Splitting field of a polynomial

Question

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

Answer 1

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 $42$?