I am considering axioms of incidence and axioms of order for plane geometry by Hilbert:
I1: Two points determine the unique line.
I2: Each line contains (at least) two points.
I3: There are three noncollinear points.
O1: If $A$ is between $B$ and $C$, then $A,B,C$ are different collinear points and $A$ is between $C$ and $B$.
O2: For two different points $A$ and $B$ there are points $C$ and $D$ such that $C$ is between $A$ and $B$ and $B$ is between $A$ and $D$.
O3: For any three different collinear points exactly one of them is between the other two.
O4: Pasch's axiom.
Now if $F$ is a field, then $F^2$ is a model to axioms I1,I2,I3. I understand that if $F$ is an ordered field, then $F^2$ is a model to axioms I1-I3, O1-O4. I am trying to prove that if for any field $F$ is $F^2$ a model to axioms I1-I3, O1-O4, then $F$ is an ordered field. I defined ordering in the field $F$ as follows:
On the line $y=0$ take points $O = (0,0)$ and $E=(1,0)$.
Then for $a,b\in F$, $a\ne b$ set $a<b$ iff $O$ is between $(a-b,0)$ and $E$.
Then the trichotomy of the relation $<$ in the field $F$ follows easily, the same holds for the compatibility with addition (i.e. $a<b$ implies $a+c<b+c$). But how can one prove the transitivity of $<$ and the compatibility with multiplication (i.e. $a<b$ and $0<c$ should imply $ac < bc$)? And is it true at all, that axioms of order in plane geometry impose an ordering in the field?
I found an answer in Hartshorne's book "Geometry, Euclid and Beyond" (a very good book, btw). And yes, it is true, the axioms of order in $F^2$ impose an ordering on $F$ (Proposition 15.3 there). For, the $x$-axis is split by $(0,0)$ into two rays (so-called line separation property). The interior points of one of the rays are told to be positive numbers, the interior points of the other ray are negative numbers. The sum and the product of positive numbers is then again a positive number. So the ordering of the points on $x$-axis corresponds to ordering on $F$.