I am reading the Neukirch's Algebraic number theory, p.142, (5.8) Corollary and stuck at some statements
Let $K$ be a local field ; i.e., a field which is complete with respect to a discrete valuation and have a finite residue field. We denote $v_{\mathfrak{p}}$ by the normalized exponential valuation, and denote $| \ |_{\mathfrak{p}}$ by the absolute value normalized by
$$ |x|_{\mathfrak{p}}= q ^{-v_{\mathfrak{p}}(x)},\tag{1}$$
where $q$ is the cardinality of the residue class field.
(5.8) Corollary. If the natural number $n$ is not divisible by the characteristic of $K$, then one finds the following indices for the subgroups of $n$-th powers $K^{*n}$ and $U^n$ in the multiplicative group $K^*$ and in the unit group $U$ :
$$ (K^* : K^{*n}) = n (U: U^n) = \frac{n}{|n|_{\mathfrak{p}}} \#\mu_n(K)$$
Q. What is $U$ and $\mu_n(K)$? I guess that $U$ is the group $\mathcal{O}^{*}$ of units of the valuation ring $\mathcal{O}$ of the $K$ ( c.f. his book p.120~122 ). And $\mu_n(K)$ is $\mu_n \cap K^{*}$, where $\mu_n$ is the group of $n$-th roots of unity ( c.f. his book p.317 ). True?
Proof : The first equality is a consequence of $K^* = (\pi) \times U$, where $\pi$ is a prime element of $K$ ( c.f. his book, proof of II-(5.3) Proposition ). By (5.7), we have
$$U \cong \mu(K) \times \mathbb{Z}^{d}_{p}, \ \operatorname{resp.} U \cong \mu(K) \times \mathbb{Z}_p^{\mathbb{N}} \tag{2}$$
when $\operatorname{char}{K}=0$, resp. $p>0$.
Q.1. I don't understand these statements. Here, $(5.7)$ Proposition is,
Note that $(\pi) \cong \mathbb{Z}$ and $K^* = (\pi) \times U$ as I mentioned above. Perhaps,
$\mu(K) \cong \mathbb{Z}/(q-1)\mathbb{Z} \oplus \mathbb{Z}/p^a\mathbb{Z}$ when $\operatorname{char}K=0$ and $\mu(K) \cong \mathbb{Z}/(q-1)\mathbb{Z}$ when $\operatorname{char}K =p $ ? If so, then we can show the above isomorphism.
( Continuing proof ) From the exact sequence
$$ 1 \rightarrow \mu_n(K) \to \mu(K) \xrightarrow{n} \mu(K) \to \mu(K)/\mu(K)^{n} \to 1, $$
one has $\# \mu_n(K) = \# \mu(K)/\mu(K)^n$. When $\operatorname{char}(K)= 0$, this gives
$$ ( U : U^n) = \# \mu_n(K) \#(\mathbb{Z}_p/n\mathbb{Z}_p)^d=\#\mu_n(K)p^{dv_p(n)} = \# \mu_n(K)/|n|_{\mathfrak{p}}, \tag{3} $$
and when $\operatorname{char}(K)=p$ one gets simply
$$(U:U^n)= \# \mu_n(K) =\#\mu_n(K)/|n|_{\mathfrak{p}} \tag{4}$$
because $(n,p)=1$, ; i.e., $n\mathbb{Z}_p = \mathbb{Z}_p$. QED.
I'm trying to understand the statements (3) and (4).
Q.2. For the first equalites in $(3)$ and $(4)$, we may use the $(2)$. At this time, perhaps,
$$ \mathbb{Z}_p^{d} / ( \mathbb{Z}^d_{p})^n \cong ( \mathbb{Z}_p/n \mathbb{Z}_p)^d,$$
and $$ \mathbb{Z}_p^{\mathbb{N}} \cong (( \mathbb{Z}_p)^{\mathbb{N}})^n $$
? If so, then through some calculation we may deduce the first equalities in $(3)$ and $(4)$.
Q.3. Why the other equalities in $(3)$ and $(4)$ are true? That is, why
$ \#(\mathbb{Z}_p/n\mathbb{Z}_p)^d= p^{dv_p(n)}$ ( EDIT : This is solved : $\# (\mathbb{Z}_p/n\mathbb{Z}_p)=p^{v_p(n)}$, where $v_p$ is the p-adic exponential valuation? ) and $p^{dv_p(n)} = (|n|_{\mathfrak{p}})^{-1}$ are true when $\operatorname{char}(K)=0$ ( EDIT : This seems also solved but there are some technical issues that I don't understand : see answer of Tob Ernak, A question in Neukirch's ANT book and can anyone see my comment for his answer ? ) and
why $|n|_{\mathfrak{p}} =1$ when $\operatorname{char}(K)=p$ ?
What are exact defitnitions of $v_p$ and $| \ |_{\mathfrak{p}}$? What relationship between $v_p$ and $| \ |_{\mathfrak{p}}$ can we use? What should I catch?
Can anyone help?
Best Answer
I'll take a shot at answering.
$Q_0$: True, those are the definitions.
$Q_1$: Also true. To see a proof of the stated isomorphisms, look here Units of p-adic integers.
$Q_2$: Again, both isomorphisms are true. In the first one, you should be careful what you mean with exponentiation. $U^n$ means $\lbrace u^n\mid u\in U\rbrace$, which is a subgroup of $U$, but $(\Bbb{Z}_p/n\Bbb{Z}_p)^d$ means $d$ copies of $\Bbb{Z}_p/n\Bbb{Z}_p$, so stay away from mixing the two. The essence is that the former is multiplicative, and the latter is additive. Then I think you can convince yourself of the first isomorphism. The second isomorphism is not true in general, but uses the critical assumption that $(n,p)=1$, meaning that it has valuation $\nu_p(n)=0$, such that $n\Bbb{Z}_p=\Bbb{Z}_p$ as we discussed in your other question. Then $n(\Bbb{Z}_p^\Bbb{N})=(n\Bbb{Z}_p)^\Bbb{N} = \Bbb{Z}_p^\Bbb{N}$.
$Q_3$: The remainder of this question will be solved when you have a clear understanding of what $|\cdot|_\frak{p}$ is. Note that there is a choice to be made whenever you define the $\frak{p}$-adic norm, but it will always have the form $|\alpha|_\mathfrak{p} = p^{-c\nu_{\mathfrak{p}}(\alpha)}$ for some constant $c$. Neukirch has the convention of picking $c=[K:\Bbb{Q}_p]$, which he calls $d$ above. Also, it is clear that $\nu_\mathfrak{p} = e\nu_p$ where $e$ is the ramification index of the extension $K/\Bbb{Q}_p$.
Edit: I will adress your comments to this question.
You are correct in what Neukirch's convention is, it is equivalent to what I wrote. Now, $\nu_\mathfrak{p} = e\nu_p$ basically by definition: We have that $\mathfrak{p}^e = (p)$ in $\mathcal{O}_K$ by definition of $e$. If we let $\alpha\in \Bbb{Q}_p$, we have that $(\alpha)=(p)^{\nu_p(\alpha)}$ by definition, and then we see that $$\mathfrak{p}^{\nu_\mathfrak{p}(\alpha)}=(\alpha)=(p)^{\nu_p(\alpha)} = (\mathfrak{p}^e)^{\nu_\mathfrak{p}(\alpha)} = \mathfrak{p}^{e\nu_p(\alpha)}.$$ I believe this holds in all characteristics. Hopefully this answers a couple of questions. (Edit 2) The leftmost equality is again the definition of $\nu_\mathfrak{p}$. The above of course implies $\nu_\mathfrak{p}(\alpha)=e\nu_p(\alpha)$ by unique prime ideal factorisation, which holds in any Dedekind domain, in particular any PID. I do all of this using ideals as that is the neatest way of doing it, because you skip talking about units, but you can of course do this using usual prime element factorisation. If $\pi$ is any uniformizer, by definition $\pi^e=p\varepsilon$ for some unit $\varepsilon\in\mathcal{O}_K$. You can continue the rest of the story from there.
$|n|_\mathfrak{p}=1$ is true if and only if $n$ is a unit in $\mathcal{O}_K$, and for this it is sufficient to be a unit in $\Bbb{F}_p((t))$, and since $(n,p)=1$, $n$ is a unit even in $\Bbb{F}_p$. (Edit 2) $\nu_p$ is not a priori defined on $\mathcal{O}_K$, so the extension is literally by defining it to be $\nu_p=\frac{1}{e}\nu_\mathfrak{p}$. Remember that any ring homomorphism, in particular embeddings, preserve units.
(Third and final edit) I have rolled back your 8 edits to this answer. For transparency, your edits consisted of deleting some of my main points, adding references to Neukirch, adding other tiny details and giving an attempt to fill out the proof of $\nu_\mathfrak{p}=e\nu_p$ using elements.
These things are what I expect you to do for yourself. Do not ever replace what an answerer is saying with your own thoughts. If you are dissatisfied with my answer in its current form, wait for another answer, or ask a new question.