ProofOfNormAndTraceOfAlgebraicNumber 2006-06-10 18:34:56 2008-02-26 01:14:23 Proof norm and trace of algebraic number 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem*{lem*}{Lemma} \newtheorem{thm}{Theorem} \begin{thm} \, 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 $T(a) = na$. \end{thm} Before proving this theorem, a lemma will be stated and proven. \begin{lem*} \, 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 \PMlinkname{absolute norm}{AbsoluteNorm} and \PMlinkname{absolute trace}{AbsoluteTrace} of $\alpha$, respectively. \, Then $d$ divides $n$, $\displaystyle N(\alpha)=(N^*(\alpha))^{\frac{n}{d}},$ and $\displaystyle T(\alpha)=\frac{n}{d}T^*(\alpha)$. \end{lem*} \begin{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{center} $\begin{array}{ll} \displaystyle N(\alpha) & \displaystyle =\prod_{\sigma \text{ emb. of }K} \sigma(\alpha) \\ \\ & \displaystyle =\prod_{\sigma \text{ emb. of }K} \left. \sigma \right|_{\mathbb{Q}(\alpha)} (\alpha) \\ \\ & \displaystyle =\left( \prod_{\tau \text{ emb. of }\mathbb{Q}(\alpha)} \tau(\alpha) \right)^{\frac{n}{d}} \\ \\ & \displaystyle =\left(N^*(\alpha) \right)^{\frac{n}{d}} \\ \\ \end{array}$ \end{center} and \begin{center} $\begin{array}{ll} \\ \displaystyle T(\alpha) & \displaystyle =\sum_{\sigma \text{ emb. of }K} \sigma(\alpha) \\ \\ & \displaystyle =\sum_{\sigma \text{ emb. of }K} \left. \sigma \right|_{\mathbb{Q}(\alpha)} (\alpha) \\ \\ & \displaystyle =\frac{n}{d} \sum_{\tau \text{ emb. of }\mathbb{Q}(\alpha)} \tau(\alpha) \\ \\ & \displaystyle =\frac{n}{d} T^*(\alpha). \end{array}$ \end{center} \end{proof} Now, the above theorem will be proven. \emph{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{center} $\begin{array}{ll} \displaystyle N(\alpha \beta) & \displaystyle =\prod_{\sigma \text{ emb. of }K} \sigma(\alpha \beta) \\ \\ & \displaystyle =\prod_{\sigma \text{ emb. of }K} \sigma(\alpha) \sigma(\beta) \\ \\ & \displaystyle =\left( \prod_{\sigma \text{ emb. of }K} \sigma(\alpha) \right) \left( \prod_{\sigma \text{ emb. of }K} \sigma(\beta) \right) \\ \\ & \displaystyle =N(\alpha)N(\beta) \\ \\ \end{array}$ \end{center} and \begin{center} $\begin{array}{ll} \\ \displaystyle T(\alpha + \beta) & \displaystyle =\sum_{\sigma \text{ emb. of }K} \sigma(\alpha + \beta) \\ \\ & \displaystyle =\sum_{\sigma \text{ emb. of }K} \sigma(\alpha)+\sigma(\beta) \\ \\ & \displaystyle =\prod_{\sigma \text{ emb. of }K} \sigma(\alpha)+\sum_{\sigma \text{ emb. of }K} \sigma(\beta) \\ \\ & \displaystyle =T(\alpha)+T(\beta). \end{array}$ \end{center} $\qedsymbol$ \begin{thm} An algebraic integer $\varepsilon$ is a unit if and only if its \PMlinkescapetext{absolute norm} $N^*(\varepsilon)=\pm 1,$. \, Thus, \PMlinkescapetext{the constant term} in the minimal polynomial of an algebraic unit is always \,$\pm 1$. \end{thm} \begin{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 $N^*(\varepsilon) = \pm 1$. \end{proof} \begin{thebibliography}{9} \bibitem{marcus} Marcus, Daniel A. {\em Number Fields}. New York: Springer-Verlag, 1977. \end{thebibliography}