## 数学代写|数论作业代写number theory代考|MA218

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## 数学代写|数论作业代写number theory代考|Application to Algebraic Congruences

The CRT is a convenient tool for reducing algebraic congruences, $f(x) \equiv 0$ $(\bmod n)$, modulo composite $n$, to the case of, e.g., prime powers. We will deal with that special case later.
The following result will be used in the Chapter on Primitive Roots.
B.IV.1 Lemma. Let $n=n_1 n_2$ where the factors are relatively prime, and $\geq 3$. The congruence $x^2 \equiv 1(\bmod n)$ then has at least four solutions modulo $n$, i.e., the solution set consists of at least four different residue classes modulo $n$.
Proof. Consider the four different congruence pairs
$$\begin{array}{ll} x \equiv \pm 1 & \left(\bmod n_1\right) \ x \equiv \pm 1 & \left(\bmod n_2\right) . \end{array}$$
Each of the pairs is uniquely solvable modulo $n$, by the CRT, producing four different residue classes modulo $n$. In each case the solutions $x$ satisfy
$$\begin{array}{cc} x^2 \equiv 1 & \left(\bmod n_1\right) \ x^2 \equiv 1 & \left(\bmod n_2\right), \end{array}$$
i.e., $x^2-1$ is divisible by both $n_1$ and $n_2$. By the Second Divisibility Theorem, this implies that $x^2-1$ is divisible by their product, i.e. $x^2 \equiv 1(\bmod n)$.

## 数学代写|数论作业代写number theory代考|Linear Congruences

The number $n$, as usual, is a positive integer.
In this short section we study linear congruences, $a x \equiv b(\bmod n)$. No really new theory is required.

B.V.1 Theorem. The congruence $a x \equiv b(\bmod n), a \nmid n$, is solvable if and only if $(a, n) \mid b$.

The solution, in this case, is unique modulo $n /(a, n)$. In other words, the solution set is a residue class modulo $n /(a, n)$, or, equivalently, is made up of $(a, n)$ different residue classes modulo $n$.

As usual, we express this by saying that the congruence has $(a, n)$ solutions modulo $n$.

Proof. The congruence $a x \equiv b(\bmod n)$ is equivalent to the existence of an integer $y$ with $a x-b=n y ; b=a x-n y$. Thus the congruence is solvable if and only if there are $x, y$ satisfying $b=a x-n y$. This, as we have noted many times before, is equivalent to $(a, n) \mid b$. That takes care of the existence part.

For the uniqueness part, assuming the condition of the Theorem, we get an equivalent congruence by dividing everything by $(a, n)$. Letting $a^{\prime}=a /(a, n)$, $b^{\prime}=b /(b, n), n^{\prime}=n /(a, n)$, our congruence is equivalent to
$$a^{\prime} x \equiv b^{\prime} \quad\left(\bmod n^{\prime}\right) .$$

Now note that $\left(a^{\prime}, n^{\prime}\right)=1$ (Lemma A.V.16), i.e., $a^{\prime}$ is invertible modulo $n^{\prime}$. Letting $r^{\prime}$ represent the inverse class of $a^{\prime}+\left(n^{\prime}\right), r^{\prime} a^{\prime} \equiv 1\left(\bmod n^{\prime}\right)$, this last congruence is equivalent to
$$x \equiv r^{\prime} a^{\prime} x \equiv r^{\prime} b^{\prime} \quad\left(\bmod n^{\prime}\right),$$
so the solution set is indeed the residue class $r^{\prime} b^{\prime}+\left(n^{\prime}\right)=r^{\prime} b^{\prime}+(n /(a, n))$.

# 数论作业代写

CRT是一个简化代数同余的方便工具，$f(x) \equiv 0$$(\bmod n)，模复合n，例如，质数幂的情况。我们稍后再处理这个特殊情况。 下面的结果将在关于原始根的章节中使用。 b.iv.1引理。设因子相对质数为n=n_1 n_2，和\geq 3。则同余式x^2 \equiv 1(\bmod n)至少有四个以n为模的解，即解集由至少四个以n为模的不同的剩余类组成。 证明。考虑四个不同的同余对$$ \begin{array}{ll} x \equiv \pm 1 & \left(\bmod n_1\right) \ x \equiv \pm 1 & \left(\bmod n_2\right) . \end{array} $$每个对都是唯一可解的模n，通过CRT，产生四个不同的剩余类模n。在每种情况下，解x都满足$$ \begin{array}{cc} x^2 \equiv 1 & \left(\bmod n_1\right) \ x^2 \equiv 1 & \left(\bmod n_2\right), \end{array} $$也就是说，x^2-1可以被n_1和n_2整除。根据第二可整除定理，这意味着x^2-1可以被它们的乘积整除，即x^2 \equiv 1(\bmod n)。 ## 数学代写|数论作业代写number theory代考|Linear Congruences 数字n，像往常一样，是一个正整数。 在这个简短的部分，我们学习线性同余，a x \equiv b(\bmod n)。不需要新的理论。 b.v.1定理。同余a x \equiv b(\bmod n), a \nmid n，当且仅当(a, n) \mid b可解。 在这种情况下，解是唯一模n /(a, n)。换句话说，解集是一个剩余类模n /(a, n)，或者，等价地，由(a, n)个不同的剩余类模n组成。 像往常一样，我们说同余式有(a, n)个解以n为模。 证明。同余a x \equiv b(\bmod n)等价于整数y与a x-b=n y ; b=a x-n y的存在性。因此当且仅当存在x, y满足b=a x-n y时，同余是可解的。正如我们之前多次提到的，这相当于(a, n) \mid b。这就解决了存在性部分。 对于唯一性部分，假设定理的条件，我们通过除以(a, n)得到等价的同余。让a^{\prime}=a /(a, n)$$b^{\prime}=b /(b, n), n^{\prime}=n /(a, n)$，我们的同余等于
$$a^{\prime} x \equiv b^{\prime} \quad\left(\bmod n^{\prime}\right) .$$

$$x \equiv r^{\prime} a^{\prime} x \equiv r^{\prime} b^{\prime} \quad\left(\bmod n^{\prime}\right),$$

## 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|数论作业代写number theory代考|MATH0034

statistics-lab™ 为您的留学生涯保驾护航 在代写数论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。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

## 数学代写|数论作业代写number theory代考|Euler’s Phi Function Revisited

In this Section we prove a property that allows us to compute $\phi(n)$ (A.V.1) whenever a full prime factorization of $n$ is known. The CRT plays a decisive role in the proof. The following Lemma reformulates an earlier observation:
B.II.1 Lemma. Suppose the integers $n_1, n_2>0$ are relatively prime. Let $0 \leq a_1<n_1, 0 \leq a_2<n_2$ and let $0 \leq x<n_1 \cdot n_2$ be the unique solution to the congruence pair
$$\begin{array}{ll} x \equiv a_1 & \left(\bmod n_1\right) \ x \equiv a_2 & \left(\bmod n_2\right) \end{array}$$
Then the class $x+\left(n_1 n_2\right)$ is invertible if and only if the classes $a_1+\left(n_1\right)$ and $a_2+\left(n_2\right)$ are.
Proof. As $a_1+\left(n_1\right)=x+\left(n_1\right)$, and $a_2+\left(n_2\right)=x+\left(n_2\right)$ we are stating that the class $x+\left(n_1 n_2\right)$ is invertible if and only if $x+\left(n_1\right)$ and $x+\left(n_2\right)$ are, or equivalently:
$$\left(x, n_1 n_2\right)=1 \Longleftrightarrow\left(x, n_1\right)=\left(x, n_2\right)=1 .$$
We proved the left arrow in the course of proving the CRT, B.I.5.
The right arrow is trivial.

An immediate consequence of the Lemma is that we have a bijection between invertible classes $x+\left(n_1 n_2\right)$ and pairs of invertible classes $\left(a_1+\left(n_1\right), a_2+\left(n_2\right)\right)$, whenever $\left(n_1, n_2\right)=1$. As there are $\phi\left(n_1 n_2\right)$ of the former, and $\phi\left(n_1\right) \phi\left(n_2\right)$ of the latter, we have proved the following Theorem:

B.II.2 Theorem (Multiplicativity of $\phi$ ). Let $n_1, n_2$ be positive integers satisfying $\left(n_1, n_2\right)=1$. Then:
$$\phi\left(n_1 n_2\right)=\phi\left(n_1\right) \phi\left(n_2\right) .$$

## 数学代写|数论作业代写number theory代考|General CRT

We now proceed to proving the CRT in its full generality. We prepare the proof by generalizing the $\mathrm{lcm}$ and gcd to include the case of more than two numbers, and by formally introducing multiplicities of prime factors. For non-zero numbers $m_1, m_2, \ldots, m_n$ we naturally let $\left(m_1, m_2, \ldots, m_n\right)$ denote the greatest (positive) number dividing all the $m_j$. Obviously,
$$\left(m_1, m_2, \ldots, m_n\right)=\left(m_1,\left(m_2, \ldots, m_n\right)\right) .$$
And for non-zero numbers $m_1, m_2, \ldots, m_n$ we let $\left[m_1, m_2, \ldots, m_n\right]$ denote the least number divisible by all the $m_j$. Obviously,
$$\left[m_1, m_2, \ldots, m_n\right]=\left[m_1,\left[m_2, \ldots, m_n\right]\right] .$$
We now turn our attention to multiplicities.
B.III.1 Definition. For any non-zero integer $n$, and any prime number $p$, we denote by $v_p(n)$ the largest exponent $e \geq 0$ such that $p^e$ divides $n$. It is called the multiplicity of $p$ in (the factorization of) $n$.

We record a few elementary observations:
B.III.2 Lemma.
a) The positive integers $m, n$ are equal if and only if $v_p(m)=v_p(n)$ for all prime numbers $p$.
b) $m$ divides $n$ if and only if $v_p(m) \leq v_p(n)$ for all prime numbers $p$.
c) For non-zero integers $m, n$, and all primes $p, v_p(m n)=v_p(m)+v_p(n)$.
We will be concerned with the multiplicities of primes entering the gcd and lcm of two numbers.

B.III.3 Lemma. Let $m, n$ be non-zero integers. Then, for all prime numbers $p$ :
a) $v_p((m, n))=\min \left(v_p(m), v_p(n)\right)$,
b) $v_p([m, n])=\max \left(v_p(m), v_p(n)\right)$.
Proof. For the first part, note that $p^e$ divides $(m, n)$ if and only if $p^e$ divides both $m$ and $n$, i.e., if and only if $e \leq$ both $v_p(m)$ and $v_p(n)$. This proves that $v_p((m, n))$ must equal the smaller of these two numbers.

For the second part, note that $p^e$ is divisible by $[m, n]$ if and only if $p^e$ is divisible by both $m$ and $n$, i.e., if and only if $e \geq \operatorname{both} v_p(m)$ and $v_p(n)$. This proves that $v_p([m, n])$ must equal the greater of these two numbers.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Euler’s Phi Function Revisited

b.ii.1引理。假设整数$n_1, n_2>0$是相对素数。设$0 \leq a_1<n_1, 0 \leq a_2<n_2$和$0 \leq x<n_1 \cdot n_2$是同余对的唯一解
$$\begin{array}{ll} x \equiv a_1 & \left(\bmod n_1\right) \ x \equiv a_2 & \left(\bmod n_2\right) \end{array}$$

$$\left(x, n_1 n_2\right)=1 \Longleftrightarrow\left(x, n_1\right)=\left(x, n_2\right)=1 .$$

b.ii.2定理($\phi$的乘法性)。设$n_1, n_2$是满足$\left(n_1, n_2\right)=1$的正整数。然后:
$$\phi\left(n_1 n_2\right)=\phi\left(n_1\right) \phi\left(n_2\right) .$$

## 数学代写|数论作业代写number theory代考|General CRT

$$\left(m_1, m_2, \ldots, m_n\right)=\left(m_1,\left(m_2, \ldots, m_n\right)\right) .$$

$$\left[m_1, m_2, \ldots, m_n\right]=\left[m_1,\left[m_2, \ldots, m_n\right]\right] .$$

b.iii.1定义。对于任何非零整数$n$和任何质数$p$，我们用$v_p(n)$表示最大的指数$e \geq 0$，使$p^e$除$n$。它被称为$n$(分解)中$p$的多重性。

b.iii.2引理。
a)对于所有质数$p$，正整数$m, n$当且仅当$v_p(m)=v_p(n)$相等。
B) $m$除$n$当且仅当$v_p(m) \leq v_p(n)$对于所有质数$p$。
c)对于非零整数$m, n$和所有质数$p, v_p(m n)=v_p(m)+v_p(n)$。

b.iii.3引理。设$m, n$为非零整数。然后，对于所有质数$p$:
A) $v_p((m, n))=\min \left(v_p(m), v_p(n)\right)$;
B) $v_p([m, n])=\max \left(v_p(m), v_p(n)\right)$。

## 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|数论作业代写number theory代考|MATH-UA248

statistics-lab™ 为您的留学生涯保驾护航 在代写数论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。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

## 数学代写|数论作业代写number theory代考|A Brief Account of RSA

In this Section we briefly discribe the RSA public key cryptographic scheme. As there are many excellent comprehensive accounts in the literature (Buchmann, Trappe-Washington) we dwell on the number theory involved, leaving most of the practical issues aside.

The mathematics behind RSA is summed up in the following two Lemmas. The first is a special case of a general Theorem to be proved in the next Chapter.

A.VI.1 Lemma. Let $n=p q$ be the product of two different prime numbers. Then $\phi(n)=(p-1)(q-1)$.

Proof. We can prove this the same way we did in the special case $n=$ $15=3 \cdot 5$. There are $n=p q$ classes modulo $p q$, represented by the numbers $m$ with $0 \leq m \leq n-1$. The non-invertible classes are represented by those $m$ with $(m, p q)>1$, i.e., those divisible by $p$ or $q$. There are $q$ and $p$ of these, respectively. Only $m=0$ is divisible by both $p$ and $q$ : by our Second Divisibility Theorem (A.II.2) a number divisible by both must be divisible by their product, as $(p, q)=1$.

So we subtract $q$ and $p$ classes and put back the zero class, which we subtracted twice. Therefore:
$$\phi(p q)=p q-q-p+1=(p-1)(q-1) .$$

The second Lemma is sometimes overlooked in the literature, probably because the probability of randomly choosing an $a$ with $(a, p q)>1$ is very small when $p, q$ are large.

A.VI.2 Lemma. Still assuming $n=p q$. For all integers $a$, and positive integers $k$, we have
$$a^{k \phi(n)+1} \equiv a \quad(\bmod n),$$
whether $(a, n)=1$ or not.

Proof. If $(a, n)=1$, Euler’s Theorem (A.V.12) states that $a^{\phi(n)} \equiv 1$ $(\bmod n)$, so the result follows on raising both members to the power $k$, and multiplying them by $a$.

Next consider the case where $(a, n)>1$. This means that $a$ is divisible by $p$ or $q$. If $a$ is divisible by both, it is divisible by their product $n$, and the result is trivial in this case. So we can assume that $p \mid a$ and $q \nmid a$.
Consider the difference
$$b=a^{k \phi(n)+1}-a .$$

## 数学代写|数论作业代写number theory代考|The Chinese Remainder Theorem

This is as good a place as any to introduce the least common multiple of two integers.

B.I.1 Definition. Let $m, n$ be non-zero integers. The least common multiple of $m$ and $n$, denoted $[m, n]$, or $\operatorname{lcm}(m, n)$, is the smallest (positive) integer divisible by both $m$ and $n$.
B.I.2 Example. $[3,4]=[3,-4]=[4,6]=12$.
The example shows that the lcm of two positive numbers may be their product or not. The following Theorem gives the full story.
B.I.3 Theorem. The lcm of two positive integers is given by
$$[m, n]=\frac{m \cdot n}{(m, n)} .$$
It therefore equals their product if and only if $m$ and $n$ are relatively prime, $(m, n)=1$. Furthermore, any common multiple of $m$ and $n$ is divisible by their least common multiple.

Proof. Let $e$ be any common multiple, $m|e, n| e$. This is clearly equivalent to
$$\frac{m}{(m, n)}\left|\frac{e}{(m, n)}, \quad \frac{n}{(m, n)}\right| \frac{e}{(m, n)} .$$
As $m /(m, n)$ and $n /(m, n)$ are relatively prime (Lemma A.V.16), the two conditions are equivalent to $e /(m, n)$ being divisible by their product, according to the Second Divisibility Theorem, (A.II.2). That is, to:
$$\frac{m}{(m, n)} \cdot \frac{n}{(m, n)} \mid \frac{e}{(m, n)} .$$
Multiplying by $(m, n)$ we therefore see that $e$ is a common multiple of $m, n$ if and only if
$$\frac{m \cdot n}{(m, n)} \mid e$$
which proves both parts of the Theorem.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|A Brief Account of RSA

RSA背后的数学可以总结为以下两个引理。第一个是下一章将要证明的一般定理的一个特例。

a.vi.1引理。设$n=p q$是两个不同质数的乘积。然后$\phi(n)=(p-1)(q-1)$。

$$b=a^{k \phi(n)+1}-a .$$

## 数学代写|数论作业代写number theory代考|The Chinese Remainder Theorem

b.i.1定义。设$m, n$为非零整数。$m$和$n$的最小公倍数，记为$[m, n]$或$\operatorname{lcm}(m, n)$，是能被$m$和$n$整除的最小(正)整数。
b.i.2示例:$[3,4]=[3,-4]=[4,6]=12$。

b.i.3定理。两个正整数的lcm由
$$[m, n]=\frac{m \cdot n}{(m, n)} .$$

$$\frac{m}{(m, n)}\left|\frac{e}{(m, n)}, \quad \frac{n}{(m, n)}\right| \frac{e}{(m, n)} .$$

$$\frac{m}{(m, n)} \cdot \frac{n}{(m, n)} \mid \frac{e}{(m, n)} .$$

$$\frac{m \cdot n}{(m, n)} \mid e$$

## 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|数论作业代写number theory代考|MATH453

statistics-lab™ 为您的留学生涯保驾护航 在代写数论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。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

Since the ring $O_p$ has no zero divisors (Corollary 2 of Theorem 2), it can be embedded in a field, using the standard construction of a field from an integral domain. Application of this construction to our situation leads to consideration of fractions of the form $\alpha / p^k$, where $\alpha$ is some $p$-adic integer, and $k \geqslant 0$. The fractions considered here could more suitably be written as pairs $\left(\alpha, p^k\right)$.

Definition. A fraction of the form $\alpha / p^k, \alpha \in O_p, k \geqslant 0$, determines a fractional $p$-adic number, or, more simply, a $p$-adic number. Two fractions, $\alpha / p^k$ and $\beta / p^m$, determine the same $p$-adic number if and only if $\alpha p^m=\beta p^k$ in $O_p$.
The set of all $p$-adic numbers will be denoted by $R_p$.
A $p$-adic integer determines an element $\alpha / 1=\alpha / p^0$ in $R_p$. It is clear that distinct $p$-adic integers determine distinct elements of $R_p$. Hence we shall assume that $O_p$ is a subset of the set $R_p$.
Addition and multiplication are defined in $R_p$ by the rules
\begin{aligned} \frac{\alpha}{p^k}+\frac{\beta}{p^m} & =\frac{\alpha p^m+\beta p^k}{p^{k+m}}, \ \frac{\alpha}{p^k} \frac{\beta}{p^m} & =\frac{\alpha \beta}{p^{k+m}} . \end{aligned}

## 数学代写|数论作业代写number theory代考|Convergence in the Field of p-Adic Numbers

In Section 3.1 we noted the analogy between $p$-adic integers and real numbers, in that both are determined by sequences of rational numbers.
Just as every real number is the limit of any sequence of rational numbers which determines it, it would be natural to conjecture that the same fact should hold for $p$-adic numbers, if the correct definition of the concept of convergence is given. The definition of limit for real or rational numbers can be based, for example, on the notion of nearness; two real or rational numbers being near if the absolute value of their difference is small. For the definition of convergence for $p$-adic numbers we thus must decide under what conditions two $p$-adic numbers are to be considered close to one another.

In the example of the first section, we spoke of the $p$-nearness of two $p$-adic integers $x$ and $y$, meaning by this that the difference of $x$ and $y$ should be divisible by a high power of $p$. It was under this definition of nearness that the analogy between the definitions of real numbers and of $p$-adic integers became apparent. If we use the concept of the $p$-value $v_p$, then the $p$-nearness of $x$ and $x$ will be characterized by the value of $v_p(x-y)$. Thus we may speak of two $p$-adic numbers $\xi$ and $\eta$ (not necessarily integers) as being near when the value of $v_p(\xi-\eta)$ is sufficiently large. Thus “small” $p$-adic numbers are characterized by the large value of their $p$-value.
After these remarks we turn to precise definitions.
Definition. The sequence
$$\left{\xi_n\right}=\left{\xi_0, \xi_1, \ldots, \xi_n, \ldots\right}$$
of $p$-adic numbers converges to the $p$-adic number $\xi$ (we denote this by $\lim {n \rightarrow \infty} \xi_n=\xi$ or $\left.\left{\xi_n\right} \rightarrow \xi\right)$ if $$\lim {n \rightarrow \infty} v_p\left(\xi_n-\xi\right)=\infty .$$
A singular feature of this definition (which distinguishes it from the usual definition of convergence for real numbers) is that the convergence of $\left{\xi_n\right}$ to $\xi$ is determined by the sequence of rational integers $v_p\left(\xi_n-\xi\right)$, which must converge to infinity. We can put the definition in a more familiar form if, instead of $v_p$, we consider another nonnegative real-valued function on the field $R_p$, which will converge to zero as $v_p$ goes to infinity. Namely, choose some real number $\rho$, satisfying $0<\rho<1$, and set
$$\varphi_p(\xi)= \begin{cases}\rho^{v_p(\xi)} & \text { for } \xi \neq 0 \ 0 & \text { for } \quad \xi=0\end{cases}$$

# 数论作业代写

\begin{aligned} \frac{\alpha}{p^k}+\frac{\beta}{p^m} & =\frac{\alpha p^m+\beta p^k}{p^{k+m}}, \ \frac{\alpha}{p^k} \frac{\beta}{p^m} & =\frac{\alpha \beta}{p^{k+m}} . \end{aligned}

## 数学代写|数论作业代写number theory代考|Convergence in the Field of p-Adic Numbers

$$\left{\xi_n\right}=\left{\xi_0, \xi_1, \ldots, \xi_n, \ldots\right}$$

$$\varphi_p(\xi)= \begin{cases}\rho^{v_p(\xi)} & \text { for } \xi \neq 0 \ 0 & \text { for } \quad \xi=0\end{cases}$$

## 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|数论作业代写number theory代考|MATH3240

statistics-lab™ 为您的留学生涯保驾护航 在代写数论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。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

## 数学代写|数论作业代写number theory代考|Sums of Powers

We now apply the general method of the preceding section to the case when the polynomial $F$ is equal to a sum of powers of the variables, i.e.,
$$F\left(x_1, \ldots, x_n\right)=a_1 x_1^{r_1}+\cdots+a_n x_n^{r_n}, \quad a_i \not \equiv 0(\bmod p) .$$
We shall assume that $n \geqslant 3$, since for $n=1$ and $n=2$ the number of solutions of the congruence $F \equiv 0(\bmod p)$ can be found by an elementary method.
By formula (2.4) the number $N$ of solutions to the congruence $a_1 x_1{ }^{r_1}+\cdots$ $+a_n x_n{ }^{r_n} \equiv 0(\bmod p)$ is given by the expression
$$N=p^{n-1}+\frac{1}{p} \sum_x^{\prime} \sum_{x_1 \ldots . . x_n} \zeta^{x\left(a_1 x_1 r_1+\cdots+a_n x_n n_n\right)},$$
which can be written in the form
$$N=p^{n-1}+\frac{1}{p} \sum_x^{\prime} \prod_{i=1}^n \sum_{x_i} \zeta^{a_i x x_i r_i} .$$
Hence we must investigate sums of the form
$$\sum_y \zeta^{a y^r}(a \not \equiv 0(\bmod p))$$

Clearly,
$$\sum_y \zeta^{a y^r}=\sum_x m(x) \zeta^{a x},$$
where $m(x)$ is the number of solutions to the congruence $y^r \equiv x(\bmod p)$. It is clear that $m(0)=1$. We shall find an explicit formula for $m(x)$ when $x \not \equiv 0$ $(\bmod p)$.
If $g$ is a primitive root modulo $p$, then
$$x \equiv g^k(\bmod p),$$
where the exponent $k$ is uniquely determined modulo $p-1$. Let $y \equiv g^4$ $(\bmod p)$. The congruence $y^r \equiv x(\bmod p)$ is then equivalent to the congruence
$$r u \equiv k(\bmod p-1) .$$

## 数学代写|数论作业代写number theory代考|The Absolute Value of Gaussian Sums

Consider the set $\mathfrak{F}$ of all complex functions $f(x)$, defined for rational integers $x$, and satisfying the condition: $f(x)=f(y)$ if $x \equiv y(\bmod p)$. Since each function $f(x) \in \mathbb{F}$ is determined by its values on a full system of residues modulo $p, \mathfrak{F}$ is a $p$-dimensional linear space over the field of complex numbers. We introduce a Hermitian inner product on $\mathfrak{F}$ by setting
$$(f, g)=\frac{1}{p} \sum_x f(x) \overline{g(x)} \quad(f, g \in \mathfrak{F}) .$$

It is easily checked that with respect to this inner product the $p$ functions
$$f_a(x)=\zeta^{-a x} \quad(a \text { a residue }(\bmod p))$$
form an orthonormal basis for $\mathfrak{F}$. Indeed, by (2.2),
$$\left(f_a, f_{a^{\prime}}\right)=\frac{1}{p} \sum_x \zeta^{\left(a^{\prime}-a\right) x}=\left{\begin{array}{lll} 1 & \text { for } & a \equiv a^{\prime}(\bmod p), \ 0 & \text { for } & a \neq a^{\prime}(\bmod p) . \end{array}\right.$$
The functions (2.17), which satisfy
$$f_a(x+y)=f_a(x) f_a(y),$$
are called additive characters modulo $p$. We shall find the coordinates of a multiplicative character $\chi$ with respect to the basis (2.17). Let
$$\chi=\sum_a \alpha_a f_a .$$
Then
$$\alpha_a=\left(\chi, f_a\right)=\frac{1}{p} \sum_x \chi(x) \zeta^{a x}=\frac{1}{p} \tau_a(\chi) .$$

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Sums of Powers

$$F\left(x_1, \ldots, x_n\right)=a_1 x_1^{r_1}+\cdots+a_n x_n^{r_n}, \quad a_i \not \equiv 0(\bmod p) .$$

$$x \equiv g^k(\bmod p),$$

$$U\left(O_K\right) \simeq \mathbb{Z}_2 \times \mathbb{Z} .$$

$$1<\theta<\eta$$

$$\theta= \pm \eta^n$$

$$\theta=\eta^n .$$

## 数学代写|数论作业代写number theory代考|Calculating the Fundamental Unit

$$\alpha_0=\sqrt{m}$$

$$\alpha_{n+1}=\frac{1}{\alpha_n-\left[\alpha_n\right]}, n=0,1,2, \ldots$$

\begin{aligned} & \alpha_0=\sqrt{31}, \ & \alpha_1=\frac{1}{\alpha_0-\left[\alpha_0\right]}=\frac{1}{\sqrt{31}-5}=\frac{5+\sqrt{31}}{6} \text {, } \ & \alpha_2=\frac{1}{\alpha_1-\left[\alpha_1\right]}=\frac{1}{\frac{5+\sqrt{31}}{6}-1}=\frac{1+\sqrt{31}}{5} \text {, } \ & \alpha_3=\frac{1}{\alpha_2-\left[\alpha_2\right]}=\frac{1}{\frac{1+\sqrt{31}}{5}-1}=\frac{4+\sqrt{31}}{3} \text {, } \ & \alpha_4=\frac{1}{\alpha_3-\left[\alpha_3\right]}=\frac{1}{\frac{4+\sqrt{31}}{3}-3}=\frac{5+\sqrt{31}}{2} \text {, } \ & \alpha_5=\frac{1}{\alpha_4-\left[\alpha_4\right]}=\frac{1}{\frac{5+\sqrt{31}}{2}-5}=\frac{5+\sqrt{31}}{3} \text {, } \ & \alpha_6=\frac{1}{\alpha_5-\left[\alpha_5\right]}=\frac{1}{\frac{5+\sqrt{31}}{3}-3}=\frac{4+\sqrt{31}}{5} \text {, } \ & \alpha_7=\frac{1}{\alpha_6-\left[\alpha_6\right]}=\frac{1}{\frac{4+\sqrt{31}}{5}-1}=\frac{1+\sqrt{31}}{6} \text {, } \ & \alpha_8=\frac{1}{\alpha_7-\left[\alpha_7\right]}=\frac{1}{\frac{1+\sqrt{31}}{6}-1}=\frac{5+\sqrt{31}}{1} \text {, } \ & \alpha_9=\frac{1}{\alpha_8-\left[\alpha_8\right]}=\frac{1}{5+\sqrt{31}-10}=\frac{5+\sqrt{31}}{6}=\alpha_1 \text {, } \ & \alpha_{10}=\alpha_2, \alpha_{11}=\alpha_3, \ldots \ & \end{aligned}

## 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|数论作业代写number theory代考|MATH393

statistics-lab™ 为您的留学生涯保驾护航 在代写数论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。我们的专家在代写数论number theory代写方面经验极为丰富，各种代写数论number theory相关的作业也就用不着说。

## 数学代写|数论作业代写number theory代考|Factoring Primes in an Arbitrary Number Field

Theorem 10.3 .1 was actually proved by Dedekind in the following slightly stronger form. For all but at most a finite number of primes, Theorem 10.5.1 gives the factorization of a prime into prime ideals in an arbitrary algebraic number field.
Theorem 10.5.1 Let $K=\mathbb{Q}(\theta)$ be an algebraic number field with $\theta \in O_K$. Let $p$ be a rational prime. Let
$$f(x)=\operatorname{irr}_{\mathbb{Q}}(\theta) \in \mathbb{Z}[x] .$$
Let ${ }^{-}$denote the natural map $: \mathbb{Z}[x] \longrightarrow \mathbb{Z}_p[x]$, where $\mathbb{Z}_p=\mathbb{Z} / p \mathbb{Z}$. Let
$$\bar{f}(x)=g_1(x)^{e_1} \cdots g_r(x)^{e_r},$$
where $g_1(x), \ldots, g_r(x)$ are distinct monic irreducible polynomials in $\mathbb{Z}_p[x]$ and $e_1, \ldots, e_r$ are positive integers. For $i=1,2, \ldots, r$ let $f_i(x)$ be any monic polynomial of $\mathbb{Z}[x]$ such that $\bar{f}_i=g_i$. Set
$$P_i=\left\langle p, f_i(\theta)\right\rangle, i=1,2, \ldots, r .$$
If $\operatorname{ind}(\theta) \not \equiv 0(\bmod p)$ then $P_1, \ldots, P_r$ are distinct prime ideals of $O_K$ with
$$\langle p\rangle=P_1^{e_1} \cdots P_r^{e_r}$$
and
$$N\left(P_i\right)=p^{\operatorname{deg} f_i}, i=1,2, \ldots, r .$$

## 数学代写|数论作业代写number theory代考|Factoring Primes in a Cyclotomic Field

Let $m$ be a positive integer and let $\zeta_m$ be a primitive $m$ th root of unity. The cyclotomic field $\mathbb{Q}\left(\zeta_m\right)$ is denoted by $K_m$. We give (without proof) the decomposition of a rational prime $p$ into prime ideals in $O_{K_m}$.

Theorem 10.6.1 Let $m=p^r m_1$, where $r \in \mathbb{N} \cup{0}, m_1 \in \mathbb{N}$, and $p \nmid m_1$. Let $h$ be the least positive integer such that $p^h \equiv 1\left(\bmod m_1\right)$. Then $h \mid \phi\left(m_1\right)$ and
$$\langle p\rangle=\left(P_1 P_2 \cdots P_{\phi\left(m_1\right) / h}\right)^{\phi\left(p^r\right)},$$
where $P_1, P_2, \ldots, P_{\phi\left(m_1\right) / h}$ are distinct prime ideals with
$$N\left(P_i\right)=p^h, i=1,2, \ldots, \phi\left(m_1\right) / h .$$
We refer the reader to Mann’s book [6] for a proof of this theorem.
Example 10.6.1 We determine the prime ideal decomposition of $\langle 3\rangle$ in $O_{K_9}$. Here $p=3, m=9, \phi(m)=6, r=2, m_1=1$, and $h=1$ so that by Theorem 10.6.1
$$\langle 3\rangle=P^6,$$
where $P$ is a prime ideal with $N(P)=3$.
Example 10.6.2 We determine the prime ideal decomposition of $\langle 2\rangle$ in $O_{K_5}$. Here $p=2, m=5, r=0, m_1=5, \phi\left(m_1\right)=4, \phi\left(p^r\right)=1$, and $h=4$ so that by Theorem 10.6.1
$$\langle 2\rangle=P,$$
where $P$ is a prime ideal with $N(P)=2^4$.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Factoring Primes in an Arbitrary Number Field

$$f(x)=\operatorname{irr}_{\mathbb{Q}}(\theta) \in \mathbb{Z}[x] .$$

$$\bar{f}(x)=g_1(x)^{e_1} \cdots g_r(x)^{e_r},$$

$$P_i=\left\langle p, f_i(\theta)\right\rangle, i=1,2, \ldots, r .$$

$$\langle p\rangle=P_1^{e_1} \cdots P_r^{e_r}$$

$$N\left(P_i\right)=p^{\operatorname{deg} f_i}, i=1,2, \ldots, r .$$

## 数学代写|数论作业代写number theory代考|Factoring Primes in a Cyclotomic Field

$$\langle p\rangle=\left(P_1 P_2 \cdots P_{\phi\left(m_1\right) / h}\right)^{\phi\left(p^r\right)},$$
$P_1, P_2, \ldots, P_{\phi\left(m_1\right) / h}$不同的素数理想在哪里
$$N\left(P_i\right)=p^h, i=1,2, \ldots, \phi\left(m_1\right) / h .$$

$$\langle 3\rangle=P^6,$$

$$\langle 2\rangle=P,$$

## 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|数论作业代写number theory代考|Dedekind Domains

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## 数学代写|数论作业代写number theory代考|Dedekind Domains

Definition 8.1.1 (Dedekind domain) An integral domain $D$ that satisfies the following three properties:
$D$ is a Noetherian domain,
$D$ is integrally closed, and
each prime ideal of $D$ is a maximal ideal,
is called a Dedekind domain.
In view of the remarks before Definition 8.1.1, we have
Theorem 8.1.1 Let $K$ be an algebraic number field. Let $O_K$ be the ring of integers of $K$. Then $O_K$ is a Dedekind domain.

The next theorem gives another class of integral domains that are Dedekind domains.

Theorem 8.1.2 Let $D$ be a principal ideal domain. Then $D$ is a Dedekind domain.
Proof: Let $D$ be a principal ideal domain. By Theorem 3.1.2 $D$ is a Noetherian domain, so (8.1.1) holds. By Theorem 3.3.1 $\mathrm{D}$ is a unique factorization domain and thus, by Theorem 4.2.5, $D$ is integrally closed, so (8.1.2) holds. By Theorem 1.5.7 each prime ideal of $D$ is maximal so that (8.1.3) holds. Hence $D$ is a Dedekind domain.

Our main objective in this chapter is to show that every ideal $I(\neq\langle 0\rangle,\langle 1\rangle)$ of a Dedekind domain can be expressed uniquely as a product of prime ideals. We also show that every ideal of a Dedekind domain is generated by at most two elements.

## 数学代写|数论作业代写number theory代考|Ideals in a Dedekind Domain

The first step toward our objective of proving that in a Dedekind domain every proper ideal is a product of prime ideals is to show that every such ideal contains a product of prime ideals. This is actually true in a Noetherian domain.

Theorem 8.2.1 In a Noetherian domain every nonzero ideal contains a product of one or more prime ideals.

Proof: Suppose that $D$ is a Noetherian domain that possesses at least one nonzero ideal that does not contain a product of one or more prime ideals. Let $S$ be the set of all such ideals. By assumption $S$ is not empty. As $D$ is Noetherian, by Theorem 3.1.3 $S$ contains a (nonzero) ideal $A$ maximal with respect to the property of not containing a product of one or more prime ideals. Clearly $A$ itself is not a prime ideal. Hence, by Theorem 1.6.1, there exist ideals $B$ and $C$ such that
$$B C \subseteq A, B \nsubseteq A, C \nsubseteq A$$
Define the ideals $B_1$ and $C_1$ of $D$ by
$$B_1=A+B, C_1=A+C$$
Clearly
$$A \subset B_1, A \subset C_1,$$
so that $B_1 \notin S, C_1 \notin S$. Hence there exist prime ideals $P_1, \ldots, P_k$ such that
$$B_1 \supseteq P_1 \cdots P_h, C_1 \supseteq P_{h+1} \cdots P_k .$$

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Dedekind Domains

8.1.1 (Dedekind域)一个满足以下三个属性的积分域$D$:
$D$是一个诺瑟域，
$D$是全封闭的，并且
$D$的每个素理想都是一个极大理想，

## 数学代写|数论作业代写number theory代考|Ideals in a Dedekind Domain

$$B C \subseteq A, B \nsubseteq A, C \nsubseteq A$$

$$B_1=A+B, C_1=A+C$$

$$A \subset B_1, A \subset C_1,$$

$$B_1 \supseteq P_1 \cdots P_h, C_1 \supseteq P_{h+1} \cdots P_k .$$

## 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。