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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
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2 Answers

up vote 1 down vote accepted

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.

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It is probably worth mentioning that the proof (and IIRC also the truth) of my first sentence depends on the axiom of choice. So if you don't accept choice you are in trouble. –  Jyrki Lahtonen Jun 16 at 6:47
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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)

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I think that is only true when we consider a Galois extension no? –  user38268 Jun 14 '12 at 11:51
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
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