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

Let $\bar{\mathbb{Q}}$ be a (fixed) algebraic closure of $\mathbb{Q}$ and $\tau\in\bar{\mathbb{Q}},\tau\notin\mathbb{Q}.$ Let $E$ be a subfield of $\bar{\mathbb{Q}}$ maximal with respect to the condition $\tau\notin E.$ Show that every finite dimensional extension of $E$ is cyclic.

Let $K = \bar{\mathbb{Q}}$.
Since $\tau \notin E$ we can define $E=\left\{ \tau\in K \vert\alpha\left(\tau\right)\neq\tau\,\,\forall\:\alpha\in H\right\} $ be the fixed field of $H,$ a minimal closed subgroup of $\mathrm{Aut}(K/\mathbb{Q})$.

But then, such a subgroup would be generated by a single element. So $H$ is cyclic. Thus $\mathrm{Aut}(K/E)$ is a cyclic extension. Hence every finite extension, say, $M$ of $E$ is cyclic.

I am not really convinced by my argument. Hints and suggestions are very much welcomed. Thanks.

share|cite|improve this question
I can't make any sense of your solution. Firstly, $E$ is given to you, you can't just define it to be something. And anyway, your definition doesn't make any sense, with all those undefined $\sigma$ and unused $\alpha$. What you wanted to say is something like let $H=\text{Aut}(K/E)$. But after that you just jump to the conclusion that you are asked to prove with no justification. So I am not convinced by your argument either. – Alex B. Apr 16 '11 at 8:42
@Alex: sorry about the $\sigma$. Can you at least give me a hint. – Nana Apr 16 '11 at 12:45
up vote 3 down vote accepted

If $K\supset E$ is a finite extension and $K\neq E$ then $K\supset E(\tau)$. If $K\supset E$ is normal with Galois group $G$ and the group fixing $E(\tau)$ is $H\subset G$ then (by Galois correspondence) any proper subgroup of $G$ has to be contained in $H$, since for any $K\supset L\supset E$, $L\neq E$, we have $L\supset E(\tau)$. Therefore, if $g\in G$, $g\notin H$, then the cyclic subgroup generated by $g$ must be $G$, i.e. $G$ is cyclic. Hence also any $L\supset E$ with $K\supset L\supset E$ is cyclic.

share|cite|improve this answer
@User8268: Thanks very much for your answer. Can you please tell me what's wrong with my attempt. Thanks. – Nana Apr 17 '11 at 1:47
@Nana: Your attempt is nonsensical. The way to problem is set up, you cannot decide what $E$ is, so just saying "...define $E=$..." means you are already on the wrong part. Your definition makes use of $\tau$ as a free variable, which makes it at best confusing (given that $\tau$ is given and fixed). And if you meant to say that $E$ happens to be equal to what you give, you need to prove it, especially given that your definition seems to depend on the choice of $H$. – Arturo Magidin Apr 17 '11 at 2:27
@Arturo:Thanks. What I meant was since $E$ was given to be maximal, it could be defined that way...I guess I was wrong then. – Nana Apr 17 '11 at 13:27
@Nana: Whether or not it can be defined this way is something to be established. Note that you do not know whether there is a unique E with this property, so you certainly do not know that any $H$ as described will necessarily yield the E you happen to be looking at. (In fact, I'm pretty sure $E$ is not unique, so it would certainly not be equal to the field you described for any choice of $H$). – Arturo Magidin Apr 17 '11 at 13:56
@Arturo: Thanks once again. – Nana Apr 17 '11 at 14:51

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.