# 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?

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What do you mean by "explicitly determine"? Do you have in mind some way of classifying all orderings on $\mathbb{R}((x))(y)$ that doesn't work for $\mathbb{R}((x,y))$? –  SJR Dec 12 '12 at 19:53
How about a reference to what is $\mathbb R ((x,y))$ –  Gerald Edgar Dec 12 '12 at 21:20
To SJR: By "explicityly determine the orderings", I wanna to say "we know how to contruct all the orderings". I learn from an exercise in Lam's book on quadratic forms that there is a description of all orderings of k(x), when k is a real field. I really don't know whether every ordering of $\mathbb{R}((x))(y)$ extends to an ordering of $\mathbb{R}((x, y))$. –  Yong Hu Dec 12 '12 at 21:29
Hi, Gerald. $\mathbb{R}((x, y))$ is the fraction field of the ring of power seires in two variables x and y over $\mathbb{R}$. –  Yong Hu Dec 12 '12 at 21:30
This looks interesting. But can you provide some references for the three "known" cases? –  Gerald Edgar Dec 12 '12 at 23:01

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))

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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.

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Thanks, Gerald. For $\mathbb{R}(y)$, I've found the solution in Lam's book (Introduction to quadratic forms over fields, AMS, 2005, p.239) and for $\mathbb{R}((y))$, there are exactly two orderings. –  Yong Hu Dec 14 '12 at 14:16