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going over some past papers with no answers and would like a bit of help if possible..

I've shown that for p a prime number then $x^p-1 \in K[x]$ is abelian where K is a subfield of $\mathbb{C}$. I've now been asked to show that the Galois group of $x^p-a$ over K is soluble with $a \neq 0 \in K$. I know any abelian group A is soluble, since ${1} \triangleleft A$ is a subnormal series with its only factor A being abelian, so $x^p-1 \in K[x]$ is certainly soluble. Not sure where to go from here and the question is worth quite a lot so guessing there is quite a bit more to do, any help appreciated.

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    Consider the splitting field of your polynomial $q(x)=x^p-a$, call it $L/K$. This is a radical extension so soluble. I imagine the question is wanting you to prove that radical $\Rightarrow$ soluble, which is a fairly standard result. [Hint: use the Galois correspondence]2012-05-15

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Hint: Let $c$ be a root of $X^p - a$ and let $\zeta$ be a primitive $p$-th root of unity. What can you say about $c, \zeta c, \dots, \zeta^{p-1} c$?

Now you can build up the splitting field of $X^p - a$ in two steps, by first adjoining $\zeta$ and then adjoining $c$. So you get fields $K\subset K(\zeta) \subset K(\zeta, c)$ and $K(\zeta, c)/K$ is Galois.

You already know what $\mathrm{Gal}(K(\zeta)/K)$ looks like (or at least that it is abelian). You should also be able to figure out the structure of $\mathrm{Gal}(K(\zeta, c)/K(\zeta))$. (One can write it down explicitely)

What is left to do: We need to find some sort of relationship between $\mathrm{Gal}(K(\zeta, c)/K)$, $\mathrm{Gal}(K(\zeta, c)/K(\zeta))$ and $\mathrm{Gal}(K(\zeta)/K)$.

Useful facts:

  • Any subgroup of a solvable group is solvable.
  • A finite product of solvable groups is solvable.
  • If $N \lhd G$ is normal, then $G$ is solvable $\iff$ $G/N$ and $N$ are solvable.

Further Hint (although I feel this might be giving away too much):

For instance you could look at $\quad \mathrm{Gal}(K(\zeta, c)/K) \mapsto \mathrm{Gal}(K(\zeta)/K), \quad \sigma \mapsto \sigma|_{K(\zeta)} \quad $ and use some isomorphism theorem.