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This is an exercise from the book Algebraic Number Theory by Jurgen Neukirch, on page 166.

And, after solving several previous exercises, I found this to be particularly difficult to solve. I am hoping only for hints, not a complete solution. And any hint is appreciated.

Also, I have made a conjecture that the number of all nonarchimedean valuations extending one of $Q$ is either 1 or $\phi(n)+1$, as to the archimedean ones, I think they are easy to find, where $\phi(n)$ is the Euler function of $n$.
And, eventually, I stated my question once again to make it clear.
Again, please provide me only with hints so I can work it out myself:

How to determine all valuations of the field $\mathbb{K}$ where $\mathbb{K}$ is obtained by adjoining $n$th root of 2 to $\mathbb{Q}$?

Thanks.

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@awllower: If you want only hints, I would suggest instead to say so before stating the question in the body, rather than hoping people will do so because of a tag. –  Arturo Magidin Mar 6 '11 at 3:55
    
@Arturo Magidin: Thanks for your advice, I get it now. –  awllower Mar 6 '11 at 3:59
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I am confused by some of your comments in the question. Any number field $K$ has infinitely many non-Archimedean valuations: for each prime number $p$, there are between $1$ and $[K:\mathbb{Q}]$ of them lying over the $p$-adic valuation of $\mathbb{Q}$. Moreover, the valuations lying over $p$ correspond to the prime ideals lying over $p$, so (OK, this uses a serious theorem) for any number field there is a positive density set of primes p such that the number of valuations $v$ of $K$ extending $v_p$ of $\mathbb{Q}$ is exactly $[K:\mathbb{Q}]$. –  Pete L. Clark Mar 6 '11 at 6:11
    
@Pete L. Clark: I am also confused now. You are telling me that there are infinitely many valuations of any algebraic number field, but this is an exercise from the book by J. Neukirch. I can't get the point of you, any elaboration is appreciated, thanks. –  awllower Mar 7 '11 at 11:38
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This comment is perhaps a little late, but the exercise was copied incorrectly. The correct problem statement is "How many extensions to $\mathbb{Q}(\sqrt[n]{2})$ does the archimedean absolute value $| \cdot |$ of $\mathbb{Q}$ admit?" –  Brandon Carter Jan 21 '12 at 16:56

1 Answer 1

up vote 6 down vote accepted

This seems to be a nontrivial problem. Finding all valuations is equivalent to finding all prime ideals, since the archimedean valuations are easy to describe. Finding all prime ideals depends on the Galois group of the extension and would be easy if the extension were abelian, which it rarely is (there are some for $n = 4$). The problem of finding the Galois group of the normal closure of these fields is discussed by Jacobson and Velez, The Galois group of a radical extension of the rationals Manuscr. Math. 67 (1990), 271-284.

For detailed information on integral bases and the decomposition of primes, a good place to start is Ribenboim's book "Algebraic Numbers". What you should do in general is find the possible decomposition and ramification subgroups, and then draw conclusions about the splitting of primes. There are exercises like these in Marcus' Number Fields.

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Thank you very much, and I shall try then. –  awllower Jan 29 '12 at 10:56

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