### 数学代写|代数数论代写Algebraic number theory代考|Ideal Class Group and Class Number

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## 数学代写|代数数论代写Algebraic number theory代考|Ideal Class Group and Class Number

In the last section, we proved that the nonzero fractional ideals of a Dedekind domain $A$ with quotient field $K$, form an Abelian group $I$ under the operation

of multiplication of ideals. The nonzero principal fractional ideals, that is the ideals of the form $\alpha A={\alpha a \mid a \in A}$ with $\alpha \neq 0$ in $K$, form a subgroup $P$ of $I$. The quotient group $I / P$ is called the ideal class group of $K$. The elements of $I / P$ are called the ideal classes. The cardinality of the ideal class group is called the class number of $K$. We will denote the class number of $K$ by $h_{K}$. In this section we shall show that the class number of a number field is finite, in which $A=\mathcal{O}_{K}$.

Recall that for a nonzero ideal $a$ of $\mathcal{O}{K}$, its norm $N(\mathrm{a})$ is the cardinality of the quotient ring $\mathcal{O}{K} / a$. We have seen that this cardinality is finite.

Theorem 3.67. Suppose $(\alpha)=\alpha \mathcal{O}{K}$ is a principal ideal of $\mathcal{O}{K}$. Then we have
$$N((\alpha))=\left|N_{K / Q}(\alpha)\right|$$
Proof. If $\alpha=0$ there is nothing to prove, Otherwise, write $\mathcal{O}{K}=\mathbb{Z} \alpha{1} \oplus$ $\cdots \oplus \mathbb{Z} \alpha_{n}$, where $n=[K: \mathbb{Q}]$. By Proposition $3.42$, we can also write $((\alpha))=$ $\mathbb{Z} \beta_{1} \oplus \ldots \oplus \mathbb{Z} \beta_{n}$, where
$$\beta_{i}=\sum_{j=i}^{n} a_{j i} \alpha_{j}$$
with $a_{i i}>0$. By Theorem 3.47, $N((\alpha))=a_{11} \ldots a_{n n}$. On the other hand,
$$(\alpha)=\mathbb{Z} \alpha \alpha_{1} \oplus \cdots \oplus \mathbb{Z} \alpha \alpha_{n}$$
which shows that $\left{\alpha \alpha_{1}, \ldots, \alpha \alpha_{n}\right}$ is a $\mathbb{Z}$-basis of $(\alpha)$. The transition matrix $U$ from $\left{\beta_{1}, \ldots, \beta_{n}\right}$ to $\left{\alpha \alpha_{1}, \ldots, \alpha \alpha_{n}\right}$ is unimodular. If for $i<j$ we let $a_{i j}=0$ and put $M=\left(a_{i j}\right)$, then
$$\alpha\left(\begin{array}{c} \alpha_{1} \ \vdots \ \alpha_{n} \end{array}\right)=U M\left(\begin{array}{c} \alpha_{1} \ \vdots \ \alpha_{n} \end{array}\right)$$
Therefore,
$$\left|N_{K / \mathbb{Q}}(\alpha)\right|=|\operatorname{det}(U M)|=|\operatorname{det}(M)|=\left|N\left(\alpha \mathcal{O}_{K}\right)\right|$$

## 数学代写|代数数论代写Algebraic number theory代考|Arithmetic in Relative Extensions

Throughout this chapter, $K$ will denote a number field and $k$ a subfield of $K$. The extension $K / k$ will be called a relative extension of number fields. We put $\mathfrak{o}=\mathcal{O}{k}$ and $\mathcal{O}=\mathcal{O}{K}$.

Theorem 4.1. $\mathcal{O}{K}={\alpha \in K \mid f(\alpha)=0$ for a monic polynomial $f(x)$ in o $[x]}$ Proof. We only have to show that any $\alpha$ in $K$ which satisfies a monic polynomial $f(x)$ in $o[x]$ is an algebraic integer. Let $$f(x)=a{0}+a_{1} x+\cdots+a_{n-1} x^{n-1}+x^{n}$$
with $a_{j}$ in 0 . Since $a_{j}$ are algebraic integers,
$$M=\mathbb{Z}\left[a_{0}, a_{1}, \ldots, a_{n-1}\right]$$
is a finitely generated $\mathbb{Z}$-module, and so is
$$\mathbb{Z}\left[a_{0}, a_{1}, \ldots, a_{n-1}, \alpha\right]=M+M \alpha+\cdots+M \alpha^{n-1}$$
Since $\mathbb{Z}[\alpha]$ is a submodule of a finitely generated $\mathbb{Z}$-module, $\mathbb{Z}[\alpha]$ is also a finitely generated ZZ-module, which shows that $\alpha$ is an algebraic integer.

Remark 4.2. This theorem allows us to regard $K / \mathbb{Q}$ as a special case of the relative extension $K / k$ of number fields with $k=\mathbb{Q}$.
EXERCISE
Let $k \subseteq K \subseteq L$ be number fields with $[K: k]=n$ and $[L: K]=m$. Let $\sigma_{1}, \ldots, \sigma_{n}$ be the distinct $k$-isomorphisms of $K$ into $\mathbb{C}$. Show that each $\sigma_{i}$ extends to $m$ distinct $k$-isomorphisms $\sigma_{i j}: L \rightarrow \mathbb{C}$.

Hint: Let $L=K(\theta)$ and $\tau_{1}, \ldots, \tau_{m}$ be the $m$ distinct $K$-isomorphisms of $L$ into $\mathbb{C}$. If we write $\alpha$ in $L$ as
$$\alpha=a_{0}+a_{1} \theta+\cdots+a_{m-1} \theta^{m-1}$$
with coefficients in $K$, put
$$\sigma_{i j}(\alpha)=\sigma_{i}\left(a_{0}\right)+\sigma_{i}\left(a_{1}\right) \tau_{j}(\theta)+\cdots+\sigma_{i}\left(a_{m-1}\right) \tau_{j}\left(\theta^{m-1}\right)$$
Recall the following.

## 数学代写|代数数论代写Algebraic number theory代考|Criterion for Ramification

We now start preparing to show that if $K / k$ is an extension of number fields, the number of primes which ramify in $K$ is finite. In fact, we shall point out exactly which primes in $k$ ramify in $K$.
Definition 4.11. The complementary set of $\mathcal{O}$ relative to $o$ is the set
$$\mathcal{O}^{\prime}=\left{\alpha \in K \mid \operatorname{Tr}_{K / k}(\alpha \mathcal{O}) \subseteq \mathfrak{o}\right}$$
Theorem 4.12. The complementary set $\mathcal{O}^{\prime}$ is a fractional ideal and contains O.

Proof. It is obvious from the properties of the trace map and definition of $\mathcal{O}^{\prime}$ that $\mathcal{O}^{\prime}$ is an $\mathcal{O}$-module and that $\mathcal{O} \subseteq \mathcal{O}^{\prime}$. All we have to do is to produce a nonzero element $d$ of $\mathfrak{o}$ such that $d \mathcal{O}^{\prime} \subseteq \mathcal{O}$.

Fix a basis $\alpha_{1}, \ldots, \alpha_{n}$ of $K$ over $k$, consisting of elements of $\mathcal{O}$. If $\alpha \in K$, write
$$\alpha=a_{1} \alpha_{1}+\cdots+a_{n} \alpha_{n} \quad\left(a_{j} \in k\right) .$$
Then for each $i=1, \ldots, n$,
$$\operatorname{Tr}{K / k}\left(\alpha \alpha{i}\right)=\sum_{j=1}^{n} a_{j} \operatorname{Tr}{K / k}\left(\alpha{i} \alpha_{j}\right)=b_{i} \in \mathfrak{0}$$

or in the matrix notation
$$\left(\operatorname{Tr}{K / k}\left(\alpha{i} \alpha_{j}\right)\right)\left(\begin{array}{c} a_{1} \ \vdots \ a_{n} \end{array}\right)=\left(\begin{array}{c} b_{1} \ \vdots \ b_{n} \end{array}\right)$$
Since the matrix ( $\left.\operatorname{Tr}{K / k}\left(\alpha{i} \alpha_{j}\right)\right) \in G L(n, o)$, solving by Cramer’s rule, we see that $d a_{j} \in 0$, where $d=\operatorname{det}\left(\operatorname{Tr}{K / k}\left(\alpha{i} \alpha_{j}\right)\right)$ is a nonzero element of $\mathfrak{o}$.
Definition 4.13. The integral ideal
$$\mathfrak{D}{K / k}=\mathcal{O}^{\prime-1}$$ is called the different of $K / k$. Definition 4.14. The ideal $\mathfrak{d}{K / k}=N_{K / k}\left(\mathfrak{D}{K / k}\right)$ of $\mathfrak{o}$ is called the discriminant of $K / k$. The following is also clear from the proof of Theorem 4.12. Theorem 4.15. The ideal $\mathfrak{o}{K / Q}$ is generated by the discriminant $d_{K}$ of $K$, that is, $\mathfrak{o}{K / Q}=d{K} \mathbb{Z}$.

The rest of the chapter is devoted to prove that a prime $\mathfrak{F}$ of $K$ is ramified if and only if $\mathfrak{P} \mid \mathfrak{D}{K / k}$, and a prime $\mathfrak{p}$ of $k$ is ramified if and only if $\mathfrak{p} \mid d{K / k}$. In particular, there are only finitely many primes of $\mathbb{Q}$ which ramify in a number field $K$. These are exactly the primes which appear in the unique factorization of $d_{K}$.

## 数学代写|代数数论代写Algebraic number theory代考|Arithmetic in Relative Extensions

F(X)=一个0+一个1X+⋯+一个n−1Xn−1+Xn

σ一世j(一个)=σ一世(一个0)+σ一世(一个1)τj(θ)+⋯+σ一世(一个米−1)τj(θ米−1)

## 数学代写|代数数论代写Algebraic number theory代考|Criterion for Ramification

\mathcal{O}^{\prime}=\left{\alpha \in K \mid \operatorname{Tr}_{K / k}(\alpha \mathcal{O}) \subseteq \mathfrak{o}\right }\mathcal{O}^{\prime}=\left{\alpha \in K \mid \operatorname{Tr}_{K / k}(\alpha \mathcal{O}) \subseteq \mathfrak{o}\right }

Tr⁡ķ/ķ(一个一个一世)=∑j=1n一个jTr⁡ķ/ķ(一个一世一个j)=b一世∈0

(Tr⁡ķ/ķ(一个一世一个j))(一个1 ⋮ 一个n)=(b1 ⋮ bn)

Dķ/ķ=○′−1被称为不同的ķ/ķ. 定义 4.14。理想dķ/ķ=ñķ/ķ(Dķ/ķ)的○被称为判别式ķ/ķ. 从定理 4.12 的证明中也可以清楚地看到以下内容。定理 4.15。理想○ķ/问由判别式生成dķ的ķ， 那是，○ķ/问=dķ从.

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