1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
2
$\begingroup$

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

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 $(4,5) = 1$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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