Bourbaki

Exercise 4, § 2

Algebra, VI, p. 37

a) Let K be a field, and P a subset of K satisfying conditions (AP_I), (AP_{II}) and (AP_{III}), and such that K2⊂PK^2 \subset P (where K2K^2 denotes the set of squares of elements of K); show that if x>0x > 0 in the ordering defined by P, then x−1>0x^{-1} > 0; deduce that the set K+∗K^*_+ of elements >0> 0 in K is a subgroup of the multiplicative group of K.

b) In the field Q\mathbf{Q}, the only set PP which satisfies the conditions of a) is the set of rational numbers ⩾0\geqslant 0 (in the usual ordering).

c) \* Let PP be a subset of KK satisfying (AP_I), (AP_{II}) and (AP_{III}), such that 1∈P1 \in P and such that x>0x > 0 implies x−1>0x^{-1} > 0 in the ordering defined by PP. Show that y2⩾z2y^2 \geqslant z^2 implies y⩾zy \geqslant z, for positive yy and zz. Deduce that, for arbitrary y>0y > 0, we have

12(y+y−1)⩾1−n−1 \frac{1}{2} (y + y^{-1}) \geqslant 1 - n^{-1}

for every integer n>0n > 0 (notice that (y+y−1)2m⩾(2mm)(y + y^{-1})^{2m} \geqslant \binom{2m}{m} for every integer m>0m > 0). Deduce that, if the ordered additive group structure defined by PP is archimedean (VI, p. 35, Ex. 31), then (y−z)′⩾0(y - z)' \geqslant 0 for every pair of elements y,zy, z of PP; if K′K' is the subfield of KK generated by PP then x2⩾0x^2 \geqslant 0 for all x∈K′x \in K'.

d) \* Let K=R(X)K = R(X) be the field of rational functions in one indeterminate over RR; let PP be the set consisting of 0 and the rational functions u∈Ku \in K such that u(t)u(t) is defined and >0> 0 for every real number tt. Show that PP satisfies the conditions of c) and generates KK, but that there exist elements v∈Kv \in K such that v2∉Pv^2 \notin P.

Solution

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