Timeline for Centraliser of the complex conjugation in the absolute Galois group
Current License: CC BY-SA 3.0
8 events
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| Feb 8, 2013 at 2:05 | comment | added | user28172 | Dear Andreas: Aha, indeed the uniqueness of the order structure in the case of the real algebraic numbers is completely elementary, so the whole thing is elementary as you say (after replacing mention of the real numbers with the real algebraic numbers). Very good. | |
| Feb 7, 2013 at 21:54 | comment | added | Andreas Thom | What really makes a difference here is that $\mathbb Q$ is dense in the real algebraic numbers (as well as the real numbers) in the order-sense. If this is not the case (i.e. if the real closed field is more general and contains infinitesimal elements etc.), one has to find indeed different arguments and use Artin-Schreier theory. | |
| Feb 7, 2013 at 21:46 | comment | added | Andreas Thom | You do not need to extend (if you do not want to), it is still just the same argument. If the real algebraic numbers have only one ordering, then any automorphism must preserve it; and then any automorphism must fix any real algebraic number, since any real number is determined by the set of rational numbers below it (no theory needed for this). It is true that I was secretly thinking about the complex numbers (in fact I misread the question), but it does not make any difference. | |
| Feb 7, 2013 at 21:07 | comment | added | user28172 | Dear Andreas: I agree that the main idea is the same. But I think your argument has a genuine gap: you don't explain why the given abstract automorphism of $\overline{\mathbf{Q}}$ admits an extension to $\mathbf{C}$ commuting with complex conjugation (and thus preserving the real field). You just assert this (implicitly) in your first sentence. The Artin-Schreier theory of real closed fields provides an algebraic version of that justification. The A-S theory was developed precisely to address "algebraic" aspects of complex conjugation, so it is a natural tool for such questions. | |
| Feb 7, 2013 at 19:22 | comment | added | Andreas Thom | The argument is just the same. An element is non-negative in the ordering if and only if it is a square - this is obviously invariant under automorphisms. No sophisticated theory is needed at this point. | |
| Feb 7, 2013 at 19:17 | comment | added | user28172 | When we extend the given automorphism $g$ to an automorphism $g'$ of the abstract field $\mathbf{C}$ there's no reason why $g'$ should commute with $c$, so no reason why $g'$ should preserve $\mathbf{R}$. Instead, let $R$ be the subfield of algebraic numbers in $\mathbf{R}$, so $\overline{\mathbf{Q}} = R(i)$. The condition on $g$ is that it preserves $R$, so we just have to show ${\rm{Aut}}(R)=1$. This requires genuine input from algebra: the Artin-Schreier theory of real closed fields, which gives the uniqueness of the ordering on $R$. That rescues the "continuity" idea used above. | |
| Feb 7, 2013 at 17:17 | vote | accept | G Gonzalez | ||
| Feb 7, 2013 at 15:44 | history | answered | Andreas Thom | CC BY-SA 3.0 |