# 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 ProofOfNormAndTraceOfAlgebraicNumber 2013-03-22 15:58:53 2013-03-22 15:58:53 Wkbj79 (1863) Wkbj79 (1863) 27 Wkbj79 (1863) Proof msc 11R04