## 数学代写|代数数论代写Algebraic number theory代考|Analytic Methods

statistics-lab™ 为您的留学生涯保驾护航 在代写代数数论Algebraic number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写代数数论Algebraic number theory代写方面经验极为丰富，各种代写代数数论Algebraic number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|代数数论代写Algebraic number theory代考|Analytic Methods

In this chapter, we shall prove Dedekind’s famous formula of 1877 for the class number $h_{K}$ of a number field $K$, namely
$$\lim {s \rightarrow 1+}(s-1) \zeta{K}(s)=h_{K} \cdot \kappa,$$
where $\zeta_{K}(s)$ is the Dedekind zeta function
$$\zeta_{K}(s)=\sum \frac{1}{N(a)^{s}}$$
The summation is over all nonzero integral ideals a of $\mathcal{O}{K}$. We shall show that the series on the right of (6.2) converges absolutely for real $s$ in the open interval $11$, $$\zeta{K}(s)=\prod_{p}\left(1-\frac{1}{N(p)^{s}}\right)^{-1}$$
where the product is over all the nonzero prime ideals $p$ of $K$.
Proof. We only sketch the proof, leaving the details to be filled in by the reader.
First, because $N(\mathfrak{p})^{s}>1$,
$$\left(1-\frac{1}{N(p)^{s}}\right)^{-1}=1+\frac{1}{N(p)^{s}}+\frac{1}{N(p)^{2 s}}+\cdots$$
We formally multiply these series, one for each prime $\mathfrak{p}$, to obtain
$$\prod_{p}\left(1-\frac{1}{N(p)^{s}}\right)^{-1}=\sum \frac{1}{N\left(p_{1}^{c_{1}} \cdots p_{g}^{c_{g}}\right)^{s}}$$
In the summation, each product $p_{1}^{c_{1}} \ldots p_{g}^{e_{g}}$ occurs exactly once. Therefore, by Dedekind’s unique factorization theorem for ideals,
$$\sum \frac{1}{N\left(p_{1}^{e_{1}} \cdots p_{g}^{c_{g}}\right)^{s}}=\sum \frac{1}{N(a)^{s}}$$the summation being over all nonzero integral ideals of $K$. Hence, the Euler product formula follows.

## 数学代写|代数数论代写Algebraic number theory代考|Preliminaries

In Chapter 5, we proved that the group $\mathcal{O}{K}^{\times}$of units (of the ring of integers) of a number field $K$ is isomorphic to $W{K} \times \mathbb{Z}^{r}$. Here $W_{K}$ is the group of roots of unity in $K$ and $r=r_{1}+r_{2}-1$. Recall that $r_{1}$ (resp. $r_{2}$ ) is the number of real (resp. pairs of complex) $\mathbb{Q}$-isomorphisms of $K$ into $\mathbb{C}$, so that $[K: \mathbb{Q}]=r_{1}+2 r_{2}$. Let $u_{1}, \ldots, u_{r}$ be a fundamental system of units in $K$, that is to say, any $u$ in $\mathcal{O}{K}^{\times}$can be uniquely expressed as $$u=\eta u{1}^{a_{1}} \ldots u_{r}^{a_{r}},$$
with $\eta$ in $W_{K}$ and $a_{1}, \ldots, a_{r}$ in $\mathbb{Z}$. We now use the set $\left{u_{1}, \ldots, u_{r}\right}$ of fundamental units in $K$ to define an important invariant of $K$, called its regulator, which is intimately related to its class number $h_{K}$.
The hyperplane
$$V=\left{\boldsymbol{v}=\left(\lambda_{1}, \ldots, \lambda_{r_{1}+r_{2}}\right) \in \mathbb{R}^{r_{1}+r_{2}} \mid \lambda_{1}+\cdots+\lambda_{r_{1}+r_{2}}=0\right}$$
is a $r$-dimensional subspace of $\mathbb{R}^{r_{1}+r_{2}}$. Let $\sigma_{1}, \ldots, \sigma_{r_{1}} ; \sigma_{r_{1}+1}, \bar{\sigma}{r{1}+1}, \ldots, \sigma_{r_{1}+r_{2}}$, $\bar{\sigma}{r{1}+r_{2}}$ be all the Q-isomorphisms of $K$ into $\mathbb{C}$. In Chapter 5 , we defined a map $\lambda: K^{\times} \rightarrow \mathbb{R}^{r_{1}+r_{2}}$ by $\lambda(\alpha)=\left(\log \left|\sigma_{1}(\alpha)\right|, \ldots, \log \left|\sigma_{r_{1}}(\alpha)\right|, \log \left|\sigma_{r_{1}+1}(\alpha)\right|^{2}, \ldots\right.$ $\left.\log \left|\sigma_{r_{1}+r_{2}}(\alpha)\right|^{2}\right)$, which is a group homomorphism from the multiplicative group $K^{\times}$into the additive group $\mathbb{R}^{r_{1}+r_{2}}$. We proved that $\lambda\left(\mathcal{O}{K}^{\times}\right)$is a full lattice in the $r$-dimensional subspace $V$ of $\mathbb{R}^{r{1}+r_{2}}$, defined above. To define the regulator, we need to compute the $r$-dimensional volume $\mu\left(\lambda\left(\mathcal{O}{K}^{\times}\right)\right)$of a fundamental parallelepiped of $\lambda\left(\mathcal{O}{K}^{\times}\right)$.

## 数学代写|代数数论代写Algebraic number theory代考|The Regulator of a Number Field

We now state and prove a theorem that leads to the definition of regulator. For $u$ in $\mathcal{O}{K}^{\times}$, let $\lambda{j}(u)$ denote the $j$-th component of the vector $\lambda(u)$ of $\mathbb{R}^{r_{1}+r_{2}}$.
Theorem 6.2. The $r$-dimensional volume $\mu\left(\lambda\left(\mathcal{O}{K}^{\times}\right)\right)$of any fundamental parallelepiped of the (full) lattice $\lambda\left(\mathcal{O}{K}^{\times}\right)$in $V$ is given by
$$\mu\left(\lambda\left(\mathcal{O}{K}^{\times}\right)\right)=\sqrt{r{1}+r_{2}}\left|\operatorname{det}\left(\phi_{i}\left(u_{j}\right)\right)\right|$$

where $\left{\phi_{1}, \ldots, \phi_{r}\right}$ is an arbitrarily chosen subset of $\left{\lambda_{1}, \ldots, \lambda_{r_{1}+r_{2}}\right}$ of cardinality $r$.
In particular the quantity
$$R_{K}=\left|\operatorname{det}\left(\phi_{i}\left(u_{j}\right)\right)\right|$$
depends only on $K$, and not on the choice of $\phi_{1}, \ldots, \phi_{r}$.
Definition 6.3. The regulator of a number field $K$ is the absolute value
$$R_{K}=\left|\operatorname{det}\left(\phi_{i}\left(u_{j}\right)\right)\right|$$
of the $r \times r$ determinant $\operatorname{det}\left(\phi_{i}\left(u_{j}\right)\right)$.
Proof. Consider the unit vector
$$\boldsymbol{u}=\frac{1}{\sqrt{r_{1}+r_{2}}}(1, \ldots, 1)$$
in $\mathbb{R}^{r_{1}+r_{2}}$. By the definition of $V$, the inner product $\langle\boldsymbol{u}, \boldsymbol{x}\rangle=0$ for all $\boldsymbol{x}$ in $V$. Hence $\boldsymbol{u} \perp V(\boldsymbol{u}$ is perpendicular to $V)$. If
$$L=\lambda\left(\mathcal{O}{K}^{\times}\right) \oplus \mathbb{Z} \boldsymbol{u}$$ then $L$ is a full lattice in $\mathbb{R}^{r{1}}+r_{2}$ and the $r$-dimensional volume $\mu\left(\lambda\left(\mathcal{O}{K}^{\times}\right)\right)$of $\lambda\left(\mathcal{O}{K}^{\times}\right)$is equal to the $\left(r_{1}+r_{2}\right)$-dimensional volume of $L$, which is the absolute value of the $\left(r_{1}+r_{2}\right) \times\left(r_{1}+r_{2}\right)$ determinant
$$\frac{1}{\sqrt{r_{1}+r_{2}}} \cdot\left|\begin{array}{ccc} 1 & \cdots & 1 \ \lambda_{1}\left(u_{1}\right) & \cdots & \lambda_{r_{1}+r_{2}}\left(u_{1}\right) \ \vdots & & \ \lambda_{1}\left(u_{r}\right) & \vdots & \lambda_{r_{1}+r_{2}}\left(u_{r}\right) \end{array}\right|$$
For all $u$ in $\mathcal{O}{K}^{\times}$, we have $$\sum{j=1}^{r_{1}+r_{2}} \lambda_{j}(u)=0$$

## 数学代写|代数数论代写Algebraic number theory代考|Analytic Methods

Gķ(s)=∑1ñ(一个)s

Gķ(s)=∏p(1−1ñ(p)s)−1

(1−1ñ(p)s)−1=1+1ñ(p)s+1ñ(p)2s+⋯

∏p(1−1ñ(p)s)−1=∑1ñ(p1C1⋯pGCG)s

∑1ñ(p1和1⋯pGCG)s=∑1ñ(一个)s总和是对所有非零积分理想的ķ. 因此，欧拉乘积公式如下。

## 数学代写|代数数论代写Algebraic number theory代考|Preliminaries

V=\left{\boldsymbol{v}=\left(\lambda_{1}, \ldots, \lambda_{r_{1}+r_{2}}\right) \in \mathbb{R}^{r_{ 1}+r_{2}} \mid \lambda_{1}+\cdots+\lambda_{r_{1}+r_{2}}=0\right}V=\left{\boldsymbol{v}=\left(\lambda_{1}, \ldots, \lambda_{r_{1}+r_{2}}\right) \in \mathbb{R}^{r_{ 1}+r_{2}} \mid \lambda_{1}+\cdots+\lambda_{r_{1}+r_{2}}=0\right}

## 数学代写|代数数论代写Algebraic number theory代考|The Regulator of a Number Field

μ(λ(○ķ×))=r1+r2|这⁡(φ一世(在j))|

Rķ=|这⁡(φ一世(在j))|

Rķ=|这⁡(φ一世(在j))|

1r1+r2⋅|1⋯1 λ1(在1)⋯λr1+r2(在1) ⋮ λ1(在r)⋮λr1+r2(在r)|

∑j=1r1+r2λj(在)=0

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|代数数论代写Algebraic number theory代考|Minkowski’s Lemma on Convex Bodies

statistics-lab™ 为您的留学生涯保驾护航 在代写代数数论Algebraic number theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写代数数论Algebraic number theory代写方面经验极为丰富，各种代写代数数论Algebraic number theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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|代数数论代写Algebraic number theory代考|Minkowski’s Lemma on Convex Bodies

The Dirichlet’s unit theorem asserts that, up to the roots of unity in $K$, the group $\mathcal{O}{K}^{\times}$of units of $K$ is a free Abelian group of rank $r=r{1}+r_{2}-1$. [We shall define the non-negative integers $r_{1}$ and $r_{2}$ in the next section.] It is not very difficult to show that $r \leq r_{1}+r_{2}-1$. The harder part that $r=r_{1}+r_{2}-1$ follows from the famous lemma of Minkowski on convex bodies.

A subset $X \subseteq \mathbb{R}^{n}$ is convex if for all $\boldsymbol{u}, \boldsymbol{v}$ in $X$ and all real $t$ in the interval $[0,1]$, the vector $t \boldsymbol{u}+(1-t) \boldsymbol{v}$ is in $X$. That is, the line segment joining $\boldsymbol{u}$ to $\boldsymbol{v}$ is entirely in $X$. It is easy to see that if $X$ is convex in $\mathbb{R}^{m}$ and $Y$ is convex in $\mathbb{R}^{n}$, then $X \times Y$ is convex in $\mathbb{R}^{m+n}$. We call $X \subseteq \mathbb{R}^{n}$ centrally symmetric if $\boldsymbol{v} \in X$ implies $-\boldsymbol{v} \in X$.

Let $\mu$ be the Lebesgue measure on $\mathbb{R}^{n}$, that is, the measure on $\mathbb{R}^{n}$, such that for a cube $X \subseteq \mathbb{R}^{n}$ given by
$$\begin{gathered} X=\left{\boldsymbol{x}=\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n} \mid a_{j} \leq x_{j} \leq b_{j}\right} \ \mu(X)=\operatorname{vol}(X)=\prod_{j=1}^{n}\left(b_{j}-a_{j}\right) \end{gathered}$$
Let $L$ be a full lattice with a fundamental parallelepiped $P$, as in (5.4) and (5.5). Of course, $P$ depends on the choice of the $\mathbb{Z}$-basis $\left{\boldsymbol{v}{1}, \ldots, \boldsymbol{v}{n}\right}$ of $L$. However, any two $\mathbb{Z}$-bases of $L$ are related by a unimodular matrix, that is a matrix of determinant $\pm 1$ with entries in $\mathbb{Z}$. Since $\mu(P)$ is the absolute value of the determinant, whose rows are $\boldsymbol{v}{1}, \ldots, \boldsymbol{v}{n}$, it follows that the volume $\mu(P)$ of $P$ is independent of the choice of the basis. Thus, we may denote $\mu(P)$ also by $\mu(L)$.

Theorem $5.4$ (Minkowski’s Lemma). Suppose $X \subseteq \mathbb{R}^{n}$ is a bounded, centrally symmetric convex set and $L \subseteq \mathbb{R}^{n}$ is a full lattice. If $\mu(X)>2^{n} \mu(L)$, then $X$ contains a nonzero vector of $L$.

Proof. First we show that if $Y \subseteq \mathbb{R}^{n}$ is a bounded set, such that ${\boldsymbol{v}+Y \mid \boldsymbol{v} \in$ $L}$ is a family of disjoint subsets of $\mathbb{R}^{n}$, then $\mu(Y) \leq \mu(P)$, where $P$ is a fundamental parallelepiped of $L$. This is almost immediate, because writing $Y$ as the disjoint union
$$Y=\cup_{v \in L} Y \cap(v+P),$$
we have by $(5.6), \mu(Y)=\sum_{v \in L} \mu(Y \cap(v+P))$.
Since $\mu$ is translation invariant, $\mu(Y \cap(v+P))=\mu((-v+Y) \cap P)$. Hence $\mu(Y)=\sum_{v \in L} \mu((-v+Y) \cap P) \leq \mu(P)$, because the sets $-v+Y$ are also pairwise disjoint.

## 数学代写|代数数论代写Algebraic number theory代考|Logarithmic Embedding

Suppose $K \subseteq \mathbb{C}$ is a number field of degree $n$ over $\mathbb{Q}$. Consider a ring homomorphism $\sigma: K \rightarrow \mathbb{C}$. We require that $\sigma(1)=1$. Hence $\sigma_{\mid \mathbb{Q}}=1_{\mathrm{Q}}$, the identity map on $\mathbb{Q}$. Such a $\sigma$ is clearly injective. [Its kernel Ker $(\sigma)$ is an ideal of the field $K$, which can only be ${0}$ or $K$.] Hence, we call $\sigma$ a Q-isomorphism of $K$ into $\mathbb{C}$. There are exactly $n \mathbb{Q}$-isomorphisms of $K$ into $\mathbb{C}$. To see this, write $K=\mathbb{Q}(\alpha)$. If $\sigma$ is a $\mathbb{Q}$-isomorphism of $K$ into $\mathbb{C}$, it is determined by $\sigma(\alpha)$, which is a conjugate of $\alpha$. But there are exactly $n$ conjugates of $\alpha$ over $\mathbb{Q}$.
One may regard such a $\sigma: K \rightarrow \mathbb{C}$ also an injective linear transformation of vector spaces, when $K$ and $\mathbb{C}$ are viewed as vector spaces over $\mathbb{Q}$. Unless stated to the contrary $\sigma: K \rightarrow \mathbb{C}$ will be a $\mathbb{Q}$-isomorphism.

If $\sigma(K) \subseteq \mathbb{R}$, we call $\sigma$ a real imbedding, otherwise it is a complex imbedding. If $\sigma$ is complex, the map $\bar{\sigma}: K \rightarrow \mathbb{C}$, given by $\bar{\sigma}(x)=\overline{\sigma(x)}$ is also a $\mathbb{Q}$ isomorphism. Thus, the complex $\mathbb{Q}$-isomorphisms occur in pairs. We shall denote the real $\mathbb{}$ complex ones by $\sigma_{r_{1}+1}, \overline{\sigma_{r_{1}+1}}, \ldots ; \sigma_{r_{1}+r_{2}}, \overline{\sigma_{r_{1}+r_{2}}}$. In particular, $n=r_{1}+2 r_{2}$.
Consider $\mathbb{C}$ as a vector space of dimension two over $\mathbb{R}$ with ${1, i}$ as the standard basis. If $z=x+i y \in \mathbb{C}$, the multiplication by $z$ is a linear transformation of $\mathbb{C}$ into itself over $\mathbb{R}$. Its matrix relative to the basis ${1, i}$ is easily seen to be $T=\left(\begin{array}{cc}x & y \ -y & x\end{array}\right)$ with determinant
$$\operatorname{det}(T)=x^{2}+y^{2}=|z|^{2}$$
If we identify $\mathbb{C}$, as a vector space over $\mathbb{R}$ with $\mathbb{R}^{2}$, via the map $x+i y \rightarrow\left(\begin{array}{l}x \ y\end{array}\right)$, then $\mathbb{R}^{r_{1}} \times \mathbb{C}^{r_{2}} \cong \mathbb{R}^{n}$.

## 数学代写|代数数论代写Algebraic number theory代考|Units of a Quadratic Field

Let $K=\mathbb{Q}(\sqrt{d})(d \neq 0,1$, a square-free integer $)$ be a quadratic field. We call $K$ a real quadratic field or an imaginary quadratic field according as $d>0$ or $d<0$. If $K$ is an imaginary quadratic field, then $r_{1}=0, r_{2}=1$, so $r=r_{1}+r_{2}-1=0$. In this case, $\mathcal{O}{K}^{\times}=W{K}$, the roots of unity in $K$. We leave it as an exercise to determine this finite group $W_{K}$.

For the real quadratic field, $r=1$ and the group of units is given by the following corollary.

Corollary 5.15. If $d>1$ is a square-free integer and $K=\mathbb{Q}(\sqrt{d})$, then the group
$$\mathcal{O}_{K}^{\times} \cong{\pm 1} \times \mathbb{Z} .$$
In particular, the Pell equation $x^{2}-d y^{2}=1$ has infinitely many solutions in integers.
EXERCISES

1. Determine the structure of $\mathcal{O}_{K}^{\times}$when $[K: \mathbb{Q}]=3$ and 4 .
2. Use Dirichlet’s unit theorem to find all integer solutions of $5 x^{2}-$ $5 y^{2}=y^{4}$.

## 数学代写|代数数论代写Algebraic number theory代考|Minkowski’s Lemma on Convex Bodies

\begin{聚集} X=\left{\boldsymbol{x}=\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n} \mid a_{j } \leq x_{j} \leq b_{j}\right} \ \mu(X)=\operatorname{vol}(X)=\prod_{j=1}^{n}\left(b_{j} -a_{j}\right) \end{聚集}\begin{聚集} X=\left{\boldsymbol{x}=\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n} \mid a_{j } \leq x_{j} \leq b_{j}\right} \ \mu(X)=\operatorname{vol}(X)=\prod_{j=1}^{n}\left(b_{j} -a_{j}\right) \end{聚集}

$$be the factorization of \mathfrak{p} into powers of distinct primes in \mathcal{O}. Since the ring$$
\mathcal{O} / \mathrm{pO}=\mathcal{O} / \mathfrak{P}{1}^{e{1}} \ldots \mathfrak{P}{g}^{e{g}} \cong \mathcal{O} / \mathfrak{P}{1}^{e{1}} \times \ldots \times \mathcal{O} / \mathfrak{P}{g}^{e{g}}
$$\mathfrak{p} is ramified \Leftrightarrow some e_{j}>1 \Leftrightarrow \mathcal{O} / \mathfrak{P}{j}^{\epsilon{j}} is not reduced \Leftrightarrow \mathcal{O} / \mathfrak{p O} is not reduced \Leftrightarrow \mathfrak{d}{(\mathcal{O} / p \mathcal{O}) /(0 / p)}=(0). Thus we need to show that \mathfrak{d}{(\mathcal{O} / p \mathcal{O}) /(\mathfrak{o} / \mathfrak{p})}=(0) \Leftrightarrow p \mid \mathfrak{o}_{K / k}. Let S=\mathfrak{o} \backslash \mathfrak{p}, A the localization of \mathfrak{o} at \mathfrak{p}, B=S^{-1} \mathcal{O}, \mathfrak{P}=S^{-1} \mathfrak{p}, the maximal ideal of A. Since \mathcal{O} is a finitely generated (but not necessarily free) o-module, B is a finitely generated A-module, generated by the same elements. Now since A is a principal ideal domain, B has a basis over A, easily seen to consist of n=[K: k] elements \alpha_{1}, \ldots, \alpha_{n}. Since S does not intersect any of the prime ideals of \mathcal{O} lying above \mathfrak{p}, we have the following diagram:$$
\begin{array}{ccc}
\mathcal{O} / \mathfrak{p O} & \cong & B / \mathfrak{P} B \
\mid & & \mid \
0 / \mathfrak{p} & \cong & A / \mathfrak{P}
\end{array}
$$For \beta in B, we denote by \bar{\beta} its residue class in B / \mathfrak{P B}. The dimension of \mathcal{O} / \mathfrak{p O} over o / \mathfrak{p} is n and so is the dimension of B / \mathfrak{P B} over A / \mathfrak{P}. Since \bar{\alpha}{1}, \ldots, \bar{\alpha}{n} generate B / \mathfrak{P B} over A / B, by comparing dimensions, they must form a basis of B / \mathfrak{P} B over A / \mathfrak{P}. Thus by Theorem 4.26 and the diagram above, \mathfrak{d}{(\mathcal{O} / \mathrm{pO}) /(\mathrm{o} / \mathrm{p})}=(0) if and only if \Delta\left(\alpha{1}, \ldots, \alpha_{n}\right)=0. Thus we show that \mathfrak{p} \mid \mathfrak{d}{K / k} if and only if \Delta\left(\alpha{1}, \ldots, \alpha_{n}\right) \in \mathfrak{P}. First, let \Delta\left(\alpha_{1}, \ldots, \alpha_{n}\right) \in \mathfrak{P}. If \left{\beta_{1}, \ldots, \beta_{n}\right} is a basis of K over k consisting of elements in \mathcal{O}, then$$
\beta_{i}=\sum_{j=1}^{n} a_{i j} \alpha_{j} \quad\left(a_{i j} \in A\right)
$$which shows that \Delta\left(\beta_{1}, \ldots, \beta_{n}\right)=\operatorname{det}\left(\operatorname{Tr}\left(\alpha_{i} \alpha_{j}\right)\right) \cdot\left(\operatorname{det}\left(a_{i j}\right)\right)^{2} \in \mathcal{O} \cap \mathfrak{P}=\mathfrak{p}. Hence, \mathfrak{o}{K / k} \subseteq \mathfrak{p}, i.e. \mathfrak{p} \mid \mathfrak{d}{K / k}. Conversely, suppose \mathfrak{p} \mid \mathfrak{v}{K / k}. If \alpha{1}, \ldots, \alpha_{n} is a basis of B over A, write each \alpha_{j}=\beta_{j} / s with \beta_{j} in \mathcal{O} and s in S. Then$$
\Delta\left(\alpha_{1}, \ldots, \alpha_{n}\right)=\operatorname{det}\left(\operatorname{tr}\left(\alpha_{i} \alpha_{j}\right)\right)=\frac{1}{s^{2 n}} \operatorname{det}\left(\operatorname{tr}\left(\beta_{i} \beta_{j}\right)\right)

## 数学代写|代数数论代写Algebraic number theory代考|What Is Number Theory

。定理 1.1。有无穷多个素数。

n=p1和1…prCr(r≥1)

3,5;5,7;11,13;17,19;29,31;…

## 数学代写|代数数论代写Algebraic number theory代考|Methods of Proving Theorems in Number Theory

X3=是2+2=(是+−2)(是−−2).

1=b(3一个2−2b2),是=一个3−6一个b2.

1. 解析方法
欧拉开创了我们所说的解析数论。对无穷级数（分析）的研究可以在数论中产生有趣的结果。让我们回顾一下欧拉关于素数无穷大的证明。撇开收敛问题不谈，通过形式上的无限级数相乘，可以看到
∑n=1∞1n=∑n=11p1和1…pr和r=∏p(1+1p+1p2+⋯), IE  1n=∏p∞(1−1p)−1
取所有素数的乘积（称为欧拉乘积）p. 请注意，第一个等式是唯一因式分解 (1.1) 的结果。

## 数学代写|代数数论代写Algebraic number theory代考|Techniques from Algebraic Geometry

X2+是2=1

X=一个2−b2,是=2一个b,和=一个2+b2,

## 有限元方法代写

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## MATLAB代写

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