Proposition 3

Let K be an ordered field and f an irreducible polynomial over K which changes sign between a and b in K. Then the extension $E = K[X]/(f)$ of K admits the structure of an ordered extension.

We will argue by induction on the degree n off. For $n = 1$ the proof is trivial. Suppose the result holds for degrees $\leq n - 1$, and let us prove it for degree n by contradiction; by Th. 1 we are thus assuming a relation of the form

Without loss of generality we may suppose that the $f_i$ have degrees $\leq n - 1$ (IV, p. 11, Cor). Then

where $h \neq 0$ has degree at most $n - 2$. Replacing X by a and b in the above inequality, we see that $h(a) f(a) > 0$ and $h(b) f(b) > 0$. Since f changes sign between a and b by hypothesis, we conclude that $h(a) h(b) < 0$. Then we have a similar inequality for one of the irreducible factors $g(X)$ of $h(X)$: that is $g(a)\ g(b) < 0$. But $1 + \sum_i P_i f_i^2(X) \equiv 0 \pmod{g(X)}$, which shows that the field $K[X]/(g)$ cannot be an ordered extension of $K$ (Th. 1), contrary to the induction hypothesis.