I'm reading Jurgen Neukirch's Algebraic Number theory, p.317~318 ( proof of class field axiom for local fields) and some question arises.
Let $L|K$ be a finite (possibly cyclic) extension of local field. Here local field means, a field which is complete with respect to a discrete valuation and have a finite residue class field. Or equivalent to finite extensions of the $\mathbb{Q}_p$ and $\mathbb{F}_p((t))$.
Let $v_L$ and $v_K$ be the (normalized) discrete valuation on $L$, $K$, respectively.
My question is, $v_L$ extends $v_K$?
- For the case that $L$, $K$ are of $p$-adic number fields ; i.e., finite extensions of the fields of $p$-adic numbers $(\mathbb{Q}_p , v_p)$, it seems that it can be proved that $v_L$ extends $v_K$.
Because, in this case, from the proof of his(Neukirch's) book p.135, (5.2)-proposition and p.133, the exponential valuation $w_L$ and $w_K$ on $L$, $K$ are given by
$$ w_L(\alpha)=(1/n_L)v_p(N_{L|\mathbb{Q}_p}(\alpha)), n_L := [L : \mathbb{Q}_p]$$
$$ w_K(\beta)=(1/n_K)v_p(N_{K|\mathbb{Q}_p}(\beta)), n_K := [K : \mathbb{Q}_p]$$
Then for $\beta \in K$, we can show that $w_L(\beta)=w_K(\beta)$ by using
(1) separability of $L|K$ as being extension of $\mathbb{Q}_p$ ,
(2) his book p.10, (2.7)-Corollary,
(3) his book p.9, (2.6)-proposition (iii),
(4) Hungerford, Algebra, p.286, Prop.6.12
(5) $1/n_K = (1/n_L)[L:K]$.
; i.e., for $\beta \in K$,
$$ w_L(\beta) := (1/n_L)v_p(N_{L|Q_p}(\beta)) = (1/n_L)v_p(N_{K|Q_p} \circ N_{L|K}(\beta))
= (1/n_L)v_p(N_{K|Q_p}(\prod_{\sigma \in Hom_{K}(L,\bar{K})}(\sigma \beta))
= (1/n_L)v_p(N_{K|Q_p}(\beta^{[L:K]}))
= (1/n_L)v_p((N_{K|Q_p}(\beta))^{[L:K]})= (1/n_L)[L:K]v_p(N_{K|Q_p}(\beta))
= (1/n_K)v_p(N_{K|Q_p}(\beta))$$.
- Next Question is, for the case that $L$, $K$ are finite extensions of $\mathbb{F}_p((t))$(another local field case), is it also true that $v_L$ extends $v_K$? Perhaps we may apply similar strategy above ?
Best Answer
I assume your notation means that $v_K$ is the normalization of $w_K$. You have given a valid proof that $w_L$ extends $w_K$, but this does not imply that $v_K$ extends $v_L$ (and indeed this is not true in general).
Here's an example: take $K=\mathbb Q_p$ so $v_K=w_K$ is the usual $p$-adic valuation on $\mathbb Q_p$, and take $L=\mathbb Q_p(p^{1/n})$. We have $n\,w_L(p^{1/n})=w_L(p)=w_K(p)=1$ and therefore $w_L(p^{1/n})=1/n$. Now the issue is that when you change from $w_L$ to $v_L$, i.e. you normalize, you define it so that $v_L(p^{1/n})=1$ since $p^{1/n}$ is a uniformizer of $L$, but then $v_L(p)=n\neq 1=v_K(p)$.
The result you're asking about is true when $L|K$ is unramified: in this case the ramification indices $e(L|\mathbb Q_p)$ and $e(K|\mathbb Q_p)$ are equal, call this value $e$, and then $v_L=e\,w_L=e\,w_K=v_K$.
Also for your question 2, yes everything we've argued works analogously when $L$ and $K$ are extensions of $\mathbb F_p((t))$, the only thing to be careful of is that you have the right definition of $N_{K|\mathbb F_p((t))}$, as many authors define this in a way that is only correct for separable extensions.