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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|编码理论代写Coding theory代考|ALGEBRAIC STRUCTURE OF FINITE FIELDS

We begin our discussion of the algebraic structure of finite fields by considering the additive properties of the field’s unit element, 1. Since the field contains 1 , it must contain $\sum_{i=1}^{2} 1=1+1$ and $\sum_{i=1}^{3} 1=1+1+1$ and $\sum_{i=1}^{n} 1=1+1+\cdots+1$. If these field integers are not all distinct, then there exist two rational integers $m$ and $n$ such that $m>n$ and $\sum_{i=1}^{m} 1=\sum_{i=1}^{n} 1$ in the field, or $\sum_{i=1}^{m-n} 1=0$ in the field.

Definition 4.401 The least positive rational integer $c$ for which $\sum_{i=1}^{c} 1=0$ in the field is called the characteristic of the field.

If $\sum_{i=1}^{n} 1$ is nonzero for every integer $n$, then the field is said to have characteristic $\infty$ (or, in the older terminology still used by some writers, characteristic 0$)$.
The field integers are closed under multiplication, because $\dagger$ $\left(\sum_{i=1}^{n} 1\right)\left(\sum_{i=1}^{m} 1\right)=\left(\sum_{i=1}^{n m} 1\right)$
If $\left(\sum_{i=1}^{m n} 1\right)=0$ and $\left(\sum_{i=1}^{n} 1\right) \neq 0$, then we can multiply by $1 /\left(\sum_{i=1}^{n} 1\right)$ and obtain $\sum_{i=1}^{m} 1=0$. Hence, if $\sum_{i=1}^{m n} 1=0$, then either $\sum_{i=1}^{m} 1=0$ or $\sum_{i=1}^{n} 1=0$. The conclusion is Theorem $4.402$.

## 数学代写|编码理论代写Coding theory代考|EXAMPLES

As a straightforward illustration of these theorems, we shall investigate the factorization of the binary polynomial $x^{4}-x$.
\begin{aligned} x^{4}-x &=x\left(x^{3}-1\right) \ &=x Q^{(1)}(x) Q^{(3)}(x) \ &=x(x-1)\left(x^{2}+x+1\right) \end{aligned}
This cyclotomic factorization is valid over any field. Over the binary field, + and – are interchangeable, and we have
$$x^{4}-x=x(x+1)\left(x^{2}+x+1\right)$$
The factor $x^{2}+x+1$ is the irreducible binary quadratic. In order to factor this irreducible polynomial, we must extend the field to include its roots. If we let $\xi$ denote one of its roots, then
$$\xi^{2}+\xi+1=0 \quad \text { or } \quad \xi^{2}=\xi+1$$
The other root of this quadratic is the binary conjugate of $\xi$, namely, $\xi^{2}$, which we denote by $\partial$. In GF (4), we then have the complete factorization,
$$x^{4}-x=x(x+1)(x+\xi)(x+\partial)$$
Here
$$\partial=\xi+1=\xi^{2}$$
and
$$\xi=\partial+1=\partial^{2}$$
The four elements of the field may be represented as the four binary polynomials of degree $<2$ in $\xi$ (or $\partial$ ); the three nonzero field elements may be represented as the successive powers of $\xi$ (or $\partial$ ).

## 数学代写|编码理论代写Coding theory代考|ALGEBRAIC CLOSURE

Since $k$ ! divides $j$ ! whenever $k<j, \operatorname{GF}(p) \subset \operatorname{GF}\left(p^{2 !}\right) \subset \mathrm{GF}\left(p^{3 !}\right) \subset$ $\mathrm{GF}\left(p^{4 !}\right) \subset \ldots$ I define $\mathrm{GF}\left(p^{\infty}\right)$ by the rule $\xi \in \mathrm{GF}\left(p^{\infty !}\right)$ iff $\xi \in \mathrm{GF}\left(p^{n !}\right)$ for all sufficiently large $n$. Every element of $G F\left(p^{\infty 1}\right)$ has finite order, even though the order of $\operatorname{GF}\left(p^{\infty}\right)$ is $\infty$.

Unlike the finite fields of characteristic $p, \operatorname{GF}\left(p^{\infty 1}\right)$ is algebraically closed, as indicated by Theorem $4.61$.

Theorem 4.61 Every polynomial of degree $d$ over $\mathrm{GF}\left(p^{\infty \infty}\right)$ has $d$ roots in $\mathrm{GF}\left(p^{\infty !}\right)$

Proof If $f(x)$ is a polynomial over $\mathrm{GF}\left(p^{\infty 1}\right)$, then there exists a $k$ such that all coefficients of $f(x)$ lie in $\mathrm{GF}\left(p^{k}\right)$. If $g(x)$ is an irreduc-ible factor of $f(x)$ over GF $\left(p^{k}\right)$ and $\operatorname{deg} g(x)=i$, then all roots of $f(x)$ lie in $\mathrm{GF}\left(p^{k i}\right)$, which is a subfield of $\mathrm{GF}\left(p^{n !}\right)$ for all sufficiently large $n$. Hence all roots of $f(x)$ lie in $\mathrm{GF}\left(p^{\infty}\right)$.
Q.E.D.

## 数学代写|编码理论代写Coding theory代考|EXAMPLES

$$x^{4}-x=x\left(x^{3}-1\right) \quad=x Q^{(1)}(x) Q^{(3)}(x)=x(x-1)\left(x^{2}+x+1\right)$$

$$x^{4}-x=x(x+1)\left(x^{2}+x+1\right)$$

$$\xi^{2}+\xi+1=0 \quad \text { or } \quad \xi^{2}=\xi+1$$

$$x^{4}-x=x(x+1)(x+\xi)(x+\partial)$$

$$\partial=\xi+1=\xi^{2}$$

$$\xi=\partial+1=\partial^{2}$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师