6
$\begingroup$

Is it known what is the centralizer of the complex conjugation in the absolute Galois group (i.e. the Galois group of the field of complex algebraic numbers over the rationals)? and, what would be a good reference for this question?

$\endgroup$

1 Answer 1

14
$\begingroup$

If some element centralizes the complex conjugation, then it must preserve the real numbers as a set. Now, since any automorphism of the real numbers preserves the set of squares, it must preserve the order; and hence be continuous. Since $\mathbb Q$ is fixed, this implies that the real numbers are fixed pointwise. It follows that any element which centralized the complex conjugation must be the identity or the complex conjugation itself.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – user28172
    Feb 7, 2013 at 19:17
  • $\begingroup$ 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. $\endgroup$ Feb 7, 2013 at 19:22
  • $\begingroup$ 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. $\endgroup$
    – user28172
    Feb 7, 2013 at 21:07
  • $\begingroup$ 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. $\endgroup$ Feb 7, 2013 at 21:46
  • $\begingroup$ 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. $\endgroup$ Feb 7, 2013 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.