13$ to conclude that there is a maximal idcal $\mathfrak{n} \subset R$ with $a \subset \mathfrak{n}$. Neccssarily, $\mathfrak{n}$ must coincidc with $\mathfrak{m}$. Thercforc the complement $R-\mathfrak{m}$ consists of units and we get (ii). Conversely, if (ii) holds, every proper ideal $\mathfrak{\subsetneq} \subsetneq R$ will be contained in $\mathrm{m}$, since proper ideals cannot contain units. In particular, $\mathfrak{m}$ is a unique maximal ideal in $R$ and we get (i).
Next assume (ii) again. Then $m$ is a maximal ideal by (i), and we see for every $m \in \mathfrak{m}$ that $1+m$ cannot be contained in $\mathfrak{m}$ since $1 \notin \mathfrak{m}$. Thus, by our assumption, $1+m$ is a unit and we have (iii). Conversely, assume (iii) and let $x \in R-\mathrm{m}$. Since $\mathfrak{m}$ is a maximal ideal, $x$ and $\mathfrak{m}$ will generate the unit ideal in $R$. Hence, there exists an equation
$$1=a x-m$$
with elements $a \in R$ and $m \in \mathfrak{m}$. Then $a x=1+m$ is a unit by (iii) and the same is true for $x$ so that (ii) holds.

Every field $K$ is a local ring with maximal ideal $0 \subset K$. Further examples of local rings are provided by discrete valuation rings, which can be viewed as principal ideal domains $R$ containing just one prime element $p \in R$ (up to multiplication by units). In such a ring, $(p) \subset R$ is the only maximal ideal.
To give an explicit example of a discrete valuation ring $R$, fix a prime $p \in \mathbb{N}$ and consider
$$\mathbb{Z}{(p)}=\left{\frac{m}{n} \in \mathbb{Q} ; m, n \in \mathbb{Z} \text { with } p \nmid n\right} \subset \mathbb{Q}$$ as a subring of $\mathbb{Q}$. Then $\mathbb{Z}{(p)}$ is an integral domain, and we claim that $\mathbb{Z}{(p)}$ is, in fact, a principal ideal domain. To show that any ideal $a \subset \mathbb{Z}{(p)}$ is principal, look at its restriction $\mathfrak{a}^{\prime}=\mathfrak{a} \cap \mathbb{Z}$, which is an ideal in $\mathbb{Z}$. As $\mathbb{Z}$ is principal, there is an as the set of numerators of fractions $\frac{m}{n} \in \mathfrak{a}$ where $p \nmid n$. Next we want to show that $\mathbb{Z}{(p)}$ contains precisely one maximal ideal and that the latter is generated by $p$. To justify this, observe that $\frac{1}{p}$ does not belong to $\mathbb{Z}{(p)}$ and, hence, that $p$ is not invertible in $\mathbb{Z}{(p)}$. Therefore $p \mathbb{Z}{(p)}$ is a proper ideal in $\mathbb{Z}_{(p)}$, and we claim that its complement $\mathbb{Z}{(p)}-p \mathbb{Z}{(p)}$ consists of units in $\mathbb{Z}{(p)}$. Any element in $\mathbb{Z}{(p)}$ can be written as a fraction $\frac{m}{n}$ with $p \nmid n$, and such a fraction satisfies $p \nmid m$ if it does not belong to $p \mathbb{Z}{(p)}$. But then $\left(\frac{m}{n}\right)^{-1}=\frac{n}{m} \in \mathbb{Z}{(p)}$ and $\frac{m}{n}$ is a unit. Therefore all elements of $\mathbb{Z}{(p)}-p \mathbb{Z}{(p)}$ are invertible, and it follows from Proposition 2 (ii) that $\mathbb{Z}{(p)}$ is a local ring with maximal ideal $p \mathbb{Z}{(p)}$. In particular, $p$ is a prime element in $\mathbb{Z}{(p)}$, in fact, up to multiplication by a unit the only prime element existing in $\mathbb{Z}{(p)}$. Indeed, a prime element of $\mathbb{Z}{(p)}$ cannot be invertible and, hence, must belong to $p \mathbb{Z}{(p)}$, which means that it is divisible by $p$. Looking at prime decompositions of elements in $\mathbb{Z}{(p)}$, we see that the ideals in $\mathbb{Z}{(p)}$ are precisely the ones occurring in the chain
$$\mathbb{Z}{(p)} \supset p \mathbb{Z}{(p)} \supset p^{2} \mathbb{Z}_{(p)} \supset \ldots \supset 0$$

## 数学代写|交换代数代写commutative algebra代考|Radicals

Definition 1. Let $R$ be a ring. The intersection
$$j(R)=\bigcap_{\mathrm{m} \in \mathrm{S}{p} m} \mathfrak{m}$$ of all maximal ideals in $R$ is called the Jacobson radical of $R$. As an intersection of ideals, the Jacobson radical $j(R)$ is an ideal in $R$ again. If $R$ is the zero ring, it makes sense to put $j(R)=R$, since an empty intersection of ideals in a ring $R$ equals $R$ by convention. Let us consider some further examples. Clearly, a ring $R$ is local if and only if its Jacobson radical $j(R)$ is a maximal ideal. Furthermore, we claim that the Jacobson radical of a polynomial ring in finitely many variables $X{1}, \ldots, X_{n}$ over a field $K$ is trivial,
$$j\left(K\left[X_{1}, \ldots, X_{n}\right]\right)=0$$
This is a special case of Hilbert’s Nullstellensatz; see $3.2 / 5$ or $3.2 / 6$. To give a simplẽ argument for this at thē prêsent stāge, lèt $\bar{K}$ bee an algebraic closure of $K$. Then, for a point $x=\left(x_{1}, \ldots, x_{n}\right) \in \bar{K}^{n}$, we may consider the ideal
$$\mathfrak{m}{x}=\left{f \in K\left[X{1}, \ldots, X_{n}\right] ; f(x)=0\right}$$
which is the kernel of the substitution homomorphism
$$K\left[X_{1}, \ldots, X_{n}\right] \longrightarrow \bar{K}, \quad g \longmapsto g(x) .$$
As the latter is surjective onto the field $K\left(x_{1}, \ldots, x_{n}\right)$ generated over $K$ by the components of $x$, we see that $K\left[X_{1}, \ldots, X_{n}\right] / \mathfrak{m}{x}$ is a field. Therefore it follows from $1.1 / 8$ (ii) that $\mathfrak{m}{x}$ is a maximal ideal in $K\left[X_{1}, \ldots, X_{n}\right]$.

Now let $f \in j\left(K\left[X_{1}, \ldots, X_{n}\right]\right)$. Then $f \in \mathfrak{m}_{x}$ and, hence, $f(x)=0$ for all $x \in \bar{K}^{n}$. From this we can conclude by induction on $n$ that $f$ is the zero polynomial. Indeed, the case $n=1$ is clear since any algebraically closed field contains infinitely many elements and, hence, $f$ is a polynomial in one variable that has an infinite number of zeros. On the other hand, if $n>1$ we can write $f$ as a polynomial in $X_{n}$ with coefficients in $K\left[X_{1}, \ldots, X_{n-1}\right]$, say $f=\sum_{i=0}^{d} f_{i} X_{n}^{i}$. Fixing an arbitrary point $x^{\prime} \in \bar{K}^{n-1}$, the polynomial
$$f\left(x^{\prime}, X_{n}\right)=\sum_{i=0}^{d} f_{i}\left(x^{\prime}\right) X_{n}^{i} \in \bar{K}\left[X_{n}\right]$$
vanishes at all points $x_{n} \in \bar{K}$ and, as before, must have coefficients $f_{i}\left(x^{\prime}\right)$ that are trivial. But then, varying $x^{\prime}$, the polynomials $f_{i}$ will vanish on all points of $\bar{K}^{n-1}$ so that the $f_{i}$ must be trivial by the induction hypothesis. In particular, $f=0$ and this shows that the Jacobson radical of $K\left[X_{1}, \ldots, X_{n}\right]$ is trivial, as claimed. For a different method of proof see Exercise 5 below.

## 数学代写|交换代数代写commutative algebra代考|Modules

Definition 1. Let $R$ be a ring. An $R$-module consists of a set $M$ together with an inner composition law $M \times M \longrightarrow M,(a, b) \longmapsto a+b$, called addition, and an external composition law $R \times M \longrightarrow M,(\alpha, a) \longmapsto \alpha \cdot a$, called scalar multiplication, such that:
(i) $M$ is abelian group with respect to addition.
(ii) $(\alpha+\beta) \cdot a=\alpha \cdot a+\beta \cdot a$ and $\alpha \cdot(a+b)=\alpha \cdot a+\alpha \cdot b$ for all $\alpha, \beta \in R$, $a, b \in M$, i.e. addition and scalar multiplication satisfy distributivity.
(iii) $(\alpha \cdot \beta) \cdot a=\alpha \cdot(\beta \cdot a)$ for all $\alpha, \beta \in R, a \in M$, i.e. the scalar multiplication is associative.
(iv) $1 \cdot a=a$ for the unit element $1 \in R$ and all $a \in M$.
Modules should be seen as a natural generalization of vector spaces. In particular, a $K$-module over a field $K$ is just a $K$-vector space. On the other hand, any idcal $a$ of a ring $R$ can be considercd as an $R$-modulc. Just vicw a as a group with respect to the addition of $R$ and use the multiplication of $R$ in order to define a scalar multiplication of $R$ on a. In particular, $R$ is a module over itself. Also note that there is a $(1: 1)$-correspondence between abelian groups and $\mathbb{Z}$-modules.

Definition 2. A map $\varphi: M \longrightarrow N$ between $R$-modules $M$ and $N$ is called $a$ morphism of $R$-modules or an $R$-module homomorphism (or just an $R$-homomorphism if the context of modules is clear ) if
(i) $\varphi(x+y)=\varphi(x)+\varphi(y)$ for all $x, y \in M$,
(ii) $\varphi(r x)=r \varphi(x)$ for all $r \in R$ and $x \in M$.
Mono-, epi-, iso-, endo-, and automorphisms of $R$-modules are defined as usual.

## 数学代写|交换代数代写commutative algebra代考|Local Rings and Localization of Rings

(i)R是具有极大理想的局部环米.
(ii) 每一个元素R−米是一个单位R.
㈢米是一个最大理想和类型的每个元素1+米和米∈米是一个单位R.

1=一个X−米

\mathbb{Z}{(p)}=\left{\frac{m}{n} \in \mathbb{Q} ; m, n \in \mathbb{Z} \text { with } p \nmid n\right} \subset \mathbb{Q}\mathbb{Z}{(p)}=\left{\frac{m}{n} \in \mathbb{Q} ; m, n \in \mathbb{Z} \text { with } p \nmid n\right} \subset \mathbb{Q}作为一个子环问. 然后从(p)是一个积分域，我们声称从(p)实际上，是一个主要的理想域。表明任何理想一个⊂从(p)是本金，看它的限制一个′=一个∩从，这是一个理想的从. 作为从是主要的，有一个作为分数分子的集合米n∈一个在哪里p∤n. 接下来我们要证明从(p)恰好包含一个最大理想，而后者是由p. 为了证明这一点，请注意1p不属于从(p)因此，p不可逆从(p). 所以p从(p)是一个适当的理想从(p), 我们声称它的补码从(p)−p从(p)由单位组成从(p). 中的任何元素从(p)可以写成分数米n和p∤n, 这样的分数满足p∤米如果它不属于p从(p). 但是之后(米n)−1=n米∈从(p)和米n是一个单位。因此所有元素从(p)−p从(p)是可逆的，从命题 2 (ii) 可以得出从(p)是具有极大理想的局部环p从(p). 尤其是，p是一个素数元素从(p), 事实上, 乘以一个单位是唯一存在的素数从(p). 确实，一个主要元素从(p)不能可逆，因此必须属于p从(p)，这意味着它可以被p. 查看元素的主要分解从(p)，我们看到理想在从(p)正是发生在链中的那些

## 数学代写|交换代数代写commutative algebra代考|Radicals

j(R)=⋂米∈小号p米米的所有最大理想R被称为 Jacobson 根R. 作为理想的交汇点，雅各布森激进派j(R)是一个理想的R再次。如果R是零环，放j(R)=R，因为理想在环中的空交集R等于R按照惯例。让我们考虑一些进一步的例子。明明是戒指R当且仅当它的 Jacobson 激进时是局部的j(R)是一个极大理想。此外，我们声称在有限多个变量中多项式环的 Jacobson 根X1,…,Xn在一个领域ķ微不足道，

j(ķ[X1,…,Xn])=0

\mathfrak{m}{x}=\left{f \in K\left[X{1}, \ldots, X_{n}\right] ; f(x)=0\右}\mathfrak{m}{x}=\left{f \in K\left[X{1}, \ldots, X_{n}\right] ; f(x)=0\右}

ķ[X1,…,Xn]⟶ķ¯,G⟼G(X).

F(X′,Xn)=∑一世=0dF一世(X′)Xn一世∈ķ¯[Xn]

## 数学代写|交换代数代写commutative algebra代考|Modules

(i)米是关于加法的阿贝尔群。
(二)(一个+b)⋅一个=一个⋅一个+b⋅一个和一个⋅(一个+b)=一个⋅一个+一个⋅b对所有人一个,b∈R, 一个,b∈米，即加法和标量乘法满足分布性。
㈢(一个⋅b)⋅一个=一个⋅(b⋅一个)对所有人一个,b∈R,一个∈米，即标量乘法是关联的。
(四)1⋅一个=一个对于单位元素1∈R和所有一个∈米.

（i）披(X+是)=披(X)+披(是)对所有人X,是∈米,
(ii)披(rX)=r披(X)对所有人r∈R和X∈米.

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## MATLAB代写

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