Ordered Fields

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 25 January 2010

Notation

We will use the following notation. We will write

1. $a<b$, if $a\le b$ and $a\ne b$,
2. $a\ge b$, if $a\nleqq b$ or $a=b$, and
3. $a>b$, if $a\nleqq b$,

where $a,b\in S$. Note that these definitions correspond to the usual conventions.

Ordered fields

An ordered monoid is a commutative monoid $G$ with an ordering $\le$ such that

if $x,y,z\in G$ and $x\le y$ then $x+z\le y+z$.

An ordered group is an abelian group $G$ with an ordering $\le$ such that

if $x,y,z\in G$ and $x\le y$ then $x+z\le y+z$.

An ordered ring is a commutative ring $A$ with an ordering $\le$ such that

1. $A$ is an ordered group under $+$, and
2. if $x,y\in A$ and $x\ge 0$ and $y\ge 0$ then $xy\ge 0$.

An ordered field is a field $𝔽$ with a total ordering $\le$ such that $𝔽$ is an ordered ring.

Let $G$ be an ordered group and let $x\in G$. The element $x$ is positive if $x\ge 0$. The element $x$ is negative if $x\le 0$. The element $x$ is strictly positive if $x>0$. The element $x$ is strictly negative if $x<0$.

Let $G$ be a lattice ordered group. If $x\in G$ define

$x^{+}=\sup \left\{x,0\right\}\text{,}$ and $x^{-}=\sup \left\{-x,0\right\}$

Let $G$ be a lattice ordered group and let $x\in G$. the absolute value of $x$ is

$\left|x\right|=\sup \left\{x,-x\right\}$.

Let $G$ be an ordered group. Let $x,y\in G$. The elements $x$ and $y$ are coprime if $\inf \left(x,y\right)=0$.

Let $G$ be an ordered group. Let $x\in G$. The element is irreducible if it is a minimal element of a set of strictly positive elements of $G$.

Let $𝔽$ be an ordered field. If $x\in 𝔽$ define $$\mathrm{sgn}\left(x\right)=\left\{\begin{array}{ll}1\text{,}& \text{if }x>0\text{,}\\ -1\text{,}& \text{if }x<0\text{,}\\ 0\text{,}& \text{if }x=0\text{.}\end{array}\right.$$

The nonnegative integers ${\mathbb{Z}}_{\ge 0}$ with the ordering defined by

$x\le y$ if there is an $n\in {\mathbb{Z}}_{\ge 0}$ with $y=x+n$,

is an ordered momoid. There is a unique extension of this ordering to $\mathbb{Z}$ so that $\mathbb{Z}$ is an ordered group. There is a unique extension of this ordering to $\mathbb{Q}$ so that $\mathbb{Q}$ is an ordered field.

We still need the proper characterisation of $ℝ$ as an ordered field that contains $\mathbb{Q}$ and satisfies the least upper bound property. What is the proper uniqueness statement? Should we put Dedikind cuts here?

Let $S$ be an ordered fields and $x,y\in S$ with $x\ge 0$ and $y\ge 0$, then $$x\le y\text{if and only if}{x}^2\le y^2\text{.}$$

		Proof.
	Assume $x,y\in S$ and $x\ge 0$ and $y\ge 0$. To show: If $x\le y$ then $x^2\le y^2$. If $x^2\le y^2$ then $x\le y$. We have: Assume $x^2\le y^2$. Then $y^2+\left(-x^2\right)\ge x^2+\left(-x^2\right)=0$. So $y^2-x^2\ge 0$. So $\left(y-x\right)\left(y+x\right)\ge 0$. Since $x\ge 0$ and $y\ge 0$ then $x+y\ge 0$ and ${\left(x+y\right)}^{-1}>0$ (or $x=0$ and $y=0$). So $\left(y-x\right)\left(y+x\right){\left(x+y\right)}^{-1}\ge 0$. So $y-x\ge 0$. Assume $y\ge x$. Then $y-x\ge 0$ Then $\left(y-x\right)\left(y+x\right)\ge 0\left(x+y\right)$. So $y^2-x^2\ge 0$. So $y^2\ge x^2$. Note: The proof above slips lots of steps. For example, the assertion that $y^2+\left(-x^2\right)=\left(y+\left(-x\right)\right)\left(y+\mathrm{x}\right)$ (which relies on the claim that $\left(-x\right)y=-\left(xy\right)$) requires some justification.

Exercises

Show that if $x,y\in \mathbb{Q}$ then $x\le y$ if and only if $y-x\ge 0$.

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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