### 数学代写|数论代写Number theory代考|MATH2O88

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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|数论代写Number theory代考|DEFINITIONS AND BASIC PROPERTIES

We begin with a definition which was given in a slightly different, but equivalent, form in Chapter $1 .$
Definition 3.1. If the complex number $\alpha$ is a root of a polynomial
$$a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{1} z+a_{0}$$
with rational coefficients and $a_{n} \neq 0$, then $\alpha$ is said to be algebraic. Any (complex) number which is not algebraic is called a transcendental number. More generally, if $\alpha$ is a root of a polynomial such as (3.1) with coefficients in a field $\mathbb{F}$, then $\alpha$ is said to be algebraic over $\mathbb{F}$; if there is no such polynomial then $\alpha$ is transcendental over $F$.

Lemma 3.1. Properties of algebraic numbers.

• A complex number $\alpha$ is algebraic (over $\mathbb{Q}$ ) if and only if it is a root of a non-zero polynomial with integral coefficients.
• If $\alpha$ is algebraic, then there exists a unique monic polynomial $f_{\alpha}$ having rational coefficients and smallest possible degree, such that $f_{\alpha}(\alpha)=0$. If $g$ is any polynomial with rational coefficients such that $g(\alpha)=0$, then $g$ is a multiple of $f_{\alpha}$.
• The polynomial $f_{\alpha}$ is irreducible over $\mathbb{Q}$. That is, $f_{\alpha}$ cannot be factorised as the product of two polynomials with rational coefficients and degree smaller than that of $f_{\alpha}$.

Proof. To prove the first assertion, just multiply a polynomial with rational coefficients by a common denominator for its coefficients. In the second statement the existence of $f_{\alpha}$ is clear; if $g(\alpha)=0$ then dividing $g$ by $f_{\alpha}$ gives
$$g(z)=f_{\alpha}(z) q(z)+r(z)$$
with $r(\alpha)=0$. But $r$ has smaller degree than $f_{\alpha}$, so $r$ is the zero polynomial and hence $g$ is a multiple of $f_{\alpha}$. The uniqueness of $f_{\alpha}$ follows since if there are two polynomials with the minimal-degree property then each is a factor of the other. To prove irreducibility note that if there is a proper factorisation $f_{\alpha}=g h$ then either $g$ or $h$ has $\alpha$ as a root, thus contradicting the minimality of $f_{\alpha}$.

## 数学代写|数论代写Number theory代考|Proving polynomials irreducible

The next two results are often useful for proving irreducibility of polynomials.
Lemma 3.4. Eisenstein’s Lemma. Let $f$ be a polynomial with integral coefficients,
$$f(z)=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{1} z+a_{0}$$

suppose that there is a prime $p$ such that $p$ is a factor of $a_{0}, a_{1}, \ldots, a_{n-1}$ but not of $a_{n}$, and such that $p^{2}$ is not a factor of $a_{0}$. Then $f$ is irreducible over the field $Q$ of rational numbers.

Proof. By Gauss’ Lemma, we need only show that $f$ cannot be written as the product of two polynomials with integer coefficients and degree less than $n$. Suppose that there are two such polynomials; without loss of generality we may assume that they have the same number of terms, say
$$f(z)=g(z) h(z)=\left(b_{m} z^{m}+\cdots+b_{1} z+b_{0}\right)\left(c_{m} z^{m}+\cdots+c_{1} z+c_{0}\right)$$
with $m<n$. Looking at the constant terms, $b_{0} c_{0}=a_{0}$ is divisible by $p$ but not by $p^{2}$; by symmetry, we may assume that $p \mid b_{0}$ and $p \nmid c_{0}$. Multiplying out all the other coefficients shows that $p$ is a factor of $b_{1}, b_{2}, \ldots, b_{m}$ too. But then $p$ is a factor of every $a_{k}$, including $a_{n}$, and this is contrary to our initial assumption. So $f$ is irreducible.
Examples.

• The polynomial $z^{3}-12 z^{2}+345 z-6789$ is irreducible since 3 is a prime factor of 12 , of 345 and of 6789 but not of the leading coefficient, and since $3^{2}$ is not a factor of 6789 .
• A slightly more subtle application of Eisenstein’s Lemma simplifies the proof of irreducibility for
$$f(z)=\frac{z^{5}-1}{z-1}=z^{4}+z^{3}+z^{2}+z+1 .$$
It is not hard to see that we can factorise $f(z)$ if and only if we can factorise
$$f(z+1)=\frac{(z+1)^{5}-1}{z}=z^{4}+5 z^{3}+10 z^{2}+10 z+5$$
but Eisenstein’s Lemma with $p=5$ shows immediately that $f(z+1)$ is irreducible, and hence so is $f(z)$.
• In fact, if $p$ is any prime then $f(z)=z^{p-1}+z^{p-2}+\cdots+z+1$ can be proved irreducible by the same method. Exercise. Give the details!

## 数学代写|数论代写Number theory代考|Closure properties of algebraic numbers

Next we shall sketch proofs that the set of (complex) algebraic numbers forms a subfield of $\mathbb{C}$, and that the algebraic integers form an integral domain. These proofs require a certain acquaintance with basic properties of vector spaces and abelian groups; however, the level required is probably too much to summarise in an appendix. Therefore, on this occasion only, we invite the interested reader to refer to other sources for background material. Two of many possibilities are Axler $[8]$ for linear algebra, Stewart and Tall [62] for groups. The reader who prefers to continue with the main topics of this book can safely proceed to section $3.2$ after noting carefully the results of Theorem 3.10, Corollary $3.11$ and Theorem 3.12.
Lemma 3.8. Let $S=\left{\alpha_{k} \mid k \in K\right}$ be a set of complex numbers. Then

• the set of linear combinations
$$\sum r_{k} \alpha_{k}$$
with finitely many terms and rational coefficients $r_{k}$ is a vector space over the field $\mathbb{Q}$;
• the set of linear combinations
$$\sum m_{k} \alpha_{k}$$
with finitely many terms and integer coefficients $m_{k}$ is an abelian group under addition.

Lemma 3.9. Finiteness criteria for algebraic numbers. Let $\alpha \in \mathbb{C}$; in the previous lemma take $S=\left{1, \alpha, \alpha^{2}, \ldots\right}$. Then

• $\alpha$ is algebraic if and only if the vector space of rational linear combinations of $S$ is finite-dimensional;
• $\alpha$ is an algebraic integer if and only if the group of integer linear combinations of $S$ is finitely generated.

Sketch of proof. If $\alpha$ is algebraic of degree $n$, then all powers of $\alpha$ can be written as linear combinations of $\left{1, \alpha, \alpha^{2}, \ldots, \alpha^{n-1}\right}$, so the vector space has a spanning set (in fact, a basis) with $n$ elements, and so is finite dimensional. Conversely, if the vector space has dimension $n$, then $\left{1, \alpha, \alpha^{2}, \ldots, \alpha^{n}\right}$ is a linearly dependent set, and this yields a polynomial identity satisfied by $\alpha$.
If $\alpha$ is an algebraic integer of degree $n$, then every power of $\alpha$ can be written as an integral linear combination of $\left{1, \alpha, \alpha^{2}, \ldots, \alpha^{n-1}\right}$, and so this set generates the group. Conversely, suppose that the group is generated by $n$ elements $p_{1}, p_{2}, \ldots, p_{n}$. Each of these is an integer linear combination of powers of $\alpha$; therefore so are $\alpha p_{1}, \alpha p_{2}, \ldots, \alpha p_{n}$, and we can write equations
$$\alpha p_{k}=m_{k 1} p_{1}+m_{k 2} p_{2}+\cdots+m_{k n} p_{n} \quad \text { for } \quad k=1,2, \ldots, n$$

## 数学代写|数论代写Number theory代考|DEFINITIONS AND BASIC PROPERTIES

• 一个复数一个是代数的（超过问) 当且仅当它是具有整数系数的非零多项式的根。
• 如果一个是代数的，则存在唯一的一元多项式F一个具有有理系数和最小可能度数，使得F一个(一个)=0. 如果G是任何具有有理系数的多项式，使得G(一个)=0， 然后G是的倍数F一个.
• 多项式F一个是不可约的问. 那是，F一个不能被分解为两个有理系数和次数小于的多项式的乘积F一个.

G(和)=F一个(和)q(和)+r(和)

## 数学代写|数论代写Number theory代考|Proving polynomials irreducible

F(和)=一个n和n+一个n−1和n−1+⋯+一个1和+一个0

F(和)=G(和)H(和)=(b米和米+⋯+b1和+b0)(C米和米+⋯+C1和+C0)

• 多项式和3−12和2+345和−6789是不可约的，因为 3 是 12 、 345 和 6789 的质因数，但不是主要系数，并且因为32不是 6789 的因数。
• 爱森斯坦引理的稍微更微妙的应用简化了不可约性的证明
F(和)=和5−1和−1=和4+和3+和2+和+1.
不难看出我们可以分解F(和)当且仅当我们可以分解
F(和+1)=(和+1)5−1和=和4+5和3+10和2+10和+5
但是爱森斯坦的引理p=5立即表明F(和+1)是不可约的，因此也是F(和).
• 事实上，如果p是任何素数F(和)=和p−1+和p−2+⋯+和+1可以用同样的方法证明不可约。锻炼。给个细节！

## 数学代写|数论代写Number theory代考|Closure properties of algebraic numbers

• 线性组合的集合
∑rķ一个ķ
具有有限多项和有理系数rķ是场上的向量空间问;
• 线性组合的集合
∑米ķ一个ķ
具有有限多项和整数系数米ķ是加法下的阿贝尔群。

• 一个是代数的当且仅当小号是有限维的；
• 一个是一个代数整数当且仅当整数线性组合的群小号是有限生成的。

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## MATLAB代写

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