Clarification of an old question: Galois Groups of Finite Extensions of Fixed Fields The question is:

Let $\overline{K}$ be an algebraic closure of $K$, $\sigma\in G(\overline{K}:K)$ and $E=\{x\in\overline{K}:\sigma(x)=x\}$. Prove that every finite extension $L\mid E$ is a Galois extension and that $G(L:E)$ is cyclic.

The accepted answer of this old question is the next hint:
"If $x\in\overline{K}$, the set $\{\sigma^i (x):i\in\mathbb{N}\}$ is finite. It follows that a suitable polynomial of the form:
$$\prod_{i=0}^{n-1}(t-\sigma^i (x))\in E[t]$$
Using this you can hopefully prove that if $N$ is a normal closure of $L\mid E$, then $G(N:E)=\langle\sigma\rangle$."
Now, I do not know why a suitable polynomial of the form $\prod_{i=0}^{n-1}(t-\sigma^{i}(x))\in E[t]$, and assuming that, how could you prove $G(N:E)=\langle\sigma\rangle$?
A different way to solve the problem would be appreciated too.
 A: Since $x\in\overline{K}$, there exists $p\in K[t]$ such that $p(x)=0$, and then $p(\sigma(x))=\sigma(p(x))=0$. Hence the set $\{\sigma^i(x):i\in\mathbb{N}\}$ is contained in the set of roots of $p$ and so it is finite. Since $\sigma$ is an automorphism, there exists $n$ such that $\sigma^n(x)=x$ and $\sigma^i(x)\neq\sigma^j(x)$ for $i,j< n$, $i\neq j$.
Now consider the polynomial
$$q(t)=\prod_{i=0}^{n-1}(t-\sigma^i(x))\in \overline{K}[t]$$
Since $\sigma^n(x)=x$,
$$\sigma(q(t))=\prod_{i=0}^{n-1}(t-\sigma^{i+1}(x))=q(t)$$
and hence $q(t)\in E[t]$. Since $\sigma^i(x)\neq\sigma^j(x)$ for $i,j< n$ and $i\neq j$, $q(t)$ is separable and we have shown that every algebraic extension of $E$ is separable. If $L/E$ is finite, take $N/E$ a normal closure. Since, by hypothesis, $N^\sigma=E$, we have $<\sigma>=\operatorname{Gal}(N/N^\sigma)=\operatorname{Gal}(N/E)$ by the fundamental theorem of Galois theory. Since $\operatorname{Gal}(N/E)$ is abelian, all subextensions of $N/E$ are Galois: in particular $L/E$ is normal and $L=N$.
