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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.

Attempt
Let $K = \bar{\mathbb{Q}}$.
Edit
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.

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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
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1 Answer

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.

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@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
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