Are there fields and ordered fields of every infinite cardinality?

Question

On the top of my head, I cannot think of any fields of cardinality more than that of the reals. (It is known that the process of algebraic closure does not increase the cardinality of an infinite field.)

What is the simplest way to give an example of a field (and an ordered field) of a specific cardinality $\alpha$?

I see there is the "Field" of surreal numbers, but it is a proper class rather than a set (and hence do not have a cardinality as such). However, there seems to be some modified construction which gives proper fields with the cardinality of some strongly inaccessible cardinal.

Answer 1

What about the field $\mathbf{Q}\left( \{ T_{i}\;|\; i\in \alpha \} \right)$, where the $T_i$ are independant formal variables indexed by $\alpha$ ?

Answer 2

Yes, this follows from the upward Löwenheim-Skolem theorem (see wikipedia or any good book on mathematical logic). The axioms for a field form a finite set of axioms of a first order nature.

Answer 3

The Lowenheim-Skolem theorem applies to fields, ordered fields, groups, rings, monoids, lattices, and other algebraic structures that are axiomatized by "first-order" axioms. It shows that, as soon as there is a countably infinite model of one of these sets of axioms, there are models of all infinite cardinalities.

The proof, intuitively, is to adjoin a large number of "new" elements and then to place as few restrictions (i.e. algebraic identities) as necessary on the "new" elements to obtain a model of the desired theory - much like adjoining transcendentals to a field.