The field Q/p(x) contains the rationals as a subfield; it also contains the element x (assuming deg(p)>1, of course). Because we’ve assumed that this field is ordered, we can now compare x to all rational numbers to determine its real value (this is the Dedekind cut); so the ordering determines the value of x.
Conversely, evaluation of a formal polynomial in the field at one of the (real, because we need to end up in the ordered field R) roots of p(x) allows determination of the ordering. So knowing the order is equivalent to knowing the value of the root.
(Does that make any more sense? or is there something more specific that’s bothering you about it?)
I just can’t leave this thread alone…
Consider a field generated by Q and an arbitrary transcendental x (such as e or pi). Because x is transcendental, this is just the field Q(x) of rational functions of an indeterminate value x with coefficients in Q. (That is, there are no nontrivial equivalences among these rational functions, because that would give a polynomial with the transcendental as root.) So any choice of transcendental number will give a field isomorphic to Q(x), but (as argued previously) each such choice gives its own ordering. So all such fields have uncountably many possible orderings.
So, there are simple extensions Q(x) of Q with…[ul][]no orderings: e.g., Q[(-2)[sup]1/2[/sup]][]one ordering: e.g., Q and Q[2[sup]1/3[/sup]][]two orderings: e.g., Q[2[sup]1/2[/sup]] []three orderings: e.g., Q where p(x)=8x[sup]3[/sup]-6x-1=0 (p is an irreducible polynomial with the three real roots cos 20°, cos 100°, cos 140°)[](etc., using higher-order irreducible polynomials)[]uncountably many orderings: e.g., Q(e).[/ul]