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Any extending absolute value is also non-Archimedean. Indeed its values on integers are the given base-field values, bounded by one, so the integer criterion for a non-Archimedean absolute value applies. For the criterion's sufficiency, the binomial theorem gives $|x+y{|}^n\le (n+1)\max (|x|,|y|{)}^n$; taking $n$th roots yields the ultrametric inequality.

Fix a $K$-basis of $L$ and its coordinate maximum norm. Every extending absolute value is a vector-space norm satisfying $\Vert av\Vert =|a|\Vert v\Vert$. The finite-dimensional non-Archimedean norm equivalence over a complete field makes it equivalent to that coordinate norm. Here is the relevant proof: induct on dimension. The span $W$ of all but the last basis vector $e$ is complete by the induction hypothesis, hence closed in the given norm. Therefore $\delta ={\inf }_{w\in W}\Vert e-w\Vert >0$. For $v=w+ae$, one has $|a|\le \Vert v\Vert /\delta$ and $\Vert w\Vert \le \max (\Vert v\Vert ,|a|\Vert e\Vert )$. The induction bound on $W$ bounds all its coordinates by a constant times $\Vert v\Vert$. The reverse bound follows from the triangle inequality. This proof applies to every complete base field here, without a local compactness assumption.

Thus two extending absolute values satisfy $c|x{|}_1\le |x{|}_2\le C|x{|}_1$ for fixed positive constants. Apply this to $x^n$ and take $n$th roots:Letting $n\to \infty$ proves equality. The bounded comparison forces equality of multiplicative absolute values argument gives the required unique extension of an absolute value to a finite extension, without proving existence.

An extending absolute value on a field is non-Archimedean by Question 1(a). We prove finite-dimensional non-Archimedean norm equivalence over a complete field, using completeness rather than local compactness.

Fix a basis $e_1,\dots ,e_d$ of a finite-dimensional vector space over $K$, and let $\Vert \cdot \Vert$ be a norm homogeneous for the given absolute value on a field and satisfying the ultrametric inequality. Write $\Vert \sum a_i{e}_i{\Vert }_0={\max }_i|a_i|$. The upper boundis immediate. For the lower bound use induction on $d$. The one-dimensional case is homogeneity. For the induction step, the span $W$ of $e_1,\dots ,e_{d-1}$ is complete for its restricted norm by the induction comparison with the coordinate norm and completeness of $K$. A complete vector subspace of a metric space is closed, soFor $v=w+a{e}_d$ with $a\ne 0$, scaling by $a$ gives $\Vert v\Vert \ge |a|\delta$; the same coefficient bound is trivial when $a=0$. AlsoThe induction bound controls every coordinate of $w$ by a constant times $\Vert w\Vert$. Together with the bound on $a$, this proves $\Vert v{\Vert }_0\le C^{\prime }\Vert v\Vert$.

Two extending absolute values on a field on $L$ are norms of this kind on its finite-dimensional vector space over $K$. Hence there is $B>0$ with $|x{|}_2\le B|x{|}_1$ for every $x\in L$. Apply this to $x^N$ and use multiplicativity:Letting $N\to \infty$ gives $|x{|}_2\le |x{|}_1$. Interchanging the two norms gives equality. ThusThe proof covers the trivial base absolute value on a field too. It proves uniqueness, without assuming an existence theorem or compactness of the coordinate sphere $\{v:\Vert v{\Vert }_0=1\}$.