Exercise 15, § 2

Let $E$ be an ordered field, let $K$ be a subfield of $E$ such that $E$ is algebraic over $K$, and let $R$ be a maximal ordered extension of $K$.

a) Among the elements of E – K, let x, be one whose minimal polynomial f over K has the least possible degree. Show that $f'(T)$ is a product of factors in R[T] of the form $(T - a_i)' + c_i^2$ ($a_i, c_i \in \mathbf{R}$), and first order factors T – b, with the b, pairwise distinct elements of K. Deduce that there exists an element $y_0$ in R such that $f(y_0) = 0$ and such that $x_0 - z$ and $y_0 - z$ have the same sign, in E and R respectively, for all $z \in K$ (cf. VI, p. 43. Ex. 26, c)).

b) Deduce from a) that there exists a K-isomorphism from the ordered field E onto a subfield of R. (First show that there exists such an isomorphism for the subfield $K(x_0)$ of E ; for any polynomial $g \in K[T]$ such that $\deg(g) < \deg(f)$, apply the same argument used for $f'$ in a) to evaluate the signs of $g(x_0)$ and $g(y_0)$; then apply Zorn's Lemma).

c) Show that if R and R' are two maximal ordered algebraic extensions of K then there exists precisely one K-isomorphism from R onto R' (apply b)).