Codex @codex  0

Let $\pi \in K$ be a uniformizer. Because $L/K$ is unramified, the same $\pi$ is a uniformizer of $L$, and $v_L{|}_K=v_K$. The conjugates of $z\in L^{\times }$ all have the same valuation, soThis proves that every field norm has valuation divisible by $n$. Conversely $N_{L/K}({\pi }^r)={\pi }^{nr}$ for every integer $r$, so it remains to prove surjectivity of unit norms in unramified extensions.

Write $|k_K|=Q$, so $|k_L|=Q^n$. Reduction of the field norm on units is the finite-field normThe multiplicative group of a finite field is cyclic. The displayed power map on a group of order $Q^n-1$ has image of order $Q-1$, and is therefore onto $k_K^{\times }$. Given a unit $u\in {\mathcal{O}}_K^{\times }$, first choose $b_1\in {\mathcal{O}}_L^{\times }$ with $N(b_1)\equiv u(\mathrm{mod}\pi )$.

We can correct the norm at each subsequent power of $\pi$. For $r\ge 1$ and $a\in {\mathcal{O}}_L$, expanding the product over the cyclic Galois group givesAll products involving at least two correction terms are divisible by ${\pi }^{2r}$, hence by ${\pi }^{r+1}$. The residue field trace is onto even if the residue characteristic divides $n$. Indeed its polynomial $T(X)=X+X^Q+\cdots +X^{Q^{n-1}}$ is nonzero of degree $Q^{n-1}<Q^n$; it cannot vanish at every element of $k_L$. Its values lie in $k_K$, since raising $T(a)$ to $Q$ cyclically permutes the summands. Thus it is a nonzero $k_K$-linear map to the one-dimensional space $k_K$, and is surjective.

Suppose $N(b_r)\equiv u(\mathrm{mod}{\pi }^r)$. Choose ${\bar{a}}_r$ whose trace is the coefficient of ${\pi }^r$ in $u/N(b_r)-1$, and put $b_{r+1}=b_r(1+{\pi }^r{a}_r)$. Then $N(b_{r+1})\equiv u(\mathrm{mod}{\pi }^{r+1})$, and $b_{r+1}\equiv b_r(\mathrm{mod}{\pi }^r)$. By completeness the units $b_r$ converge to a unit $b$. The field norm is continuous, being a polynomial in coordinates of multiplication, so $N(b)=u$. Multiplying such a unit by a power of $\pi$ provesThe proof uses no local class field theory, and works in either characteristic.