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

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

Let $A$ be a commutative ring with 1 . Suppose $M \neq{0}$ is an $A$-module. We say that $M$ is a free $A$ – module of rank $n$ ( $n$ being an integer $\geq 1$ ) if there are $n$ elements $\alpha_{1}, \ldots, \alpha_{n}$ in $M$ such that every element $\alpha$ of $M$ can be uniquely written as
$$\alpha=a_{1} \alpha_{1}+\cdots+a_{n} \alpha_{n}$$
with $a_{j}$ in $A$. We write it as
$$M=A \alpha_{1} \oplus \ldots \oplus A \alpha_{n}$$
The set $\left{\alpha_{1}, \ldots, \alpha_{n}\right}$ is called a basis of $M$ over $A$. If the elements of a basis are taken in a fixed order, it is called an ordered basis. In this section, we shall prove that for a number field $K$, its ring of integers $\mathcal{O}_{K}$ is a free $\mathbb{Z}$-module of rank $[K: k]$. We recall some basic facts needed from linear algebra and Galois theory.

Suppose $x=\left(x_{i j}\right)$ is in $M(n, A)$, that is $x$ is an $n$ by $n$ matrix with entries in $A$.

Definition 3.13. The trace $\operatorname{tr}(x)$ of $x$ is the sum $x_{11}+\cdots+x_{n n}$ of the diagonal entries of $x$.
The following theorem is obvious.
Theorem 3.14. Let $x, y$ be in $M(n, A)$ and $a$ in $A$. Then
(1) $\operatorname{tr}(x+y)=\operatorname{tr}(x)+\operatorname{tr}(y)$.
(2) $\operatorname{tr}(a x)=a \operatorname{tr}(x)$
(3) $\operatorname{tr}(x y)=\operatorname{tr}(y x)$.
Now suppose $M$ a free $A$-module of rank $n$ over $A$. Let $\lambda: M \rightarrow M$ be a homomorphism of $A$-modules, or simply an A-homomorphism. We associate a matrix $L$ over $A$ to $\lambda$ with respect to an ordered basis $\left{\alpha_{1}, \ldots, \alpha_{n}\right}$ of $M$ over $A$ in the same way as to a linear transformation. If $L_{1}$ and $L_{2}$ are the matrices of $\lambda$ with respect to two ordered bases, then $L_{2}=P^{-1} L_{1} P$ for some $P$ in $G L(n, A)$, that is, for a matrix $P$ over $A$ whose determinant has multiplicative inverse in $A$.

For the rest of the section, let $K / k$ be an extension of number fields. Since the dimension $\operatorname{dim}{k}(K)$ cannot be more than $\operatorname{dim}{\mathrm{Q}}(K)$, it is clear that $K / k$ is a finite extension. We may regard $K$ as a $k$-module of rank $n=[K: k]$. For $\alpha$ in $K$, the multiplication by $\alpha$ is a $k$-homomorphism $m_{\alpha}: K \rightarrow K$. Let $L_{\alpha}$ be the matrix of $m_{\alpha}$ with respect to an ordered basis of $K$ over $k$.

A number field $K$ is a quadratic field if the degree $[K: \mathbb{Q}]=2$. By Theorem $3.18, K=\mathbb{Q}(\alpha)$, where $\alpha$ is a root of an irreducible polynomial $f(x)=a x^{2}+$ $b x+c$ of degree 2 over $\mathbb{Q}$. Since $\alpha$ is not a rational number, the discriminant $D=b^{2}-4 a c$ of $f(x)$ cannot be zero or a perfect square. Write $D=d m^{2}$, with the integer $d \neq 0,1$, having no square factor larger than 1. From the quadratic formula for solving quadratic polynomial equations, it is clear that $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{d})$. We summarize this as

Proposition 3.30. A quadratic field $K=\mathbb{Q}(\sqrt{d})$ for a square-free integer $d \neq 0,1$.

The following theorem exhibits an integral basis of the ring of integers of a quadratic field explicitly.
Theorem 3.31. Suppose $d \neq 0,1$ is a square-free integer. Put
$$\omega= \begin{cases}\sqrt{d} & \text { if } d \equiv 2,3 \quad(\bmod 4) \ \frac{1+\sqrt{d}}{2} & \text { if } d \equiv 1 \quad(\bmod 4)\end{cases}$$
Then ${1, \omega}$ is an integral basis of $K=\mathbb{Q}(\sqrt{d})$.
Proof. First we show that $\mathcal{O}{K} \supseteq \mathbb{Z}+\mathbb{Z}$. For this, all we need to show is that in case of $d \equiv 1(\bmod 4), \omega=(1+\sqrt{d}) / 2$ is a root of a monic polynomial of degree 2 over $\mathbb{Z}$. It is easy to see that $x^{2}-\operatorname{tr}{K / Q}(\omega) x+N_{K / k}(\omega) \in \mathbb{Z}[x]$ is such a polynomial. Next we show that $\mathcal{O}{K} \subseteq \mathbb{Z}+\mathbb{Z}$. Suppose $\alpha=a+b \sqrt{d} \in \mathcal{O}{K}$ with $a, b \in \mathbb{Q}$. We know that $n=N_{K / k}(\alpha)=a^{2}-d b^{2}, m=\operatorname{tr}_{K / k}(\alpha)=2 a \in \mathbb{Z}$. Now if $m$ is even, then $a \in \mathbb{Z} \Rightarrow d b^{2} \in \mathbb{Z}$. Since $d$ is square-free, this implies that $b \in \mathbb{Z}$. This shows that $\alpha \in \mathbb{Z}+\mathbb{Z} \omega$. If $m$ is odd, then $d b^{2}-\frac{1}{4} \in \mathbb{Z}$. Since $d$ is square-free, $b=c / 2$ with $c$ odd. This gives $\omega=\frac{1+\sqrt{d}}{2}$ and $d \equiv 1$ $(\bmod 4)$.

Corollary 3.32. The discriminant of the quadratic field $K=\mathbb{Q}(\sqrt{d})$, where $d \neq 0,1$ is square-free, is given by
$$d_{K}= \begin{cases}4 d & \text { if } d \equiv 2,3 \quad(\bmod 4) \ d & \text { if } d \equiv 1 \quad(\bmod 4)\end{cases}$$
Proof. The two $\mathbb{Q}$-homomorphisms $\sigma_{i}: K \rightarrow \mathbb{C}$ are the identity $\sigma_{1}=1_{K}$ and the conjugation $\sigma_{2}$ defined by $\sigma_{2}(x+y \sqrt{d})=x-y \sqrt{d}$. Let $\left{\alpha_{1}, \alpha_{2}\right}$ be the integral basis of $K$ given by Theorem 3.31. Using $d_{K}=\left(\operatorname{det}\left(\sigma_{i}\left(\alpha_{j}\right)\right)\right)^{2}$, a short calculation is all we need.

## 数学代写|代数数论代写Algebraic number theory代考|Unique Factorization Property for Ideals

Let $A$ be a commutative ring with 1. We recall the definition of an ideal of $A$. Suppose $a$ is a nonempty subset of $A$. We say that a is an ideal of $A$, if for every $a \in A$ and $x, y \in a, a x$ and $x+y \in a$. Every ideal contains 0 , the zero element of $A$. The whole ring $A$ itself is an ideal. An ideal a of $A$ is a proper ideal if $A \geqslant a$. If $a \geqslant{0}$, then we call a nonzero ideal.

Theorem 3.34. If $\mathfrak{A}$ is a nonzero ideal of $\mathcal{O}_{K}$, then $a=\mathfrak{2} \cap \mathbb{Z}$ is a nonzero ideal of $Z$.

Proof. If $0 \neq \alpha \in \mathfrak{A}$, then $\alpha$ satisfies a nonzero monic polynomial over $\mathbb{Z}$, i.e.
$$a_{0}+a_{1} \alpha+\cdots+\alpha^{n}=0$$
with $a_{j}$ in $\dddot{Z}$ and $a_{0} \neq 0$. Using the defining properties of an ideal, we see that $a_{0}=-a_{1} \alpha-\cdots-a_{n} \alpha^{n} \in \mathfrak{a} \cap \mathbb{Z}=a$.

Let $a$ be an ideal. The relation $x \sim y \Leftrightarrow x-y \in a$ is an equivalence relation which partitions $A$ into disjoint sets of the form $x+a={x+a \mid a \in a}$, called the cosets of a in $A$. This set of cosets is a ring, called the quotient of $A$ by a and is denoted by $A / a$. The ring operations on $A /$ a are defined in an obvious way, namely,
$$(x+\mathbf{a})+(y+\mathbf{a})=(x+y)+\mathbf{a},(x+\mathbf{a})(y+\mathbf{a})=x y+\mathfrak{a}$$
Remark 3.35. Let a be an ideal of $A$. The notation $x \equiv y$ (mod a) means that $x-y \in \mathfrak{a}$.

Definition 3.36. Suppose $m$ is a proper ideal of $A$. We call $m$ a maximal ideal if for no other proper ideal a, we can have $m \varsubsetneqq a$. We call a proper ideal $\mathfrak{p}$ a prime ideal, if $a, b \in A, a b \in \mathfrak{p}$ implies that either $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$.
Theorem 3.37. Suppose a is an ideal of A. Then

1. $a$ is maximal if and only if $A / a$ is a field.
2. $a$ is prime if and only if $A / a$ is an integral domain.

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

α=a1α1+⋯+anαn

M=Aα1⊕…⊕Aαn

（一）tr⁡(x+y)=tr⁡(x)+tr⁡(y).
(2) tr⁡(ax)=atr⁡(x)
(3) tr⁡(xy)=tr⁡(yx).

ω={d if d≡2,3(mod4) 1+d2 if d≡1(mod4)

dK={4d if d≡2,3(mod4) d if d≡1(mod4)

## 数学代写|代数数论代写Algebraic number theory代考|Unique Factorization Property for Ideals

a0+a1α+⋯+αn=0

(x+a)+(y+a)=(x+y)+a,(x+a)(y+a)=xy+a

1. a是最大的当且仅当A/a是一个字段。
2. a是素数当且仅当A/a是一个积分域。

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