Show that formally real fields can be totally ordered into ordered fields A field $F$ is said to be formally real if the only solution to $\sum_{i=1}^{n}{x_{i}^{2}}=0$ is the trivial solution $x_{i}=0$, for any $n \in \mathbb{N}$.
How do I show that there exists a total ordering $\geq$ on a formally real field $F$ so that $(F,+,\cdot,\geq)$ becomes an ordered field?
 A: SKETCH: First show that if $F$ is formally real, then the equation $-1=\sum_{i=1}^nx_i^2$ has no solution in $F$ for any $n\in\Bbb Z^+$. Then let $\mathscr{P}$ be the set of prepositive cones in $F$. That is, $\mathscr{P}$ is the set of $P\subseteq F$ such that:


*

*for all $x,y\in P$, $x+y,xy\in P$:  

*for all $x\in F$, $x^2\in P$; and  

*$-1\notin P$.


Show that $\mathscr{P}\ne\varnothing$. HINT: Consider the set of all sums of squares of elements of $F$. Then use Zorn’s lemma to get $P\in\mathscr{P}$ that is maximal with respect to $\subseteq$. Finally, show that $P$ is a positive cone for $F$, i.e., that $F=P\cup(-P)$.
Added: It will be helpful to show that if $P\in\mathscr{P}$, and $x\in F\setminus P$, then either
$$P\cup\{x\}\subseteq P+Px\in\mathscr{P}\;,$$
or
$$P\cup\{-x\}\subseteq P+P(-x)\in\mathscr{P}\;,$$
where $P+Py=\{p+qy:p,q\in P\}$. The hardest part of this is showing that $-1$ cannot belong to both $P+Px$ and $P+P(-x)$. For this a preliminary result is useful: if $y\ne 0$, then at least one of $y$ and $-y$ does not belong to $P$. 
To prove this, suppose that $y,-y\in P$. Then $y(-y)=-y^2\in P$. But by hypothesis $(y^{-1})^2\in P$, so $-1=(-y^2)(y^{-1})^2\in P$, which is impossible.
Now if $-1\in P+Px$ and $-1\in P+P(-x)$, then there are $p,q,r,s\in P$ such that $p+qx=-1=r-sx$. And $1\in P$ (why?), so if $p'=p+1$ and $r'=r+1$, then $p',r'\in P$. Moreover, $p'+qx=0=r'-sx$, so
$$\frac{-p'}q=x=\frac{r'}s\;.$$
But then $-p's=r'q\in P$, and $p's\in P$, contradicting the previous result.
