# Is the Euclidean norm of an algebraic number contained an a different extension?

I know that for every $\alpha \in \mathbb{Q}$, the Euclidean norm $\|\alpha\|=(\alpha\,\bar{\alpha})^{1/2} = \alpha\ge0$ where $\bar{\alpha}$ is the complex conjugate of $\alpha$. An obvious observation is that, $0\le\|\alpha\|\in\mathbb{Q}$. Does this observation carry over to any algebraic number field extension $K$ of $\mathbb{Q}$? That is, if $K=\mathbb{Q}(\vartheta)$ where $\vartheta$ is some algebraic number, then is it true that $\forall\gamma\in K, \|\gamma\| = (\gamma\,\bar{\gamma})^{1/2}\in K$? If not please give a counter-example and in addition give a constructive proof on how one can obtain a minimal extension $L$ of $K$ that contains $\|\gamma\|$.

For your first question, consider $K=\mathbb{Q}(i)$. $1+i\in K$, but $||1+i||=\sqrt{2}\not\in K$.
I have to make a slight guess about what you are asking next. Are you saying that you want a minimal extension such that $||\gamma||\in L$ for all $\gamma\in K$? This might be difficult! Take the same example. All primes of the form $p\equiv 1\mod 4$ occur as a norm in $K$. That means $L$ will be an infinite extension of $K$!
• Oops my bad. I meant to say a minimal extension $L$ of $\mathbb{Q}$ and not $K$. I hope that is clear now. – saint Apr 20 '17 at 13:16
• This example would still give an infinite extension of $\mathbb{Q}$, but it is not suggestive of a general approach. – gobucksmath Apr 20 '17 at 13:21
• Note that by the Primitive Elemement Theorem the problem is solved. For instance, consider the example given by @gobucksmath above with $K=\mathbb{Q}(i)$ and minimal polynomial $p(x)=x^{2}+1$. Note that $\sqrt{2} \notin K$. Thus we have the minimal polynomial $q(x)=x^{2}-2$ and the number field $L=\mathbb{Q}/\langle q(x)\rangle$. By the primitive element theorem, one can find a extension containing containing both K and L. – saint Apr 20 '17 at 14:10