Let \(K = \QQ[\sqrt 2 + \sqrt 5]\).
Find \([K: \QQ]\).
Show that \(K/\QQ\) is Galois, and find the Galois group \(G\) of \(K/\QQ\).
Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \(\QQ\) and \(K\).
Solution
Put \[ \theta=\sqrt2+\sqrt5. \] Then \[ \theta^2=7+2\sqrt{10}, \qquad \sqrt{10}=\frac{\theta^2-7}{2}\in\mathbb Q(\theta). \] Moreover \[ \theta\sqrt{10}=5\sqrt2+2\sqrt5, \] so subtracting \(2\theta\) gives \[ 3\sqrt2=\theta(\sqrt{10}-2). \] Hence \(\sqrt2\in\mathbb Q(\theta)\), and then \(\sqrt5=\theta-\sqrt2\in\mathbb Q(\theta)\). Thus \[ K=\mathbb Q(\theta)=\mathbb Q(\sqrt2,\sqrt5). \] Since \(\sqrt5\notin\mathbb Q(\sqrt2)\), the degree is \[ [K:\mathbb Q]=4. \]
The field contains all four conjugates \[ \pm\sqrt2\pm\sqrt5 \] of \(\theta\), so \(K/\mathbb Q\) is Galois. Its automorphisms independently change the signs of \(\sqrt2\) and \(\sqrt5\), giving \[ G\cong C_2\times C_2. \] Let \(\sigma_2\) change the sign of \(\sqrt2\) and fix \(\sqrt5\), and let \(\sigma_5\) change the sign of \(\sqrt5\) and fix \(\sqrt2\). Then the Galois correspondence is \[ \begin{array}{c|c} H&K^H\\ \hline G&\mathbb Q\\ \langle\sigma_2\rangle&\mathbb Q(\sqrt5)\\ \langle\sigma_5\rangle&\mathbb Q(\sqrt2)\\ \langle\sigma_2\sigma_5\rangle&\mathbb Q(\sqrt{10})\\ \{1\}&K. \end{array} \] These are all subgroups of \(G\) and hence all intermediate fields.