Galois group of \(\mathbb{Q}(\sqrt{2+\sqrt{2}})\) (D&F 14.2.14)

Show that \(\mathbb{Q}(\sqrt{2+\sqrt{2}})\) is a cyclic quartic field.

Let \(a_+ = \sqrt{2+\sqrt{2}}\) and \(a_- = \sqrt{2-\sqrt{2}}\). It can be seen that \(a_+\) is a root of the minimal polynomial \(p(x) = (x^2-2)^2-2 = x^4-4x^2+2\) (being irreducible by Eisenstein), and solving for all roots of \(p(x)\) gives \(\pm a_+, \pm a_-\). Also note that since \({a_+}^2-2 = \sqrt{2}\) and \({a_+}{a_-} = \sqrt{2}\ (*)\), it follows that \(a_- \in \mathbb{Q}(a_+)\), and so \(\mathbb{Q}(a_+)\) contains all roots of \(p(x)\). Since \(\mathbb{Q}(a_+)\) has degree 4 over \(\mathbb{Q}\), it is in fact the splitting field of \(p(x)\) and is thus Galois.

We need only find 4 automorphisms in \(\text{Gal}(\mathbb{Q}(a_+)/\mathbb{Q})\) and show that they form a cyclic group. Note that the automorphisms are fully determined by their action on \(a_+\) (due to the starred expression above). Let \(\varphi\) be such an automorphism; then \(\varphi(a_+)\varphi(a_-) = \varphi(a_+)^2-2\), so \(\varphi(a_-) = (\varphi(a_+)^2-2)/\varphi(a_+)\). If \(\varphi(a_+) = a_+\), then clearly \(\varphi(a_-) = a_-\). If \(\varphi(a_+) = a_-\), then \(\varphi(a_-) = -a_-\). If \(\varphi(a_+) = a_-\), then a manual calculation gives \(\varphi(a_-) = -a_+\). Lastly if \(\varphi(a_+) = -a_-\), then we similarly get \(\varphi(a_-) = a_+\).

Associate \(a_+, -a_+, a_-, -a_-\) with the numbers \(1,2,3,4\) in order. Then the automorphisms listed above correspond to the permutations 1, \((1\ 2)(3\ 4)\), \((1\ 3\ 2\ 4)\) and \((1\ 4\ 2\ 3)\). It can be checked that \((1\ 3\ 2\ 4)\) generates this set and thus the conclusion follows.