# Cyclotomic Field Extension over arbitrary Field

Let $K$ be an arbitrary field (for example $\mathbb{Q}_p$) and $\zeta_n$ the $n$-th primitive root of unity.

My question is why is the automorphism group $\operatorname{Aut}(K(\zeta_n) \mid K)$ of the field extension $K(\zeta_n) \mid K$ always independent of "base" field $K$ (especially equals cyclic group $(\Bbb Z/n\Bbb Z)^*)$?

Or equivalently why each $\alpha \in \operatorname{Aut}(K(\zeta_n) \mid K)$ can map only $\zeta_n \mapsto \zeta_n ^k$ for $1 \le k \le n-1$?

There are many problems in your question.

You are making two statements that are not really equivalent but calling them so. And you claim a group is cyclic which it is not (for example, when $n=8$, you get a non-cyclic group of order 4. the Klein four-group.)

First statement is false. The automorphism group depends on $K$. For $K$ the field of rational numbers the automorphism group is the multiplicative group of invertible elements modulo $n$, which has order $\phi(n)$ (Euler phi-function).

Now in this case take $F$ to be an intermediate field between rationals and the $n$-th cyclotomic field, then Aut$\,(F(z_n)/F)$ will be a a proper subgroup of the above group.

The second statement uses one field-theoretic result and one group theoretic result. The number of roots for a polynomial equation with coefficients in a field is at most the degree.

Group theoretic-result needed: an automorphism of a group preserves the order of an element.

A field automorphism is also an automorphism of its multiplicative group of non-zero elements. A root of unity of order $n$ is exactly an element of order $n$ in the multiplicative group, which are found among $z_n^k$ for $1\le k\le n-1$. (Choose the $k$ to be coprime to $n$).