15

Attempt to answer: Every set up to cardinality admits a ring structure, (except the empty set). Let $S$ be a set. If $S$ if finite, we may assume it's $ \mathbb{Z} / n \mathbb{Z}$. Otherwise, construct the free abelian ring $\mathbb{Z}[S]$ with $|S| = |\mathbb{Z}[S]|$? Then find a bijection from $S$ to $\mathbb{Z}[S]$. I don't think this works, unfortunately. I know we can give group structure to $S$, we can then give it the structure of the group ring over the integers. Or we could make it abelian and construct the ring of endomorphisms? I don't think any of these preserve cardinality, otherwise proving that every set is a group is trivial. In the proof that every set is a group, I don't think that group admits a multiplication with a unit. (Rings should have a 1).

Phthalo Johnson
  • 151
  • 3
    Note: the axiom of choice is equivalent in ZF to the assertion that every nonempty set carries a group structure. https://mathoverflow.net/a/12988/1946 – Joel David Hamkins Jul 11 '23 at 17:41
  • 14
    I'd be curious about a field structure on a set with six elements. – Najib Idrissi Jul 11 '23 at 17:42
  • Just a side note: “…up to cardinality…” is redundant here, and essentially anywhere in mathematics one might say it. It means “…up to bijection…”, but algebraic structure obviously transfers along bijections, so it’s just the same as asking whether every set admits a field (etc) structure. // Relatedly, I’ve VTC’d: this question is off-topic here. It’s a good question, and would be fine on math.stackexchange, but it’s really not research-level — it could fit well as homework in a first or second course in logic. – Peter LeFanu Lumsdaine Jul 12 '23 at 19:25
  • @PeterLeFanuLumsdaine I agree with your main point, but arguably, the phrase "up to cardinality" plays the role of communicating that the axiom of choice is being tacitly assumed. – Timothy Chow Jul 12 '23 at 22:59
  • 1
    @TimothyChow How does "up to cardinality" convey AC? I suppose it would if you think of cardinalities as well-ordered, but ZF has a perfectly robust notion of cardinality without AC. And this is the notion that Peter is using. – Joel David Hamkins Jul 13 '23 at 22:54
  • @JoelDavidHamkins It doesn't convey it explicitly, but I suspect that someone who says "every set up to cardinality" rather than "every set up to isomorphism" is probably thinking of the cardinal numbers $\aleph_0, \aleph_1, \ldots$ – Timothy Chow Jul 14 '23 at 12:50
  • It can quite important for such questions to specify the definition of a ring (associative? commutative? unital?) – YCor Jul 14 '23 at 12:57

1 Answers1

24

In ZFC, every nonempty cardinality is the cardinality of a ring. For finite cardinalities, we have $\mathbb{Z}/n\mathbb{Z}$ as you mentioned. For infinite cardinalities, this is an immediate consequence of the Löwenheim-Skolem theorem. So there are rings and fields in every infinite cardinality.

Without the axiom of choice, it is not true even that every infinite set carries a group structure. See the answer by Ashutosh; the result is due originally to Hajnal and Kertész.

It follows:

Corollary. The following are equivalent over ZF set theory:

  1. Every infinite set carries a group structure.
  2. Every infinite set carries an abelian group structure
  3. Every infinite set carries a ring structure.
  4. Every infinite set carries a field structure.
  5. The axiom of choice holds.

Proof. Statement 5 implies statement 4 by the Löwenheim-Skolem theorem. And clearly 4 implies 3 implies 2 implies 1. And 1 implies 5 by the argument in the Hajnal/Kertész result, in the answer of Ashutosh. $\Box$

Note that in 5 implies 4, one can specify any desired characteristic.

(For finite fields, of course they have to be prime powers.)

Joel David Hamkins
  • 224,022
  • 12
    Fwiw you don't need anything as sophisticated as Lowenheim-Skolem, you can just take say $\mathbb F_2[S]$, where you mean the free associative $\mathbb F_2$-algebra generated by elements of $S$. This is evidently a countable union of finite powers of $S$, so under choice has the same cardinality as $S$. – Kevin Casto Jul 11 '23 at 17:57
  • 15
    Kevin, thanks for the more specific construction. I agree. However, my view also is that the Löwenheim-Skolem theorem is valuable because it avoids the need for ad hoc constructions realizing different kinds of (first-order) mathematical structure in different cardinalities. It is an elementary result in model theory, but sweeping, in that it answers all such cardinality questions at once, for any desired kind of mathematical structure. It is part of the framework of my understanding of how mathematics works. – Joel David Hamkins Jul 11 '23 at 18:19
  • 6
    I would also quibble with the statement that Lowenheim-Skolem is "sophisticated". It's a basic result which, I suspect, is presented in any book/course on model theory (I learned it as an undergraduate in a course using Monk's book). – Lee Mosher Jul 12 '23 at 14:02
  • @KevinCasto: I think both versions are enlightening, in slightly different ways. As Joel says, Löwenheim–Skolem fits it most clearly into a big general picture. On the other hand your example (and similar ones for other theories) give concrete structures, algebraically natural and tractable, with the invocation of choice confined to the construction of a bijection with the original set. – Peter LeFanu Lumsdaine Jul 12 '23 at 19:22
  • 1
    Since you're listing algebraic structures, left-cancellative magma is enough for the proof to get choice. – Asaf Karagila Jul 13 '23 at 22:44