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Homeproof of norm and trace of algebraic number

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# 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 and absolute trace 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 absolute norm $N^{*}(\varepsilon)=\pm 1,$. Thus, the constant term 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.

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## Comments

## TeX problem

In the middle of this entry, I have the code:

\begin{proof}[Proof of theorem 1]

According to the PM Content and Style Guide, this is supposed to make "Proof of theorem 1." appear italicized, which is what I want. Instead, what is appearing is "Proof." italicized, then "[Proof of theorem 1]". How do I fix this?

## Re: TeX problem

It looks ok in page images mode, so maybe it's a PM problem. Perhaps rerendering might help.

## Re: TeX problem

Rerendering doesn't help, but it looks the way that I want it to in page images mode. Hmmm....

## Re: TeX problem

I forced the "Proof of theorem 1" to appear the way that I want it to, but now, in html mode only, there's some gibberish that appears just before the bibliography. Any ideas on how to fix this?

## Re: TeX problem

I don't know, but you might try removing the empty line between the last line of your proof and \end{proof}. At least this would mean the square would appear in the right place, and it might even do something to the weird text.

## Re: TeX problem

It put the square in the right spot (I didn't even notice that!), but the weird text is still there. :-( Any other advice?

## Eureka! QED

I finally figured it out! Since I made the "Proof of theorem 1" outside of the proof environment, I had put a \qed at the end of the proof. In order to avoid the "ARRAY&$!*#" stuff, I had to put \qed in math mode.

In any case, I'm so happy it's fixed now! Thanks silverfish for giving me suggestions.

$\qed$

Warren