Exercise 9, § 2

Let q $\geq 0$ be a non-square natural number; let A denote the set $\{ 1, \sqrt{q} \}$ in the field $K = \mathbf{Q}(\sqrt{q})$: show that K is A-orderable. Show that there exists an extension E of K such that E is pythagorean, orderable, and that every odd degree polynomial over E has a root in E, but that E does not admit the structure of a maximal ordered field (consider a maximal A-orderable extension of K).