There is a general result which holds for the rational numbers $ \mathbb Q $ (as well as number fields in general):



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The nontrivial part is to prove that the norm thus defined satisfies the triangle inequality. $ K $ is either $ \mathbb R $ or $ \mathbb Q\_p $ for a prime $ p $, and for the latter case one can argue using Hensel's lemma that there is an equivalence between having norm $ \leq 1 $ and being integral over the $ p $-adic integers, which combined with the ultrametric property of the $ p $-adic norm is sufficient to prove the claim. I'm satisfied with this argument in the sense that it seems to give a "moral" explanation for why we should expect the claim to be true.


However, the argument for $ K = \mathbb R $ is simply to note that the claim is obviously true for the only nontrivial finite extension of $ \mathbb R $, which is $ \mathbb C/\mathbb R $. While this clearly is sufficient for a proof, I don't understand why one should expect this to be true in advance. Is there a better motivation for why we should expect this result to hold, perhaps an argument which treats all completions symmetrically instead of having one argument for the finite primes and another for the prime at infinity?



Why does the field norm on the field extension $ \mathbb C/\mathbb R $ induce a vector space norm?
The map $|N(\cdot)|^{1/n}$ is a continuous multiplicative extension of $|\cdot|$.
By a multiplicative function I mean a function $\chi:L\to [0,\infty)$
such that $\chi(0)=0$, $\chi(1)=1$ and for every $x,y\in L$, $\chi(xy)=\chi(x)\chi(y)$.
A multiplicative function which satisfies for every $x,y\in L$, $\chi(x+y)\leq \chi(x)+\chi(y)$ is called an absolute value.


**Theorem:** Let $K$ be a field and $|\cdot|$ an absolute value on $K$.
Assume $K$ is complete with respect to the induced metric.
Then for every finite field extension of $K$, every continuous multiplicative extension of $|\cdot|$ is an absolute value.
In fact, there exists a unique such continuous multiplicative extension of $|\cdot|$, which is $|N(\cdot)|^{1/n}$.


The uniqueness follows at once from the fact that all norms are equivalent on finite dimensional vector spaces over complete fields: for multiplicative functions $\chi\_1,\chi\_2$, the function $\chi\_1/\chi\_2$ (defined on the multiplicative group) is multiplicative too, hence must be unbounded or trivial.
The less trivial part of the theorem is its first part.


In my other answer I gave a proof of this fact which indeed holds in the generality of complete fields. This post is to give an easy proof I found, under the assumption that the fields are locally compact.


We consider a local field $K$, endowed with an absolute value $|\cdot|$,
a finite field extension $L$ of $K$ and a continuous multiplicative extension $\chi$ of $|\cdot|$. We argue to show that $\chi$ is an absolute value on $L$.


Consider $L$ as a $K$-vector space and consider the corresponding space $\Omega$ consisting of $K$-norms on $L$.
Consider the multiplicative group $L^\*$ as a locally compact group (see below for justification) and let it act on $\Omega$ by $\|\cdot\|\mapsto \chi(x)^{-1} \|x\cdot\|$
for $x\in L^\*$. Note that for $x\in K^\*$,
$$ \chi(x)^{-1} \|x\cdot\|=\chi(x)^{-1} |x|\|\cdot\|=\|\cdot\|,$$
thus the $L^\*$-action on $\Omega$ factors via $L^\*/K^\*$,
which is a compact group, as it is homeomorphic to a projective space.
This action admits a fixed point. Indeed, for every norm $\|\cdot\|\in \Omega$,
the map
$$ L \ni v \mapsto \int\_{L^\*/K^\*} \chi(x)^{-1}\|xv\|~ \text{dHaar}\_{L^\*/K^\*}(x) \in [0,\infty)$$
is easily seen to be an $L^\*$-fixed norm on $L$.
We let $\|\cdot\|$ be such a fixed point which is normalized to satisfy $\|1\|=1$. Then for every $x\in L^\*$,
$$ \|x\|=\|x\cdot 1\|=\chi(x)\cdot \chi(x)^{-1}\|x\cdot 1\|=
\chi(x)\cdot \|1\|=\chi(x). $$
Thus $\|\cdot\|=\chi$ and in particular, we conclude that $\chi$ is indeed a norm.
This finishes the proof.




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To see that $L^\*$ is a topological group we need to verify the continuity of the inversion map $x\mapsto x^{-1}$ on $L^\*$.
Its continuity on $K^\*$ follows by a standard argument from the existence of the absolute value $|\cdot|$.
From the continuity of the inversion on $K^\*$ we get its continuity on $\text{GL}\_n(K)$, as inversion is polynomial in the matrix entries and $\det(\cdot)^{-1}$, thus also its continuity on $L^\*$.


