# 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.

• @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. Mar 6, 2011 at 3:55
• @Arturo Magidin: Thanks for your advice, I get it now. Mar 6, 2011 at 3:59
• 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 $[K:\mathbb{Q}]$. Mar 6, 2011 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. Mar 7, 2011 at 11:38
• 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?"
– bzc
Jan 21, 2012 at 16:56

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.
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 $\mathbb{Q}$.