Let $F$ be a local field, and $K/F$ a finite extension, then there is at most one norm on $K$ extending the norm on $F$.

Question

Let $F$ be a local field, that is to say, a field endowed with an absolute value $\lvert\cdot\rvert_F:F\to\mathbb{R}_{\ge0}$ making it into a locally compact topological space, and let $K/F$ be a finite field extension. Note that any two absolute values $\lvert\cdot\rvert_1$ and $\lvert\cdot\rvert_2$ on $K$ that extend $\lvert\cdot\rvert_F$ must be equivalent as norms, by local compactness. But how do we show that $\lvert\alpha\rvert_1 = \lvert\alpha\rvert_2$ for all $\alpha\in K$?

Answer 1

Here's a nice trick that I learnt from Koblitz's book on $p$-adic numbers. Suppose $|\alpha|_1<|\alpha|_2$. By equivalence of norms, there is $C>0$ such that $|x|_2\le C|x|_1$ for all $x$. But $C|\alpha^n|_1 =C(|\alpha|_1/|\alpha|_2)^n|\alpha^n|_2<|\alpha^n|_2$ for large enough $n$, contradiction.