The group $G=\{ \sigma \in Aut(K) : \sigma|_\mathbb{Q} = id_\mathbb{Q} \}$ of $K=\mathbb{Q}(7^\frac{1}{5})$

Question

I'm trying to find the group $G=\{ \sigma \in Aut(K): \sigma|_\mathbb{Q} = id_\mathbb{Q} \}$ for $K=\mathbb{Q}(7^\frac{1}{5})$ and here is my reasoning:

Since $\mathbb{Q}\subseteq K$ is a finite extension and $\textrm{char}(\mathbb{Q})=0$ the extension is separable, hence $|S|=[K:\mathbb{Q}]_s=[K:\mathbb{Q}]=\textrm{deg}(f)=5$ where $f=x^5-7$ is the minimal polynomial of $\alpha=7^\frac{1}{5}$ and $S=\{\sigma:K\rightarrow \overline{\mathbb{Q}} :\sigma \hspace{0.2cm} \textrm{is an homomorphism and} \hspace{0.2cm} \sigma|_\mathbb{Q}=id_\mathbb{Q} \}$. Clearly, $G\subseteq S$.

Note that for each $j=1,...,4$, $\alpha\beta^j$ is a root of $f$, where $\beta=e^\frac{2\pi i}{5}$ and together with $\alpha$ they are all the roots of $f$. Hence $K\simeq \mathbb{Q}[x]/(f) \simeq \mathbb{Q}(\alpha\beta^j)$ for each $j=1,...,4$ by the isomorphism $\sigma_j: K\rightarrow \mathbb{Q}(\alpha\beta^j)$, $\sigma_j(\alpha)=\alpha\beta^j$, $\sigma_j|_\mathbb{Q}=id_\mathbb{Q}$. Since $\mathbb{Q}(\alpha\beta^j)\subseteq \overline{\mathbb{Q}}$ and there are $4$ distinct $\sigma_j$ we arrive that $S=\{id_K, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \}$. Because $\alpha\beta^j \notin K$ we get that $\sigma_j \notin G$. Therefore: $G=\{id_K\}$

If anyone could verify my reasoning and give me feedback would help me very much. Also any alternative proof is appreciated.

Answer 1

Looks good. A couple of comments on your proof. Any field automorphism of $K$ is automatically going to fix $\mathbb Q$ pointwise, so you can just write $G = \operatorname{Aut}K$.

Your proof contains a lot of fluff and can be shorted. There is no need to mention anything about separability (since we are in characteristic zero as you point out) or about the fields $K(\alpha \beta^j)$ or about embeddings of $K$ into $\overline{\mathbb Q}$.

The main idea is that any automorphism $\sigma$ of $K$ is going to send $\alpha = \sqrt[5]{7}$ to a root of $x^5= 7$. Since $K$ doesn't contain any such roots except for $\alpha$ itself, this means that $\sigma$ must fix $\alpha$ and hence $\sigma = \operatorname{id}_K$.