# Finite separable extension over fixed field $K=L^\sigma$ is normal?

I'm trying to prove the following:

Let $L$ be an algebraically closed field and let $\sigma \in \mathrm{Aut}(L)$. I want to show that for any separable $\alpha \in L$ over $K = L^\sigma$ ($K$ is the fixed field under $\sigma$), we have that $K(\alpha)/K$ is a normal extension.

I've been stuck on this for quite some time now and started to wonder whether it is even true?

Basically what I managed to figure out is the following: If $n \in \mathbb N$ is chosen maximally such that $\alpha, \sigma(\alpha), \dots, \sigma^{n-1}(\alpha)$ are pairwise distinct, the minimal polynomial $f$ of $\alpha$ in $K[X]$ must divide

$$P = \prod_{i=0}^{n-1} (X - \sigma^i(\alpha)) \in K[X]$$

($P$ is in $K[X]$ because it is invariant under $\sigma$)

This implies that if $K(\alpha)$ is normal, we must have $K(\alpha) = K(\alpha, \sigma(\alpha), \dots, \sigma^{n-1}(\alpha))$, which is the case iff $K(\alpha)$ is invariant under $\sigma$.

I think that it might be possible to conlude $\sigma^i(\alpha) \in K(\alpha)$ by investigating the coefficients of $P$, but it seems a bit complicated: One could start out by observing

\begin{align*} \sigma(\alpha) \cdots \sigma^{n-1}(\alpha) \in K(\alpha) &\implies \sum_{i=1}^{n-1} \sigma(\alpha) \cdots \widehat{\sigma^i(\alpha)}\cdots \sigma^{n-1}(\alpha) \in K(\alpha)\\ \sum_{i=0}^{n-1} \sigma^i (\alpha ) \in K(\alpha) &\implies \sum_{} \sigma^i(\alpha) \sigma^j(\alpha)\in K(\alpha) \end{align*}

But I doubt that there is no other way. I would very much appreciate good hints rather than a complete solution.

Thank you =)

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The answer to the question in the title is certainly "no", since $\mathbf Q$ is perfect and there are lots of finite extensions of $\mathbf Q$ that aren't Galois. But I don't yet see how it connects with the body. – Dylan Moreland Sep 21 '11 at 0:52
@Dylan: Could you give an example of a finite extension of $\mathbb Q$ that is not normal? – Sam Sep 21 '11 at 0:54
@Sam: $\mathbb{Q}(2^{1/3})$ – Hans Parshall Sep 21 '11 at 0:55
Ah, right. Thanks to you both. Then - assuming that the claim in the body is right - there cannot be a $\sigma \in \mathrm{Aut}(\mathbb C)$ such that $\mathbb Q = \mathbb C ^\sigma$ (I thought there had to be such a $\sigma$ by some application of Zorn's lemma, but didn't really think it through properly...) – Sam Sep 21 '11 at 1:02
I have changed the title after the above comments, since it didn't really relate to the body. – Sam Sep 21 '11 at 1:27

The answer is yes if $\sigma$ has finite order. You have Artin's Theorem (Theorem V.2.15 in Hungerford's Algebra):
If $F$ is a field, $G$ a finite group of automorphisms of $G$ with fixed field $K$, then $L/K$ is finite Galois with group $G$.
Now, in your situation, $G$ is the subgroup generated by $\sigma$, which is finite if $\sigma$ has finite order. Artin's Theorem tells you that $L$ is finite Galois over $K$ with group $G$. Since $G$ is abelian, being cyclic, any subextension of $L/K$ is Galois over $K$.
Thanks. I think I've got it now: If $M/K$ is a finite (separable) extension, we can consider the normal closure $M'$ of $M$. Then we have that $M'/K$ is a finite Galois extension and $K = L^\sigma = M'^{\sigma'}$, where $\sigma' = \sigma|_{M'}$. Thus $\mathrm{Gal}(M'/K) = \langle \sigma' \rangle$ is cyclic, and this implies that $M \subseteq M'$ is normal over $K$. Indeed, $M/K$ is seen to be a cyclic Galois extension. – Sam Sep 21 '11 at 2:17