# Galois Groups of Finite Extensions of Fixed Fields

I am trying to prove the following proposition:

Let $L$ be an algebraically closed field, $g \in Aut(L)$ and $K=\{x \in L \; | \; g(x)=x\}$. Show that every finite extensions $E/K$ is a cyclic Galois extension.

So far my thoughts have led me to the following:

If $E=K$, then clearly $E/K$ is Galois and $Gal(E/K)=Aut(K/K)=i$, where $i$ is the identity automorphism, so it is trivially cyclic. Assume $[E:K]=n$ for some $n>1$. Since $E/K$ is a finite extension, it is algebraic. Note that if $\beta \in E\setminus F$ then we have the minimal polynomial $m_{\beta}(X)$ over $K[x]$. Since $K$ is fixed by $g$, we know that $g(\beta)$ is also a root of $m_{\beta}(X)$. We can also conclude that $g(\beta)\not \in K$ since $\beta \not \in K$.

Where to take any of these observations, I have yet to determine. I feel I must be missing something elementary, for I cannot see how to conclude this is a Galois extension from here.

Any tips would be greatly appreciated, especially over full solutions.

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If $g$ has finite order, then $L/K$ would be cyclic, and hence so would all intermediate extensions. It is possible to generalize this to the case of infinite order by recalling basic facts of infinite Galois theory. Yet something simpler is probably out there! (I deleted an earlier version of this comment because it had IMHO disturbing typos). –  Jyrki Lahtonen Jun 2 '12 at 19:09

Hint: Assume that $x(\in L)$ is algebraic over $K$. The set $\{g^i(x)\mid i\in\mathbf{N}\}$ is then finite, as $x$ has only finitely many conjugates over $K$. It follows that a suitable polynomial of the form $$\prod_{i=0}^{n-1}(T-g^i(x))\in K[T].$$ Using this you can hopefully prove that if $N$ is a normal closure of $E/K$, then $Gal(N/K)=\langle g\rangle$. The rest is easy.
I am sorry, but I don't get it. Why is a suitable polynomial of the form $\prod_{i=0,...,n-1}(T-g^{i}(x))\in K[T]$? And assuming that, How could you prove $G(N/K)=⟨g⟩$? Thank you. –  Gauloises Jun 18 at 17:12