Galois group of \(x^4-14x^2+9 \in \mathbb{Q}[x]\) (D&F 14.2.12)

Determine the Galois group of the splitting field over \(\mathbb{Q}\) of \(x^4-14x^2+9\).

We can solve for the roots explicitly: they are \(\pm\sqrt{7 \pm \sqrt{40}}\). For convenience, let \(a_+ = \sqrt{7+\sqrt{40}}\) and \(a_- = \sqrt{7-\sqrt{40}}\), so that the roots are \(\pm a_+, \pm a_-\). Since \(x^4-14x^2+9\) is separable, the splitting field is Galois over \(\mathbb{Q}\), i.e. there are exactly 4 automorphisms fixing \(\mathbb{Q}\).

Observe that for any automorphism \(\varphi\), we have \(\varphi(a_+)\varphi(a_-) = \varphi(\sqrt{49-40}) = 3\). Hence the action of \(\varphi\) on \(a_+\) determines its action on \(a_-\), and hence the remaining roots. In particular, the automorphisms induced by mapping \(a_+\) to each of the 4 roots exhaust the Galois group of the splitting field. It can then be verified that the group of automorphisms is isomorphic to \(V_4\).