Compute the Galois group of the polynomial \(f(x) = x^4 - 3\) over \(\mathbb{Q}\), and describe its action on the roots.
Solution
Let \(\alpha=\sqrt[4]{3}\). The roots are \(\alpha,i\alpha,-\alpha,-i\alpha\), so the splitting field is \[ K=\mathbb Q(\alpha,i). \] Eisenstein at \(3\) gives \([\mathbb Q(\alpha):\mathbb Q]=4\). Since \(\mathbb Q(\alpha)\subset\mathbb R\), we have \(i\notin\mathbb Q(\alpha)\), hence \([K:\mathbb Q]=8\).
Define \[ r(\alpha)=i\alpha,\qquad r(i)=i, \] and let \(s\) be complex conjugation: \[ s(\alpha)=\alpha,\qquad s(i)=-i. \] Then \(r\) has order \(4\), \(s\) has order \(2\), and \[ srs=r^{-1}. \] Thus \(\langle r,s\rangle\cong D_4\) has order \(8\), so it is the whole Galois group. On the ordered roots \((\alpha,i\alpha,-\alpha,-i\alpha)\), \[ r=(1\ 2\ 3\ 4),\qquad s=(2\ 4). \] Therefore \[ \operatorname{Gal}(K/\mathbb Q)\cong D_4 \] (the dihedral group of order \(8\)).