# How does one show that $v(N_{L/K}L^{\times})=f_{L/K}\cdot v(K^{\times})$?

Let $L/K$ be a finite extension of $\mathfrak{p}$-adic number fields. Let us denote by $\mathfrak{p}$ the maximal ideal in the valuation ring $\mathcal{O}_{K}$ and $v$ the associated $\mathfrak{p}$-adic valuation.

How does one prove that the image $v(N_{L/K}L^{\times})$ is equal to $f_{L/K}\cdot v(K^{\times})$, where $f_{L/K}$ is the inertia degree of $L/K$?

I've tried splitting $L/K$ looking at the maximal unramified extension $L\cap\widetilde{K}/K$, but I can use multiplicativity of the norm only for elements in the domain. In short, it is just not working out.

Is there a known reference for this? How does one prove this equality? Thanks.

• Is this not just because up to sign, $N_{L/K}(\pi_L) = \pi_K^{[\mathbb F_L:\mathbb F_K]}=\pi_K^f$? – Mathmo123 Jan 17 '17 at 23:07

Write $\alpha = u\cdot \pi_L^n$ then $N(\alpha) = u'\pi_K^{nf}$ where here $u, u'$ are units in $L, K$ respectively and $\pi_L, \pi_K$ are the respective uniformizers in each ring. But then as $v(\pi_L), v(\pi_K)$ generate the valuation subgroups of $\Bbb Z$ in their respective rings and because $v(\alpha) = v(\pi_L^n)=n$ and similarly if we represent an arbitrary $\beta=\bar{u}\pi_K^m\in K^\times$ then $v(\beta) = v(\pi_K^m)=m$ we see that
$$v(N_K^L(L^\times)) = \langle v(\pi_L)\rangle \cong \langle f\rangle\le\Bbb Z\cong v(\langle\pi_K\rangle)=v(K^\times)$$