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Fix an algebraic closure $\bar{\mathbf{Q}}$ of $\mathbf{Q}$.

Let $B\subset \mathbf{P}^1_{\bar{\mathbf{Q}}}$ be a closed subscheme of finite cardinality.

Let $K$ be a number field such that $B$ can be defined over $K$. Let $B_K$ be a closed subscheme of $\mathbf{P}^1_K$ such that the base change to $\bar{\mathbf{Q}}$ is $B$.

Is the orbit of $B_K$ under the action of the absolute Galois group Gal$(\bar{\mathbf{Q}}/\mathbf{Q})$ finite? Is it a closed subscheme?

The answer is trivially yes if $K=\mathbf{Q}$. I expect the cardinality of the orbit to be less or equal to $[K:\mathbf{Q}]\cdot$#$B$ in general. But why?

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You expect something and are asking us why? (last line of the question) maybe you could tell us that better. =P –  Patrick Da Silva Jul 29 '11 at 20:46
    
The title asks about the orbit of an algebraic number, but the body asks about the orbit of a closed subscheme (of ${\bf P}^1_K$, etc.). If these are not the same, perhaps some editing is in order to bring title and body into alignment. –  Gerry Myerson Jul 29 '11 at 23:11
    
@Patrick. By "I expect that " I mean that "I would not be surprised if"...In any case, I might be wrong. I just thought it would be nice to add some personal intuition. –  Oen Jul 30 '11 at 0:33
    
@Gerry. There's not a big difference in the two questions. I just tried to minimize the length of the title. –  Oen Jul 30 '11 at 0:34
    
@Oen: I don't know any arithmetic geometry, but could you explain why you are considering the projective line $\mathbf{P}^1_{\overline{\mathbf{Q}}}$ instead of just, say, $\operatorname{Spec} \overline{\mathbf{Q}}[x]$? Does the presence of points at infinity simplify some issue under consideration? –  Zhen Lin Jul 30 '11 at 9:21

1 Answer 1

I don't know from closed subschemes, but I know something about algebraic numbers, and OP assures me in the comments that there's not much difference, so I'll answer the question in the title. A finite set of algebraic numbers generates a finite extension of the rationals, which has a normal closure that is also a finite extension of the rationals, and every element of the absolute Galois group must take each of the finitely many algebraic numbers to one of its finitely many conjugates in this normal closure, so, yes, the orbit of a finite set of algebraic numbers under the action of the absolute Galois group is finite.

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Ok. So set-theoretically you just add all the conjugates. –  Oen Jul 30 '11 at 9:38
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@Oen, I'd say the orbit of any one algebraic number is the set of all of its conjugates, and if the orbit of a set means the union of the orbits of the members of the set, then we're talking about the union of all the sets of conjugates. –  Gerry Myerson Jul 30 '11 at 12:38

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