Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 18H (Galois Theory)

The automorphisms satisfy ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$. The $n$ rotations $z\mapsto {\zeta }_n^{j}z$ and $n$ inversion maps $z\mapsto {\zeta }_n^j/z$ are all distinct, so the dihedral group $G$ has order $2n$, including the case $n=1$ when it has order two. The invariantis fixed by both generators.

Use two standard field-theoretic results: the fixed-field theorem for a finite group of automorphisms gives $[L:L^G]=|G|$ and makes $L/L^G$ Galois with group $G$; and a nonconstant rational function $P(z)/Q(z)$ in lowest terms satisfies $[\mathbb{C}(z):\mathbb{C}(P/Q)]=\max (\mathrm{deg}P,\mathrm{deg}Q)$. Here $w=(z^{2n}+1)/z^n$ has coprime numerator and denominator and degree $2n$. Thus $[L:\mathbb{C}(w)]=[L:L^G]=2n$. Since $\mathbb{C}(w)\subseteq L^G$, the tower law proves the dihedral fixed field in one rational variable:

For odd prime $n$, every subgroup is one of $\{1\}$, $⟨\sigma ⟩$, a reflection subgroup $H_j=⟨{\sigma }^j\tau ⟩$ for $0\le j<n$, or $G$. Indeed, a subgroup containing a nonidentity rotation contains all rotations; with any reflection it is then $G$. A subgroup with no nonidentity rotation contains at most one reflection, since the product of distinct reflections is a nonidentity rotation.

The Galois correspondence gives the following fixed fields. The trivial subgroup corresponds to $L=\mathbb{C}(z)$, the rotation subgroup to $\mathbb{C}(z^n)$, and $G$ to the base $\mathbb{C}(w)$. Since composition of the stated automorphisms gives $({\sigma }^j\tau )(z)={\zeta }_n^{-j}/z$, the reflection subgroup corresponds toThis rational invariant has degree two, hence generates the fixed field of its order-two subgroup, by the same degree argument as part (i). These fields have degrees respectively $2n,2,n,1$ over $\mathbb{C}(w)$.

An intermediate field is Galois over the base exactly when its corresponding subgroup is normal in $G$. The trivial subgroup, the rotations and $G$ are normal; no reflection subgroup is normal because $n$ is odd and greater than two. Hence the Galois intermediate extensions and groups areThe $n$ reflection fixed fields are not Galois over the base.