Sppliting field and Galois theory and its Automorphism group

Question

I'm studying elementary Galois theory and came across these two questions:

1. If $L = Gal(x^n-1, \mathbb{Q})$ then $Aut_{\mathbb{Q}} L$ is abelian.

This question is followed by

2. If $ L = Gal(x^{n}-a, K)$ and $K$ contains a primitive root of unity, then $Aut_{K} L$ is abelian.

My approach to the first question:

Let $\xi \neq 1$ such that $\xi^n = 1$, and $\xi_k = \cos \frac{2k\pi}{n} + i\sin \frac{2 k\pi}{n}$, where $k = 0,1,\ldots, n-1$. We notice that $L=\mathbb{Q}[\xi]$. There exist isomorphisms $$\sigma_k : \mathbb{Q}[\xi] \rightarrow \mathbb{Q}[\xi_k]$$

such that $\sigma_k(\xi) = \xi_k = \xi^j$, for some $0\leq j \leq n-1$ and $\sigma_k\Big|_{\mathbb{Q}} = Id$. Thus $\sigma_k = \sigma^j$, for every $\sigma_k \in Aut_{\mathbb{Q}}L$, where $$\sigma : \mathbb{Q}[\xi] \rightarrow \mathbb{Q}[\xi]$$

is defined by $\sigma (\xi)= \xi$. We may conclude that $Aut_{\mathbb{Q}}L = \langle \sigma\rangle $ is cyclic, therefore is abelian.

If you could tell me if this approach is valid or not I would really appreaciate it. Any ideas where I could use the first question on the second?

Thank you for your time.

Answer 1

In your reasoning you are ignoring the fact, that not every root of $x^n-1$ generates the splitting field $L$. The chain of arguments could be like this:

1. The $n$-th roots of unity form a cyclic group,
2. Every generator $\xi$ of this group is a primitive element of $L$,
3. An automorphism $\sigma$ of $L/\mathbb{Q}$ is uniquely determined by the image $\sigma(\xi)=\xi^k$ for some $k$,
4. Take two automorphisms $\sigma,\tau$. Then $\sigma(\xi)=\xi^k$ and $\tau(\xi)=\xi^\ell$, hence $(\sigma\tau)(\xi)=\xi^{k+\ell}=(\tau\sigma)(\xi)$. Thus $\sigma\tau=\tau\sigma$ because automorphisms are uniquely determined by the image of $\xi$.