I have $f(x)=x^7-6$ $\in \mathbb Q[x]$ I can see the roots are $e^{2\pi ik/7}\times6^{1/7}$ with $k$ from $0$ to $6$. How can I show the splitting field $N$ has: $[N:\mathbb Q]=42$?

  • 2
    $\begingroup$ Use this result for the composite of the cyclotomic extension $\Bbb{Q}[\zeta_7]$ and $\Bbb{Q}[\sqrt[7]{6}]$. $\endgroup$ – sharding4 Jul 4 '18 at 13:03

Let $\zeta_7 = e^{2\pi i / 7}$. Then $[\Bbb Q(\zeta_7):\Bbb Q] = 6$. We have that $N$ contains $\zeta_7$, because the quotient of two adjacent roots of the polynomial is $\zeta_7$. So $\Bbb Q(\zeta_7)\subseteq N$, and therefore $[N:\Bbb Q] = [N:\Bbb Q(\zeta_7)]\cdot [\Bbb Q(\zeta_7):\Bbb Q] = 6[N:\Bbb Q(\zeta_7)]$ is divisible by $6$.

We also have $[\Bbb Q(\sqrt[7]6):\Bbb Q] = 7]$. And $N$ contains $\sqrt[7]6$, since that's a root of the polynomial. Therefore $\Bbb Q(\sqrt[7]6)\subseteq N$, and $[N:\Bbb Q] = [N:\Bbb Q(\sqrt[7]6)]\cdot [\Bbb Q\sqrt[7]6):\Bbb Q] = 7[N:\Bbb Q(\sqrt[7]6)]$ is divisible by $7$.

So $[N:\Bbb Q]$ is divisible by both $6$ and $7$, and must therefore be divisible by $42$. It can't be $0$, because $N$ as a vector space over $\Bbb Q$ isn't the trivial space. Can you convince yourself that $[N:\Bbb Q]$ can't be larger than $42$?

  • $\begingroup$ No need for a complete answer, but why can't It be larger? I have been thinking why, but do not know for sure $\endgroup$ – Daniel Moraes Jul 4 '18 at 13:28
  • 1
    $\begingroup$ @user528821 You first need to show that $N = \Bbb Q(\zeta_7, \sqrt[7]6) = \Bbb Q(\zeta_7)(\sqrt[7]6)$, which should be quite easy. Now, can $[\Bbb Q(\zeta_7)(\sqrt[7]6):\Bbb Q(\zeta_7)]$ possibly be larger than $[\Bbb Q(\sqrt[7]6):\Bbb Q] = 7$? $\endgroup$ – Arthur Jul 4 '18 at 13:36

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.