How to find out the type of automorphism of this field extension of Q? Let $\alpha$ = $2^{1/5}$ $\in\mathbb R$ and $\xi$ = $e^{2\pi i/5}$. Let $K=\mathbb Q[\alpha , \xi]$. Pick the correct statements from below:

*

*There exists a field automorphism $\sigma$ of $\mathbb C$ such that $\sigma (K)=K$ and $\sigma \not= id$.

*There exists a field automorphism $\sigma$ of $\mathbb C$ such that $\sigma(K)\not=K$.

*There exists a finite extension $E$ of $\mathbb Q$ such that $K\subseteq E$ and $\sigma(K)\subseteq E$ for every field automorphism $\sigma$ of $E$.

*For all field automorphisms $\sigma$ of $K$, $\sigma(\alpha\xi)$ = $\alpha\xi$.

I am preparing for an exam so I was solving previous year questions. I studied field theory but I did not study this part. So when this question came to me, I studied about the field automorphisms and the field extensions of $\mathbb Q$  through some research papers on google and I got to know that there exist only identity automorphisms about rationals. But I could not find out about the similar or this particular example. I have seen the example of the $\mathbb Q[2^{1/3}, e^{2\pi i/3}]$ but that wasn't in much detail so I couldn't get the idea for this question. Any help or hints would be appreciated.
 A: 1)) As noted in the comments, the complex conjugation $\sigma$ is a non-identity field automorphism of $\Bbb C$. We have $\sigma(\alpha)=\alpha\in K$, $\sigma(\xi)=1/xi\in K$, so $\sigma(K)\subset K$. Then $K=\sigma(\sigma(K))\subset \sigma(K)$, so $K=\sigma(K)$.
2)) Let $\sigma$ be any field automorphism of $\Bbb C$. Then $\sigma(1)=1$. Let $n$ be any natural number. Since $n=1+1+\dots+1$ (the sum of $n$ $1$’s), $\sigma(n)= \sigma(1)+ \sigma(1)+\dots+\sigma(1)= 1+1+\dots+1=n$. Since $0=\sigma(0)=\sigma(n-n)= \sigma(n)+ \sigma(-n)$, we have $\sigma(-n)=-\sigma(n)=-n$. Let $m$ be any natural number. Then  $\sigma(n/m)\sigma(m)=\sigma(n)$, so $\sigma(n/m)= \sigma(n)/ \sigma(m)=n/m$. That is $\sigma(q)=q$ for each $q\in\Bbb Q$.
Since $\xi\in K$ and $K$ is a field, $K$ contains all powers of $\xi$, which are all complex roots of an equation $x^5=1$. Since
$1=\sigma(1)=\sigma(\xi^5)= \sigma(\xi)^5$,  $\sigma(\xi)$  is such root, so $\sigma(\xi)\in K$. Since $\alpha\in K$ and
all complex roots of the equation $x^5=1$ are contained in $K$, $K$ contains all complex roots of the equation $x^5=\alpha^5=2$.
Since $2=\sigma(2)=\sigma(\alpha^5)= \sigma(\alpha)^5$,  $\sigma(\alpha)$  is such root, so $\sigma(\xi)\in K$.
Since $K=\Bbb Q[\alpha,\xi]$, and $\sigma(\Bbb Q)\subset \Bbb Q\subset K$, $\sigma(\alpha)\in K$, and  $\sigma(\xi)\in K$, we have $\sigma(K)\subset K$. Then $K=\sigma(\sigma(K))\subset \sigma(K)$, so $K=\sigma(K)$.
3)) Since $K=\Bbb Q[\alpha,\xi]$ is an extension of $\Bbb Q$ by finitely many elements, algebraic over $\Bbb Q$, $K$ is a finite extension of $\Bbb Q$. Thus, by the answer to the previous question, it suffices to put $E=K$.
4)) If $\sigma$ is  complex conjugation then $\sigma(\alpha\xi)=\alpha/\xi\ne\alpha\xi$.
