I am studying real closd fields for quantifier elimination of RCF and proved the following theorem.
Let $F$ be a field. Then the following properties are equivalent:
- The field $F$ is real closed.
- There is a unique ordering of $F$ whose positive cone is the set of squares of $F$ and every polynomial of $F[X]$ of odd degree has a root in $F$.
- The ring $F[i]=F[X]/(X^2+1)$ is an algebraically closed field.
In the case of $\mathbb{Q}$, I can see without a problem that it is not real closed using all of the properties, but I cannot figure out how the cone $\Sigma \mathbb{Q}^2 = \{\Sigma_iq_i^2:q_i \in \mathbb{Q} \}$ looks like. I've tried setting up an arbitrary equation of the form: $$ \Sigma_i q_i^2=p, \ p\in \mathbb{Q} $$ Intuitively, I would just multiply everything by the lcm of the denominators and reduce everyhing mod the numerator of p. This should lead to a solution since we can keep adding "ones". Is this correct or am I missing anything?