# Galois Group of $x^n - a$

Homework problem: If the field F contains a primitive nth root of unity, prove that the Galois group of $x^n - a$, for $a \in F$, is abelian.

I'm not really sure where to start here and I'm confused about the wording of the problem. I would appreciate hints.

Here are a few steps to get you started. Let $\omega$ be a primitive $n$th root of unity in $F$, and let $u$ satisfy $u^n=a$. The roots of $x^n-a$ are then $\omega^iu$ for $i=0,\ldots,n-1$. The Galois group $G$ of $x^n-a$ permutes these roots, and any element of $G$ is completely defined by where it sends $u$ (which can be chosen arbitrarily). That is, for any $i=0,\ldots,n-1$ there is a field automorphism $\sigma_i$ which maps $u$ to $\omega^iu$. See if you can take it from here.
• +1 and a minor nitpick. If $x^n-a$ is not irreducible, then not all the $\sigma_i$ are ok (think examples like $x^6-4$ over the rationals). But the remaining $\sigma_i$:s still commute, so the conclusion is valid. – Jyrki Lahtonen Mar 31 '14 at 6:42
• @JyrkiLahtonen: Of course. This was an overlook on my part. Thanks for catching it. To the OP, the point is that the Galois group permutes the roots of the irreducible factors of $x^n-a$, so that the only $\sigma_i$ in the Galois group are those for which $u$ and $\omega^iu$ are roots of the same irreducible factor of $x^n-a$. As Jyrki points out, you can still show that these viable $\sigma_i$ all commute with one another. – Jared Mar 31 '14 at 6:49