Question regarding finding fixed field of extension This question was part of an assignment which i am solving of a university in which I am not  student.

Let $\alpha = {{2}^{1/5}} \in \mathbb{R}$ and e =  $exp([2\pi i]/S)$ . Let $K=\mathbb{Q}(\alpha e)$. Then pick Correct statements.

1 There exists a field automorphisms $\sigma$ of $\mathbb{C}$ such that $\sigma(K)= K$ and $\sigma \neq id$.


*There exists a field automorphisms $\sigma$ of $\mathbb{C}$ such that $\sigma(K) \neq  K$
3.For all field automorphisms $\sigma$ of K , $\sigma(\alpha e)=\alpha e$.
4 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.
I was confused what S is but then I took S as 5 to solve the problem as I really wanna learn how to do it. I think taking S doesn't makes sense.
I have proved 4 correct. But rest i am unable to do. They are related to finding field to K in $\mathbb{Q}$ and I am unable to find them . SO, can you please tell how to do that .
thank you !!
 A: I am sure that $S=5$, it is the most likely answer to that conundrum.
Note that $(\alpha e)^5=2$, so the minimal polynomial of this particular element is $x^5 - 2$ (shown irreducible by Eisenstein on the shifted polynomial) i.e. the extension $\mathbb Q(\alpha e)$ has degree $5$.
Furthermore, the other roots of $x^5-2$ are $\alpha,\alpha e^2,\alpha e^3$ and $\alpha e^4$.
The key ideas used in the proofs of all the above questions is : every automorphism sends a root of a polynomial, to another root of the same polynomial.

1
Suppose that $\sigma(K) = K$ and $\sigma \neq id$. Since $\sigma(\mathbb Q) = \mathbb Q$, we cannot have $\sigma(\alpha e) = \alpha e$ since the set fixed by $\sigma$ would then be the whole of $K$ , since it is a field containing $\mathbb Q$ and $\alpha e$.
Therefore, $\sigma(\alpha e) \neq \alpha e$, but it must be a root of $x^5-2=0$, hence is one of $e,\alpha e^2,\alpha e^3,\alpha e^4$. We show that none of these can belong in $K$.
Indeed, suppose that $\alpha e^n$ belongs in $K$ for $n \not \equiv 1 \mod 5$. Then, we know that $\alpha e \in K$, so taking the quotient, $e^l \in K$ for some $l$ not a multiple of $5$. By Bezout's lemma there is a $d$ with $dl \equiv 1 \mod 5$, so $(e^{l})^d = e^{ld} = e$ is also in $K$.
But then, $K$ is an extension of $\mathbb Q(e)$ so $[K:\mathbb Q]$ must be a multiple of $[\mathbb Q(e) : \mathbb Q]$. The former is $5$, and the latter is $4$, by the minimal polynomial of $e$ being $x^4+x^3+x^2+x+1$, so this is not true.
Consequently, no other root of $x^5-2$ belongs in $K$, therefore $\sigma(K) \neq K$ for any $\sigma \neq id$.

2
This one is pretty clear : any field homomorphism from $K$ is defined by what it does with $\alpha e$ because it always fixes the rational numbers. So simply define $\sigma : K \to \mathbb Q(\alpha)$ by $\alpha e \to \alpha$. This is well defined and extends to $\mathbb C$, via fixing all other elements of $\mathbb C$, and this qualifies as an automorphism of $\mathbb C$ which sends $K$ to a field not equal to $K$.

3
The image of $\sigma : K \to \mathbb C$ must contain a root of $x^5-2 = 0$. If the image equals $K$ then this root can only be $\alpha e$ as shown in part $1$. Consequently $\sigma(\alpha e) = \alpha e$.

4
This one is easy : consider the field consisting of all roots of $x^5-2 = 0$ i.e. $K(\alpha,\alpha e,\alpha e^2,\alpha e^3,\alpha e^4)$. Any automorphism must send $\alpha e$ to a root of $x^5-2 = 0$ but this extension contains all the roots, so it will contain the image of $K$ under any automorphism of $\mathbb C$.
