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Here is a problem in charpter $2$ section $5$ in Algebraic Number Theory written by Jiirgen Neukirch.

The problem is

Let $K$ be a $p-adic$ number field, $v_p$ the normalized exponential valuation of $K$, and $dx$ the Haar measure on the locally compact additive group $K$, scaled so that $\int_O dx = 1$. Then one has $v_p(a) = \int_{aO} dx$.

I would like to ask why the last equation holds,I think it should be $\shortmid a\shortmid _p =\int_{aO} dx$.

Please help me with it,thank you very much.

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Dear curiosity, For those without Neukirch's book on hand, could you briefly explain what is the normalized exponential valuation. (E.g. is it an absolute value, taking products to products, or a valuation in the ring-theoretic sense, taking products to sums?) In other words, what is the difference between $v_p(a)$ and $|a|_p$? Regards, – Matt E May 26 '11 at 18:29
I am sorry for making you confused, here $v_p$ means taking products to products while the other means the usual valuation in the ring-theoretic sense. – curiosity May 27 '11 at 4:41
up vote 3 down vote accepted

The question is equivalent to asking for the ratio of the measures $\mu(a \mathcal{O})/\mu(\mathcal{O})$, where $\mathcal{O}$ is the valuation ring and $a \in K^{\times}$. But in a $p$-adic field, whenever you have one ball centered at zero contained in another ball centered at $0$, the larger ball is simply a disjoint union of translates of the smaller ball. Since the Haar measure is translation invariant, the ratio of the volumes just comes out to be the number of translates. If $a \in \mathcal{O}$, then the number of translates of $a \mathcal{O}$ in $\mathcal{O}$ is nothing else than the cardinality of the quotient ring $\mathcal{O}/a \mathcal{O}$.

I leave it to you to work out the problem according to the notation in the book (especially since I think your comment trying to clarify the notation may be mistaken). But for instance in $\mathbb{Q}_p$, the ratio of the volume of $\mathcal{O}$ to $p \mathcal{O}$ is equal to $p$, because $\# \mathcal{O}/p\mathcal{O} = \# \mathbb{Z} / p \mathbb{Z} = p$. In general this ratio will be multiplicative in $a$, not additive (so it should be a $K$-adic norm, not a $K$-adic valuation).

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@Peter: Thank you,I will take a try. – curiosity May 27 '11 at 12:19
@Peter:I hane worked it out,thank you very much. – curiosity May 30 '11 at 17:41

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