Prove that the splitting field of $x^{p}-q$ for prime numbers $p,q$ is an extension of degree $p(p-1)$ in $\mathbb{Q}$.

I know that the degree of the splitting field is bounded by $p!$, but I don't know how to prove that it equals $p(p-1)$ exactly. Any help would be appreciated.

  • $\begingroup$ You can find the splitting field exactly. First find the splitting field of $x^p-1$ and then $x^p-q$ over that field... $\endgroup$ – Asvin May 1 '15 at 6:34
  • $\begingroup$ And how are the splitting fields of $x^p-1$ and $x^p-q$ related? Sorry, I'm a little confused. $\endgroup$ – user3566523 May 1 '15 at 6:37
  • $\begingroup$ The splitting field you want has to contain all the p- roots of unity. Try looking at the difference of two roots of your polynomial. Then build your splitting field as a tower of roots of unity+whatever's missing. $\endgroup$ – Asvin May 1 '15 at 6:43
  • $\begingroup$ It might help a bit to fix p and q. Maybe 3 and 2. $\endgroup$ – Asvin May 1 '15 at 6:44

Let $\zeta = e^{2\pi i/p}$ and $\omega = q^{1/p}$. Note that for $0 \leq j \leq p-1$, $\zeta^{j}\omega$ is a root of $x^p-q$. Since these are $p$ distinct roots and the polynomial is monic of degree $p$, we must have \begin{gather*} x^p-q=\prod_{j=0}^{p-1}(x-\zeta^{j}\omega). \end{gather*} From this, it is pretty easy to see that $\mathbb{Q}(\omega, \zeta)$ is a splitting field of the polynomial. You can show $[\mathbb{Q}(\omega):\mathbb{Q}]=p$ with Eisenstein's criterion and then note that $\zeta$ is a root of $x^{p-1}+ \cdots + x+1$ which is irreducible by the standard trick of applying Eisenstein's criterion to $(x+1)^{p-1}+\cdots + (x+1) + 1$, so that $[\mathbb{Q}(\zeta):\mathbb{Q}]=p-1$. Since gcd$(p,p-1)=1$, the degrees of the extensions multiply.

  • $\begingroup$ This is very helpful. I do have one question. How would you use Eisenstein's criterion show that $[\mathbb{Q}(\omega) : \mathbb{Q}]=p$. $\endgroup$ – user3566523 May 1 '15 at 7:09
  • $\begingroup$ To show that $x^p-q$ is irreducible over $\mathbb{Q}$, so that it is the minimal polynomial of $\omega$ over $\mathbb{Q}$. $\endgroup$ – Noah Olander May 1 '15 at 7:16

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