# Characterization of an element being algebraic over $\mathbb{Q}$.

Let $Aut(\mathbb{C}/\mathbb{Q})$ be the set of field automorphisms of $\mathbb{C}$ over $\mathbb{Q}$ (in short, all field automorphisms of $\mathbb{C}$). Let $x$ be an element of $\mathbb{C}$ such that the set $\{\sigma(x)|\sigma \in Aut(\mathbb{C}/\mathbb{Q})\}$ is finite. Is it true then that $x$ is algebraic over $\mathbb{Q}$ (and if so, why?) ?

It's being used in the following paper about elliptic curves to prove that a elliptic curve with complex multiplication has modular invariant $j$ which is algebraic over $\mathbb{Q}$: http://www.math.tifr.res.in/~eghate/cm.pdf

Any help would be appreciated.

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Note that this is only true if the axiom of choice is assumed: otherwise $Aut(\mathbb C)$ might, for example, be finite, hence every $x\in \mathbb C$ would satisfy your property. – Generic Human Jun 14 '12 at 11:55

All the permutations of any transcendence basis can be lifted to an automorphism $\sigma$ of $\mathbb{C}$. The cardinality of any transcendence basis of $\mathbb{C}/\mathbb{Q}$ is uncountable. So if $x$ is transcendental over $\mathbb{Q}$, it can be included to a transcendence basis, and hence the set $$\{\sigma(x)\mid\sigma\in Aut(\mathbb{C})\}$$ would be uncountable. The claim follows.
The polynomial $f(y)=\prod (y-\sigma(x))$ is invariant under $Aut(\mathbb{C}/\mathbb{Q})$, hence its coeffecients must lie in $\mathbb{Q}$. (provided I remember my Galois theory correctly)
As Generic Human comments, AC (or Zorn's lemma) is needed to prove that for each number $z\in\mathbb{C}\setminus\mathbb{Q}$ there is an automorphism $\sigma$ of $\mathbb{C}$ such that $\sigma(z)\neq z$. – Jyrki Lahtonen Jun 14 '12 at 12:17