# 数学代写|数论作业代写number theory代考|The Discriminant of a Set of Elements in an Algebraic Number Field

#### Doug I. Jones

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## 数学代写|数论作业代写number theory代考|The Discriminant of a Set of Elements in an Algebraic Number Field

Let $K$ be an algebraic number field of degree $n$. Let $\omega_1, \omega_2, \ldots, \omega_n$ be any $n$ elements of $K$. An important quantity defined in terms of $\omega_1, \omega_2, \ldots, \omega_n$ and their conjugates relative to $K$ is the discriminant $D\left(\omega_1, \ldots, \omega_n\right)$. As we shall see the discriminant has some very nice properties. For example, $D\left(\omega_1, \ldots, \omega_n\right)$ is always a rational number, which is nonzero if and only if $\omega_1, \ldots, \omega_n$ are linearly independent over $\mathbb{Q}$. Moreover, if $\omega_1, \ldots, \omega_n$ are all algebraic integers then $D\left(\omega_1, \ldots, \omega_n\right)$ is a rational integer.

Definition 6.4.1 (Discriminant of $n$ elements in an algebraic number field of degree $n$ ) Let $K$ be an algebraic number field of degree $n$. Let $\omega_1, \ldots, \omega_n$ be $n$ elements of the field $K$. Let $\sigma_k(k=1,2, \ldots, n)$ denote the $n$ distinct monomorphisms $: K \longrightarrow \mathbb{C}$. For $i=1, \ldots, n$ let
$$\omega_i^{(1)}=\sigma_1\left(\omega_i\right)=\omega_i, \omega_i^{(2)}=\sigma_2\left(\omega_i\right), \ldots, \omega_i^{(n)}=\sigma_n\left(\omega_i\right)$$
denote the conjugates of $\omega_i$ relative to $K$. Then the discriminant of $\left{\omega_1, \ldots, \omega_n\right}$ is
$$D\left(\omega_1, \ldots, \omega_n\right)=\left|\begin{array}{cccc} \omega_1^{(1)} & \omega_2^{(1)} & \cdots & \omega_n^{(1)} \ \omega_1^{(2)} & \omega_2^{(2)} & \cdots & \omega_n^{(2)} \ \vdots & \vdots & \cdots & \vdots \ \omega_1^{(n)} & \omega_2^{(n)} & \cdots & \omega_n^{(n)} \end{array}\right|^2$$

## 数学代写|数论作业代写number theory代考|Basis of an Ideal

We now use our knowledge of the properties of the discriminant to show that every ideal in the ring $O_K$ of integers of an algebraic number field $K$ has a finite basis considered as an Abelian group, that is, as a $\mathbb{Z}$-module. Thus $O_K$ is Noetherian.
We first prove a preliminary result.
Theorem 6.5.1 Let $K$ be an algebraic number field with $[K: \mathbb{Q}]=n$. Let $I$ be $a$ nonzero ideal in $O_K$. Then there exist $\eta_1, \ldots, \eta_n \in I$ such that
$$D\left(\eta_1, \ldots, \eta_n\right) \neq 0$$
Proof: By Theorem 6.1.2 we have $K=\mathbb{Q}(\theta)$ for some $\theta \in O_K$. By Theorem 6.4.3 $D(\theta) \neq 0$. Further, by Theorem 6.1.7, as $I$ is a nonzero ideal of $O_K$, there exists $c \in I \cap \mathbb{Z}$ with $c \neq 0$. Hence, as $I$ is an ideal of $O_K$,
$$\eta_1=c, \eta_2=c \theta, \ldots, \eta_n=c \theta^{n-1} \in I$$
and
$$D\left(\eta_1, \ldots, \eta_n\right)=D\left(c, c \theta, \ldots, c \theta^{n-1}\right)=c^{2 n} D\left(1, \theta, \ldots, \theta^{n-1}\right)=c^{2 n} D(\theta) \neq 0 .$$
We are now in a position to prove that every ideal of the ring of integers of an algebraic number field has a finite basis.

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## 数学代写|数论作业代写number theory代考|The Discriminant of a Set of Elements in an Algebraic Number Field

$$\omega_i^{(1)}=\sigma_1\left(\omega_i\right)=\omega_i, \omega_i^{(2)}=\sigma_2\left(\omega_i\right), \ldots, \omega_i^{(n)}=\sigma_n\left(\omega_i\right)$$

$$D\left(\omega_1, \ldots, \omega_n\right)=\left|\begin{array}{cccc} \omega_1^{(1)} & \omega_2^{(1)} & \cdots & \omega_n^{(1)} \ \omega_1^{(2)} & \omega_2^{(2)} & \cdots & \omega_n^{(2)} \ \vdots & \vdots & \cdots & \vdots \ \omega_1^{(n)} & \omega_2^{(n)} & \cdots & \omega_n^{(n)} \end{array}\right|^2$$

## 数学代写|数论作业代写number theory代考|Basis of an Ideal

$$D\left(\eta_1, \ldots, \eta_n\right) \neq 0$$

$$\eta_1=c, \eta_2=c \theta, \ldots, \eta_n=c \theta^{n-1} \in I$$

$$D\left(\eta_1, \ldots, \eta_n\right)=D\left(c, c \theta, \ldots, c \theta^{n-1}\right)=c^{2 n} D\left(1, \theta, \ldots, \theta^{n-1}\right)=c^{2 n} D(\theta) \neq 0 .$$

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