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# 数学代写|数论代写Number Theory代考|MATH25 Integral Basis and Discriminant of Cyclotomic Fields

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## 数学代写|数论代写Number Theory代考|Integral Basis and Discriminant of Cyclotomic Fields

We first find the discriminant and an integral basis of cyclotomic fields generated by $p$ th root of unity for a prime $p$. Recall that if $\zeta$ is a primitive $n$th root of unity, then the degree of the $n$th cyclotomic field $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ is $\phi(n)$ (cf. Theorem A.40).
Theorem 2.25 Let $\zeta$ be a primitive pth root of unity, p an odd prime. Then ${1, \zeta, \ldots$, $\left.\zeta^{p-2}\right}$ is an integral basis of $K=\mathbb{Q}(\zeta)$ and $d_K=(-1)^{\frac{p-1}{2}} p^{p-2}$.

Proof Let $\Phi(X)$ denote the minimal polynomial of $\zeta$ over $\mathbb{Q}$ of degree $p-1$. Using Lemma $2.6$ and Corollary 2.16, we have
$$D_{K / \mathbb{Q}}\left(1, \zeta, \ldots, \zeta^{p-2}\right)=(-1)^{\frac{p-1}{2}} N_{K / \mathbb{Q}}\left(\Phi^{\prime}(\zeta)\right)=(\text { ind } \zeta)^2 d_K .$$
So the theorem is proved in view of above equation once it is shown that $N_{K / \mathbb{Q}}\left(\Phi^{\prime}(\zeta)\right)$ equals $p^{p-2}$ and $p$ does not divide ind $\zeta$. The polynomial $\Phi(X)$ satisfies the relation
$$\Phi(X)(X-1)=X^p-1 .$$
On differentiating both sides of above equation with respect to $X$ and then substituting $X=\zeta$, we see that
$$\Phi^{\prime}(\zeta)=\frac{p}{(\zeta-1) \zeta} .$$

## 数学代写|数论代写Number Theory代考|An Algorithm for Computing Integral Basis

We now prove a theorem which gives an algorithm for computing integral bases of algebraic number fields.

Theorem 2.34 Let $K=\mathbb{Q}(\theta)$ be an algebraic number field of degree $n$ with $\theta$ an algebraic integer. Then $K$ has an integral basis of the form $\left{\xi_0, \xi_1, \ldots, \xi_{n-1}\right}$, where
$$\xi_0=1 \text { and } \xi_i=\frac{c_{i 0}+c_{i 1} \theta+\cdots+c_{i(i-1)} \theta^{i-1}+\theta^i}{d_i}$$
with $c_{i j}, d_i$ belonging to $\mathbb{Z}$ and $d_i>0$ dividing $d_{i+1}$ for $1 \leq i \leq n-1$. The numbers $d_i$ are uniquely determined by $\theta$. In particular, $\left[\mathcal{O}K: \mathbb{Z}[\theta]\right]=\prod{i=1}^{n-1} d_i$.

Proof Let $r$ denote the index of $\mathbb{Z}[\theta]$ in $\mathcal{O}K$. Then by Lagrange’s theorem for finite groups $r \mathcal{O}_K \subset Z[\theta]$. So $\mathcal{O}_K$ is a subgroup of the free abelian group $\frac{1}{r} \mathbb{Z}[\theta]$ which has basis $\left{\frac{1}{r}, \frac{\theta}{r}, \ldots, \frac{\theta^{n-1}}{r}\right}$. By Lemma $2.12$ and Remark 2.13, there exists a $\mathbb{Z}$ – basis of $\mathcal{O}_K$ of the type $\left{\xi_0, \xi_1, \ldots, \xi{n-1}\right}$, where for $0 \leq i \leq n-1$
$$\xi_i=\frac{a_{i 0}+a_{i 1} \theta+\cdots+a_{i i} \theta^i}{r},$$
with $a_{i j}$ belonging to $\mathbb{Z}$ and $a_{i i}>0$ for each $i$. Observe that the positive rational number $\xi_0=\frac{a_{00}}{r}$ being an element of a $\mathbb{Z}$-basis of $\mathcal{O}_K$ must be 1 .

We claim that $a_{i i}$ divides $r$ as well as $a_{i j}$ when $0 \leq j<i, 1 \leq i \leq n-1$ and that $d_i$ divides $d_{i+1}$ for $1 \leq i \leq n-1$ where $d_i=\frac{r}{a_{i i}}$. The claim together with the fact that $\xi_0=1$ immediately shows that $\left{\xi_0, \ldots, \xi_{n-1}\right}$ is an integral basis of $K$ of the desired form once we set $c_{i j}=\frac{a_{i j}}{a_{i j}}$ and $d_i=\frac{r}{a_{i j}}, 0 \leq j<i, 1 \leq i \leq n-1$.

## 数学代写|数论代写数论代考|回旋场的积分基与判别

$$D_{K / \mathbb{Q}}\left(1, \zeta, \ldots, \zeta^{p-2}\right)=(-1)^{\frac{p-1}{2}} N_{K / \mathbb{Q}}\left(\Phi^{\prime}(\zeta)\right)=(\text { ind } \zeta)^2 d_K .$$

$$\Phi(X)(X-1)=X^p-1 .$$

$$\Phi^{\prime}(\zeta)=\frac{p}{(\zeta-1) \zeta} .$$

## 数学代写|数论代写数论代考|一种计算积分基的算法

$$\xi_0=1 \text { and } \xi_i=\frac{c_{i 0}+c_{i 1} \theta+\cdots+c_{i(i-1)} \theta^{i-1}+\theta^i}{d_i}$$
， $c_{i j}, d_i$属于$\mathbb{Z}$, $d_i>0$除$d_{i+1}$为$1 \leq i \leq n-1$。数字$d_i$是由$\theta$唯一确定的。特别是$\left[\mathcal{O}K: \mathbb{Z}[\theta]\right]=\prod{i=1}^{n-1} d_i$ .

$$\xi_i=\frac{a_{i 0}+a_{i 1} \theta+\cdots+a_{i i} \theta^i}{r},$$
with $a_{i j}$ 属于 $\mathbb{Z}$ 和 $a_{i i}>0$ 对于每一个 $i$。观察正有理数 $\xi_0=\frac{a_{00}}{r}$ 的一个元素 $\mathbb{Z}$-基础 $\mathcal{O}_K$

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