I have encountered this argument while reading Tent and Ziegler's "Course in model theory", and I don't know why it is justified. It arises during the proof that every ordered field has a real closure.
Let $R$ be an ordered field, and let $f\in R[X]$ be a polynomial of odd degree without a zero. Now let $\alpha$ be a zero for $f$ in the algebraic closure of $R$ and put $L=R(\alpha)$. It follows from prior arguments that $L$ cannot be ordered (as $R$ was constructed to be maximal with regard to this property), and so $-1$ is a sum of squares in $L$. Why does it follow that there are polynomials $g_i \in R[X]$ of degree less than $n$ such that $f$ divides $1+\sum g_i^2$?
Thanks
The elements of $L$ are of the form $g(\alpha)$ with $g(x)\in R[x]$, $\deg g<n$. If $$ -1=\sum_{i=1}^kg_i(\alpha)^2 $$ for some $g_i(x)\in R[x], i=1,2,\ldots,k,$ then $\alpha$ is a zero of the polynomial $$ p(x)=1+\sum_i g_i(x)^2\in R[x]. $$ Because $R$ is ordered, $p(x)$ cannot be the constant polynomial zero (think about the coefficient of degree $2\max\{\deg g_i\mid i=1,2,\ldots,k\}$). Therefore $p(x)$ must be divisible by the minimal polynomial of $\alpha$, i.e. $f(x)$