Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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] – Edward Hughes May 15 '12 at 13:55
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.