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Is there an upper bound on the cardinality of a field?

The "biggest" fields I know are the field of real numbers, or the field of complex numbers. Is there a field with cardinality greater than continuum?

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migrated from Jul 15 '14 at 21:29

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Have you heard of a transcendence basis? – blue Jul 15 '14 at 21:32
No, I have not... – Marco Flores Jul 16 '14 at 3:11

3 Answers 3

up vote 5 down vote accepted

Given any set $X$, you can come up with an integral domain $\mathbb Q[X]$, of polynomials with each $x\in X$ an indeterminate. This ring is an integral domain, so it has a field of fractions, and that field of fractions has at least $X$ elements.

If $X$ is infinite, I believe the field of fractions will have the same cardinality as $X$, but don't quote me.

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It will be, at least if we assume the axiom of choice, which I assume we are. :-) – Asaf Karagila Jul 15 '14 at 21:53

Yes, There are arbitrarily large fields. This follows from a result in logic, namely the Upward Löwenheim–Skolem theorem (see

To be more concrete, consider the countable algebriacly closed field, $\bar{\mathbb{Q}}$. Every element of this field is algebraic (meaning it is a root of a polynomial). However, we can begin adding "transcendental" elements, like $\pi$ and $e$. These don't affect the "field properties" of the field. Then, we can add any cardinality of transcendentals to construct a field of that particular size.

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Well there are only at most $\beth_1$ transcendentals to add – John Fernley Jul 15 '14 at 21:45
@John: The term "Transcendental" is not taken in the context of real/complex numbers here. It is taken in the context of adding elements which are not realizations of types over the model. Or in simpler words, not algebraic over a given field. – Asaf Karagila Jul 15 '14 at 21:50

No, surreal numbers form a 'field' which is a proper class, it contains all ordinals as a subclass for example. It also has subfields of every cardinality, see

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