2
$\begingroup$

Choose an embedding $\overline{\mathbf{Q}}\rightarrow \mathbf{C}$ from an algebraic closure of the field of rationals to the field of complex numbers.

Question 1: Is it true that $\mathbf{C}$ is isomorphic to $\overline{\mathbf{Q}}(T_{i\in I})$ a pure transcendental extension of degree the cardinality of some uncountable set $I$ ?

Question 2 Is there a concrete description of the (profinite ?) group $\mathrm{Aut}_{\overline{\mathbf{Q}}}(\mathbf{C})$ ?

Improve this question
$\endgroup$
9
  • 9
    $\begingroup$ For your question question 1: no this field is never algebraically closed ($T_i$ cannot have a square root). $\endgroup$
    – Simon Henry
    Commented Apr 11, 2016 at 21:53
  • 6
    $\begingroup$ Question 2. Without Choice, the group is trivial. With Choice, the group is huge but there is not a concrete description of single element (otherwise you wouldn't need Choice to show it exists). $\endgroup$
    – zeno
    Commented Apr 11, 2016 at 23:04
  • $\begingroup$ @SimonHenry thanks for your remark, you are right! $\endgroup$
    – Ofra
    Commented Apr 11, 2016 at 23:06
  • $\begingroup$ @zeno What do you mean by "Choice" ? $\endgroup$
    – Ofra
    Commented Apr 11, 2016 at 23:06
  • 4
    $\begingroup$ @zeno It may look pedantic, but I believe one should say that "without choice, the group may be trivial". Does one know what this group may look like without assuming choice? $\endgroup$
    – ACL
    Commented Apr 12, 2016 at 8:05

0

Reset to default

You must log in to answer this question.