Bourbaki

Exercise 3, § 3

General Topology, IV, 388

Bourbaki marks this one as harder than the rest.

a) If ff is an isomorphism of the field R\mathbf{R} onto a subfield of R\mathbf{R} show that ff must be the identity mapping of R\mathbf{R} onto itself. [Remark that all rational numbers are invariant under ff and that

f(x2)=(f(x))2⩾0, f(x^2) = (f(x))^2 \geqslant 0,
so that ff is increasing].

b) Let K0K_0 be the field R(X)\mathbf{R}(X) of rational fractions in one indeterminate over R\mathbf{R}, ordered by taking the polynomials >0> 0 to be those whose leading coefficient is >0> 0. Let KK be the maximal ordered algebraic extension of K0K_0. Show that there is an order-preserving automorphism ff of KK such that f(ξ)=ξf(\xi) = \xi for all ξ∈R\xi \in \mathbf{R} and f(X)=X2(∗)f(X) = X^2(*).

Solution

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Transcribed from the printed volume. Not translated.