Is every infinite set equipotent to a field?

Question

For example, $\mathbb N$ is equipotent to $\mathbb Q$ which is a field.

$\mathbb R$ is equipotent to itself, which is a field.

But what about $\mathbb R^{\mathbb R}$, $P(\mathbb R^{\mathbb R})$ etc.?

Well, $\Bbb R^{\Bbb N}$ is equipotent to $\Bbb R$, so perhaps that's a bad example. — Ben Grossmann, Feb 29 '16 at 16:13
Or what about $\aleph_1$, which is the cardinality of the set of all countable ordinals? (If $\aleph_1=2^{\aleph_0}$ as Cantor conjectured and as was later shown to be a question not answered by the usual axioms, then the answer is clear, but can you explicitly make the set of all countable ordinals into a field somehow?) $\qquad$ — Michael Hardy, Feb 29 '16 at 16:17
@MichaelHardy it's things like this that make me realize how nice assuming the continuum hypothesis can be for sanity — Ben Grossmann, Feb 29 '16 at 16:21
I think this answer is in some ways better than any relying on the Löwenheim–Skolem theorem, since it gives an explicit construction of a field of arbitrary infinite cardinality. $\qquad$ — Michael Hardy, Feb 29 '16 at 16:33
@Stefan : What do you mean by $\mathbb Q(\omega_1)$? Usually $\mathbb Q(x)$ means the smallest field that includes $\mathbb Q$ and contains some additional element called $x$, so it's essentially the field of all rational functions of $x$ with coefficients in $\mathbb Q$. But that can't be what you mean. $\qquad$ — Michael Hardy, Feb 29 '16 at 16:36
@MichaelHardy By $\mathbb Q(\omega_1)$ I mean the quotient field of the polynomial ring $\mathbb Q[X_\alpha \mid \alpha < \omega_1]$ with $\omega_1$ many variables. — Stefan Mesken, Feb 29 '16 at 16:49
@Omnomnomnom: If you like niceness, then how about assuming $V=L$? It solves lots of problems, and you can always 'justify' your assumption by saying that no one can ever describe a set that isn't in $L$, so every universally quantified statement that you prove using your assumption really applies to every set that anyone can give you. — user21820, Mar 01 '16 at 00:14
@user21820 I can describe many set that are not in $L$... (Obviously I can't prove their existence in $\operatorname{ZFC}$ - unless I can prove its inconsistency.) — Stefan Mesken, Mar 01 '16 at 00:29
@Stefan: Let me rephrase my statement. Nobody can ever describe in ZFC a set that isn't in $L$. To be completely precise, there is no first-order definable object over ZFC that can be proven to be not in $L$. Since ZFC is supposed to be the foundation theory, it does not make much sense to say that one can describe a set that isn't in L if one cannot prove in ZFC that such a set even exists. Or do you mean something else? — user21820, Mar 01 '16 at 06:03

GEdgar · Accepted Answer · 2016-03-03T01:58:41.663

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Is it true that for any infinite set $E$, the cardinal of the field $\mathbb Q(E)$ is equal to the cardinal of $E$?

added
What is $\mathbb Q(E)$?
Here $E$ is considered to be a set of "indeterminates". Or, if that is not convenient, invent a set of indeterminates perhaps with some notation like $X_e$, one for each element $e \in E$. Then $\mathbb Q[E]$ is the set of polynomials in these indeterminates, with coefficients in $\mathbb Q$. (Of course each polynomial involves only finitely many of the elements of $E$.)

So $\mathbb Q(E)$ is a ring. It is an algebra over $\mathbb Q$. It is an integral domain.

And $\mathbb Q(E)$ is the set of rational functions in these indeterminates. In other words, formulas of the type $f/g$, where $f,g \in \mathbb Q[E]$ and $g \ne 0$.

So $\mathbb Q(E)$ is a field. It has the set $E$ (or a set $\{X_e : e \in E\}$ identified with $E$) of mutually algebraically independent elements that generate it (as a field over $\mathbb Q$).

To compute the cardinality of $\mathbb Q(E)$, compute in turn: how many monomials in $E$ are there; how many linear combinations of monomials (i.e. polynomials); how many quotients of those (rational functions).

edited Mar 03 '16 at 01:58

answered Feb 29 '16 at 16:25

GEdgar

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My first reaction was to downvote and flag to delete as "not an answer" (fortunately I withheld both in time). I don't think phrasing your answers as questions is a good idea... – Najib Idrissi Feb 29 '16 at 16:33
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I think your answer can be phrased in a "non-rhetorical" fashion. – user2902293 Feb 29 '16 at 16:50
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Plus: easier than the L-S theorem. – GEdgar Feb 29 '16 at 17:18
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A remark: This theorem relies on the axiom of choice (in this case, in the cardinal arithmetic needed to establish $|\Bbb Q(E)|=|E|$), and is in fact equivalent to the axiom of choice, since it implies the related theorem that every set is equipotent to a group. – Mario Carneiro Feb 29 '16 at 20:26
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So you are saying that not only this proof, but the result itself is equivalent to AC. – GEdgar Feb 29 '16 at 22:46
@GEdgar Excuse me, what is ℚ(E) exactly? Link to an article about it? – Alexandre Khoury Mar 02 '16 at 19:47

score 7 · Answer 2 · answered Feb 29 '16 at 16:16

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The field axioms are a countable set over a countable language. Since there is an infinite model of the field axioms, there is (by Lowenheim-Skolem) a model of every infinite cardinality.

answered Feb 29 '16 at 16:16

user21820

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Note that the cardinal arithmetic required here is the same as that required in GEdgar's (Gregory Grant's) answer, since one can easily see that the size of the model built by adding $κ$ new constants axiomatized to be all distinct is going to be between $κ$ and $|\mathbb{Q}| \cdot κ$. – user21820 Mar 01 '16 at 00:11

score 3 · Answer 3 · answered Feb 29 '16 at 16:18

For every infinite set $X$ there is a field $\mathbb F$ which is equipotent to $X$. In fact, much more is true. This is an immediate consequence of the Upward Löwenheim-Skolem Theorem. Fix a countable field $K$ (e.g. $\mathbb Q$). By the Löwenheim-Skolem Theorem there is some $\mathbb F$ such that $\mathbb F$ is equipotent to $X$ and $K \prec \mathbb F$. This implies that $K$ and $\mathbb F$ have the same theory and since $K$ is a field, $\mathbb F$ is a field as well.

score 3 · Answer 4 · edited Apr 13 '17 at 12:21

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I think this answer by Gregory Grant is in some ways better than any relying on the Löwenheim–Skolem theorem, since it gives an explicit construction of a field of arbitrary infinite cardinality.

edited Apr 13 '17 at 12:21

Community

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answered Feb 29 '16 at 16:37

Michael Hardy

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Since this link refers to a post on math.stackexchange, I don't think that the "This is a link-only answer" argument for closure applies as it will (most likely) remain accessible. – Stefan Mesken Mar 01 '16 at 02:07

Is every infinite set equipotent to a field?

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