Timeline for Automorphisms of $\mathbb{C}$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 7, 2012 at 10:11 | comment | added | Emil Jeřábek | @Bird: Why is $\mathrm{Aut}(\overline{\mathbb Q})$ a subgroup of $\mathrm{Aut}(\mathbb C)$ as a group? That’s the question no one here was able to answer so far. | |
| Apr 6, 2012 at 19:05 | comment | added | expmat | @Emerton: you are right that the simpler question was my original one but, nonetheless, the other question is also interesting... ;) | |
| Apr 6, 2012 at 1:16 | answer | added | Pat Devlin | timeline score: 14 | |
| Apr 5, 2012 at 2:57 | comment | added | Emerton | The natural map is from $Aut(\mathbb C)$ to $G_{\mathbb Q}$ (given by restriction), not the other way around. As has now been noted in several comments and answers, this map is surjective. On the other hand, this does not answer your first question, as to whether $G_{\mathbb Q}$ is a subgroup of $Aut(\mathbb C)$. In fact, you probably don't actually care about this question; the surjectivity is what you seemed interested in. Nevertheless, I imagine that the answer is no, in the sense that the surjection $Aut(\mathbb C) \to G_{\mathbb Q}$ presumably doesn't split. Regards, | |
| Apr 4, 2012 at 21:46 | answer | added | Daniel Miller | timeline score: 14 | |
| Apr 4, 2012 at 19:46 | comment | added | Emil Jeřábek | I see, that’s a different Zorn lemma argument than I had in mind. | |
| Apr 4, 2012 at 18:12 | comment | added | user20421 | @Emil: Sorry but I don't really understand. For any automorphism $g$ of $\bar{\mathbb Q}$, we could construct partially ordered set $S$={ ($L$,$g_L$) | $L$ subfield of ${\mathbb C}$, $g_L$ automorphism of $L$ and $g_L|_{\bar{\mathbb Q}}=g$ }, then use Zorn's lemma? | |
| Apr 4, 2012 at 17:16 | comment | added | user20421 | For the "simpler" one, I think "Zorn's lemma" argument can work. | |
| Apr 4, 2012 at 17:08 | comment | added | Ralph | That's in essential Theorem V.2.8 in Lang's algebra book. "in essential", because you first have to choose a transcendence base $T$ of $\mathbb{C}|\bar{\mathbb{Q}}$, extend your automorphism $f$ to $\bar{\mathbb{Q}}(T)$ by $f|T = id_T$ and then you can apply the cited theorem. | |
| Apr 4, 2012 at 16:58 | comment | added | Emil Jeřábek | See e.g. en.wikipedia.org/wiki/Transcendence_degree#Applications . | |
| Apr 4, 2012 at 16:26 | comment | added | expmat | Can you point me to a reference for this fact? | |
| Apr 4, 2012 at 16:22 | comment | added | Emil Jeřábek | Yes, any automorphism of $\overline{\mathbb Q}$ can be extended to an automorphism of $\mathbb C$, hence $\mathrm{Aut}(\overline{\mathbb Q})$ is a quotient of $\mathrm{Aut}(\mathbb C)$. This holds for any extension pair of algebraically closed fields. | |
| Apr 4, 2012 at 16:11 | history | edited | expmat | CC BY-SA 3.0 |
added 122 characters in body
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| Apr 4, 2012 at 14:19 | history | asked | expmat | CC BY-SA 3.0 |