I am try to determine

What is the degree of the splitting field of $x^5 − 7$ over $\mathbb{Q}$?

I concluded the degree of $\mathbb{Q}[\sqrt[5]{7}]$ over $\mathbb{Q}$ is 5 because the polynomial is irreducible in $\mathbb{Q}$. And that the solutions to the polynomial are $\alpha = \sqrt[5]{7}$ times the 5th roots of unity $\beta = e^{2\pi i/5}$, that is $\alpha$, $\alpha\beta$, $\alpha\beta^2$, $\alpha\beta^3$, and $\alpha\beta^4$. Furthermore the degree of $\mathbb{Q}[\beta]$ over $\mathbb{Q}$ is 4. Thus the degree of the splitting field is $[\mathbb{Q}[\alpha,\beta]:\mathbb{Q}]$ = 20, because 4 and 5 have no common divisors.

Is this correct?

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    $\begingroup$ How do you know that the degree of $\mathbb{Q} [ \beta]$4 over $\mathbb{Q}$ is $4$? Also, You need to show that $\alpha$ and $\beta$ generate the whole splitting field of that polynomial (which I guess is clear) $\endgroup$ – Pawel Feb 13 '17 at 4:49
  • $\begingroup$ @Paquarian I know the degree is 4 because it's cyclic, right? $\endgroup$ – wogsland Feb 13 '17 at 4:59
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    $\begingroup$ If you have in mind $x^4-1$, then $\beta$ is not a root of that polynomial. $\endgroup$ – Pawel Feb 13 '17 at 5:11
  • $\begingroup$ Correct indeed. $\endgroup$ – Andreas Caranti Feb 13 '17 at 13:43
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    $\begingroup$ @Paquarian, what makes you think that OP has the wrong polynomial $x^{4} -1$ in mind, instead of the correct $x^{4} + x^{3} + x^{2} + x + 1$? $\endgroup$ – Andreas Caranti Feb 13 '17 at 13:44

$\newcommand{\rt}{\sqrt[5]7}$ $\newcommand{\Q}{\mathbb{Q}}$

Yes, you are right.

Let $\zeta := \exp(2 \pi i / 5)$, then the splitting field of your polynomial is $\mathbb{Q}(\rt, \zeta \rt, \dots, \zeta^4\rt) = \Q(\zeta, \rt)$. The degree of $\Q(\zeta)/\Q = 4$, because either you know that $Gal(\Q(\zeta)/\Q) \cong (\mathbb{Z}/5 \mathbb{Z})^x$ (where $\mathbb Z / 5 \mathbb Z$ is a field, thus the only non-unit is $0$) or you can also easily show that $\zeta$ is a root of $\Phi_5(X) = X^4+X^3+X^2+X+1$ and that this polynomial is irreducible.

$[\Q(\rt): \Q] = 5$, as your given polynomial is irreducible (Eisenstein).

Then, as you have already mentioned, the degree of the big extension must be $20$ as $(4,5) = 1$

  • $\begingroup$ Thanks for this. I think you've fleshed out some of the places where my reasoning was a little holy. $\endgroup$ – wogsland Feb 13 '17 at 23:05

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