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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: $[N:\mathbb Q]=42$?

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    $\begingroup$ 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
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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 $42$?

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  • $\begingroup$ 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
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    $\begingroup$ @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

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