Exercise 8, § 2

Let A be a nonempty subset of a commutative field K of characteristic $\neq 2$; then K is said to be A-orderable if no element of A is a sum of squares in K. We say that K is orderable if there exists a total ordering of K which is compatible with its ring structure. a) Show that if K is A-orderable then it is orderable (Ex. 7), and hence of characteristic (0). b) Show that pure extensions and algebraic extensions of odd degree of an A-orderable field K are A-orderable (argue as in Prop. 3 of VI, p. 23). c) Let K be an A-orderable field and let b be an element of K not of the form $ca - d$, where $a \in A$ and c and d are sums of squares in K. Show that the field $K(\sqrt{b})$ is A-orderable. d) An A-orderable field K is called maximal if no proper algebraic extension of K is A-orderable. Show that a maximal A-orderable field K has the following properties: 1. K is pythagorean, in other words every sum of squares is a square; 2. no element of A is a square; 3. every element of K which is not a square has the form $c^2u - d^2$, where $u \in A$; 4. every odd degree polynomial in $K[X]$ has at least one root in K (use b) and c)). e) Show that if K satisfies the conditions of d) then every element algebraic over K has degree $2^q$ over K for some q (argue as in Prop. 8 of VI, p. 26, by induction on the exponent of 2 in the degree of the element under consideration). Show that on the other hand no quadratic extension of K is A-orderable. Deduce that K is a maximal A-orderable field (use Galois theory and Prop. 12 of I, p. 77). f) Let K be an A-orderable field and $\Omega$ an algebraically closed extension of K. Show that there exists a maximal A-orderable field contained in $\Omega$ and containing K.