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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.

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

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  • $\begingroup$ I should not have missed that. Is there a natural way to make it an ordered field? $\endgroup$
    – N Unnikrishnan
    Commented Jan 17, 2015 at 8:13
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    $\begingroup$ Actually, if $k$ is an orderer field, then so is $k(T)$ as there is on it a unique order such that $T$ is strictly positive and strictly smaller than any strictly positive element of $k$. By induction it follow that if $k(I)$ denotes the field in my answer, this field is ordered if $I$ is countable. And by transfinite induction, it follows that it is always countable, as if $I$ is non-empty, there's at least one (good) order on $I$. $\endgroup$
    – Olórin
    Commented Jan 17, 2015 at 10:49
  • $\begingroup$ I mean "Always orderable", not "always countable", of course. By the way, orderability of $I$ is used to order the various $X_i$'s, akd therefore $K(I)$. $\endgroup$
    – Olórin
    Commented Jan 17, 2015 at 10:57
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    $\begingroup$ The simple idea behind that is that if you know how to order $k(T)$ (use the order I gave) then modulo comparing $T$ and $S$, you know how to order $k(T,S) = k(T)(S)$ in the same way. $\endgroup$
    – Olórin
    Commented Jan 17, 2015 at 11:00
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    $\begingroup$ It works as $2\times 3$ and $3^{-1}$. Anyway, I suggest you find information about this somewhere else, or browse math.stackexchange with the seach field about this, and in case the question has neven being ask (which I really doubt), ask the question. $\endgroup$
    – Olórin
    Commented Aug 9, 2015 at 16:17
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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.

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  • $\begingroup$ For example, if we add two new transcendental elements $\alpha$ and $\beta$ to a field, we will need to add the identity $\alpha\beta = \beta\alpha$ to satisfy the commutativity axiom of fields. $\endgroup$
    – Carl Mummert
    Commented Jan 11, 2015 at 13:38
  • $\begingroup$ I am in a confusion regarding accepting an answer, since, while you and @HennoBrandsma answer the existential question, there is a subquestion about the simplest way to give an example of a field that Robert Green has answered. While I would prefer the LST deduction as far as the existential question is concerned, for a general reader, Green's explicit construction answers all questions (for fields, that is). $\endgroup$
    – N Unnikrishnan
    Commented Jan 17, 2015 at 8:22
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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.

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