The Galois group of $x^4-2$ over $\QQ$

Let \[ \alpha=2^{1/4}. \] The roots are \[ \alpha,-\alpha,i\alpha,-i\alpha, \] so the splitting field is \[ L=\QQ(\alpha,i). \]

The polynomial \(x^4-2\) is Eisenstein at \(2\), hence \[ [\QQ(\alpha):\QQ]=4. \] Since \(\QQ(\alpha)\subset\RR\) but \(i\notin\RR\), \[ [L:\QQ]=8. \] Thus the Galois group has order \(8\).

Define automorphisms \[ r(\alpha)=i\alpha, \qquad r(i)=i, \] and \[ s(\alpha)=\alpha, \qquad s(i)=-i. \] Then \[ r^4=s^2=1, \] and \[ srs=r^{-1}. \] Moreover, \(r\) has order \(4\), \(s\) has order \(2\), and together they generate all \(8\) automorphisms.

Hence \[ \Gal(L/\QQ) \cong \langle r,s\mid r^4=s^2=1,\ srs=r^{-1}\rangle, \] the dihedral group of order \(8\).