### 数学代写|交换代数代写commutative algebra代考|MATH3033

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|Background and Overview

The main theme of the present chapter is to discuss the process of coefficient extension for modules and its reverse, called descent. For example, imagine a ring $R$ and an $R$-module $M$ whose structure seems to be difficult to access. Then one can try to replace the coefficient domain $R$ by a bigger ring $R^{\prime}$ over which the situation might become easier to handle. In other words, we would select a certain extension homomorphism $R \longrightarrow R^{\prime}$ and use it in order to derive from $M$ a best possible $R^{\prime}$-module $M^{\prime}$ extending the $R$-module structure we are given on $M$. In particular, $M^{\prime}$ will respect all relations that are already present in $M$. The technical frame for such a construction is given by the socalled tensor product. Passing from $M$ to the tensor product $M^{\prime}=M \otimes_{R} R^{\prime}$ we say that $M^{\prime}$ is obtainèd from $M$ via coeefficient extension with respeect to $R \longrightarrow R^{\prime}$. Of course, the extension homomorphism $R \longrightarrow R^{\prime}$ must be chosen in an intelligent way so that the results obtained for $M^{\prime}$ can be descended to meaningful information on $M$.

Let us discuss an example from Linear Algebra. We consider a quadratic matrix with coefficients from $\mathbb{R}$, say $A \in \mathbb{R}^{n \times n}$ where $n>0$, and look at the $\mathbb{R}$-linear map
$$\mathbb{R}^{n} \longrightarrow \mathbb{R}^{n}, \quad x \longmapsto A \cdot x .$$
Recall that an element $\lambda \in \mathbb{R}$ is called an eigenvalue of $A$ if there exists an associated eigenvector, i.e. a vector $z \in \mathbb{R}^{n}-{0}$ such that $A z=\lambda z$. Note that the eigenvalues of $A$ are precisely the zeros of the characteristic polynomial $\chi_{A}(X)=\operatorname{det}(X \cdot$ id $-A)$ : Since the field $\mathbb{R}$ is nnt algebrairally rlnsed; it is possible that the set of eigenvalues of $A$ is empty.

However, if we assume $A$ to be symmetric, then the characteristic polynomial $\chi_{A}(X)$ decomposes completely into linear factors over $\mathbb{R}$ and, hence, the set of eigenvalues of $A$ cannot be empty. We want to explain how this result can be derived by means of coefficient extension from $\mathbb{R}$ to $\mathbb{C}$. Viewing $\mathbb{R}^{n}$ as an $\mathbb{R}$-vector space, a canonical candidate for its coefficient extension via $\mathbb{R} \longrightarrow \mathbb{C}$ is of course the $\mathbb{C}$-vector space $\mathbb{C}^{n}$. So we look at the $\mathbb{C}$-linear map
$$\mathbb{C}^{n} \longrightarrow \mathbb{C}^{n}, \quad x \longmapsto A \cdot x .$$
Furthermore, consider the canonical Hermitian form on $\mathbb{C}^{n}$ given by $\langle x, y\rangle=x^{t} \cdot \bar{y}$ for column vectors $x, y \in \mathbb{C}^{n}$, where $x^{t}$ means the transpose of $x$ and $\bar{y}$ the

complex conjugate of $y$. Then, since $A$ is a symmetric matrix with real entries, we get
$$\langle A \cdot x, y\rangle=(A \cdot x)^{t} \cdot \bar{y}=x^{t} \cdot \overline{A \cdot y}=\langle x, A \cdot y\rangle$$
for $x, y \in \mathbb{C}^{n}$. Now use the fact that the field $\mathbb{C}$ is algebraically closed. Therefore the characteristic polynomial $\chi_{A}(X)$ admits a zero $\lambda \in \mathbb{C}$ and there is a corresponding eigenvector $z \in \mathbb{C}^{n}-{0}$. Since $\langle z, z\rangle \neq 0$, the equation
$$\lambda\langle z, z\rangle=\langle\lambda z, z\rangle=\langle A \cdot z, z\rangle=\langle z, A \cdot z\rangle=\langle z, \lambda z\rangle=\bar{\lambda}\langle z, z\rangle$$
shows $\lambda=\bar{\lambda}$. Hence, all zeros of the characteristic polynomial $\chi_{A}(X)$ must be real and we are done. In our argument we can rely on the fact that the characteristic polynomial $\chi_{A}(X)$ is the same for $A$ as a matrix in $\mathbb{C}^{n \times n}$ or in $\mathbb{R}^{n \times n}$. This makes the descent from $\mathbb{C}$ to $\mathbb{R}$ particularly easy.

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

Let $M$ and $N$ be $R$-modules. Recall that a map $\Phi: M \times N \longrightarrow E$ to some $R$-module $E$ is called $R$-bilinear if, for all $x \in M$ and $y \in N$, the maps
$$\begin{array}{ll} \Phi(x, \cdot): N \longrightarrow E, & z \longmapsto \Phi(x, z), \ \Phi(\cdot, y): M \longrightarrow E, & z \longmapsto \Phi(z, y), \end{array}$$
are $R$-linear, by which we mean that they define morphisms of $R$-modules.
Definition 1. A tensor product of $M$ and $N$ over $R$ consists of an $R$-module $T$ together with an $R$-bilinear map $\tau: M \times N \longrightarrow T$ such that the following universal property holds:

For each $R$-bilinear map $\Phi: M \times N \longrightarrow E$ to some $R$-module $E$, there is a unique $R$-linear map $\varphi: T \longrightarrow E$ such that $\Phi=\varphi \circ \tau$, i.e. such that the diagram

Remark 2. Tensor products are uniquely determined by the defining universal property, up to canonical isomorphism.

The proof consists of a well-known standard argument which we would like to repeat once more. Let
$$\tau: M \times N \longrightarrow T, \quad \tau^{\prime}: M \times N \longrightarrow T^{\prime}$$
be tensor products of $M$ and $N$ over $R$. Then there is a diagram
with $R$-linear maps $\varphi, \psi$, where the existence of $\varphi$ satisfying $\tau^{\prime}=\varphi \circ \tau$ follows from the universal property of $\tau: M \times N \longrightarrow T$ and, likewise, the existence of $\psi$ satisfying $\tau=\psi \circ \tau^{\prime}$ from the universal property of $\tau^{\prime}: M \times N \longrightarrow T^{\prime}$. Then we have
$$\mathrm{id}_{T} \circ \tau=\tau=\psi \circ \tau^{\prime}=(\psi \circ \varphi) \circ \tau$$

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

Given two morphisms of $R$-modules $\varphi: M \longrightarrow M^{\prime}$ and $\psi: N \longrightarrow N^{\prime}$, their tensor product over $R$ is defined as the $R$-linear map
$$\varphi \otimes \psi: M \otimes_{R} N \longrightarrow M^{\prime} \otimes_{R} N^{\prime}, \quad x \otimes y \longmapsto \varphi(x) \otimes \psi(y),$$
which is well-defined, due to the fact that the map $M \times N \longrightarrow M^{\prime} \otimes_{R} N^{\prime}$, $(x, y) \longmapsto \varphi(x) \otimes \psi(y)$, is $R$-bilinear in $x$ and $y$. In particular, we can consider the tensor product
$$\varphi \otimes \mathrm{id}{N}: M \otimes{R} N \longrightarrow M^{\prime} \otimes_{R} N$$
of an $R$-linear map $\varphi: M \longrightarrow M^{\prime}$ with the identity map $\mathrm{id}_{N}: N \longrightarrow N$ on any $R$-module $N$. Thereby we tensor $\varphi$ with $N$ over $R$, as we will say. In the same way, we can tensor sequences of $R$-linear maps with $N$. As is easily seen, the process of tensoring $R$-linear maps with an $R$-module $N$ commutes with the composition of such maps.
Proposition 1. Let
$$M^{\prime} \stackrel{\varphi}{\longrightarrow} M \stackrel{\psi}{\longrightarrow} M^{\prime \prime} \longrightarrow 0$$
be an exact sequence of $R$-modules. Then, for any $R$-module $N$, the sequence

$$M^{\prime} \otimes_{R} N \stackrel{\varphi \otimes \mathrm{id}{N}}{\longrightarrow} M \otimes{R} N \stackrel{\psi \otimes \operatorname{id}{N}}{\longrightarrow} M^{\prime \prime} \otimes{R} N \longrightarrow 0$$
obtained by tensoring with $N$ is exact. The property is referred to as the right exactness of tensor products.
Proof. First observe that $\operatorname{im}\left(\varphi \otimes \mathrm{id}{N}\right) \subset \operatorname{ker}\left(\psi \otimes \mathrm{id}{N}\right)$ since
$$\left(\psi \otimes \mathrm{id}{N}\right) \circ\left(\varphi \otimes \mathrm{id}{N}\right)=(\psi \circ \varphi) \otimes \mathrm{id}{N}=0 .$$ ‘herefore $\psi \otimes \mathrm{id}{N}$ admits the factorization
and we see:
\begin{aligned} \operatorname{im}\left(\varphi \otimes \operatorname{id}{N}\right)=\operatorname{ker}\left(\psi \otimes \operatorname{id}{N}\right) & \Longleftrightarrow \bar{\Psi} \text { is injective } \ \psi \otimes \operatorname{id}_{N} \text { is surjective } & \Longleftrightarrow \bar{\Psi} \text { is surjective } \end{aligned}

## 数学代写|交换代数代写commutative algebra代考|Background and Overview

Rn⟶Rn,X⟼一个⋅X.

Cn⟶Cn,X⟼一个⋅X.

⟨一个⋅X,是⟩=(一个⋅X)吨⋅是¯=X吨⋅一个⋅是¯=⟨X,一个⋅是⟩

λ⟨和,和⟩=⟨λ和,和⟩=⟨一个⋅和,和⟩=⟨和,一个⋅和⟩=⟨和,λ和⟩=λ¯⟨和,和⟩

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

τ:米×ñ⟶吨,τ′:米×ñ⟶吨′

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

(ψ⊗一世dñ)∘(披⊗一世dñ)=(ψ∘披)⊗一世dñ=0.’因此ψ⊗一世dñ承认因式分解
，我们看到：

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