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# 数学代写|代數數論代写Algebraic Number Theory  代考|MATH223 Local and global different

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## 数学代写|代數數論代写Algebraic Number Theory代考|Local and global different

DEFINITION 176. A lattice $\Lambda<\mathbb{R}^n$ is discrete and cocompat subgroup, equivalently a free $\mathbb{Z}$-submodule generating by a basis.
LEMMA 177. The two definitions are equivalent.
Corollary 178. (Of the proof of the Lemma) Let $\Lambda=\oplus_i \mathbb{Z}i$. Then $\mathcal{F}=\left{\sum{i=1}^n a_i vi \mid a_i \in[0,1]\right}$ is a bounded fundamental domain. DEFINITION 179. The covolume of $\Lambda$ is $\operatorname{vol}(\mathcal{F})=\operatorname{det}\left(\cdots v_i \cdots\right)=\sqrt{\operatorname{det}\left(\left\langlev_i, v_j\right\rangle\right){i, j}}$
PROPOSITION 180. # ${\lambda \in \Lambda \cap B(0, R)} \sim \frac{\operatorname{vol} B(R)}{\operatorname{vol}(\mathcal{F})}$.
Proof. The set $\bigcup_{\lambda \in \Lambda \cap B(0, R)}(\mathcal{F}+\lambda)$ has volume # ${\lambda \in \Lambda \cap B(0, R)} \operatorname{vol}(\mathcal{F})$ and its symmetric difference with $B(R)$ is contained in a spherical shell of radius $R$ and constant thickness, hence has volume $O\left(R^{-1} \operatorname{vol}(B(R))\right)$.

COROLLARY 181. $\operatorname{vol}(\mathcal{F})$ is indpendent of the choice of $\mathcal{F}$, and will be denoted $\operatorname{vol}\left(\mathbb{R}^n / \Lambda\right)$ or $\operatorname{covol}(\Lambda)$

THEOREM 182 (Minkowski). Let $\Lambda<\mathbb{R}^n$ be a lattice (discrete and cocompat subgroup, equivalently a free $\mathbb{Z}$-submodule generating by a basis). Let $X \subset \mathbb{R}^n$ be convex, bounded and symmetic aboud the origin. Suppose that $\operatorname{vol}(X) \geq 2^n \operatorname{vol}\left(\mathbb{R}^n / \Lambda\right)$. Then there is $0 \neq \lambda \in X \cap \Lambda$.

Proof. Suppose first that $\operatorname{vol}(X)>2^n \operatorname{vol}\left(\mathbb{R}^n / \Lambda\right)$. Assume by contradiction that $X \cap \Lambda={0}$. Then all translates of $\frac{1}{2} X$ are disjoint: if there are $x, y \in X$ and a non-zero $\lambda \in \Lambda$ such that $\frac{1}{2} x=$ $\frac{1}{2} y+\lambda$ then $\lambda=\frac{1}{2} x+\frac{1}{2}(-y) \in X$ by assumption. Now let $r=\operatorname{diam}(\mathcal{F})$. Then for all $R>0$,
$$\bigcup_{\lambda \in \Lambda \cap B(0, R)}\left(\frac{1}{2} X+\lambda\right) \subset B(0, R+r) .$$
Since the union on the left is disjoint, we have
$$#{\lambda \in \Lambda \cap B(0, R)} \frac{\operatorname{vol}(X)}{2^n} \leq \operatorname{vol}\left(B_{\mathbb{R}^n}(R+r)\right) .$$

## 数学代写|代數數論代写Algebraic Number Theory代考|Discriminant bounds

LEMMA 183. For a number field $K$ the image of $\mathcal{O}K$ in $K{\infty}$ is a lattice of covolume $2^{-s} \sqrt{\left|d_K\right|}$. REMARK 184. The lemma implicitely depends on the choice of inner product on $K_{\infty}$ made in its proof.

Proof. Let $T \subset \operatorname{Hom}{\mathbb{Q}}(K, \mathbb{C})$ be a set of representatives for the infinite places of $K$, say $T=T{\mathbb{R}} \sqcup T_{\mathbb{C}}$, and let $\imath: K \rightarrow K_{\infty}=\prod_{\tau \in T} K_\tau$ be the embedding. We have seen (Theorem 122) that $\iota$ induces an isomorphism $K \otimes_{\mathbb{Q}} \mathbb{Q}{\infty} \rightarrow K{\infty}$, and in particular $\iota$ maps every $\mathbb{Q}$-basis of $K$ to an $\mathbb{R}$-basis of $K_{\infty}$. Since $\mathcal{O}K$ is the $\mathbb{Z}$-span of an integral basis, it follows that its image is a lattice in $K{\infty}$. On $K_{\infty}$ we take the Hermitian product $\left\langle\left(x_\tau\right),\left(y_\tau\right)\right\rangle=\sum_{i=1}^n x_\tau \bar{y}\tau$. Then for $\omega_i, \omega_j \in \mathcal{O}_K$ we have $\left\langle l\left(\omega_i\right), l\left(\omega_j\right)\right\rangle=\sum{\tau \in T} \tau\left(\omega_i\right) \bar{\tau}\left(\omega_j\right)$

THEOREM 185. Let $K$ be a number field. There are at most finitely many extensions of degree $n$ having a given discriminant.

Proof. Can easily reduce to the case $K=\mathbb{Q}$ and counting extensions $L$ such that $i \in L$ (the discriminant of $L(i)$ diffes by a constant). Thus we are counting totally complex $L$. Fix an infinite place $v_0$ and let
$$X=\left{\left(x_v\right) \in L_{\infty}|| \mathfrak{I} x_{v_0}|\leq C \sqrt{|D|},| \Re x_{v_0}|<1,| x_v \mid<1 v \neq v_0\right} .$$

## 数学代写|代數數論代写ALGEBRAIC NUMBER THEORY代 考|DISCRIMINANT BOUNDS

.REMARK184.Thelemmaimplicitelydependsonthechoiceofinnerproductonk{infty}\$在其证明中作出。 证明。让$T \subset \operatorname{Hom} \mathbb{Q}(K, \mathbb{C})$是无限地方的一组代表$K$，说$T=T \mathbb{R} \sqcup T_{\mathbb{C}}$，然后让$\imath: K \rightarrow K_{\infty}=\prod_{\tau \in T} K_\tau$是嵌入。我们已经看到$T h e o r e m 122$那$\iota$诱导同构$K \otimes_{\mathbb{Q}} \mathbb{Q} \infty \rightarrow K \infty$, 特别是$\iota$映射每个$\mathbb{Q}$-基础$K$到一个$\mathbb{R}$-基础$K_{\infty}$. 自从$\mathcal{O} K$是个$\mathbb{Z}$– 一个整数基础的跨度，因此它的图像是一个格子$K \infty$. 上$K \infty$我们采用厄米特产 品$\left\langle\left(x_\tau\right),\left(y_\tau\right)\right\rangle=\sum_{i=1}^n x_\tau \bar{y} \tau$. 然后为$\omega_i, \omega_j \in \mathcal{O}_K$我们有$\left\langle l\left(\omega_i\right), l\left(\omega_j\right)\right\rangle=\sum \tau \in T \tau\left(\omega_i\right) \bar{\tau}\left(\omega_j\right)$定理 185. 让$K$是一个数字字段。度的扩展至多有限$n$有给定的判别式。 证明。可以很容易地减少到情况$K=\mathbb{Q}$和计数扩展$L$这样$i \in L$thediscriminantof$\$L$ (idiffesbyaconstant). Thuswearecountingtotallycomplex大号

## 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。