Multiply ordered fields

I’m sure you know about ordered fields. The rationals, the reals, etc. But those fields can only be ordered in one way. Now consider the field Q[sqrt(2)]. That field has two possible orderings: one where sqrt(2) is positive and one where it’s negative.

What about a field with 0 possible orderings? Easy – any finite field, any complex field, insert third example here. Many to choose from.

A field with 4 orderings is also pretty easy to construct: Q[sqrt(2)][sqrt(3)] is an example – the sqrt(2) and sqrt(3) can be positive or negative as they choose.

How about a field with continuum-many orderings? I can think of two very different examples: Q[sqrt(2)][sqrt(3)][sqrt(5)][sqrt(7)] etc, and the field of rational functions. The first has continuum-many via infinitely many countable choices, and the second has continuum-many via x taking on any real value.¹ Unlike all other examples on this list, these ordered fields are not all isomorphic.

What if we want exactly 3 orderings? I struggled for a while but I think I found one. Extend the field Q with an element we call k, whose cube is one less than its triple. There are three orderings of this field.

That setup for 3 orderings can of course be generalized for higher odd numbers, and the 4 orderings setup lets you double any number of orderings that can be achieved. Altogether, that lets you get any finite number of orderings.

Any amount of orderings that can be achieved by a field can also be achieved by a commutative ring and vice-versa: any field is a commutative ring, and any commutative ring with more than 0 orderings has the same number of orderings as its field of fractions.

And, of course, the obligatory unsolved question: Is there a field with countably many orderings?

1. If x is algebraic, there are two orderings depending on whether the corresponding polynomial is positive or negative. There are also two more orderings for x being positive or negative infinity. ↩︎