My question
If $E$ is a formally real extension of an ordered field $F$, does $E$ always admit an ordering compatible with $F$?
Less ambitious: What if $E$ is a formally real simple algebraic extension?
Background
I'm reading Section 6.2 on real fields in the book Abstract Algebra by Pierre Antoine Grillet. My attention is caught by the following:
Proposition 2.1. If $F$ is a formally real field and $\alpha^2\in F$, $\alpha^2>0$, then $F(\alpha)$ is formally real.
Proposition 2.2. If $F$ is a formally real field, then every finite extension of $F$ of odd degree is formally real.
The proofs are done by contradiction: If $-1$ is a sum of squares in a simple extension $F(\alpha)$, then… See my imitated proofs below for more details.
My efforts
I know that $E$ and $F$ don't necessarily have compatible orderings. As a simple example, there are two different ways to make $\mathbb{Q}(\sqrt{2})$ an ordered field: either $\sqrt{2}>0$ or $-\sqrt{2}>0$. Each is a simple extension of the other, but their orderings are not compatible.
For simplicity, let's call $E$ an order extension of an ordered field $F$ iff $E$ is an ordered field that extends $F$, and their orderings are compatible.
I made a slight change to the proof that a field is formally real iff $-1$ is not a sum of squares. Combined with the original proofs to the two propositions above, I was able to make some improvements. Most parts of the lengthy arguments below are probably standard to anyone familiar with the subject. I include them only for completeness.
If $F$ is an ordered field, and if $\alpha^2\in F$, $\alpha^2>0$, then $F(\alpha)$ can be made an order extension of $F$.
Without loss of generality, assume that $\alpha\notin F$. Let $S$ be the set of all nonzero members of $F(\alpha)$ that can be written in the form
\begin{align}
a_1z_1^2+\cdots+a_nz_n^2,
\end{align}
where $a_1,\dots,a_n$ are positive members of $F$, and $z_1,\dots,z_n$ are nonzero members of $F(\alpha)$. All nonzero suqares in $F(\alpha)$ are members of $S$. If $-1\in S$, then
\begin{align}
-1=\sum_{i=1}^{n}a_i(x_i+y_i\alpha)^2=\sum_{i=1}^na_i(x_i^2+\alpha^2y_i^2)+2\alpha\sum_{i=1}^na_ix_iy_i
\end{align}
for some $x_i,y_i\in F$. Since each $a_i(x_i^2+\alpha^2y_i^2)$ is in $F$, the sum of all $x_iy_i$ must be zero. But then $-1$ is a sum of positive members of $F$, impossible. Hence $-1\notin S$ (in fact, this already proves $F(\alpha)$ to be formally real).
The set $S$ contains every positive $a\in F$, because
\begin{align}
a=\frac{a}{\alpha^2}\alpha^2.
\end{align}
It is a multiplicative group. Indeed, $S$ contains $1\in F$, $S$ is closed under multiplication, and
\begin{align}
\frac{1}{z}=\frac{z}{z^2}\in S
\end{align}
for all $z\in S$. This also shows that $S$ is closed under addition: If $z,w\in S$, then $1/w\in S$, so $z+w\neq 0$. Otherwise, $z/w=-1\in S$.
The rest is very similar to the proof that a field is formally real iff $-1$ is not a sum of squares: Let $\mathscr{A}$ be the collection of all subsets $A$ of $F(\alpha)-\{0\}$ with the following properties:
- $-1\notin A$, but $S\subseteq A$ (in particular, all nonzero squares in $F(\alpha)$ are in $A$);
- $A$ is a multiplicative group;
- $A$ is closed under addition.
Then $\mathscr{A}$ is nonempty, and $\subseteq$ defines a partial ordering on it. If $\mathscr{C}$ is a chain in $\mathscr{A}$, then $\bigcup\mathscr{C}$ is an upper bound of $\mathscr{C}$. By Zorn's lemma, $\mathscr{A}$ has a maximal member $P$. Given nonzero $z\in F(\alpha)$, if $-z\notin P$, then put
\begin{align}
P'=\{a+bz:\text{$a,b\in P\cup\{0\}$, $a$ and $b$ are not both zero}\}.
\end{align}
It's not hard to show that $P'$ is a multiplicative group closed under addition, $P\subseteq P'$, and $-1\notin P'$. By maximality, $P'=P$. Hence $z\in P$. Since $z$ is arbitrary, $F(\alpha)-\{0\}$ is partitioned by $P$, $-P$ and $\{0\}$. We can define an ordering on $F(\alpha)$ by letting $P$ be the set of all positive members. Since all positive members of $F$ are in $S\subseteq P$, this ordering is compatible with the original one on $F$. This completes the proof.
This $S$ can also be used to imrove Proposition 2.2. The only difference from Proposition 2.1 is in showing that $-1\notin S$.
Suppose that $F$ is an ordered field, $E$ is a finite extension of $F$, and $[E:F]$ is odd. Then $E$ can be made an order extension of $F$.
We prove by induction on the degree of extension. It suffices to consider simple extensions $F(\alpha)$ of odd degrees $m$. The base case $m=1$ is trivial. If $m>1$, then let $S$ be as before. Put $p(X)=\mathrm{Irr}(\alpha:F)$, so the degree of $p(X)$ is $m$. If
\begin{align}
-1=a_1f_1(\alpha)^2+\cdots+a_nf_n(\alpha)^2
\end{align}
for some positive $a_1,\dots,a_n\in F$ and polynomials $f_1(X),\dots,f(X)\in F[X]$ of degree less than $m$, then
\begin{align}
1+a_1f_1(X)^2+\cdots+a_nf_n(X)^2=p(X)g(X)
\end{align}
for some $g(X)\in F[X]$. The leading coefficient of each $f_i(X)^2$ is positive, so the degree of $p(X)g(X)$ is an odd number less than $2m$. It follows that $g(X)$ is of an odd degree less than $m$. Let $\beta$ be a root to $g(X)$. Then
\begin{align}
1+a_1f_1(\beta)^2+\cdots+a_nf_n(\beta)^2=p(\beta)g(\beta)=0.
\end{align}
But according to the inductive hypothesis, $F(\beta)$ is formally real. A contradiction.
Combining the results above, by induction I think it's safe to say that:
If $F$ is an ordered field, and if $\alpha^n\in F$, $\alpha^n>0$, then $F(\alpha)$ can be made an order extension of $F$.
It is only natural for me then to conjecture that:
If $F$ is an ordered field, $\alpha$ is algebraic over $F$, and $F(\alpha)$ is formally real, then $F(\alpha)$ can be made an order extension of $F$.
If the degree of $p(X)=\mathrm{Irr}(\alpha:F)$ is less than $5$, then we have root formula by radicals. I can't think of much for $\mathrm{deg}\,p(X)=2m\geq6$.
If this holds, then a standard argument of Zorn's lemma shows that:
If $F$ is an ordered field, and $E$ is a formally real algebraic extension of $F$, then $E$ can be made an order extension of $F$.
Best Answer
I think that the answer is negative. Order $\Bbb Q(x)$ in a nonarchimedean way and consider the extension $\Bbb R/\Bbb Q(x)$, induced by choosing any transcendental number such as $\pi$. Note that the ordering of $\Bbb R$ is unique and archimedean.
This can be modified to give an algebraic counterexample: Take $F$ to be the real closure of $\Bbb Q(\pi)$ with the ordering induced by the embedding $\Bbb Q(\pi) \hookrightarrow \Bbb R$. Then $F$ is real closed (hence has a unique ordering) and archimedean. But as an abstract field $F$ is an algebraic extension of $\Bbb Q(x)$, which can be ordered in a non-archimedean way.
Here's an explanation of the non-archimedean ordering on $\Bbb Q(x)$: define a non-zero rational function $f/g$ to be positive if $\ell(f)/\ell(g)$ is positive, where $\ell$ denotes the leading coefficient.