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

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• 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

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