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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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    @Alex: sorry about the $\sigma$. Can you at least give me a hint.2011-04-16

1 Answers 1

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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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    @Arturo: Thanks once again.2011-04-17