Show that the field extension \(\mathbb Q\subseteq\mathbb Q\left( \sqrt{2+\sqrt2}\right)\) is Galois and determine its Galois group.
Solution
Let \(\alpha = \sqrt{2+\sqrt 2}\), and let \(K = \QQ(\alpha)\).
Minimal polynomial: Squaring gives \(\alpha^2 = 2 + \sqrt 2 \implies \alpha^2 - 2 = \sqrt 2 \implies (\alpha^2 - 2)^2 = 2 \implies \alpha^4 - 4\alpha^2 + 2 = 0\). Let \(f(x) = x^4 - 4x^2 + 2 \in \QQ[x]\). By Eisenstein’s criterion at \(p=2\), \(f(x)\) is irreducible over \(\QQ\). Thus \([K : \QQ] = 4\).
Roots of \(f(x)\): The four roots of \(f(x)\) are: \[ \pm \alpha = \pm \sqrt{2+\sqrt 2}, \qquad \pm \beta = \pm \sqrt{2-\sqrt 2}. \] Notice that: \[ \alpha \beta = \sqrt{(2+\sqrt 2)(2-\sqrt 2)} = \sqrt{4 - 2} = \sqrt 2 = \alpha^2 - 2 \in \QQ(\alpha). \] Therefore, \(\beta = \frac{\alpha^2 - 2}{\alpha} \in K\). Hence, all four roots \(\pm \alpha, \pm \beta\) lie in \(K = \QQ(\alpha)\). This proves that \(K\) is the splitting field of the separable polynomial \(f(x)\) over \(\QQ\), so \(K/\QQ\) is Galois of degree 4.
Galois group: Any \(\sigma \in \Gal(K/\QQ)\) is determined by where it sends \(\alpha\). Since \(f(x)\) is irreducible and \(K\) contains all its roots, there exists an automorphism \(\sigma \in \Gal(K/\QQ)\) sending \(\alpha \mapsto \beta\). Then: \[ \sigma(\sqrt 2) = \sigma(\alpha^2 - 2) = \beta^2 - 2 = (2 - \sqrt 2) - 2 = -\sqrt 2. \] Now compute \(\sigma(\beta)\): \[ \sigma(\beta) = \sigma\left(\frac{\sqrt 2}{\alpha}\right) = \frac{-\sqrt 2}{\beta} = -\alpha. \] Applying \(\sigma\) again: \[ \sigma^2(\alpha) = \sigma(\beta) = -\alpha, \qquad \sigma^3(\alpha) = \sigma(-\alpha) = -\beta, \qquad \sigma^4(\alpha) = \sigma(-\beta) = \alpha. \] Thus \(\sigma\) has order 4, which generates the group. Therefore, \(\Gal(K/\QQ) \cong \ZZ_4\) (the cyclic group of order 4).