Unique extension of a nonarchimedean absolute value

Put $n=[L:K]$ and $v(x)=\mathrm{\mid }{N}_{L\mathrm{/}K}(x){\mathrm{\mid }}^{1\mathrm{/}n}$. The determinant interpretation shows that v vanishes exactly at zero and $v(a)=\mathrm{\mid }a\mathrm{\mid }$ for $a\in K$, since multiplication by a is a scalar n by n matrix. Norm multiplicativity gives $v(xy)=v(x)v(y)$.

For $v(x)\le 1$, let $E=K(x)$ and m=[E:K]. The tower law and determinant of the scalar E-linear action give $N_{L\mathrm{/}K}(x)=N_{E\mathrm{/}K}(x{)}^{[L:E]}$. The companion matrix of multiplication by x shows that its norm is $(-1{)}^{m}f(0)$ for the monic minimal polynomial f. Thus $\mathrm{\mid }f(0)\mathrm{\mid }\le 1$, so all coefficients lie in the valuation ring. It follows that $\mathrm{\mid }f(-1)\mathrm{\mid }\le 1$, and the same companion-matrix calculation for x+1 gives $v(x+1)\le 1$.

For y nonzero and $v(x)\le v(y)$, apply the preceding step to x/y to obtain $v(x+y)\le v(y)$. If y=0 there is nothing to prove; interchange x,y when needed. This proves the strong triangle inequality. The uniqueness lemma gives uniqueness and completeness. Its proof applies in arbitrary characteristic; no conjugate count or separability was used. For the trivial base value the norm formula is identically one on nonzero elements.