This is a basic question about Galois Extension, but I want some details about it.

Let $F$ be a splitting field over $\mathbb Q$ the polynomial $x^8-5\in\mathbb Q[x]$. Recall that $F$ is the subfield of $\mathbb C$ generated by all roots of this polynomial.

(1) Find the degree $[F:\mathbb Q]$ of the number of the number field $F$.

(2) Determine the Galois group $Gal(F/\mathbb Q)$.

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    $\begingroup$ The field is generated by the (positive) 8th root of 5 and a square root of $i$ --- that should get you started. See what you can do with that information. $\endgroup$ – Gerry Myerson Feb 15 '14 at 5:49
  • $\begingroup$ @GerryMyerson $x^8-5=\prod_{k=1}^{8}(x-5^{\frac{1}{8}}e^{\frac{i}{8}k})$ Then how we find the basis in $\{e^{\frac{i}{8}k}\}$? $\endgroup$ – gaoxinge Feb 16 '14 at 3:53
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    $\begingroup$ I don't believe your formula, and I don't understand your question. Do you understand my comment about generators of the extension? $\endgroup$ – Gerry Myerson Feb 16 '14 at 4:11
  • $\begingroup$ @GerryMyerson Do you mean $F=\mathbb Q(e^{\frac{2\pi i}{8}},...,e^{\frac{2\pi i}{8}\times 8})$? $\endgroup$ – gaoxinge Feb 16 '14 at 4:47
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    $\begingroup$ No. Look at what I wrote: "the (positive) 8th root of 5" --- what happened to that? "a square root of $i$" --- just one square root of $i$ is all you need/want. $\endgroup$ – Gerry Myerson Feb 16 '14 at 7:52

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