# Is every infinite set equipotent to a field? [duplicate]

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. Feb 29, 2016 at 16:13
• @Omnomnomnom thanks Feb 29, 2016 at 16:15
• Also, tip for LaTeXing: \Bbb R gives you $\Bbb R$ Feb 29, 2016 at 16:16
• Relevant: Fields of arbitrary cardinality. Feb 29, 2016 at 16:17
• @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. Feb 29, 2016 at 16:49

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

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).

• 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... Feb 29, 2016 at 16:33
• I think your answer can be phrased in a "non-rhetorical" fashion. Feb 29, 2016 at 16:50
• Plus: easier than the L-S theorem. Feb 29, 2016 at 17:18
• 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. Feb 29, 2016 at 20:26
• So you are saying that not only this proof, but the result itself is equivalent to AC. Feb 29, 2016 at 22:46

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.

• 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 $|\mathbb{Q}| \cdot κ$. Mar 1, 2016 at 0:11

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.

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.

• 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. Mar 1, 2016 at 2:07