Remark 26

a) Let K be a commutative field, S a subset of K such that K is a maximal S-orderable field (chap. VI, § 2, exerc. 8). Let $f$ be a polynomial of $K[X]$, L its field of roots. Show that if, for every (total) order structure on K compatible with its ring structure, L admits a structure of an ordered extension of K, then necessarily $L = K$. (Argue by contradiction: proceeding as in exerc. 8 e) of chap. VI, § 2, reduce to the case where one would have $[L : K] = 2$. Conclude by observing that if $b \in K$ is not a square, there exists a total order structure on K, compatible with its ring structure, for which $b < 0$ (cf. chap. VI, § 2, no. 3, lemma of th. 1)).

b) Extend the results of No. 3 (with the exception of prop. 6) to the case where K is a maximal S-orderable field (). (Begin by proving the analogue of prop. 5, using a). Establish next the analogue of cor. 2 of th. 2 for the case where the commutative algebra B consists of hermitian endomorphisms, using prop. 10 of chap. VIII, § 9, no. 4. Pass to the case of a normal endomorphism u for A = K(i), observing that one can then write $u = v + i w$, where v and w are hermitian and permutable. Finally, if A is the field of quaternions over K, E0 the space E considered as a vector space of dimension 2n over K(i), and if one sets $\Phi(x, y) = \Phi_1(x, y) + \Phi_2(x, y)i$, where $\Phi_1$ and $\Phi_2$ take their values in K(i), observe that if u is normal for $\Phi$, it is also normal for $\Phi_1$.)