# Fields of arbitrary cardinality

So I took an introductory abstract algebra course a few semesters ago, and we were shown that groups and rings can both be made into products, i.e. if I have some group $G$ (resp. ring $R$) and some indexing set $I$, then I can make a group $G^{I}$ (resp. ring $R^{I}$) with the appropriate cardinality. However, it was also demonstrated that for fields, you cannot keep the multiplicative inverse under a product structure, i.e. if I have fields $F_{1}, F_{2}$, then $F_{1} \times F_{2}$ will still be a ring, but you will not (not just "not in general", but won't) have multiplicative inverses for all non-zero (i.e. not $(0, 0)$) elements, as you could pick $(0, x)$, where $x \neq 0$. According to Wikipedia, an ultraproduct will preserve field structure, but I'm not sure what the cardinality of $\prod_{i \in I} F_{i} / \mathscr{U}$ would be in general.

I saw another post that said you could make a field of arbitrarily large cardinality by extending a given field through throwing in a whole bunch of "transcendental" elements. The example given was that if I had $\mathbb{Q}$, I could start by dumping in the complexes, but then I suppose after that I'd just start throwing in dogs and cats or something; I'm really not sure what that responder meant, and ceteris paribus, my dogs tend to stay inside, so I'd rather keep them out of my fields, particularly the large ones the thread was interested in.

Moreover, the thread mentioned earlier seemed more concerned with just making the fields big. My question is a bit more nuanced: Given any cardinality $\kappa > 1$, can I generally construct a field $F$ of a cardinality $\kappa$? Moreover, how vague do I have to be about it (i.e. do I have to use some really choicy methods, or can I make it a bit more straightforward)?

Thanks.

EDIT: Sorry. I am aware that for finite fields, you are limited to powers of primes. I meant for infinite cardinals.

• For this to be provable without AC would imply a uniform construction of a group structure (the invertibles in the field) on any set, by adding a point to the set, putting a field structure on the new set, then identifying its unit group with the old set. The last step does not use AC. I'm pretty sure "any set has an abelian group structure" is a form of AC.
– zyx
May 21 '16 at 21:34

Let $F$ be a finite field, and $X$ an infinite set, let $\hat X$ be the set of all finite strings of things in $X$ where two strings are equal if they differ only by their order (e.g. $x_1x_3x_2=x_1x_2x_3$). Then let $A$ be the set of formal sums $\{\sum_{i=1}^n f_ix_i\mid n\in\Bbb N,\ f_i\in F\ \forall i,\ x_i\in \hat X\ \forall i\}$. Then $A$ is a (commutative) ring and its field of fractions is a field with cardinality equal to $X$.

• Wow! Now that is some domain! Endorsed!!! May 24 '15 at 16:55
• Hm, but wouldn't multiplication be noncommutative here? I guess taking the polynomial ring with all elements of $X$ as variables, i.e. using maps $X\to\mathbb N_0$ with finite support in place of finite strings might be better. May 24 '15 at 16:58
• Is it "choicy"? Here $X$ need not be well-orderable. But without AC, how do you know your result has the same cardinal as $X$? For example, won't it in fact have cardinal ${}\ge |X^2|$ ?? May 24 '15 at 17:04
• It seems to me, waif that I am, that demonstrating this set has cardinality $|X|$ still has "choicy bits". Clearly it has cardinality $\leq |X|^{\omega}$. May 24 '15 at 17:04
• Sorry to revive a dead problem, but if there exists a bijection $\phi: X \to \hat{X}$, then there exists a bijection $X \to X^{2}$. With Schroeder-Bernstein in mind, if we fix $x_{0} \in X$, then $x \mapsto [ x x_{0} ]$ is an injection, and the map $x \mapsto \begin{cases} \phi(x) & \operatorname{length}(\phi(x)) = 2 \\ [x_{0}] & \textrm{otherwise} \end{cases}$ is a surjection. This implies Tarski's theorem of choice (en.wikipedia.org/wiki/Tarski%27s_theorem_about_choice), which if Wikipedia is to be believed implies the Axiom of Choice. So this is actually still using the full AC.
– AJY
Feb 29 '16 at 17:27

On the question of whether choice is required: as Gregory Grant's answer shows, the following is a theorem of ZF:

$$\text{Suppose \kappa is a cardinality. Then there is a field F with \vert F\vert \ge \kappa - in particular, \kappa injects into F.}$$

Moreover, if $\kappa$ is equinumerous with $\kappa^{<\omega}$ (the cardinality of finite sequences from $\kappa$), $\vert F\vert=\kappa$.

What if we drop the hypothesis $\kappa=\kappa^{<\omega}$? Well, the only well-orderable cardinalities which don't satisfy this are the finite ones, so for well-orderable cardinalities nothing changes. Moreover, it is consistent that there are sets which are not well-orderable but are in bijection with their finite sequences - for example, $\mathbb{R}$. So this applies to a broader class of cardinalities than just the well-orderable ones.

NOTE: this is different from saying that the global theorem doesn't imply AC. There are principles which don't imply AC "locally" but do "globally" - for example, GCH (http://math.bu.edu/people/aki/7.pdf) for which as far as I know the question of "local implication" is still open. Still, I suspect that the statement is strictly weaker than choice.

However, it is consistent that there are sets which are infinite but not in bijection with their set of finite sequences. A strong example of such a set is an amorphous set - this is a set which cannot be written as the disjoint union of two infinite sets.

This leads to a really interesting question:

Is it consistent with $ZF$ that there is a field of amorphous cardinality?

In general, the question of what structure amorphous sets can have has been thoroughly studied on the combinatorial side; see the answers to https://mathoverflow.net/questions/86654/what-sort-of-structure-can-amorphous-sets-support. However, on the algebraic and model-theoretic side much less is known; the only paper I'm aware of is http://www.researchgate.net/profile/Agatha_Walczak-Typke/publication/38338949_The_first-order_structure_of_weakly_Dedekind-finite_sets/links/0c960524ac2860b857000000.pdf.

However, this paper proves a remarkable result: among other things, it shows that a complete theory with no finite models has an amorphous model if and only if it is $\aleph_0$-categorical (exactly one countably infinite theory up to isomorphism) and strongly minimal (in every model of the theory, every set which is definable from parameters is either finite or co-finite). So the above question boils down to:

Is there a theory, extending the theory of infinite fields, which is $\aleph_0$-categorical and strongly minimal?

I believe the answer is "yes," but I don't know.

EDIT: Some partial results:

• Say an amorphous set $A$ is bounded if there is some $n$ such that, given any partition of $A$ into infinitely many nonempty pieces, each piece has size $<n$. Then no bounded amorphous set can carry a group structure, let alone a field, by a simple coset argument.

• Meanwhile, there are unbounded amorphous sets with group structures: indeed there can be amorphous vector spaces! (Over a finite field, of course.) Axiom of choice and automorphisms of vector spaces

There are of course analogous questions for other finite-ish cardinalities (generally called "Dedekind-finite").

• Impressive! The devil is in the details, eh? May 24 '15 at 17:21
• I think that I saw somewhere a result about amorphous fields. May 24 '15 at 17:37
• Perhaps the theory of algebraically closed fields of characteristic 0 with a countable infinity of constants that are are all algebraically independent over $\mathbb Q$? It is at least $\aleph_0$-categorical and it also feels like it could be proved strongly minimal by some kind of quantifier elimination. May 24 '15 at 21:20
• Can't allow characteristic zero - then $\{1\cdot n: n\in\omega\}$ defines a countable infinite subset of the field, which in turn implies that the field is not amorphous. May 24 '15 at 22:53
• @NoahSchweber: Ah, that makes sense. But then I can prove that amorphous fields don't exist; see my answer. May 25 '15 at 13:29

The claim that there are fields of every infinite cardinality depends essentially on "choicy methods".

Without the Axiom of Choice, it is consistent that there may exist an amorphous set, that is, an infinite set that does not have two disjoint infinite subsets. Such a set cannot be given a field structure.

An amorphous field cannot contain any nonzero element of infinite multiplicative order, because the odd and even powers of such an element would be two disjoint infinite subsets of the field.

On the other hand if every nonzero element has finite order, then the field is a countable disjoint union of finite sets: $$F = \bigcup_{n\in\mathbb N}\{ x\in F \mid x\text{ has order }n \}$$ (because a field has at most $n$ elements of order $n$), and a countable disjoint union of finite sets cannot be amorphous.

This approach was suggested by Noah Schweber's answer.

If $\kappa$ is a finite cardinality, this is only possible if $\kappa =p^n$ for some prime $p$ and whole number $n$.

If $\kappa$ is infinite this is possible by Zorn Lemma.

Indeed pick a set $A$ or cardinality $\kappa$ [we need this to be sure that we work inside a set, otherwise we might run into the set of all sets].

Define $$\mathcal F := \{ (B, +, \cdot) \mid (B, +, \cdot) \mbox{ is a field and } B \subset A \}$$

Now on $\mathcal F$ consider the order "subfield". Prove that the Zorn's lemma hold's, and thus $\mathcal F$ has a maximal element.

I claim that any maximal element $M$ must have cardinality $\kappa$. Indeed if that is not the case, as $M$ is infinite $M(x)$ and $M$ have the same cardinality.

Moreover, as $M$ has cardinality strictly less than $A$, it follows that $A \setminus M$ has cardinality bigger than $M$, which is the same as $M(X) \setminus M$. Therefore, you can find a one to one function $$f: M(X) \setminus M \to A \setminus M$$ and you can use this to construct a field isomorphic to $M(X)$ on $B=M \cup f( M(X) \setminus M)$ which is an extension of $M$. This contradicts the maximality.

• As this answer shows, yes, you have to use "choicy methods". May 24 '15 at 16:51
• @DavidWheeler this isn't true - as the answer by Gregory Gant shows, you do not need any amount of choice (as long as the cardinality $\kappa$ you're trying is well-orderable) to build a field of cardinality $\kappa$. Just because you need choice for some argument doesn't mean you need choice for the theorem. May 24 '15 at 16:58
• @user28111 I thought that given an infinite set $X$ to prove that the set of strings of finite length in $X$ has the same cardinality one needs choice. Are you sure you don't? May 24 '15 at 17:02
• @N.S. Sorry, bad communication on my part - all I meant was that the argument that we need choice isn't valid. It may well be that choice is required; all I know offhand is that, if we restrict to $\kappa$ satisfying $\kappa^{<\omega}=\kappa$, we don't need any choice; I'm pretty sure this is strictly weaker than "$\kappa$ is well-orderable." May 24 '15 at 17:05
• I think Gregory Cant's construction is perfect for a set of cardinality strictly greater than $\omega$. Then, we can reduce the problem of creating fields "of any cardinality" to the continuum hypothesis, in essence, since we have $\Bbb Q$ and $\Bbb R$. This obviates the need for AC altogether, but replaces it with a slightly subtler problem. May 24 '15 at 17:11

A lot of info was scattered about, so here's a condensed bit of the developments in Gregory Grant's answer and the comments. The final solution was that the existence of a field structure on every infinite set $$X$$ is equivalent to the full axiom of choice.

Firstly, to show choice implies the existence, we borrow Gregory Grant's construction. Let $$X$$ be an infinite set, and let $$X^* = \bigcup_{n \in \mathbb{N}} X^n$$ denote the set of all finite strings on $$X$$. Let $$\sim$$ be an equivalence relation on $$X^*$$ where $$(a_{1}, \ldots, a_{n}) \sim (b_{1}, \ldots, b_{m})$$ if and only if $$(\forall x \in X)( \# \{k : a_{k} = x \} = \# \{ k : b_{k} = x \} )$$, i.e. the strings are equal up to rearranging the terms of the strings. Set $$\hat{X} = X^{*} / \sim$$, and let $$R$$ denote the set of formal sums over $$X$$. Finally, let $$F$$ denote the field of fractions of $$R$$. Then $$F$$ is a field equitotient with $$X$$.

In the other direction, if every infinite set $$X$$ can be given a field structure, then a fortiori every infinite set can be given a group structure, so we can use this argument.