Let \(\alpha = \sqrt{2 + \sqrt{2}}\) in \(\mathbb{C}\) and let \(f\) be the minimal polynomial of \(\alpha\) over \(\mathbb{Q}\). Let \(E\) be the splitting field for \(f\) over \(\mathbb{Q}\). Determine the Galois group \(\operatorname{Gal}(E/\mathbb{Q})\).
Solution
Let \[ \alpha=\sqrt{2+\sqrt2}, \qquad \beta=\sqrt{2-\sqrt2}. \]
1. The minimal polynomial of \(\alpha\) over \(\mathbb Q\) is\[ f(x)=x^4-4x^2+2. \]
Since \[ \alpha^2=2+\sqrt2, \] we have \[ (\alpha^2-2)^2=2, \] so \(f(\alpha)=0\). The polynomial \(f\) is Eisenstein at \(2\): every nonleading coefficient is divisible by \(2\), and the constant term \(2\) is not divisible by \(4\). Hence \(f\) is irreducible over \(\mathbb Q\). Therefore \(f\) is the minimal polynomial of \(\alpha\) and \[ [\mathbb Q(\alpha):\mathbb Q]=4. \]
2. The roots of \(f\) are\[ \pm\alpha, \qquad \pm\beta. \]
The equation \(f(x)=0\) is equivalent to \[ (x^2-2)^2=2, \] so \[ x^2=2\pm\sqrt2. \] Taking square roots gives precisely the four displayed roots.
3. The field \(\mathbb Q(\alpha)\) already contains \(\beta\).
We have \[ \sqrt2=\alpha^2-2\in\mathbb Q(\alpha). \] Also \[ \alpha\beta =\sqrt{(2+\sqrt2)(2-\sqrt2)} =\sqrt2, \] where the positive square root is taken because \(\alpha,\beta>0\). Thus \[ \beta=\frac{\sqrt2}{\alpha}\in\mathbb Q(\alpha). \]
4. The splitting field is\[ E=\mathbb Q(\alpha). \]
By 2 and 3, \(\mathbb Q(\alpha)\) contains all four roots \(\pm\alpha,\pm\beta\) of \(f\). Hence it is the splitting field of \(f\). Since \(f\) is separable over the characteristic-zero field \(\mathbb Q\), the extension \(E/\mathbb Q\) is Galois. By 1, \[ |\operatorname{Gal}(E/\mathbb Q)|=[E:\mathbb Q]=4. \]
5. There is an automorphism \(\sigma\in\operatorname{Gal}(E/\mathbb Q)\) satisfying\[ \sigma(\alpha)=\beta, \]and it has order \(4\).
Because \(\beta\) is another root of the irreducible polynomial \(f\), the assignment \(\alpha\mapsto\beta\) defines a \(\mathbb Q\)-embedding \[ \mathbb Q(\alpha)\longrightarrow E. \] Since \(E=\mathbb Q(\alpha)\) is finite over \(\mathbb Q\), this embedding is an automorphism; call it \(\sigma\).
Now \[ \sqrt2=\alpha^2-2, \] so \[ \sigma(\sqrt2)=\beta^2-2=(2-\sqrt2)-2=-\sqrt2. \] Using \(\beta=\sqrt2/\alpha\), \[ \sigma(\beta) =\frac{\sigma(\sqrt2)}{\sigma(\alpha)} =\frac{-\sqrt2}{\beta} =-\alpha. \] Therefore \[ \alpha\xmapsto{\sigma}\beta \xmapsto{\sigma}-\alpha \xmapsto{\sigma}-\beta \xmapsto{\sigma}\alpha. \] Thus \(\sigma\) has order \(4\).
6. Consequently\[ \operatorname{Gal}(E/\mathbb Q)\cong C_4. \]
By 4 the Galois group has order \(4\), and by 5 it contains an element of order \(4\). Hence the group is cyclic of order \(4\).