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# 数学代写|抽象代数代写Abstract Algebra代考|Math417 Subfields of a Finite Field

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## 数学代写|抽象代数代写Abstract Algebra代考|Subfields of a Finite Field

Theorem $21.1$ gives us a complete description of all finite fields. The following theorem gives us a complete description of all the subfields of a finite field. Notice the close analogy between this theorem and Theorem 4.3, which describes all the subgroups of a finite cyclic group.
Theorem 21.11 Subfields of a Finite Field
For each divisor $m$ of $n, \operatorname{GF}\left(p^n\right)$ has a unique subfield of order $p^m$. Moreover, these are the only subfields of $\mathrm{GF}\left(p^n\right)$.

PROOF To show the existence portion of the theorem, suppose that $m$ divides $n$. Then, since
$$p^n-1=\left(p^m-1\right)\left(p^{n-m}+p^{n-2 m}+\cdots+p^m+1\right),$$
we see that $p^m-1$ divides $p^n-1$. For simplicity, write $p^n-1=\left(p^m-1\right) t$. Let $K=\left{x \in \operatorname{GF}\left(p^n\right) \mid x^{p^m}=x\right}$. We leave it as an easy exercise for the reader to show that $K$ is a subfield of $\operatorname{GF}\left(p^n\right)$ (Exercise 37). Since the polynomial $x^{p^m}-x$ has at most $p^m$ zeros in $\operatorname{GF}\left(p^n\right)$, we have $|K| \leq p^m$. Let $\langle a\rangle=\operatorname{GF}\left(p^n\right)^*$. Then $\left|a^t\right|=p^m-1$, and since $\left(a^t\right)^{p^{m-1}}=1$ , it follows that $a^t \in K$. So, $K$ is a subfield of $\operatorname{GF}\left(p^n\right)$ of order $p^m$.

The uniqueness portion of the theorem follows from the observation that if $\operatorname{GF}\left(p^n\right)$ had two distinct subfields of order $p^m$, then the polynomial $x^{p^m}-x$ would have more than $p^m$ zeros in $\operatorname{GF}\left(p^n\right)$. This contradicts Theorem 16.3.

## 数学代写|抽象代数代写Abstract Algebra代考|Historical Discussion of Geometric Constructions

The ancient Greeks were fond of geometric constructions. They were especially interested in constructions that could be achieved using only a straightedge without markings and a compass. They knew, for example, that any angle can be bisected, and they knew how to construct an equilateral triangle, a square, a regular pentagon, and a regular hexagon. But they did not know how to trisect every angle or how to construct a regular seven-sided polygon (heptagon). Another problem that they attempted was the duplication of the cube-that is, given any cube, they tried to construct a new cube having twice the volume of the given one using only an unmarked straightedge and a compass. Legend has it that the ancient Athenians were told by the oracle at Delos that a plague would end if they constructed a new altar to Apollo in the shape of a cube with double the volume of the old altar, which was also a cube. Besides “doubling the cube,” the Greeks also attempted to “square the circle”- to construct a square with area equal to that of a given circle. They knew how to solve all these problems using other means, such as a compass and a straightedge with two marks, or an unmarked straightedge and a spiral, but they could not achieve any of the constructions with a compass and an unmarked straightedge alone. These problems vexed mathematicians for over 2000 years. The resolution of these perplexities was made possible when they were transferred from questions of geometry to questions of algebra in the 19 th century.
The first of the famous problems of antiquity to be solved was that of the construction of regular polygons. It had been known since Euclid that regular polygons with a number of sides of the form $2^k, 2^k \cdot 3,2^k \cdot 5$, and $2^k \cdot 3 \cdot 5$ could be constructed, and it was believed that no others were possible. In 1796, while still a teenager, Gauss proved that the 17-sided regular polygon is constructible. In 1801, Gaussbio]Gauss, Carl asserted that a regular polygon of $n$ sides is constructible if and only if $n$ has the form $2^k p_1 p_2 \cdots p_i$, where the $p$ ‘s are distinct primes of the form $2^{2^s}+1$. We provide a proof of this statement in Theorem 31.5.

## 数学代写|抽象代数代写Abstract Algebra代考|Subfields of a Finite Field

$$p^n-1=\left(p^m-1\right)\left(p^{n-m}+p^{n-2 m}+\cdots+p^m+1\right),$$

\left 缺少或无法识别的分隔符 . 我们把它作为一个简单的练习留给读者来证明 $K$ 是一个子 字段 $G F\left(p^n\right)$ (练习 37) 。由于多项式 $x^{p^m}-x$ 至多有 $p^m$ 归零 $\mathrm{GF}\left(p^n\right)$ ，我们有 $|K| \leq p^m$. 让
$\langle a\rangle=\operatorname{GF}\left(p^n\right)^*$. 然后 $\left|a^t\right|=p^m-1$ ，并且因为 $\left(a^t\right)^{p^{m-1}}=1$ ，它遭循 $a^t \in K$. 所以， $K$ 是一个子字段 $\mathrm{GF}\left(p^n\right)$ 秩序 $p^m$. $\mathrm{GF}\left(p^n\right)$. 这与定理 $16.3$ 相矛盾。

## 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。