Norm restricted to $\mathbb Q$ Consider an algebraic extension $K$ of $\mathbb Q$ and suppose that on $K$ we have a non trivial norm $||\cdot||$. Is it possible to have that the restriction $||\cdot||_{\mathbb Q}$ is the trivial norm over $\mathbb Q$?
A norm is trivial when $||x||=1$ for all $x\neq 0$.
Thank you in advance.
 A: This is not possible. Suppose $\|\cdot\|$ is such a norm on $K$. If it is archimedean, then by definition, there exists some $n\in\mathbb Z$ with $\|n\|>1$, so we can assume that $\|\cdot\|$ is non-archimediean, and satisfies the ultra-metric inequality: $\forall x,y\in K$,
$$\|x+y\|\le\max(\|x\|,\|y\|).$$
Let $\alpha\in K^\times$ be some element of non-trivial norm. Replacing $\alpha$ with $\alpha^{-1}$ if necessary, we can assume that $\|\alpha\| < 1$.
Since $K$ is an algebraic extension, $\alpha$ is the root of a polynomial in $f(X)\in \mathbb Q[X]$ - say $$f(X) = a_0+a_1X+\cdots + a^nX^n.$$
Then 
$$\begin{align}
1&=\|a_0\|&&\text{since $\|\cdot\|$ is trivial on $\mathbb Q$}\\
&=\|a_1\alpha+\cdots+a_n\alpha^n\|&&\text{since $f(\alpha) = 0$}\\
&\le \max(\|a_1\alpha\|,\ldots,\|a_n\alpha^n\|)&&\text{by the ultra-metric inequality}\\
&=\max(\|\alpha\|,\ldots,\|\alpha\|^n)&&\text{since $\|\cdot\|$ is trivial on $\mathbb Q$}\\
&=\|\alpha\|&&\text{since $\|\alpha\|<1$}\\
&<1,
\end{align}$$
which is absurd.
Note that the assumption that $K$ was an algebraic extension was crucial to the proof. Otherwise, for example, $\mathbb Q(t)$ has an absolute value given by $$\|f(x)\| = 2^{\deg(f)},$$which is trivial on $\mathbb Q$.
