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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 $d:=[K:Q_p]$.

Tom
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    It would help if you posted the corollary and proof, given that people are unlikely to have the book next to them. – Batman Jul 05 '14 at 16:12

2 Answers2

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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.

Tob Ernack
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  • Thanks. I am trying to understand the the (5.8) Corollary. Can I ask question more? First, why the $q$ is of the form $p^f$? Is there a field homomorphism $\mathbb{F}_p \to \mathcal{O}_K / \mathfrak{p}$ regardless of whether $\operatorname{char}{K}$ is $0$ or $>0$? Second, for $(p)=p\mathcal{O}_K = \mathfrak{p}^{e}$ to be established, it seems that it will be good if the integral closure of $\mathbb{Z}_p$ (which is valuation ring of $\mathbb{Q}_p$) in $K$ is $\mathcal{O}_K$ ( C.f. Neukirch's book p.45 ) And is it true? Or is there any other route to show that $(p)= \mathfrak{p}^{e}$? – Plantation Oct 24 '23 at 09:03
  • Thrid, finally, why $v_{\mathfrak{p}} = e v_p$ ? Where can we find assoicated proof? Can you explain these? – Plantation Oct 24 '23 at 09:03
  • And.. if you have time, then can you see my question about the proof of the (5.8) Corollary that I posted ? : https://math.stackexchange.com/questions/4792835/understanding-the-neukirch-algebraic-number-theory-p-142-5-8-corollary – Plantation Oct 24 '23 at 09:04
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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.

Keenan Kidwell
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  • Thank you very much. I agree with you. I conclude that Cor II.5.8 is a mistake and its denominator $|n|_p$ should be replaced by $p^{dv_p(n)}$ where $v_p(p)=p$. – Tom Jul 06 '14 at 11:54
  • In the above, $p^{dv_p(n)}$ should be $p^{-dv_p(n)}$. – Tom Jul 06 '14 at 12:04