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Determine the Galois group of the field extension $E/\mathbb{Q}$, where $E$ is the splitting field of the polynomial $x^4-2\in \mathbb{Q}[x] $.

Here it is clear that $\Bbb Q(\sqrt[4]{2})$ is a splitting field. But I couldn't proceed further. Could somebody please give me hint? Thanks.

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    $\begingroup$ Hint: the extension $E/\Bbb Q(\sqrt{-1})$ is cyclic of degree $4$. $\endgroup$ – Ángel Valencia Mar 28 '16 at 23:39
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    $\begingroup$ The splitting field is $\mathbf Q[\sqrt[4]2,\mathrm i]$. It has degree $8$ over $\mathbf Q$, hence the Galois group has order $8$. $\endgroup$ – Bernard Mar 28 '16 at 23:41
  • $\begingroup$ @ Ángel Valencia.. I see but what would be its Galois group? $\endgroup$ – UserAb Mar 28 '16 at 23:48
  • $\begingroup$ Note that the Galois group of $E/\Bbb Q$ is a subgroup of $S_4$ of order 8. What group is it? $\endgroup$ – Ángel Valencia Mar 28 '16 at 23:58
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Consider the polynomial $x^4-2$. It has $\sqrt[4]{2}$ as a root; so $\Bbb Q(\sqrt[4]{2})\subseteq E$. On the other hand, $x^4-2$ has two complex roots; since $\Bbb Q(\sqrt[4]{2})$ is a real field (i.e. a subfield of $\Bbb R$), you have the extension $\Bbb Q(\sqrt[4]{2},i)/\Bbb Q(\sqrt[4]{2})$, which has degree $2$, and $x^4-2$ splits completely in this field; so $\Bbb Q(\sqrt[4]{2})\subset E\subseteq\Bbb Q(\sqrt[4]{2},i)$. Since the extension $\Bbb Q(\sqrt[4]{2},i)/\Bbb Q(\sqrt[4]{2})$ has no intermediate fields, you have $E=\Bbb Q(\sqrt[4]{2},i)$, and $[E:\Bbb Q]=8$.

Now, remember that the Galois group of $E/\Bbb Q$ is a transitive subgroup of $S_4$. By group theory it is known that any subgroup of order $8$ of $S_4$ is isomorphic to the dihedral group $D_8$. Therefore, the Galois group of $E/\Bbb Q$ is $D_8$.

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    $\begingroup$ @SpamIAm You're right, I edited my answer, thank you. $\endgroup$ – Ángel Valencia Mar 29 '16 at 5:43

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