Bourbaki

Exercise 28, § 2

Algebra, VI, p. 44

It is set in small type as supplementary.

a) Let K be a subfield of R and let 0 be a real number, algebraic over K. Show that the number of structures on K(θ)\mathbf{K}(\theta) of an ordered extension of K is equal to the number of real conjugates of 0 (use Ex. 14, a) (VI, p. 40) and 26, a)). b) Let K be a subfield of R, equipped with nn distinct structures of an ordered extension of Q, all archimedean ; let PiP_i (1≤i≤n1 \leq i \leq n) be the sets of positive elements under these orderings. Show that there exists a family of nn elements bib_i of K (1≤i≤n1 \leq i \leq n) such that bi∈Pib_i \in P_i for all i and bi∉Pkb_i \notin P_k for all k≠ik \neq i. (For each pair of distinct indices i,ki, k, let aik∈Pka_{ik} \in P_k be such that aik∉Pia_{ik} \notin P_i; consider the elements

ci=∑k≠i(1+aik1−aik)2r c_i = \sum_{k \neq i} \left( \frac{1 + a_{ik}}{1 - a_{ik}} \right)^{2r}
where r is a sufficiently large integer.) c) Let K=Q(θ)K = \mathbf{Q}(\theta) be an algebraic extension of Q contained in R, and let nn be the number of real conjugates of θ\theta. Let C be the set of sums of squares of elements of K ; show that C∗=C∩K∗C^* = C \cap K^* is a subgroup of K∗K^* and that (K∗:C∗)=2n(K^*:C^*) = 2^n. (Use b) and Cor. 2 of VI, p. 23.) \*

Solution

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Transcribed from the printed volume. Not translated.