Proposition 2

— Let $A$ be a totally ordered integral domain, and let $K$ be its field of fractions. Then there exists one and only one ordering on $K$ which restricts to the given ordering on $A$ and makes $K$ an ordered field.

Every $x \in K$ can be expressed in the form $x = ab^{-1}$, with $a$ and $b$ in $A$ and $b \neq 0$. If $x$ is positive, then $a$ and $b$ have the same sign, and conversely. Thus we see that, if there exists an ordering on $K$ satisfying the prescribed conditions, then it is unique, and the set $P$ of positive elements is identical to the set of $ab^{-1}$, where $a$ and $b$ are elements of $A$ of the same sign, and $b \neq 0$. It remains to show that $P$ satisfies conditions (AP$_I$), (AP$_{II}$), (AP$_{III}$) and (AP$_{IV}$). This is obvious for (AP$_{II}$) and (AP$_{IV}$). For (AP$_I$), consider $ab^{-1} + cd^{-1}$, where we may assume that $a$, $b$, $c$ and $d$ are positive; this sum is $(ad + bc)(bd)^{-1}$, and $ad + bc$ and $bd$ are positive.

To show $(\mathrm{AP}_{III})$, consider an identity of the form $ab^{-1} = -cd^{-1}$, so that $ad + bc = 0$. If we assume that $a$ and $b$ have the same sign and that $c$ and $d$ have the same sign, then the sign rules show that $ad$ and $bc$ have the same sign; whence $ad = bc = 0$, so $a = c = 0$; hence $P$ does indeed satisfy $(\mathrm{AP}_{III})$.