3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Is this not just because up to sign, $N_{L/K}(\pi_L) = \pi_K^{[\mathbb F_L:\mathbb F_K]}=\pi_K^f$? $\endgroup$ – Mathmo123 Jan 17 '17 at 23:07
2
$\begingroup$

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

$$v(N_K^L(L^\times)) = \langle v(\pi_L)\rangle \cong \langle f\rangle\le\Bbb Z\cong v(\langle\pi_K\rangle)=v(K^\times)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.