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Let $K/F$ be an extension of fields and let $\alpha$ be an element of $K$ such that the set

$\quad X = \{ \varphi(\alpha) \; | \; \varphi \in \operatorname{Aut}(K/F) \}$

is finite. Then, if $K/F$ is a Galois extension (more generally, if the fixed field of $\operatorname{Aut}(K/F)$ is $F$), $\alpha$ is algebraic over $F$. In fact, consider the polynomial $p(t) = \prod_{\beta \in X} (t-\beta)$ (it has $\alpha$ among its roots). For every $\varphi \in \operatorname{Aut}(K/F)$ (with an abuse of notation I will also denote by $\varphi$ the extension of this map to a map $K[t] \rightarrow K[t]$), we have $\varphi(p(t)) = p(t)$. So the coefficients of $p(t)$ belongs to the fixed field of $\operatorname{Aut}(K/F)$.

Are there any counterexample to this statement if $K/F$ is not Galois?

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  • $\begingroup$ How did you reach the first assertion ? $\endgroup$ Feb 26 '15 at 17:59
  • $\begingroup$ @Fardad Pouran: I've edited my question $\endgroup$ Feb 26 '15 at 18:16
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A very easy counterexample is $\alpha \in \mathbb{R}$ any trascendental number over $\mathbb{Q}$. Since $\operatorname{Aut} (\mathbb{R} / \mathbb{Q}) = \{ 1_{\mathbb{R}}\}$, the set $\{ \varphi (\alpha) : \varphi \in \operatorname{Aut} (\mathbb{R} / \mathbb{Q}) \} = \{ \alpha \}$ is a finite set.

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