Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 18H (Galois Theory)

The cyclotomic Galois group theorem statesfor $b$ coprime to $n$, with $[\mathbb{Q}({\mu }_n):\mathbb{Q}]=\phi (n)$, where $\phi$ is the Euler totient function.

Choose $\alpha$ with ${\alpha }^n=a$. The roots of $x^n-a$ are $\alpha {\zeta }_n^j$, so $F=K(\alpha )$ contains every root. Conversely their ratios generate the roots of unity, so $F$ is precisely the splitting field of $x^n-a$ over $\mathbb{Q}$. In characteristic zero this polynomial is separable polynomial, since $a\ne 0$, and a splitting field of a separable polynomial is a Galois extension.

Each automorphism over $K$ sends $\alpha$ to $\alpha {\zeta }_n^j$. The map $\sigma \mapsto \sigma (\alpha )/\alpha$ embeds $\mathrm{Gal}(F/K)$ into the cyclic group ${\mu }_n$: it is multiplicative because the automorphisms fix roots of unity, and injective because $\alpha$ generates $F/K$. Hence this group is cyclic, of order dividing $n$.

For a Galois intermediate field $K/\mathbb{Q}$, restriction of automorphisms gives an exact sequenceSurjectivity uses the extension theorem for embeddings into a normal splitting field. Both kernel and quotient are abelian, so the commutator subgroup of $\mathrm{Gal}(F/\mathbb{Q})$ lies in the abelian kernel. Its next commutator subgroup is trivial. Thus

For $n=p$ prime, $[K:\mathbb{Q}]=p-1$ and the cyclic group $\mathrm{Gal}(F/K)$ has order either $1$ or $p$. ThereforeThese correspond respectively to $a$ being or not being a $p$th power in $K$. In fact it is a $p$th power in $K$ precisely when it is a $p$th power in $\mathbb{Q}$. To see the nontrivial implication, suppose $\alpha \in K$ and ${\alpha }^p=a$. Taking the field norm gives $b^p=a^{p-1}$ with $b=N_{K/\mathbb{Q}}(\alpha )\in {\mathbb{Q}}^{\times }$. Then $(a/b{)}^p=a$, so $a$ is already a rational $p$th power.

Both degrees occur: $a=1$ gives $p-1$, while $a=2$ is not a rational $p$th power and gives $p(p-1)$. This includes $p=2$, with possible degrees $1$ and $2$.