### 数学代写|代数数论代写Algebraic number theory代考|Factoring Rational Primes in Z

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## 数学代写|代数数论代写Algebraic number theory代考|Factoring Rational Primes in Z

Let $A$ be the ring $\mathbb{Z}[i]$ of Gaussian integers and $p=2,3,4, \ldots$ a rational prime. This $p$ may or may not be a prime element of $A$. To find exactly when it is, recall the famous theorem of Fermat on the sum of two squares, which was proved by Euler (cf. [8, p. 48]).

Theorem 2.14 (Fermat). An odd prime $p$ in $\mathbb{Z}$ is a sum of two squares $\left(p=a^{2}+b^{2}\right)$ if and only if $p=4 k+1$ for $k$ in $\mathbb{N}$.

The norm of any divisor of $\alpha=a+i b$ must be a divisor of $N(\alpha)=a^{2}+b^{2}$, and for $\alpha=\beta \gamma$ with $\beta, \gamma$ both non-units, $1<N(\beta)<N(\alpha)$ (only the units have norm 1). Therefore, if $a^{2}+b^{2}$ is a prime, then $\alpha$ has to be a prime in $\mathbb{Z}[i]$. We have thus proved the following fact:

Theorem 2.15. A prime $p$ is a sum of two squares, $p=a^{2}+b^{2} \Leftrightarrow p$ is a product $(a+i b)(a-i b)$ of two primes $a \pm i b$ in $\mathbb{Z}[i]$.

For $p=2$, its two prime factors $1+i, 1-i$ in $\mathbb{Z}[i]$ are associates: $1+i=i(1-i)$. Therefore,
$$2=i(1-i)^{2} .$$
We say that 2 ramifies in $\mathbb{Z}[i]$. By Fermat’s Theorem (Theorem 2.15), $p \equiv 1$ $(\bmod 4) \Leftrightarrow p$ is a product
$$p=\pi_{1} \pi_{2}$$
of two primes $\pi_{1}, \pi_{2}$ in $\mathbb{Z}[i]$. Moreover, $\pi_{1}$ and $\pi_{2}$ are complex conjugates of each other and hence they are distinct. This discussion can be wrapped up as follows: In order to do that, observe that ${1, i}$ is a $\mathbb{Z}$-bases of $\mathbb{Z}[i]$ and so is its conjugate ${1,-i}$. These two bases make a $2 \times 2$ matrix
$$A=\left(\begin{array}{cc} 1 & i \ 1 & -i \end{array}\right)$$
with $|\operatorname{det}(A)|=2$, called the discriminant of $\mathbb{Q}(i)$.
Theorem 2.16. Let $p$ be a prime. Then
i) $p$ ramifies in $\mathbb{Z}[i] \Leftrightarrow$ it divides the discriminant of $\mathbb{Q}(i)$,ii) $p$ factors into two distinct primes of $\mathbb{Z}[i] \Leftrightarrow p \equiv 1(\bmod 4)$, and iii) $p$ stays prime in $\mathbb{Z}[i] \Leftrightarrow p \equiv 3(\bmod 4)$.

## 数学代写|代数数论代写Algebraic number theory代考|Generalities

Let $K / k$ be a field extension and suppose $\alpha$ is an element of $K$. We say that $\alpha$ is algebraic over $k$ if $\alpha$ satisfies a nonzero polynomial over $k$. Suppose $n=\operatorname{dim}{k}(K)$ is finite and $\alpha$ is in $K$. Then the $n+1$ vectors $1, \alpha, \ldots, \alpha^{n}$ cannot be linearly independent and hence satisfy a nontrivial linear relation $$c{0}+c_{1} \alpha+\cdots+c_{n} \alpha^{n}=0$$
with $c_{j}$ in $k$. This not only shows that $\alpha$ is algebraic over $k$ but also proves that it is a root of a nonzero polynomial of degree at most $n$ over $k$. The smallest degree of a polynomial over $k$ satisfied by $\alpha$ is called the degree of $\alpha$ over $k$. It is denoted by $\operatorname{deg}_{k}(\alpha)$.

## 数学代写|代数数论代写Algebraic number theory代考|Algebraic Integers

The subject of algebraic number theory originated with Gauss, who studied the arithmetic in the ring $\mathbb{Z}[i]={x+i y \mid x, y \in \mathbb{Z}}$ of the so called Gaussian integers. We begin with a useful fact about field extensions which is true for the ones to be dealt with in this book.

Definition 3.1. A field extension $K / k$ is a simple extension if there is an element $\alpha$ in $K$ such that $K=k(\alpha)$.

Here $k(\alpha)$ is the field of all quotients of polynomials in $\alpha$ over $k$. It is the smallest field containing $k$ and $\alpha$. We say that $K$ has been obtained by adjoining $\alpha$ to $k$. We also say that $\alpha$ generates $K$ over $k$.

From now on, we shall regard $\mathbb{C}$, the field of complex numbers, as our universal domain. This essentially means that all fields, unless stated otherwise, shall be subfields of $\mathbb{C}$. A field must have at least two distinct elements, namely, 0 and 1. Therefore, a subfield of $\mathbb{C}$ must contain $\mathbb{Z}$, and hence it must be an

extension of $\mathbb{Q}$. The following is a standard result from field theory (cf. $[8, \mathrm{p}$. $72]$ ).

Theorem 3.2. If $k$ is a subfield of $\mathbb{C}$, then any finite extension $K / k$ is $a$ simple extension.

Definition 3.3. A number field is a finite extension of $\mathbb{Q}$. A number field $K$ is a quadratic field or a cubic field according as $[K: \mathbb{Q}]$ is 2 or 3 . We call a field extension $K / k$ an extension of number fields if $k$ is a subfield of the number field $K$. Clearly, $k$ is also a number field.

Definition 3.4. A complex number $\alpha$ is an algebraic number if it is algebraic over Q.

It is not hard to see that the set of all algebraic numbers is a subfield of $\mathbb{C}$. It is called the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$ and is denoted by $\overline{\mathbb{Q}}$.

Every element of a number field $K$ with $[K: \mathbb{Q}]=n$ is an algebraic number of degree at most $n$. By Theorem $3.2$, there is always an $\alpha$ in $K$ with deg $(\alpha)=$ $n$.
The following definition is crucial to what follows.

2=一世(1−一世)2.

p=圆周率1圆周率2

## 数学代写|代数数论代写Algebraic number theory代考|Generalities

C0+C1一个+⋯+Cn一个n=0

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## MATLAB代写

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