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proof of norm and trace of algebraic number
Before proving this theorem, a lemma will be stated and proven.
Note also that each of the $d$ embeddings of $\mathbb{Q}(\alpha)$ into $\mathbb{C}$ extends to exactly $\displaystyle \frac{n}{d}$ embeddings of $K$ into $\mathbb{C}$ . Thus,
and
Now, the above theorem will be proven.
Proof of theorem 1. Let $f(x) \in \mathbb{Q}[x]$ be the minimal polynomial for $\alpha$ over $\mathbb{Q}$ . Then $\operatorname{deg} f=d$ , where $d$ is as in the previous lemma. Note that $|N^*(\alpha)|$ is equal to the absolute value of the constant term of $f$ and that $T^*(\alpha)$ is equal to the opposite of the coefficient of $x^{d-1}$ of $f$ . Thus, $N^*(\alpha), T^*(\alpha) \in \mathbb{Q}$ . Therefore, $\displaystyle N(\alpha)=(N^*(\alpha))^{\frac{n}{d}} \in \mathbb{Q}$ and $\displaystyle T(\alpha)=\frac{n}{d}T^*(\alpha) \in \mathbb{Q}$ . Moreover, if $\alpha$ is an algebraic integer, then $f(x) \in \mathbb{Z}[x]$ , $N^*(\alpha), T^*(\alpha) \in \mathbb{Z}$ , $\displaystyle N(\alpha)=(N^*(\alpha))^{\frac{n}{d}} \in \mathbb{Z}$ , and $\displaystyle T(\alpha)=\frac{n}{d}T^*(\alpha) \in \mathbb{Z}$ .
If $a \in \mathbb{Q}$ , then $d=1$ , $N(a)=(N^*(a))^n=a^n$ , and $T(a)=nT^*(a)=na$ .
Finally, if $\alpha, \beta \in K$ , then
and
$\qedsymbol$
If $N^*(\varepsilon) = \pm 1$ , then let $f(x) \in \mathbb{Z}[x]$ be the minimal polynomial of $\varepsilon$ over $\mathbb{Q}$ . Let $a_1, \cdots , a_{d-1} \in \mathbb{Z}$ such that $\displaystyle f(x)=x^d+\sum_{j=1}^{d-1} a_j x^j \pm 1$ . Then $\displaystyle 0=f(\varepsilon)=\varepsilon^d+\sum_{j=1}^{d-1} a_j \varepsilon^j \pm 1$ . Thus, $\displaystyle \varepsilon \left( \varepsilon^{d-1}+\sum_{j=1}^{d-1} a_j \varepsilon^{j-1} \right) = \pm 1$ . Since $\displaystyle \varepsilon^{d-1}+\sum_{j=1}^{d-1} a_j \varepsilon^{j-1} \in \mathcal{O}_K$ , it follows that $\varepsilon$ is a unit in $\mathcal{O}_K$ .
Conversely, let $\varepsilon$ be a unit in $\mathcal{O}_K$ . Let $\upsilon \in \mathcal{O}_K$ with $\varepsilon \upsilon = 1$ . Since $N^*(\varepsilon) N^*(\upsilon) = N^*(\varepsilon \upsilon)=N^*(1)=1$ and $N^*(\varepsilon), N^*(\upsilon) \in \mathbb{Z}$ , it follows that $N^*(\varepsilon) = \pm 1$ . 
Bibliography
- 1
- Marcus, Daniel A. Number Fields. New York: Springer-Verlag, 1977.