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Does every infinite field contain a countably infinite subfield?

It's easy to see that every field $K$ contains either the rational numbers $\Bbb Q$ (when $K$ has characteristic $0$) or a finite field $\Bbb F_p$ (when $K$ has characteristic $p$). Thus, in the characteristic $0$ case, the answer is an easy "yes."

But if $K$ is infinite and has characteristic $p>0$, does the fact that $K \supset \Bbb F_p$ allow us to conclude that $K$ has a countably infinite subfield?

David Zhang
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    Note that David Ulrich's answer applies to a much broader class of structures than just fields - see e.g. the Lowenheim-Skolem theorem. – Noah Schweber Nov 07 '15 at 22:00
  • @NoahSchweber Much wider class, clearly. All the way to downward L-S? Your comment puts me in a curious state, wondering whether I've proved the L-S theorem. Heh. I don't quite see it. Unless it's true that any formula is equivalent to one in some special form, for example universal quantifiers followed by existential quantifiers followed by no quantifiers, or something... ? – David C. Ullrich Nov 07 '15 at 22:18
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    @DavidC.Ullrich You haven't, it takes more work. (Maybe I should have phrased my comment more carefully.) What you've shown is that if $S$ is a set with $\kappa$-many functions on $S$, each of finite arity, then $S$ has a subset of size at most $\kappa$ which is closed under each of those functions. The other ingredient is Skolem functions: we expand the language of the structure in question with a number of functions such that being closed under those functions implies being an elementary substructure. (cont'd) – Noah Schweber Nov 07 '15 at 22:29
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    Syntactically, what's going on is that given any prenex normal form first-order formula $\forall x_1\exists y_1\forall x_2\exists y_2 . . . \forall x_n\exists y_n\varphi(\overline{x},\overline{y})$, we can convert it to one of the form $\exists f_1. . . \exists f_n\forall x_1. . .\forall x_n\varphi(\overline{x}, f_1(x_1), f_2(x_1, x_2), . . . , f_n(x_1, x_2, . . . , x_n))$, where the existential quantifiers "$\exists f$" are second-order - that is, quantifying over functions. This provides the "special form" you mention in your previous comment. – Noah Schweber Nov 07 '15 at 22:31
  • @NoahSchweber As you were typing your first reply i had a glimmering of things I once knew - looked up Skolem normal form. Thx... – David C. Ullrich Nov 07 '15 at 22:32

3 Answers3

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Yes. If $S$ is a countably infinite subset of $K$ then the subfield generated by $S$ is countable.

Since the question got six upvotes before any answers, maybe this is not obvious. Details:

Let $S_0=S$. Let $S_{n+1}$ consist of the elements of $S_n$ together with all the $x+y$, $x-y$, $xy$ and $x^{-1}$ for $x,y\in S_n$. Then $\bigcup_{n=0}^\infty S_n$ is a countable subfield.

(Yes, the same argument shows that $K$ has a subfield of any infinite cardinality less than or equal to the cardinality of $K$.)

David C. Ullrich
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  • By "the subfield generated by $S$", you mean the smallest subfield of $K$ containing all elements of $S$, right? – David Zhang Nov 07 '15 at 19:02
  • @DavidZhang Yes. – David C. Ullrich Nov 07 '15 at 19:03
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    In that case, I'm afraid that the countability of this subfield is not obvious to me. Can you briefly elaborate on the argument? – David Zhang Nov 07 '15 at 19:05
  • Would basically the same argument lead us to conclusion that we have a subfield of any cardinality $\varkappa$, where $\aleph_0 \le \varkappa \le |K|$? – Martin Sleziak Nov 07 '15 at 19:06
  • @DavidZhang Just added details to the answer. – David C. Ullrich Nov 07 '15 at 19:07
  • @MartinSleziak Seems right to me. – David C. Ullrich Nov 07 '15 at 19:07
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    Yes.The result of closing off under countably many finitary operations gives a set of equal (infinite) cardinal. – DanielWainfleet Nov 07 '15 at 19:09
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    @DavidC.Ullrich Ah, I see! Thanks, that clears it right up. I suppose this does require the axiom of choice then, to pick a countably infinite subset. – David Zhang Nov 07 '15 at 19:10
  • "Since the question got six upvotes before any answers, maybe this is not obvious" - that and we were trying (and failing) to solve it ourselves in the chat room prior to posting. Thanks for your answer –  Nov 07 '15 at 19:15
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    @DavidZhang I believe so, yes. Heh, with no AC you can't prove that a countable union of countable sets is countable... – David C. Ullrich Nov 07 '15 at 19:17
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    @AlexClark Looks like this one is gonna make me rich... – David C. Ullrich Nov 07 '15 at 19:18
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    @DavidZhang Without some (weak) form of the axiom of choice, your question already has a negative answer for sets, much less fields. – Andrew Dudzik Nov 07 '15 at 19:19
  • @Slade That is very true, but I was thinking that since fields have so much more structure than sets, an AC-less argument might be possible. For example, I was trying to think of a way to constructively exhibit a countable subfield starting with $\Bbb F_p$. – David Zhang Nov 07 '15 at 19:23
  • @DavidZhang I predict you won't find a proof with no AC. Don't ask me to prove that.... – David C. Ullrich Nov 07 '15 at 19:27
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    @DavidZhang See for example my comments on the other answer for why I'm skeptical. Note that it could well be he's right, but it's not clear to me - in any case that sort of issue shows that avoiding AC can be awesomely difficult, it hides in corners you thought were safe. (Feel free to find that little arrow, btw... heh) – David C. Ullrich Nov 07 '15 at 20:06
  • @NateEldredge Erm. Good point - shoulda just said throw in what's needed.... thx – David C. Ullrich Nov 07 '15 at 21:55
  • @DavidZhang Actually, how's this argument: If $K$ is algebraic over $\mathbb{F}_p$, then we can prove, I think without AC, that $K$ is countable, using uniqueness of finite fields. But if there is a transcendental element $t\in K$, then $\mathbb{F}_p(t)$ is a countable subfield. – Andrew Dudzik Nov 08 '15 at 02:32
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As Ulrich points out, any field generated by a countable set of elements of $K$ will be countable, but this doesn't guarantee a proper subfield, so let me pose a different question: does every infinite field properly contain an infinite subfield? (I'll even throw out the word countable.) Over course $\Bbb Q$ is a trivial counterexample in characteristic $0$, so let's consider characteristic $p$.

The answer to that question is no.

We know the finite extensions of $\Bbb F_q$ are $\Bbb F_{q^n}$ for naturals $n$, which are unique for their sizes, with inclusions $\Bbb F_{q^d}\subseteq\Bbb F_{q^n}$ if and only if $d\mid n$ (i.e. $d$ is a divisor of $n$). But can we characterize all algebraic extensions of $\Bbb F_q$? Yes: to do so we need supernatural numbers for exponents.

A supernatural number is a formal infinite product $n=\prod_p p^e$ of prime powers, with no restrictions on the exponents, and in fact $e=\infty$ is an allowed exponent. The notion of divisibility for supernatural numbers should be obvious.

Define $\Bbb F_{q^n}$ for supernatural $n$ to be $\bigcup_{d\mid n}\Bbb F_{q^d}$ within $\overline{\Bbb F_{q}}$, where $d$ ranges over all finite, natural number divisors of $n$. Given any $K\subseteq\overline{\Bbb F_q}$, we can associate to it the supernatural number $n$ which is the LCM of all finite $d$ for which $\Bbb F_{q^d}\subseteq K$, or equivalently $n=\prod p^{e(p)}$ where $e(p)$ is maximal subject to $\Bbb F_{q^{p^e}}\subseteq K$ (or $e(p)=\infty$ if that holds true for all $e$).

Exercise: Prove $K=\Bbb F_{q^n}$ where $n$ is the supernatural number associated to $K$.

Indeed, the lattice of supernatural numbers ordered by divisibility is isomorphic to the lattice of algebraic extensions of $\Bbb F_q$ ordered by inclusion. The isomorphism is $n\mapsto \Bbb F_{q^n}$.

In particular, this means for any prime $\ell$, the infinite field $\Bbb F_{q^{\ell^\infty}}$ does not properly contain any infinite subfields. Contrast with $\overline{\Bbb F_q}$, which properly contains infinite subfields $\Bbb F_{p^n}$ precisely when $n$ is an infinite supernatural which properly divides $\prod p^\infty$. (In particular it properly contains the aforementioned infinite subfields $\Bbb F_{q^{\ell^\infty}}$s!)

There is also some Galois theory that can be discussed here. The fundamental theorem of Galois theory in this context says the Galois extensions of $\Bbb F_q$ are lattice anti-isomorphic to the lattice of closed subgroups of the absolute Galois group ${\rm Gal}(\overline{\Bbb F_q}/\Bbb F_q)\cong\widehat{\Bbb Z}\cong\prod_p\Bbb Z_p$ considered with the profinite topology (note $\Bbb Z_p$ is the additive group of $p$-adic integers, not to be confused with $\Bbb Z/p\Bbb Z$). The closed subgroups, enumerated by supernaturals $n=\prod_p p^{e}$, are of the form $\prod_p \Bbb Z_p/(p^e)$, where $\Bbb Z_p/(p^e)\cong\Bbb Z/p^e\Bbb Z$ with $0\le e<\infty$ and $\Bbb Z_p/(p^e):=\Bbb Z_p$ when $e=\infty$ (which makes sense because $p^e\to0$ in $\Bbb Z_p$'s topology as $e\to\infty$).

anon
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    Isn't $\Bbb Q$ an easy example of an infinite field with no proper infinite subfield? – Akiva Weinberger Nov 08 '15 at 01:21
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    @AkivaWeinberger Yes that is pretty trivial. I suppose I was focused on thinking about characteristic $p$ because that was the topic of discussion in the chatroom (which is what brought me here). – anon Nov 08 '15 at 01:46
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    You might want to add "of characteristic $p$" or "other than $\Bbb Q$" or similar. – Akiva Weinberger Nov 08 '15 at 01:47
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Suppose $K$ has characteristic $p$.

If $K$ has a transcendental element $\alpha$, then $\mathbf{F}_p(\alpha) \subseteq K$ is countably infinite, being isomorphic to the rational function field $\mathbf{F}_p(x)$.

Otherwise, $K \subseteq \overline{\mathbf{F}}_p$ which is countable, so you can just take $K$ itself to be the countably infinite subfield.