Exercise 27, § 2
Algebra, VI, p. 43
Bourbaki marks this one as harder than the rest.
Let K be an ordered field, let E be a purely transcendental extension of K and let be a transcendence basis For E (V, p. 109). a) \* If K is archimedean then E has a structure of an ordered extension of K for which E is comparable to K if and only if the cardinality of I is at most equal to that of a transcendence basis for R over K (where K is considered as an ordered subfield of R, cf. Ex. 26, a)); the set of such structures is then bijective with the set of injective maps from I into R such that is algebraically independent over K. ■
b) If K is not archimedean then there always exists (at least) one structure on E of an ordered extension of K such that E is comparable to K (when I contains only one element, use Ex. 24 and notice that Q does not have a supremum in K ; pass to the general case by Zorn's Lemma).
Solution
No solution has been written yet.