orderings of the field R((x, y)) I don't know much about the theory of ordered fields.  But I know that, for the real fields 
$\mathbb{R}(y)$, $\mathbb{R}((x))(y)$, and $\mathbb{R}((x))((y))$,
we can explicitly determine all the orderings of the field.
My question: Can we determine all the orderingds of the field $\mathbb{R}((x, y))$?
Can anyone give some brief explanations or a reference?  
 A: A full description of the orderings of R((x,y)) is given in the paper
Alonso, M. E.(E-MADC); Gamboa, J. M.(E-MADC); Ruiz, J. M.(E-MADC)
On orderings in real surfaces.
J. Pure Appl. Algebra 36 (1985), no. 1, 1–14
In fact this paper describes all orderings of R[[x,y]] in terms of analytic half branches at the origin and the non-algebraic ones correspond to the orderings of R((x,y))
A: Not an answer, but some commentary larger than a comment.  
My interest comes from transseries (PLUG), where an annoying problem is the lack of a good notion of transseries of two variables.  I will use terminology from there, though.  
Let's first do $\mathbb R(y)$.  One model example is the case where $y$ is large and positive.  Let's write $\mathbb R(t)$.  We want to define an ordering so that $t>r$ for all $r \in \mathbb R$.  It can be done as follows.  Any rational function $f(t)$ in $t$ can be written (using long division) as a Laurent series
$$
f(t) = \sum_{k=N}^\infty a_k t^{-k}
$$
There could be finitely many terms with $k<0$, that is with positive powers on $t$, but all but finitely many terms have $k \gt  0$, that is negative powers of $t$.  Ordering is done assuming (as I said) that $t$ is large and posiitve.  So: $f(t)\gt 0$ means the leading term is positive.  In the above expansion, if $a_N \ne 0$ then the sign of $f(t)$ is defined to be the sign of $a_N$.
And $f(t) > g(t)$ is defined to mean $f(t)-g(t)>0$.  
For general ordering on $\mathbb R(y)$ we get something isomorphic to this $\mathbb R(t)$ as follows.  If $y$ is large and positve, we are done, and we may map $y \mapsto t$.  If $y$ is large and negative, we may map $y \mapsto -t$.  Otherwise (an amazing completeness property of $\mathbb R$) we have $y \sim r$ for some $r \in \mathbb R$, meaning $-1/n \lt y-r \lt 1/n$ for all $n \in \mathbb N$.  If $y \gt r$ then $1/(y-r)$ is large and positive, and we may map $y \mapsto r+1/t$.  And similarly if $y\lt r$ we may map $y \mapsto r-1/t$.  
So for each real number $r$ we get one ordering on $\mathbb R(y)$ where $y$ is infinitesimally larger than $r$ and one where $y$ is infinitesimally smaller than $r$.  And two more orderings, one where $y$ is large and poistive, one where $y$ is large and negative.
In all cases, the ordered field $\mathbb R(y)$ is isomorphic to the model example $\mathbb R(t)$.  
Note this actually works for $\mathbb R((y))$, the set of all those Laurent series, whether they come from rational functions or not.  
Tired now, maybe more later for the other cases.
