3
$\begingroup$

Is there such an example? Or it is known that a pseudo algebraically closed field which is a finite extension of a formally real field is algebraically closed?

$\endgroup$

1 Answer 1

2
$\begingroup$

As far as I can tell, you can take your formally real field to be the field $\mathbb{Q}^{tr}$ of totally real algebraic numbers (see this paper for a description of its Galois group). Then (according to Wikipedia), adjoining a square root of $-1$ gives you a pseudo algebraically closed field.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .