Without the Axiom of Choice, does every infinite field contain a countably infinite subfield?

Earlier today I asked whether every infinite field contains a countably infinite subfield. That question quickly received several positive answers, but the question of whether those answers use the Axiom of Choice has spawned off an interesting discussion in its own right. Thus I'd like to pose that question separately:

Without using the Axiom of Choice, can it be shown that every infinite field contains a countably infinite subfield?

Note that the corresponding question for sets has a negative answer, but since fields have so much more structure than sets, it is not a priori unreasonable to me that there might exist an AC-less resolution for fields.

EDIT: For the purposes of this question, an infinite set is one that is not bijective with $\{1, \dots, n\}$ for any $n \in \Bbb N$. Thanks to @MartinSleziak for bringing this issue up in chat.

Impossible, by the result from Hickman (1978). He proves that the existence of a Dedekind-finite infinite field (that is, an infinite field does not have any countable subset) is consistent with ZF. More precisely, he proves the following theorem:

Theorem. For a prime $p$ and a sequence of natural numbers $m_0\mid m_1\mid\cdots$, it is consistent with ZF that there is an Dedekind-finite infinite field, such that the field is a union of a sequence of increasing finite fields $(F_n)_{n<\omega}$ with $|F_k| = p^{m_k}$.

(His original paper uses the terminology "medial" to denote a Dedekind-finite infinite set.)

Furthermore, he proves that every Dedekind-finite infinite rings with only a finite number of zero-divisors is a field. Thus every Dedekind-finite infinite ring is a field or has infinite number of zero-divisors.

You can check the details in his paper:

Hickman, J. L. "Dedekind-finite fields." Bulletin of the Australian Mathematical Society 19.01 (1978): 117-124.

Here's a sketch for a concrete construction of a counterexample, probably close to the result by Hickman that Hanul Jeon found:

If I remember field theory right, ZF proves that there exists:

1. A nested sequence of finite fields $$ \mathbb F_p = K_0 \subseteq K_1 \subseteq K_2 \subseteq \cdots $$ such that $K_n$ has $p^{2^n}$ elements, and the only subfields of $K_i$ are $K_j$ with $j\le i$.
2. A sequence of generators $g_0, g_1, g_2, \ldots$ such that $K_n = \mathbb F_p[g_n]$.
3. A sequence of order-2 authomorphisms $\varphi_n$ of $\cup_n K_n$ such that $\varphi_n$ fixes $K_{n-1}$ but does not fix $g_n$, and such that all the $\varphi_n$s commute.

Let $K=\cup_n K_n$. We will show (assuming that AC fails badly enough) a Dedekind-finite field that is "morally isomorphic" to $K$, except the isomorphism isn't actually going to exist.

As a start, we'll develop a strange "redundant" representation of the elements of $K$.

First for every finite $A\subseteq \mathbb N$, let $\phi_A$ be the product of $\varphi_n$ for all $n\in A$.

Let a path mean a finite (1-indexed) sequence of $0$s and $1$s. For a path $a\in\{0,1\}^n$, let $g_a = \phi_{\{n\mid a(n)=1\}}(g_n)$. This is a generator of $K_n$; different paths map to different generators. For different paths of the same length we have $$ g_a = \phi_{\{n\mid a(n)\ne b(n)\}}(g_b) $$ A recipe for an element $k\in K$ is a pair $(a,q)\in \{0,1\}^*\times \mathbb F_p[X]$, such that $k=q(g_a)$.

Define an equivalence relation $\sim$ on all recipes as the symmetric transitive closure of

• $(a,q)\sim(a',q')$ if $|a|=|a'|=n$ and $ q(g_n)=q'(\phi_{\{n\mid a(n)\ne a'(n)\}}(g_n))$.
• $(a,q)\sim(a.b, q\circ r)$, where $b\in\{0,1\}$ is arbitrary, and $a.b$ is the sequence formed by appending $b$ to the end of $a$, and $r$ is any polynomial such that $r(g_{|a|+1})=g_{|a|}$.

It should be clear that two recipes are related exactly when they represent the same element of $K$. Let $[k]$ mean the equivalence class of all recipes for of $k$.

We can calculate directly on recipe classes: It should be clear that if we have two recipe classes, we can always find a path that is represented in both of them. (Just chose an arbitrary path whose length is at least the maximum of some path from one class and some path from the other). Then,

$$ \forall (a,q)\in [k], (a,q')\in [k'] : (a,q+q')\in[k+k'] \land (a,qq')\in[kk']$$

The point of all this is that the definiton of $\sim$ -- and thus computation on represetation classes -- does not depend on the elements of a path being $\{0,1\}$ in particular, but just that there are only two choices for each position in a path, which can be compared for equality.

Now, finally assume that AC fails badly enough that there is a countable sequence $(A_n)$ of unordered pairs such that $\cup_n A_n$ is not well-orderable. This is well known to be consistent with ZF.

Redefine a "path" to mean a function $a:\{1,2,\ldots,n\}\to\cup_n A_n$ such that $a(i)\in A_i$. A recipe is still a pair of a path and a polynomial, and $\sim$ is defined as above, except that $b\in\{0,1\}$ must be replaced by $b\in A_{|a|+1}$.

The properties of the old $\sim$ transfers to the new $\sim$ because each of them speak about only finitely long paths at a time, and we can always choose a finite and sufficiently long path in $(A_i)$ to play the same role as $0^n$ while verifying each of the claims for the new $\sim$.

There is still a set of all recipes, and so the set of all recipe classes under $\sim$ stil exist. Any two recipe classes can be added, multiplied, subtracted and divided just like before. So the set of recipe classes form a field.

On the other hand, this field cannot have a countably infinite subfield. Any infinite subfield would need to contain elements that generate $K_n$ for arbitrarily large $n$, and so in particular, an infinite subfield must include a recipe class that contains $(a,X)$ for every path $a$. However, this would mean that the set of paths is countable, which would mean there is a surjection $\mathbb N\to \cup_n A_n$, which contradicts $\cup_n A_n$ not being well-orderable.