A question in Neukirch's ANT book

Question

In Corollary II.5.8, Neukirch Algebraic Number Theory(p142, line 11), why $d=v'_p(p)$ where $v'$ is normalized valuation?

EDIT In other word, let $K$ be a finite extension of $Q_p$, I.e. a local field of characteristic 0. Let $\pi$ be a prime element. Then $p=\pi^d u$ for some unit $u$? Here $d:=[K:Q_p]$.

Answer 1

It's not true that $d=[K:\mathbf{Q}_p]$. The integer $d$ (more commonly denoted $e=e(K/\mathbf{Q}_p)$) is the ramification index of $K$ over $\mathbf{Q}_p$. It satisfies $e\leq[K:\mathbf{Q}_p]$ with equality if and only if $K/\mathbf{Q}_p$ is totally ramified. As for why $e=v^\prime(p)$ (where by "normalized valuation" I assume you mean the normalized discrete valuation of $K$), what is the defining property of a uniformizer $\pi$? It is an element of normalized valuation one. So $v^\prime(p)=v^\prime(\pi^e u)=ev^\prime(\pi)+v^\prime(u)=e$ because units are those elements with valuation zero.

Answer 2

I found this while asking myself the same question after reading Corollary (5.8).

In the statement of Proposition (5.7), $d$ is defined as the degree of the extension, $d = [K : \mathbb{Q}_p]$.

At the beginning of section 5, Neukirch defines the normalized absolute value in terms of the normalized valuation: $$|x|_\mathfrak{p} = q^{-v_\mathfrak{p}(x)}$$ where $q$ is the order of the residue class field of $K$ (so $q = p^f$, where $f = [\mathcal{O}_K/\mathfrak{p}:\mathbb{F}_p]).$

Note that $v_\mathfrak{p}(x) = ev_p(x)$, where $v_p$ is the valuation of $\mathbb{Q}_p$ extended to $K$ and $e$ is the ramification degree, i.e. $(p) = \mathfrak{p}^e$.

The denominator in Corollary (5.8) is actually $|n|_\mathfrak{p}=q^{-v_\mathfrak{p}(n)}=p^{-efv_p(n)} = p^{-dv_p(n)}$ because $d = ef$ (by the Hilbert Ramification Theory section in Chapter I).

So the given expression is in fact correct.