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## avatest™帮您通过考试

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## 数学代写|数论代写Number Theory代考|Units of Norm 1

Let $m$ be a positive squarefree integer. Theorem 11.2.1 tells us that there exist positive integers $x$ and $y$ such that $x^2-m y^2=1$. Hence $\lambda=x+y \sqrt{m}$ is a unit of $O_K$, where $K=\mathbb{Q}(\sqrt{m})$, such that $\lambda>1$ and $N(\lambda)=1$. Since $\lambda^n \rightarrow \infty$ as $n \rightarrow \infty, O_K$ has infinitely many units of norm 1 , namely $\left{\lambda^n \mid n \in \mathbb{Z}\right}$. All of these units are of the form $u+v \sqrt{m}$, where $u$ and $v$ are integers such that $u^2-m v^2=1$. However, when $m \equiv 1(\bmod 4)$, there may be units in $O_K$ of the form $(u+v \sqrt{m}) / 2$, where $u$ and $v$ are both odd integers. For example $(3+\sqrt{5}) / 2$ is a unit of norm 1 in $O_{\mathbb{Q}(\sqrt{5})}$. In contrast, $O_{\mathbb{Q}(\sqrt{17})}$ does not contain any units of the form $(u+v \sqrt{17}) / 2$, where $u$ and $v$ are both odd integers, since $u^2-17 v^2= \pm 4$ cannot hold modulo 8 for odd integers $u$ and $v$.

Let $\lambda=x+y \sqrt{m}$ be a unit of $O_K(K=\mathbb{Q}(\sqrt{m}))$ of norm 1 with $x$ and $y$ both integers or possibly in the case $m \equiv 1(\bmod 4)$ both halves of odd integers. We now show how the signs of $x$ and $y$ determine to which of the four intervals $(-\infty,-1),(-1,0),(0,1)$, or $(1, \infty) \lambda$ belongs.

Theorem 11.3.1 Let $m$ be a positive squarefree integer. Let $x$ and $y$ both be integers or both halves of odd integers such that $x^2-m y^2=1$. Then
\begin{aligned} x+y \sqrt{m}>1 & \Longleftrightarrow x>0, y>0, \ 00, y<0, \ -10, \ x+y \sqrt{m}<-1 & \Longleftrightarrow x<0, y<0 . \end{aligned}

## 数学代写|数论代写Number Theory代考|Units of Norm −1

Let $m$ be a positive squarefree integer. We have already observed that the ring $O_{\mathbb{Q}(\sqrt{m})}$ of integers of the real quadratic field $\mathbb{Q}(\sqrt{m})$ may or may not contain units of norm -1 . Indeed $O_{\mathbb{Q}(\sqrt{2})}$ has units such as $1+\sqrt{2}$ of norm -1 whereas $O_{\mathbb{Q}(\sqrt{3})}$ does not contain any units of norm -1 . We suppose that $O_{\mathbb{Q}(\sqrt{m})}$ contains units of norm -1 and show that there exists a unique unit $\sigma>1$ in $O_{\mathbb{Q}(\sqrt{m})}$ of norm -1 such that all units in $O_{\mathbb{Q}(\sqrt{m})}$ of norm -1 are given by $\pm \sigma^{2 k+1}(k=0, \pm 1, \pm 2, \ldots)$ and all units in $O_{\mathbb{Q}(\sqrt{m})}$ of norm 1 are given by $\pm \sigma^{2 k}(k=0, \pm 1, \pm 2, \ldots)$.

Theorem 11.4.1 Let $m$ be a positive squarefree integer. Suppose that $O_{\mathbb{Q}(\sqrt{m})}$ contains units of norm -1 . Then there exists a unique unit $\sigma>1$ of norm -1 in $O_{\mathbb{Q}(\sqrt{m})}$ such that every unit in $O_{\mathbb{Q}(\sqrt{m})}$ is of the form $\pm \sigma^n$ for some integer $n$.

Proof: Let $\rho$ be a unit in $O_{\mathbb{Q}(\sqrt{m})}$ of norm -1 . Let $\rho^{\prime}$ denote its conjugate. Then
$$\rho \rho^{\prime}=N(\rho)=-1$$
so that
$$\rho^2 \rho^{\prime 2}=1$$
Thus $\rho^2$ is a unit of $O_{\mathbb{Q}(\sqrt{m})}$ of norm 1. Hence, by Theorem 11.3.2(b), we have
$$\rho^2= \pm \epsilon^n$$
for some integer $n$, where $\epsilon$ is the fundamental unit of $O_{\mathbb{Q}(\sqrt{m})}$ of norm 1. Clearly $\rho^2>0$ and $\epsilon^n>0$ so that
$$\rho^2=\epsilon^n .$$
If $n$ is even, say $n=2 k$, then
$$\rho^2=\epsilon^{2 k}$$
so that
$$\rho= \pm \epsilon^k$$
Hence
$$N(\rho)=N\left( \pm \epsilon^k\right)=N(\epsilon)^k=1,$$
contradicting $N(\rho)=-1$. Thus $n$ must be odd, say $n=2 l+1$, and so
$$\rho^2=\epsilon^{2 l+1} .$$
Hence
$$\epsilon=\left(\rho \epsilon^{-l}\right)^2 .$$

## 数学代写|数论代写Number Theory代考|Units of Norm 1

\begin{aligned} x+y \sqrt{m}>1 & \Longleftrightarrow x>0, y>0, \ 00, y<0, \ -10, \ x+y \sqrt{m}<-1 & \Longleftrightarrow x<0, y<0 . \end{aligned}

## 数学代写|数论代写Number Theory代考|Units of Norm −1

$$\rho \rho^{\prime}=N(\rho)=-1$$

$$\rho^2 \rho^{\prime 2}=1$$

$$\rho^2= \pm \epsilon^n$$

$$\rho^2=\epsilon^n .$$

$$\rho^2=\epsilon^{2 k}$$

$$\rho= \pm \epsilon^k$$

$$N(\rho)=N\left( \pm \epsilon^k\right)=N(\epsilon)^k=1,$$

$$\rho^2=\epsilon^{2 l+1} .$$

$$\epsilon=\left(\rho \epsilon^{-l}\right)^2 .$$

## 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。

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## avatest™帮您通过考试

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## 数学代写|数论代写Number Theory代考|Factoring Primes in a Monogenic Number Field

Let $K$ be an algebraic number field. Recall that $K$ is said to be monogenic (Definition 7.1.5) if there exists $\theta \in O_K$ such that
$$O_K=\mathbb{Z}+\mathbb{Z} \theta+\cdots+\mathbb{Z} \theta^{n-1},$$
where $[K: \mathbb{Q}]=n$. The next theorem shows how to factor $\langle p\rangle$ (with $p$ a rational prime) into prime ideals in a monogenic number field. It was originally proved by Dedekind [3] in 1878.

Theorem 10.3.1 Let $K=\mathbb{Q}(\theta)$ be an algebraic number field of degree $n$ such that
$$O_K=\mathbb{Z}+\mathbb{Z} \theta+\cdots+\mathbb{Z} \theta^{n-1} .$$
Let $p$ be a rational prime. Let
$$f(x)=\operatorname{irr}_{\mathbb{Q}} \theta \in \mathbb{Z}[x] .$$
Let ${ }^{-}$denote the natural map $: \mathbb{Z}[x] \rightarrow \mathbb{Z}_p[x]$, 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
$$P_i=\left\langle p, f_i(\theta)\right\rangle, i=1,2, \ldots, r .$$
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|数论代写Number Theory代考|Some Factorizations in Cubic Fields

Example 10.4.1 We factor $\langle 5\rangle$ as a product of prime ideals in $O_K$, where $K=$ $\mathbb{Q}(\sqrt[3]{2})$. Set $\theta=\sqrt[3]{2}$. We have seen in Example 7.1.6 that $\left{1, \theta, \theta^2\right}$ is an integral basis for $K=\mathbb{Q}(\theta)$ so that $K$ is monogenic. The minimal polynomial of $\theta$ over $\mathbb{Q}$ is $x^3-2$. We have
$$x^3-2=(x+2)\left(x^2+3 x+4\right)(\bmod 5),$$
where $x+2$ and $x^2+3 x+4$ are irreducible $(\bmod 5)$. Hence, by Theorem 10.3.1, we have
$$\langle 5\rangle=P Q,$$
where
$$P=\langle 5, \theta+2\rangle, Q=\left\langle 5, \theta^2+3 \theta+4\right\rangle$$
are distinct prime ideals with
$$N(P)=5, N(Q)=5^2=25 .$$
As a check on the calculation in Example 10.4.1 we compute $P Q$ directly.

We have
\begin{aligned} P Q & =\langle 5, \theta+2\rangle\left\langle 5, \theta^2+3 \theta+4\right\rangle \ & =\left\langle 25,5(\theta+2), 5\left(\theta^2+3 \theta+4\right), \theta^3+5 \theta^2+10 \theta+8\right\rangle \ & =\left\langle 25,5(\theta+2), 5\left(\theta^2+3 \theta+4\right), 5 \theta^2+10 \theta+10\right\rangle \ & =\langle 5\rangle\left\langle 5, \theta+2, \theta^2+3 \theta+4, \theta^2+2 \theta+2\right\rangle \ & =\langle 5\rangle \end{aligned}
as
$$1=1 \cdot 5+(2 \theta+2)(\theta+2)-2\left(\theta^2+3 \theta+4\right) .$$

## 数学代写|数论代写Number Theory代考|Factoring Primes in a Monogenic Number Field

$$O_K=\mathbb{Z}+\mathbb{Z} \theta+\cdots+\mathbb{Z} \theta^{n-1},$$

$$O_K=\mathbb{Z}+\mathbb{Z} \theta+\cdots+\mathbb{Z} \theta^{n-1} .$$

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

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

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

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

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

## 数学代写|数论代写Number Theory代考|Some Factorizations in Cubic Fields

\begin{aligned} \operatorname{ord}_P(\langle\gamma\rangle+A B) & =\min \left(\operatorname{ord}_P(\langle\gamma\rangle), \operatorname{ord}_P(A B)\right) \ & =\min \left(\operatorname{ord}_P(\gamma), 0\right) \ & =0 \ & =\operatorname{ord}_P(A) \end{aligned}

## 数学代写|数论代写Number Theory代考|Norm of a Fractional Ideal

$$A=\frac{1}{\alpha} I .$$

$$N(A)=\frac{N(I)}{N(\langle\alpha\rangle)},$$

$$A=\frac{1}{\alpha} I=\frac{1}{\beta} J$$

$$\beta I=\alpha J,$$

$$\langle\beta\rangle I=\langle\alpha\rangle J,$$

$$N(\langle\beta\rangle) N(I)=N(\langle\beta\rangle I)=N(\langle\alpha\rangle J)=N(\langle\alpha\rangle) N(J),$$

$$\frac{N(I)}{N(\langle\alpha\rangle)}=\frac{N(J)}{N(\langle\beta\rangle)} .$$

## 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。

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## avatest™帮您通过考试

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## 数学代写|数论代写Number Theory代考|Index and Minimal Index of an Algebraic Number Field

Let $K$ be an algebraic number field of degree $n$ over $\mathbb{Q}$. An element $\alpha \in O_K$ is called a generator of $K$ if $K=\mathbb{Q}(\alpha)$. By Theorem 6.4.3 $\alpha$ is a generator of $K$ if and only if $D(\alpha) \neq 0$. For a generator $\alpha$ of $K$, the index of $\alpha$ is the positive integer ind $\alpha$ given by
$$D(\alpha)=(\text { ind } \alpha)^2 d(K)$$
(see Definition 7.1.4). We now define the index $i(K)$ and minimal index $m(K)$ of the field $K$.
Definition 7.4.1 (Index of a field) The index of $K$ is
$$i(K)=\operatorname{gcd}{\text { ind } \alpha \mid \alpha \text { a generator of } K} .$$
Definition 7.4.2 (Minimal index of a field) The minimal index of $K$ is
$$m(K)=\min {\text { ind } \alpha \mid \alpha \text { a generator of } K} .$$
Clearly
$$i(K) \mid m(K) .$$

## 数学代写|数论代写Number Theory代考|Integral Basis of a Cyclotomic Field

Let $m$ be a positive integer. The number of positive integers less than or equal to $m$ that are coprime with $m$ is denoted by $\phi(m)$. The arithmetic function $\phi(m)$ is called Euler’s phi function. Let $\zeta_m$ be any primitive $m$ th root of unity. There are $\phi(m)$ primitive $m$ th roots of unity, namely $\zeta_m^r, r=1,2, \ldots, m,(r, m)=1$. Let $K_m=\mathbb{Q}\left(\zeta_m\right)$. It is easy to show that $K_m=\mathbb{Q}\left(\zeta_m^r\right)$ for any $r \in{1,2, \ldots, m}$ with $(r, m)=1$, so that $K_m$ is independent of the primitive $m$ th root of unity chosen. The field $K_m$ is called the $m$ th cyclotomic field. For odd $m$ the fields $K_m$ and $K_{2 m}$ coincide as $-\zeta_m$ is a primitive $2 m$ th root of unity. Clearly $\zeta_m$ is a root of the polynomial
$$f_m(x)=\prod_{\substack{r=1 \(r, m)=1}}^m\left(x-\zeta_m^r\right) .$$
It is known that $f_m(x) \in \mathbb{Z}[x]$ and that $f_m(x)$ is irreducible, so that
$$\operatorname{irr}_{\mathbb{Q}}\left(\zeta_m\right)=f_m(x)$$
Moreover, the degree of $f_m(x)$ is $\phi(m)$ so that
$$\left[K_m: \mathbb{Q}\right]=\phi(m) .$$
The smallest field containing both $K_m$ and $K_n$ is $K_{[m, n]}$, where $[m, n]$ denotes the least common multiple of $m$ and $n$. Also, $K_m \cap K_n=K_{(m, n)}$. If $m \not \equiv 2(\bmod 4)$ then $K_m \subseteq K_n$ holds if and only if $m \mid n$. Thus if $m$ and $n$ are distinct and not congruent to $2(\bmod 4)$ the cyclotomic fields $K_m$ and $K_n$ are distinct.
The next theorem gives an integral basis for $K_m$ as well as a formula for the discriminant $d\left(K_m\right)$.
Theorem 7.5.1 Let $m$ be a positive integer. Let $\zeta_m$ be a primitive mth root of unity. Let $K_m$ denote the cyclotomic field $\mathbb{Q}\left(\zeta_m\right)$. Then $\left{1, \zeta_m, \zeta_m^2, \ldots, \zeta_m^{\phi(m)-1}\right}$ is an integral basis for $K_m$. Further,
$$d\left(K_m\right)=(-1)^{\frac{\phi(m)}{2}} \frac{m^{\phi(m)}}{\prod_{p \mid m} p^{\frac{\phi(m)}{p-1}},}$$
where the product is over all primes $p$ dividing $m$.

## 数学代写|数论代写Number Theory代考|Index and Minimal Index of an Algebraic Number Field

$$D(\alpha)=(\text { ind } \alpha)^2 d(K)$$
(见定义7.1.4)。现在我们定义字段$K$的索引$i(K)$和最小索引$m(K)$。

$$i(K)=\operatorname{gcd}{\text { ind } \alpha \mid \alpha \text { a generator of } K} .$$

$$m(K)=\min {\text { ind } \alpha \mid \alpha \text { a generator of } K} .$$

$$i(K) \mid m(K) .$$

## 数学代写|数论代写Number Theory代考|Integral Basis of a Cyclotomic Field

$$f_m(x)=\prod_{\substack{r=1 (r, m)=1}}^m\left(x-\zeta_m^r\right) .$$

$$\operatorname{irr}{\mathbb{Q}}\left(\zeta_m\right)=f_m(x)$$ 此外，$f_m(x)$的程度是$\phi(m)$，所以 $$\left[K_m: \mathbb{Q}\right]=\phi(m) .$$ 同时包含$K_m$和$K_n$的最小字段是$K{[m, n]}$，其中$[m, n]$表示$m$和$n$的最小公倍数。还有，$K_m \cap K_n=K_{(m, n)}$。如果$m \not \equiv 2(\bmod 4)$，那么$K_m \subseteq K_n$当且仅当$m \mid n$成立。因此，如果$m$和$n$是不同的，而不等于$2(\bmod 4)$，那么切光场$K_m$和$K_n$是不同的。

$$d\left(K_m\right)=(-1)^{\frac{\phi(m)}{2}} \frac{m^{\phi(m)}}{\prod_{p \mid m} p^{\frac{\phi(m)}{p-1}},}$$

## 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。

Posted on Categories:Number Theory, 数学代写, 数论

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 数学代写|数论代写Number Theory代考|Prime Ideals in Rings of Integers

In Theorem 1.5 .6 we saw that a maximal ideal of an integral domain $D$ is always a prime ideal. We noted that the converse is not always true but that it is true in a principal ideal domain. In this section we show the important result that a prime ideal is always maximal in the ring of integers of an algebraic number field.

Theorem 6.6.1 Let $P$ be a prime ideal of the ring $O_K$ of integers of an algebraic number field $K$. Then $P$ is a maximal ideal of $O_K$.

Proof: Suppose that the assertion of the theorem is false. Then there exists a prime ideal $P_1$ of $O_K$ that is not a maximal ideal. Let $S$ be the set of all proper ideals of $O_K$ that strictly contain $P_1$. As $P_1$ is not a maximal ideal, $S$ is a nonempty set. By Theorem 6.5.3 $O_K$ is a Noetherian domain. Hence, by Theorem 3.1.3, $S$ contains a maximal element; that is, there is a maximal ideal $P_2$ such that
$$P_1 \subset P_2 \subset O_K$$
By Theorem 1.5.6 $P_2$ is a prime ideal. Since every nonzero ideal in $O_K$ contains a nonzero rational integer (Theorem 6.1.7) we see that $P_1 \cap \mathbb{Z} \neq{0}$. Hence, by Theorem 1.6.2, $P_1 \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. But $\mathbb{Z}$ is a principal ideal domain (Theorem 1.4.1) so $P_1 \cap \mathbb{Z}=\langle p\rangle$ for some $p \in \mathbb{Z}$. By Theorem $1.5 .4 p$ is a prime. Thus
$$\langle p\rangle=P_1 \cap \mathbb{Z} \subseteq P_2 \cap \mathbb{Z} \subseteq \mathbb{Z}$$
Now $P_2 \cap \mathbb{Z} \neq \mathbb{Z}$ as $1 \notin P_2$, so as $\langle p\rangle$ is a maximal ideal of $\mathbb{Z}$ (Theorem 1.5.7), we have
$$P_1 \cap \mathbb{Z}=P_2 \cap \mathbb{Z}=\langle p\rangle .$$

## 数学代写|数论代写Number Theory代考|Integral Basis of an Algebraic Number Field

A basis of the principal ideal of the ring $O_K$ of integers of an algebraic number field $K$ generated by 1 , that is, $O_K$ itself, is called an integral basis for $K$.

Definition 7.1.1 (Integral basis of an algebraic number field) Let $K$ be an algebraic number field. A basis for $O_K$ is called an integral basis for $K$.

In view of this definition the following theorem, which gives an integral basis for a quadratic field, is just a restatement of Theorem 5.4.2.

Theorem 7.1.1 Let $K$ be a quadratic field. Let $m$ be the unique squarefree integer such that $K=\mathbb{Q}(\sqrt{m})$. Then ${1, \sqrt{m}}$ is an integral basis for $K$ if $m \not \equiv 1(\bmod 4)$ and $\left{1, \frac{1+\sqrt{m}}{2}\right}$ is an integral basis for $K$ if $m \equiv 1(\bmod 4)$.

If $\left{\eta_1, \ldots, \eta_n\right}$ and $\left{\lambda_1, \ldots, \lambda_n\right}$ are two integral bases for an algebraic number field $K$ then Theorem 6.5 .4 shows that $D\left(\eta_1, \ldots, \eta_n\right)=D\left(\lambda_1, \ldots, \lambda_n\right)$, and that if $\left{\eta_1, \ldots, \eta_n\right}$ is an integral basis for $K$ and $\lambda_1, \ldots, \lambda_n \in O_K$ are such that $D\left(\lambda_1, \ldots, \lambda_n\right)=D\left(\eta_1, \ldots, \eta_n\right)$ then $\left{\lambda_1, \ldots, \lambda_n\right}$ is also an integral basis for $K$. We can therefore make the following definition.

## 数学代写|数论代写Number Theory代考|Prime Ideals in Rings of Integers

$$P_1 \subset P_2 \subset O_K$$

$$\langle p\rangle=P_1 \cap \mathbb{Z} \subseteq P_2 \cap \mathbb{Z} \subseteq \mathbb{Z}$$

$$P_1 \cap \mathbb{Z}=P_2 \cap \mathbb{Z}=\langle p\rangle .$$

## 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。

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## 数学代写|数论代写Number Theory代考|Integral Closure

Let $A$ and $B$ be integral domains with $A \subseteq B$. In Theorem 4.1 .7 we showed that the set of all elements of $B$ that are integral over $A$ is a subdomain of $B$ containing $A$. We now give this domain a name.

Definition 4.2.1 (Integral closure) Let $A$ and $B$ be integral domains with $A \subseteq B$. The integral closure of $A$ in $B$ is the subdomain of $B$ consisting of all elements of $B$ that are integral over $A$. The integral closure of $A$ in $B$ is denoted by $A^B$.
From Theorem 4.1.7 we have
Theorem 4.2.1 Let $A$ and $B$ be integral domains with $A \subseteq B$. Then the integral closure $A^B$ of $A$ in $B$ is an integral domain satisfying
$$A \subseteq A^B \subseteq B$$
Clearly $A^A=A$ for any integral domain $A$.
Our next theorem determines the integral closure of $\mathbb{Z}$ in the field $\mathbb{Q}(i)=$ ${x+y i \mid x, y \in \mathbb{Q}}$.

## 数学代写|数论代写Number Theory代考|Minimal Polynomial of an Element Algebraic over a Field

Let $K$ be a subfield of the field $\mathbb{C}$ of complex numbers. Let $\alpha \in \mathbb{C}$ be algebraic over $K$ (see Definition 4.1.3). As $\alpha$ is algebraic over $K$, there exists a nonzero polynomial $g(x) \in K[x]$ such that $g(\alpha)=0$. We let $I_K(\alpha)$ denote the set of all polynomials in $K[x]$ having $\alpha$ as a root, that is,
$$I_K(\alpha)={f(x) \in K[x] \mid f(\alpha)=0}$$
Clearly the set $I_K(\alpha)$ contains the zero polynomial. It is easy to check that $I_K(\alpha)$ is an ideal of $K[x]$. Moreover, $I_K(\alpha) \neq\langle 0\rangle$ as $g(x) \in I_K(\alpha)$.

As $K$ is a field, by Theorem 2.2.1(b) we know that $K[x]$ is a Euclidean domain and thus, by Theorem 2.1.2, a principal ideal domain. Hence there exists $p(x) \in$ $K[x]$ such that
$$I_K(\alpha)=\langle p(x)\rangle .$$
Suppose $p_1(x) \in K[x]$ is another polynomial that generates $I_K(\alpha)$, that is,
$$I_K(\alpha)=\left\langle p_1(x)\right\rangle$$
Then
$$\langle p(x)\rangle=\left\langle p_1(x)\right\rangle$$
and so, by Theorem 1.3.1, we have
$$p_1(x)=u(x) p(x),$$
where $u(x)$ is a unit in $K[x]$. However, from Example 1.1.18(c), we have
$$U(K[x])=K^*$$
so that
$$u(x) \in K^*$$

## 数学代写|数论代写Number Theory代考|Integral Closure

$$A \subseteq A^B \subseteq B$$

$$I_K(\alpha)=\langle p(x)\rangle .$$

$$I_K(\alpha)=\left\langle p_1(x)\right\rangle$$

$$\langle p(x)\rangle=\left\langle p_1(x)\right\rangle$$

$$p_1(x)=u(x) p(x),$$

$$U(K[x])=K^*$$

$$u(x) \in K^*$$

## 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。

Posted on Categories:Number Theory, 数学代写, 数论

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 数学代写|数论代写Number Theory代考|Euclidean Domains

To define a Euclidean domain we must first define a Euclidean function.
Definition 2.1.1 (Euclidean function) Let $D$ be an integral domain. A mapping $\phi: D \rightarrow \mathbb{Z}$ is called a Euclidean function on $D$ if it has the following two properties:
$\phi(a b) \geq \phi(a)$, for all $a, b \in D$ with $b \neq 0$,
if $a, b \in D$ with $b \neq 0$ then there exist $q, r \in D$
such that $a=q b+r$ and $\phi(r)<\phi(b)$.
Example 2.1.1 $\phi(a)=|a|(a \in \mathbb{Z})$ is a Euclidean function on $\mathbb{Z}$.
Example 2.1.2 Let $D=F[x]$, where $F$ is a field. $D$ is the domain of polynomials in $x$ with coefficients in $F$. Let $p(x) \in D$. Then
$$\phi(p(x))= \begin{cases}\operatorname{deg}(p(x)), & \text { if } p(x) \neq 0 \ -1, & \text { if } p(x)=0\end{cases}$$
is a Euclidean function on $D$.

In general the elements $q$ and $r$ in (2.1.2) are not uniquely determined. If $D$ is an integral domain that is not a field and that possesses a Euclidean function $\phi$ for which the quotient and remainder $r$ in (2.1.2) are always uniquely determined by $a$ and $b$ then $D=F[x]$ for some field $F$. This result is due to Rhai [14]; see also Jodeit [12].

## 数学代写|数论代写Number Theory代考|Examples of Euclidean Domains

In view of Examples 2.1.1 and 2.1.2 we have
Theorem 2.2.1
(a) $\mathbb{Z}$ is a Euclidean domain.
(b) Let $F$ be a field. Then $F[x]$ is a Euclidean domain.
From Theorems 2.1 .2 and 2.2 .1 we see that $\mathbb{Z}$ and $F[x]$ are principal ideal domains. In the remainder of this section we investigate when the integral domains $\mathbb{Z}+\mathbb{Z} \sqrt{m}(m \equiv 2,3(\bmod 4))$ and $\mathbb{Z}+\mathbb{Z}\left(\frac{1+\sqrt{m}}{2}\right)(m \equiv 1(\bmod 4))$ are Euclidean with respect to the function that maps $r+s \sqrt{m}$ to $\left|r^2-m s^2\right|$. In this section we denote this function by $\phi_m$. Later in Section 9.2 we recognize $\phi_m$ as the absolute value of the norm of the element $r+s \sqrt{m}$. Integral domains that are Euclidean with respect to the absolute value of the norm are called norm-Euclidean.

Definition 2.2.1 (Function $\phi_m$ ) Let $m$ be a squarefree integer. The function $\phi_m$ : $\mathbb{Q}(\sqrt{m}) \rightarrow \mathbb{Q}$ is defined by
$$\phi_m(r+s \sqrt{m})=\left|r^2-m s^2\right|$$
for all $r, s \in \mathbb{Q}$.
The basic properties of $\phi_m$ are given in the next lemma.
Lemma 2.2.1 Let $m$ be a squarefree integer:
(a) $\phi_m: \mathbb{Z}+\mathbb{Z} \sqrt{m} \rightarrow \mathbb{N} \cup{0}$.
(b) If $m \equiv 1(\bmod 4)$ then $\phi_m: \mathbb{Z}+\mathbb{Z}\left(\frac{1+\sqrt{m}}{2}\right) \rightarrow \mathbb{N} \cup{0}$.
(c) Let $\alpha \in \mathbb{Q}(\sqrt{m})$. Then $\phi_m(\alpha)=0 \Longleftrightarrow \alpha=0$.
(d) $\phi_m(\alpha \beta)=\phi_m(\alpha) \phi_m(\beta)$ for all $\alpha, \beta \in \mathbb{Q}(\sqrt{m})$.
(e) $\phi_m(\alpha \beta) \geq \phi_m(\alpha)$ for all $\alpha, \beta \in \mathbb{Z}+\mathbb{Z} \sqrt{m}$ with $\beta \neq 0$.
(f) If $m \equiv 1(\bmod 4)$, then $\phi_m(\alpha \beta) \geq \phi_m(\alpha)$ for all $\alpha, \beta \in \mathbb{Z}+\mathbb{Z}\left(\frac{1+\sqrt{m}}{2}\right)$ with $\beta \neq 0$.

## 数学代写|数论代写Number Theory代考|Euclidean Domains

$\phi(a b) \geq \phi(a)$，对于所有$a, b \in D$和$b \neq 0$，

$$\phi(p(x))= \begin{cases}\operatorname{deg}(p(x)), & \text { if } p(x) \neq 0 \ -1, & \text { if } p(x)=0\end{cases}$$

## 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。