# Number of finite and infinite fields of a particular characteristic

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 $$p$$.

• 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 $n$. – user208649 Aug 13 '20 at 18:50
• 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 $p^n$. – Andreas Blass Aug 13 '20 at 18:54
• @AndreasBlass I'm not taking of the cardinality of a field but rather the cardinality of the set of fields of char p – John White Aug 13 '20 at 18:55
• 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 $p$. – Andreas Blass Aug 13 '20 at 18:56
• 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. – Andreas Blass Aug 13 '20 at 19:10

In fact, there are more isomorphism classes of fields than can fit in any set! Thus, $$W_p$$ has no well-defined cardinality.
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 $$\aleph_0$$.
• Woops. I forgot about fields with order $p^k$. Is there an explicit bijection that shows this? – John White Aug 13 '20 at 19:00
• Given $k>0$ and prime $p$, all fields of order $p^k$ are isomorphic. – Somos Aug 13 '20 at 21:25
• 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 $\mathbb{F}_p$. – diracdeltafunk Aug 13 '20 at 23:48