= Solution
Let $e_1,\ldots,e_d$ be a $\mathbb Q_p$-basis of $K$, and define the coordinate norm
$$
\left\|\sum_j x_je_j\right\|_0=\max_j|x_j|_p.
$$
First an extending <absolute value on a field> is automatically non-Archimedean. Its values on integers are at most one, and the binomial theorem gives
$$
|x+y|^m\leq(m+1)\max(|x|,|y|)^m.
$$
Taking $m$th roots and letting $m\to\infty$ proves the ultrametric inequality.
Consequently
$$
|x|\leq C\|x\|_0,\qquad C=\max_j|e_j|.
$$
This bound makes $x\mapsto|x|$ continuous for the coordinate topology. The coordinate sphere $S=\{x:\|x\|_0=1\}$ is a closed subset of $\mathbb Z_p^d$, hence compact. It contains no zero, so the continuous positive function $|x|$ attains a positive minimum $c$ there. Scaling any nonzero vector by a power of $p$ to put it in $S$ proves
$$
\boxed{c\|x\|_0\leq|x|\leq C\|x\|_0.}
$$
This is the <compact-sphere proof of finite-dimensional non-Archimedean norm equivalence>.
A Cauchy sequence for $|\cdot|$ is therefore Cauchy in each coordinate. The completeness of $\mathbb Q_p$ gives a coordinate limit, and the upper bound gives convergence to it for $|\cdot|$. Thus \b[every extending absolute value makes $K$ complete].
For uniqueness, if $|\cdot|_1$ and $|\cdot|_2$ are two extensions, their respective bounds give constants $A,B>0$ with
$$
A|x|_1\leq|x|_2\leq B|x|_1.
$$
Apply this to $x^m$ and use multiplicativity:
$$
A^{1/m}|x|_1\leq|x|_2\leq B^{1/m}|x|_1.
$$
Letting $m\to\infty$ proves $\boxed{|x|_1=|x|_2}$ for every $x$. This proves the <unique extension of an absolute value to a finite extension> without assuming that an arbitrary coordinate norm is multiplicative.
Back to article page
