How to determine all valuations of the field $\mathbb{Q(\sqrt[n]{2})}$? 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.
 A: 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.
A: The minimal polynomial of the primitive element $\sqrt[n]{2}$ is $ f = x^n-2 $ over $\mathbb{Q}$.
Note that by Ostrowski’s Theorem the completion of any field w.r.t an archimedean absolute value is isomorphic to either $\mathbb{C}$ or $\mathbb{R}$.
Let $n$ be even: $f$ has 2 roots in $\mathbb{R}$ and $\frac{n-2}{2}$ pairs of complex conjugate roots in $\mathbb{C}$.
We have the isomorphism $\mathbb{Q}(\sqrt[n]{2})\otimes_\mathbb{Q}\mathbb{R} \cong \mathbb{R}^2 \times \mathbb{C}^{\frac{n-2}{2}}$ (see Neukirch Chapter II, Section 8, Prop (8.3)), which means we have $2+\frac{n-2}{2} = \frac{n+2}{2}$ extensions to the archimedean absolute value of $\mathbb{Q}$.
For $n$ odd: f has 1 root in $\mathbb{R}$ and $\frac{n-1}{2}$ pairs of complex conjugate roots in $\mathbb{C}$. Analogously to above, we get $\frac{n+1}{2}$ distinct extensions of the archimedean absolute value of $\mathbb{Q}$.
