Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 18F (Galois Theory)

Let $a=3^{1/4}>0$. The four roots are $a,ia,-a,-ia$, so $K=\mathbb{Q}(a,i)$. Eisenstein's criterion at three gives $[\mathbb{Q}(a):\mathbb{Q}]=4$. This field is real, so adjoining $i$ doubles the degree and $[K:\mathbb{Q}]=8$.

Define $r(a)=ia$, $r(i)=i$, and complex conjugation $s(a)=a$, $s(i)=-i$. The first automorphism exists because over $\mathbb{Q}(i)$ the minimal polynomial of $a$ has degree four and all four roots are in $K$; the second is immediate. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct and exhaust the eight automorphisms. Thusthe dihedral group of order eight, acting on the four roots as the symmetries of a square. On every element of $K$, the action is determined by the displayed images of $a,i$, or by extending them linearly to the basis $a^j{i}^e$ with $0\le j<4$, $0\le e<2$.

Let $a=i\sqrt{2+\sqrt{2}}$ and $b=i\sqrt{2-\sqrt{2}}$. The roots are $\pm a,\pm b$. Since $\sqrt{2}=-a^2-2$ and $ab=-\sqrt{2}$, we have $b=-\sqrt{2}/a\in \mathbb{Q}(a)$. Hence the splitting field is $K=\mathbb{Q}(a)$. The quartic is Eisenstein at two, so its degree is four.

The automorphism $r$ sending $a$ to $b$ sends $\sqrt{2}$ to $-\sqrt{2}$, and therefore sends $b=-\sqrt{2}/a$ to $\sqrt{2}/b=-a$. Its successive images of $a$ are $a,b,-a,-b,a$. Thus $r$ has order four and supplies all automorphisms:Complex conjugation is $r^2$. The images of $a$ specify the full action because $K$ is generated by that one root.