Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 =)

share|cite|improve this question
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
up vote 3 down vote accepted

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 $K$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.