proof of norm and trace of algebraic number

Theorem 1.

Let $K$ be a number field and $\alpha\in K$ . The norm $N(\alpha)$ and the trace $T(\alpha)$ of $\alpha$ in the field extension $K/\mathbb{Q}$ both are rational numbers and especially rational integers in the case $\alpha$ is an algebraic integer. If $\beta$ is another element of $K$ , then $N(\alpha\beta)=N(\alpha)N(\beta)$ and $T(\alpha+\beta)=T(\alpha)+T(\beta)$ . If $[K\!:\!\mathbb{Q}]=n$ and $a\in\mathbb{Q}$ , then $N(a)=a^{n}$ and $T(a)=na$ .

Before proving this theorem, a lemma will be stated and proven.

Lemma.

Let $K$ be a number field with $[K\!:\!\mathbb{Q}]=n$ , $\alpha\in K$ such that $[\mathbb{Q}(\alpha)\!:\!\mathbb{Q}]=d$ , and $N^{*}(\alpha)$ and $T^{*}(\alpha)$ denote the absolute norm (http://planetmath.org/AbsoluteNorm) and absolute trace (http://planetmath.org/AbsoluteTrace) of $\alpha$ , respectively. Then $d$ divides $n$ , $\displaystyle N(\alpha)=(N^{*}(\alpha))^{\frac{n}{d}},$ and $\displaystyle T(\alpha)=\frac{n}{d}T^{*}(\alpha)$ .

Proof.

Note that $d$ divides $n$ because $n=[K\!:\!\mathbb{Q}]=[K\!:\!\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha)\!:\!\mathbb% {Q}]=[K\!:\!\mathbb{Q}(\alpha)]d$ .

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,

$\begin{array}[]{ll}\displaystyle N(\alpha)&\displaystyle=\prod_{\sigma\text{ % emb. of }K}\sigma(\alpha)\\ \\ &\displaystyle=\prod_{\sigma\text{ emb. of }K}\left.\sigma\right|_{\mathbb{Q}(% \alpha)}(\alpha)\\ \\ &\displaystyle=\left(\prod_{\tau\text{ emb. of }\mathbb{Q}(\alpha)}\tau(\alpha% )\right)^{\frac{n}{d}}\\ \\ &\displaystyle=\left(N^{*}(\alpha)\right)^{\frac{n}{d}}\\ \\ \end{array}$

and

$\begin{array}[]{ll}\\ \displaystyle T(\alpha)&\displaystyle=\sum_{\sigma\text{ emb. of }K}\sigma(% \alpha)\\ \\ &\displaystyle=\sum_{\sigma\text{ emb. of }K}\left.\sigma\right|_{\mathbb{Q}(% \alpha)}(\alpha)\\ \\ &\displaystyle=\frac{n}{d}\sum_{\tau\text{ emb. of }\mathbb{Q}(\alpha)}\tau(% \alpha)\\ \\ &\displaystyle=\frac{n}{d}T^{*}(\alpha).\end{array}$

∎

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

$\begin{array}[]{ll}\displaystyle N(\alpha\beta)&\displaystyle=\prod_{\sigma% \text{ emb. of }K}\sigma(\alpha\beta)\\ \\ &\displaystyle=\prod_{\sigma\text{ emb. of }K}\sigma(\alpha)\sigma(\beta)\\ \\ &\displaystyle=\left(\prod_{\sigma\text{ emb. of }K}\sigma(\alpha)\right)\left% (\prod_{\sigma\text{ emb. of }K}\sigma(\beta)\right)\\ \\ &\displaystyle=N(\alpha)N(\beta)\\ \\ \end{array}$

and

$\begin{array}[]{ll}\\ \displaystyle T(\alpha+\beta)&\displaystyle=\sum_{\sigma\text{ emb. of }K}% \sigma(\alpha+\beta)\\ \\ &\displaystyle=\sum_{\sigma\text{ emb. of }K}\sigma(\alpha)+\sigma(\beta)\\ \\ &\displaystyle=\prod_{\sigma\text{ emb. of }K}\sigma(\alpha)+\sum_{\sigma\text% { emb. of }K}\sigma(\beta)\\ \\ &\displaystyle=T(\alpha)+T(\beta).\end{array}$

$\qed$

Theorem 2.

An algebraic integer $\varepsilon$ is a unit if and only if its $N^{*}(\varepsilon)=\pm 1,$ . Thus, in the minimal polynomial of an algebraic unit is always $\pm 1$ .

Proof.

Let $K=\mathbb{Q}(\varepsilon)$ . Since $\varepsilon$ is an algebraic integer, $d=[K\!:\!\mathbb{Q}]$ is finite. Let $\mathcal{O}_{K}$ denote the ring of integers of $K$ .

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 $N^{*}(\varepsilon)=\pm 1$ . ∎

References

1 Marcus, Daniel A. Number Fields. New York: Springer-Verlag, 1977.

Title	proof of norm and trace of algebraic number
Canonical name	ProofOfNormAndTraceOfAlgebraicNumber
Date of creation	2013-03-22 15:58:53
Last modified on	2013-03-22 15:58:53
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	27
Author	Wkbj79 (1863)
Entry type	Proof
Classification	msc 11R04