Exercise 4, § 2

An ordered field K is called Euclidean if, for all x \geq 0 in K, there exists y \in K such that x = y^2.

(a) Let K be a Euclidean ordered field and G a subfield of K such that there exists no extension of G contained in K, distinct from G and comparable with G. Show that, iff is a place of K with values in an ordered field L, which is an algebraic extension of G, and reducing to the identity on G, then f is equivalent to the canonical place of K with respect to G. (f may be considered to take its values in a maximal ordered field N which is an algebraic extension of L. Observe that G is Euclidean and that, if x \in G, y \in K and x < y, then f(y) = \infty or f(y) > x. If A and p are the ring and ideal of the placef, show successively that p \subset I(G), F(G) \subset A, I(G) \subset p and A \subset F(G), noting that N is comparable with G.)

(b) Let K be a Euclidean ordered field, f a place of K with values in a maximal ordered field L and A and p the ring and ideal off. Let G be a maximal subfield among the subfields E of K such that E \cap p = 0. Show that G \subset A and that G is algebraically closed in K and therefore Euclidean. Prove that there exists no extension of G contained in K, distinct from G and comparable with G. Moreover, the subfield of L generated by f(A) is algebraic over f(G) and f is therefore equivalent to the canonical place of K with respect to G.

(c) Let K be a maximal ordered field, L a maximal ordered subfield of K, distinct from K and such that K is comparable with L (Algebra, Chapter VI, § 2, Exercise 15). Show that there exists no place f of K with values in L reducing to the identity automorphism on L.