Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 18G (Galois Theory)

Let $a=\sqrt{2+\sqrt{5}}>0$ and $b=i/a$, so $b^2=2-\sqrt{5}$. The roots are $\pm a,\pm b$. The element $2+\sqrt{5}$ is not a square in $\mathbb{Q}(\sqrt{5})$: its field norm is $-1$, whereas the field norm of a square is a square of a rational number and hence nonnegative. Thus $[\mathbb{Q}(a):\mathbb{Q}(\sqrt{5})]=2$, and $[\mathbb{Q}(a):\mathbb{Q}]=4$. This also proves irreducibility of the quartic.

The splitting field is $K=\mathbb{Q}(a,i)$, since $b=i/a$ and $i=ab$. Because $\mathbb{Q}(a)\subset \mathbb{R}$, adjoining $i$ doubles the degree:The Galois group acts faithfully on the four roots and preserves the partition into the two opposite pairs $\{a,-a\}$ and $\{b,-b\}$, possibly interchanging the pairs. The full permutation group preserving this partition has order eight and is the dihedral group of a square. Since the Galois group has order eight, it is exactly that group. Equivalently its generators can be chosen as $r:a\mapsto b,b\mapsto -a$ and complex conjugation $s:a\mapsto a,b\mapsto -b$, with $r^4=s^2=1$ and $srs=r^{-1}$.

The positive roots can be writtenwhose squares are $2+\sqrt{3}$ and $2-\sqrt{3}$ and whose product is one. Therefore all roots lie in $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Conversely $a+b=\sqrt{6}$ and $a-b=\sqrt{2}$, giving $\sqrt{3}=\sqrt{6}/\sqrt{2}$ in the splitting field. HenceThe quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are distinct, so the compositum has this degree. Their independent sign changes are four commuting automorphisms, each nonidentity of order two. Thus