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I have a doubt. Let $\mathbb{K}$ be a finite extension of $\mathbb{Q}$. May I infer that the number of distinct restrictions of $Gal(\mathbb{C}/\mathbb{Q})$ to $\mathbb{K}$ must be finite? If yes, why?

TheWanderer
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    I find hard to understand what you really meant by "distinct restrictions". If you mean that the restricted automorphisms are different, then the obvious answer is yes, as there cannot be more than $;[\Bbb K:\Bbb Q];$ different automorphisms (or embeddings into $\Bbb C,$) fixing $;\Bbb Q;$ of $;\Bbb K;$ . But if you meant whether there are infinite different automorphisms on $;Gal(\Bbb C/\Bbb Q);$ that restrict to the same finite automorphisms (embeddings) of $;\Bbb K/\Bbb Q;$ are different then...I don't know. – DonAntonio Mar 28 '18 at 12:29
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    It is better to call the field automorphism group of $\mathbb C$ over $\mathbb Q$ something other than Gal', maybe justAut', because (of course) this is not an algebraic extension, and Galois theory does not quite apply... – paul garrett Mar 28 '18 at 12:33
  • I'm asking whether there are infinite different automorphisms on $Gal(\mathbb{C}/\mathbb{Q})$ that restrict to the same finite automorphisms of $\mathbb{K}/\mathbb{Q}$. – TheWanderer Mar 28 '18 at 12:33
  • @DonAntonio What do you mean by the sentence the restricted automorphisms are different, then the obvious answer is yes, as there cannot be more than $[\mathbb{K}:\mathbb{Q}]$ different automorphisms fixing $\mathbb{Q}$ of $\mathbb{K}$? Could you be more precise, please? Do you mean $\sigma|_{\mathbb{K}}$, for $\sigma \in \mathbb{C}$? – TheWanderer Mar 28 '18 at 12:42
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    The number of restricted automorphisms must be finite, of course. That's what I meant. – DonAntonio Mar 28 '18 at 12:55

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I hope I am not mistaking the comments above, if I got it right, the question is: Given some $\varphi\in\mathrm{Aut}_{\mathbb{Q}}(\mathbb{K})$, are there inifinitely many $\psi\in\mathrm{Aut}(\mathbb{C})$ with $\psi\vert_{\mathbb{K}}=\varphi$?

The answer to this is yes, for example theorem 7 of this paper (link is taken from this question) states:

Every automorphism of a subfield of $\mathbb{C}$ may be extended to an automorphism of $\mathbb{C}$.

So, as $\mathbb{K}$ is finite over $\mathbb{Q}$, you might choose inifinitely many algebraically independent (over $\mathbb{K}$) elements $\alpha\in\mathbb{C}\setminus\mathbb{K}$, with minimal polynomials, say, $f_\alpha\in\mathbb{K}[x]$, and independently set each of them to arbitrary roots of the $\varphi(f_\alpha)$, to get inifinetly many different extensions to $\mathrm{Aut}(\mathbb{C})$ (which by construction restrict to $\varphi$ on $\mathbb{K}$).

user103697
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