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I was wondering if there is any work on the number of distinct finite and infinite fields (upto field isomorphism) for a particular characteristic $p$.

In particular, questions of the form: If $W_p := \{F \;|\; F \text{ is a field and } \text{char}(F) = p\}$, and for any $F \neq F'$, it holds that $F \not\cong F'$. Then what is $|W_p|$?

For example, for characteristic $p = 0$, there are no finite fields. There are no fields with characteristic 1. (I'm defining fields to have $0 \neq 1$). For characteristic $p > 1$, there is only 1 finite field (upto isomorphism) for that characteristic.

So my question is two-fold:

  1. What do we know about infinite fields of characteristic $0$? Are there infinitely many of them? How about uncountably infinite? i.e. What is the cardinality of $W_0$?
  2. What do we know about fields of characteristic $p > 1$. There is only $1$ finite field but how many distinct infinite fields of char $p$ exist?

It seems to me that there should be at least $|\mathbb{R}|$ fields of char $0$ just from field extensions of $\mathbb{Q}$. But is the cardinality of $W_0$ larger than that of $\mathbb{R}$? If they are equal then can we construct a bijection?

For infinite fields with char $p > 1$, I've seen several explicit constructions (In particular, one example is the field of rational functions with coefficients in $\mathbb{F}_p$). I've only seen a handful examples though so I'm not sure there are an infinite number of infinite fields with char $p > 1$.


(Edit) I forgot about fields of order $p^k$ so clearly there are an infinite number of finite fields with char $p$.

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    $\begingroup$ If you don't limit the cardinality, there are class-many fields of characteristic $p$ - $W_p$ is not a set (neither is $W_0$). There are certainly infinitely many - an easy infinite family of countable ones is $\mathbb F_p$ adjoined $n$ indeterminates for each natural number $n$. $\endgroup$ – user208649 Aug 13 '20 at 18:50
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    $\begingroup$ A finite field of characteristic $p$ has cardinality $p^n$ for some positive integer $n$. Furthermore, for every prime $p$ and positive integer $n$, there is a unique (up to isomorphism) field of cardinality $p^n$. $\endgroup$ – Andreas Blass Aug 13 '20 at 18:54
  • $\begingroup$ @AndreasBlass I'm not taking of the cardinality of a field but rather the cardinality of the set of fields of char p $\endgroup$ – John White Aug 13 '20 at 18:55
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    $\begingroup$ Nevertheless, my comment answers your question for the case of finite fields, for it immediately gives that there are exactly $\aleph_0$ nonisomorphic finite fields of characteristic $p$. $\endgroup$ – Andreas Blass Aug 13 '20 at 18:56
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    $\begingroup$ As indicated in the answer by diracdeltafunk, there's a proper class of nonisomorphic infinite fields of any given characteristic. Instead of the Löwenheim-Skolem argument in that answer, you can start with the $p$-element field (or $\mathbb Q$) and adjoin any cardinal number of independent transcendental elements. The resulting fields are all nonisomorphic, and there are as many of them as there are cardinals, i.e., a proper class. $\endgroup$ – Andreas Blass Aug 13 '20 at 19:10
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In fact, there are more isomorphism classes of fields than can fit in any set! Thus, $W_p$ has no well-defined cardinality.

By the Löwenheim–Skolem Theorem, there exists at least one field of each possible characteristic and each infinite cardinality. If you wish to avoid this kind of abstract model-theoretic argument, it's not too hard to actually construct fields of arbitrary characteristic and arbitrary infinite cardinalities (although not super obvious – let me know if you'd like me to expand on this).

For finite fields, your understanding has a gap: there are infinitely many finite fields of each prime characteristic $p$! The cardinality of the set of isomorphism classes of finite fields of characteristic $p$ is $\aleph_0$.

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  • $\begingroup$ Woops. I forgot about fields with order $p^k$. Is there an explicit bijection that shows this? $\endgroup$ – John White Aug 13 '20 at 19:00
  • $\begingroup$ Given $k>0$ and prime $p$, all fields of order $p^k$ are isomorphic. $\endgroup$ – Somos Aug 13 '20 at 21:25
  • $\begingroup$ Yes, and to show that two finite fields of the same order are isomorphic, we just note that any finite field of order $p^k$ must be the splitting field of $x^{p^k} - x$ over $\mathbb{F}_p$. $\endgroup$ – diracdeltafunk Aug 13 '20 at 23:48

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