Why Galois group of $K[ζ_{p^r}]/K[ζ_p]$ is cyclic?

Question

Why Galois group of $K[ζ_{p^r}]/K[ζ_p]$,p is a odd prime,is cyclic? Why Galois group of $K[ζ_{2^r}]/K[ζ_4]$ is cyclic? ζ denotes a primitive root of unity.K is a number field. These facts are used in the proof of “local n-th powers for all but finitely many primes are indeed global powers when $K[ζ_{2^{ord_2(n)}]/K$ is cyclic” in Milne’s CFT Chapter VIII theorem 1.4,I’m almost sure he didn’t prove these facts before in FT,ANT or CFT.

For his notes: https://www.jmilne.org/math/CourseNotes/cft.html

Answer 1

In Chapter 6 of his ANT notes, Milne shows that $Gal(\mathbb{Q}[\zeta_{p^{r}}]/\mathbb{Q})\simeq(\mathbb{Z}/p^{r}\mathbb{Z})^{\times}$. In particular, $Gal(\mathbb{Q}[\zeta]/\mathbb{Q}{})\simeq(\mathbb{Z}/p\mathbb{Z}{})^{\times}$, and so $Gal(\mathbb{Q}[\zeta_{p^{r}}% ]/\mathbb{Q}[\zeta]\mathbb{)}$ is the kernel of $(\mathbb{Z}/p^{r}\mathbb{Z}{})^{\times}\rightarrow(\mathbb{Z}/p\mathbb{Z}{})^{\times}$. This is cyclic for $p$ odd --- see Chapter 3 of his Group Theory notes. When you replace $\mathbb{Q}$ with $K$, you replace the Galois group with a quotient [Oops, subgroup].

Answer 2

The classical proof goes on like this : first $\mathbf Q(\zeta_n)/\mathbf Q$ is Galois with group $(\mathbf Z/n)^*$, hence Gal($K(\zeta_n)/K$) is a subgroup of $(\mathbf Z/n)^*$. In the particular case where $n=p^r,(\mathbf Z/p^r)^*$ is cyclic if $p$ is odd, and is of type $(2, 2^{r-2})$ if $p=2$ and $r\ge 2$.

Just for fun, let us give a direct answer by induction to your question for $p$ odd. Denote $K_r = K(\zeta_{p^r})$, with Galois group $G_r$, and let $a$ be the maximal $r$ s.t. $K=K_r$. Obviously $K_{a+1}/K$ is cyclic of degree $p$ by Kummer theory. Suppose that $G_r$ is cyclic of order $p^{r-a}$ and let us show that $G_{r+1}$ is cyclic of order $p^{r+1-a}$. By construction, it's easy to see that $G_{r+1}$ is generated by at most 2 elements (use Burnside's "basis theorem" for $p$-groups if necessary). Then, if $G_{r+1}$ were not cyclic, it would be of type $(p, p^{r-a})$, or equivalently, there would exist $\alpha \in K^*$ s.t. $K_{r+1}$ is the compositum of $K(\sqrt [p] \alpha)$ and $K_r$. By Kummer theory above $K_r$, one would have $\zeta_r .\alpha ^{-1} \in ({K_r}^*)^p$. Norming down to $K$, one would get $\zeta_a \in {K^*}^p$ (with a coherent choice of roots of unity): contradiction.