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# 数学代写|代数数论代写Algebraic Number Theory代考|MATH661 The Norm and the Trace

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## 数学代写|代数数论代写Algebraic Number Theory代考|The Norm and the Trace

We begin by defining two important rational numbers associated with an element of an algebraic number field $K$. Recall that if $K$ is an algebraic number field, then $K$ can be viewed as a finite-dimensional vector space over $\mathbb{Q}$. Then if $\alpha \in K$, the map from $K$ to $K$ defined by $\Phi_\alpha: v \rightarrow \alpha v$ defines a linear operator on $K$. We define the trace of $\alpha$ by $\operatorname{Tr}K(\alpha):=\operatorname{Tr}\left(\Phi\alpha\right)$ and the norm of $\alpha$ by $\mathrm{N}K(\alpha):=\operatorname{det}\left(\Phi\alpha\right)$ (where $\operatorname{Tr}$ and det are the usual trace and determinant of a linear map). We sometimes also use the notation $\operatorname{Tr}{K / \mathbb{Q}}$ for $\operatorname{Tr}_K$ and $\mathrm{N}{K / \mathbb{Q}}$ for $\mathrm{N}_K$.

Thus, to find $\operatorname{Tr}K(\alpha)$, we choose any $\mathbb{Q}$-basis $\omega_1, \omega_2, \ldots, \omega_n$ of $K$ and write $$\alpha \omega_i=\sum a{i j} \omega_j \quad \forall i,$$
so $\operatorname{Tr}K(\alpha)=\operatorname{Tr} A$ and $\mathrm{N}_K(\alpha)=\operatorname{det} A$ where $A$ is the matrix $\left(a{i j}\right)$.
Lemma 4.1.1 If $K$ is an algebraic number field of degree $n$ over $\mathbb{Q}$, and $\alpha \in \mathcal{O}_K$ its ring of integers, then $\operatorname{Tr}_K(\alpha)$ and $\mathrm{N}_K(\alpha)$ are in $\mathbb{Z}$.

Proof. We begin by writing $\alpha \omega_i=\sum_{j=1}^n a_{i j} \omega_j \forall i$. Then we have
$$\alpha^{(k)} \omega_i^{(k)}=\sum_{j=1}^n a_{i j} \omega_j^{(k)} \quad \forall i, k,$$
where $\alpha^{(k)}$ is the $k$ th conjugate of $\alpha$. We rewrite the above by introducing the Kronecker delta function to get
$$\sum_{j=1}^n \delta_{j k} \alpha^{(j)} \omega_i^{(j)}=\sum_{j=1}^n a_{i j} \omega_j^{(k)},$$
where $\delta_{i j}=\left{\begin{array}{ll}0 & \text { if } i \neq j, \ 1 & \text { if } i=j .\end{array}\right.$ Now, if we define the matrices
$$A_0=\left(\alpha^{(i)} \delta_{i j}\right), \quad \Omega=\left(\omega_i^{(j)}\right), \quad A=\left(a_{i j}\right),$$

## 数学代写|代数数论代写Algebraic Number Theory代考|Existence of an Integral Basis

Let $K$ be an algebraic number field of degree $n$ over $\mathbb{Q}$, and $\mathcal{O}K$ its ring of integers. We say that $\omega_1, \omega_2, \ldots, \omega_n$ is an integral basis for $K$ if $\omega_i \in \mathcal{O}_K$ for all $i$, and $\mathcal{O}_K=\mathbb{Z} \omega_1+\mathbb{Z} \omega_2+\cdots+\mathbb{Z} \omega_n$. Exercise 4.2.1 Show that $\exists \omega_1^, \omega_2^, \ldots, \omega_n^* \in K$ such that
$$\mathcal{O}_K \subseteq \mathbb{Z} \omega_1^+\mathbb{Z} \omega_2^+\cdots+\mathbb{Z} \omega_n^* \text {. }$$
Theorem 4.2.2 Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be a set of generators for a finitely generated $\mathbb{Z}$-module $M$, and let $N$ be a submodule.
(a) $\exists \beta_1, \beta_2, \ldots, \beta_m$ in $N$ with $m \leq n$ such that
$$N=\mathbb{Z} \beta_1+\mathbb{Z} \beta_2+\cdots+\mathbb{Z} \beta_m$$
and $\beta_i=\sum{j \geq i} p_{i j} \alpha_j$ with $1 \leq i \leq m$ and $p_{i j} \in \mathbb{Z}$
(b) If $m=n$, then $[M: N]=p_{11} p_{22} \cdots p_{n n}$.

Proof. (a) We will proceed by induction on the number of generators of a $\mathbb{Z}$-module. This is trivial when $n=0$. We can assume that we have proved the above statement to be true for all $\mathbb{Z}$-modules with $n-1$ or fewer generators, and proceed to prove it for $n$. We define $M^{\prime}$ to be the submodule generated by $\alpha_2, \alpha_3, \ldots, \alpha_n$ over $\mathbb{Z}$, and define $N^{\prime}$ to be $N \cap M^{\prime}$. Now, if $n=1$, then $M^{\prime}=0$ and there is nothing to prove. If $N=N^{\prime}$, then the statement is true by our induction hypothesis.

So we assume that $N \neq N^{\prime}$ and consider $A$, the set of all integers $k$ such that $\exists k_2, k_3, \ldots, k_n$ with $k \alpha_1+k_2 \alpha_2+\cdots+k_n \alpha_n \in N$. Since $N$ is a submodule, we deduce that $A$ is a subgroup of $\mathbb{Z}$. All additive subgroups of $\mathbb{Z}$ are of the form $m \mathbb{Z}$ for some integer $m$, and so $A=k_{11} \mathbb{Z}$ for some $k_{11}$. Then let $\beta_1=k_{11} \alpha_1+k_{12} \alpha_2+\cdots+k_{1 n} \alpha_n \in N$. If we have some $\alpha \in N$, then
$$\alpha=\sum_{i=1}^n h_i \alpha_i$$
with $h_i \in \mathbb{Z}$ and $h_1 \in A$ so $h_1=a k_{11}$. Therefore, $\alpha-a \beta_1 \in N^{\prime}$. By the induction hypothesis, there exist
$$\beta_i=\sum_{j \geq i} k_{i j} \alpha_j,$$
$i=2,3 \ldots, m$, which generate $N^{\prime}$ over $\mathbb{Z}$ and which satisfy all the conditions above. It is clear that adding $\beta_1$ to this list gives us a set of generators of $N$.

## 数学代写|代数数论代写Algebraic Number Theory代考|The Norm and the Trace

$$\alpha^{(k)} \omega_i^{(k)}=\sum_{j=1}^n a_{i j} \omega_j^{(k)} \quad \forall i, k,$$

$$\sum_{j=1}^n \delta_{j k} \alpha^{(j)} \omega_i^{(j)}=\sum_{j=1}^n a_{i j} \omega_j^{(k)},$$

\alpha=\sum_{i=1}^n h_i \alpha_i
$$和 h_i \in \mathbb{Z} 和 h_1 \in A 所以 h_1=a k_{11}. 所以， \alpha-a \beta_1 \in N^{\prime}. 根据纳纳假设， 存在$$
\beta_i=\sum_{j \geq i} k_{i j} \alpha_j,

$i=2,3 \ldots, m_{｝ \text { 生成 } N^{\prime} \text { 超过 } \mathbb{Z} \text { 并且满足以上所有条件。很明显, 浾加 } \beta_1 \text { 这个列表给了我们一组生成器 } N \text {. }$

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