I have trouble understanding the proof of theorem 1.2.2. in the book 'Real Algebraic Geometry' written by Bochnak, Coste and Roy. The theorem states this:
"Let $F$ be a field. Then the following statements are equivalent.
(i) $F$ is real closed.
(ii) There is a unique ordering of $F$ whose positive cone is the set of
squares in $F$ and every polynomial of $F[x]$ of odd degree has a root in $F$.
(iii) The ring $F(i) = F[x]/(x^2+1)$ is an algebraically closed field."
When proving (iii)$\to$(i), the authors state that $F$ is a real field. But I don't see why. In first place, I think that statement (iii) of the theorem should also say that $F(i)$ is a proper algebraic extension of $F$, so we don't have $i \in F$. Assuming this, as $i \not \in F$, $-1$ is not a square in $F$.
But I don't see why can't $-1$ be a sum of squares in $F$ instead.
Best Answer
If $i\not \in F$ and $\overline{F}= F(i)$ then let $A = \{ c^2, c\in F\}$ and say that $u\ge v$ iff $u-v\in A$.
$a^2+b^2\ge 0$:
Assume the opposite, so $\overline{F}= F(\sqrt{a^2+b^2}) = F(\sqrt{a+ib}\sqrt{a-ib})$.
$a,b\in F,i\not \in F$ gives that $\sqrt{a+ib}$ is $F$-conjugate to $\sqrt{a-ib}$. But $\overline{F}/F$ is a quadratic extension so an element times its $F$-conjugate is in $F$. This is a contradiction.
From there $-1$ is not a sum of squares, $F$ is an ordered field, its positive elements have a square root, its negative elements don't, and $[\overline{F}:F]=2$ gives that every polynomial factors in a product of degree $\le 2$ poynomials, so odd degree polynomials have a root.