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How do I see that there is a group of an arbitrary cardinality? Is this also true for abelian groups? Also, given a commutative ring $R\neq 0$ how do I see that there is an $R$-module of arbitrary cardinality? I'm sure I saw this result somewhere but I can't seem to find it anywhere (books, google,...) Thanks!

ashpool
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    When $\kappa$ is an infinite cardinal, what is the cardinality of a $\mathbb{Q}$-vector space of dimension $\kappa$? – Robin Chapman Jul 18 '10 at 15:06
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    In addition to Robin's suggestion, you can consider the polynomial ring in $\kappa$ variables (for an infinite cardinal $\kappa$) over a given commutative ring. – Asaf Karagila Jul 18 '10 at 15:19
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    Continuing with this Socratic approach, is there a countable $\mathbb{R}$-vector space? – Donu Arapura Jul 18 '10 at 15:20
  • @Donu: There can't be a countable $\mathbb{R}$-vector space, as any real vector space would have a surjective map onto $\mathbb{R}$, therefore would be of greater-or-equal cardinality, which is $2^{\aleph_0}$ which in turn is strictly greater than $\aleph_0$ and therefore uncountable. – Asaf Karagila Jul 18 '10 at 15:26
  • Thanks, but I was hoping that kwan would think about it (look at the second sentence of the question). – Donu Arapura Jul 18 '10 at 15:31
  • The 0-dimensional space over any field will be countable. – Carl Mummert Jul 18 '10 at 15:48
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    I guess I should have said cardinality equal to $\aleph_0$ in my original comment. I think Asaf caught my meaning. In my second comment that should read "look at the third question". It appears that I can't count finite sets either. – Donu Arapura Jul 18 '10 at 15:53
  • Thanks guys! All of your comments were extremely helpful. – ashpool Jul 18 '10 at 15:56
  • Note that as Joel pointed out in his answer, all this relies on the axiom of choice. In a world without choice, the algebraic aspects of the constructions above all work fine, but cardinal arithmetic can go right up the creek; so (for instance) the cardinality of $\mathbb{Q}^\kappa$ may not be equal to $\kappa$. – Peter LeFanu Lumsdaine Jul 18 '10 at 16:45
  • It's certainly not true that there are always R-modules of arbitrary _finite_cardinalities, e.g. when R = F_p. – Qiaochu Yuan Jul 18 '10 at 16:48

2 Answers2

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For any algebraic theory that is expressible in first order logic in a countable language, and this includes groups, rings, fields, partial orders, lattices, etc. etc., then the basic fact is expressed by the Lowenheim-Skolem theorem, which asserts that if the theory has an infinite model, then it has models of every infinite cardinality. In general, one gets models of the theory of every cardinality above the size of the language (and this covers your $R$-module case).

One needs the Axiom of Choice to prove this, however, and this use is necessary, since the Axiom of Choice is equivalent to the assertion that every set carries a group structure, as explained in the answer to this MO question.

Community
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Joel David Hamkins
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  • Does the theory of $R$-modules get a constant per ring element? – Mariano Suárez-Álvarez Jul 18 '10 at 15:14
  • @Mariano: More precisely, one unary function per ring element. – François G. Dorais Jul 18 '10 at 15:15
  • Ah, ok. And the ring operations are codified by axioms, I guess, – Mariano Suárez-Álvarez Jul 18 '10 at 15:41
  • The Lowenheim-Skolem theorem, as described above, is also called the Lowenheim-Skolem-Tarski theorem by some. – Péter Komjáth Jul 18 '10 at 17:54
  • I suppose that one can reconcile this answer with the comments above concerning $\mathbb{R}$-vector spaces by observing that it would require an uncountable language: one constant (= 0-ary function surely) per real number. Would there be modification of Lowenhweim-Skolem for such uncountable languages? – Donu Arapura Jul 19 '10 at 15:03
  • Donu, Yes, the Lowenheim-Skolem theorem is completely general: the upward version says that every infinite model of a theory has elementary extensions to all larger cardinalities that are at least the size of the language. The downward version finds elementary submodels of any model of the theory, of any desired smaller size that is at least the size of the language. – Joel David Hamkins Jul 19 '10 at 20:02
  • @PaceNielsen The question was whether there were groups, etc. of every cardinality, so the theory here is "group theory" (or "ring theory" or whatever first-order algebraic theory you like). A model of this theory is a group, and so to say that there are models of this theory of every infinite size is to say that there are groups of every infinite size. And the same for rings, fields, etc. – Joel David Hamkins Apr 16 '16 at 15:26
  • @JoelDavidHamkins I guess my follow-up question is what makes a first-order algebraic theory different from finite sets in this respect? (I'm apparently unfamiliar with the basic fact you are using.) – Pace Nielsen Apr 16 '16 at 22:02
  • @PaceNielsen I am just noticing your question now, years later. An extremely important difference is that the class of finite sets is not characterizable by a first-order theory. Any first-order theory that is true in arbitrarily large finite structures is also true in some infinite structure. This is a consequence of the compactness theorem of first-order logic. https://en.wikipedia.org/wiki/Compactness_theorem So the class of finite sets is not an elementary class. It is defined in second-order logic, or in set theory, but not by a first-order theory. – Joel David Hamkins Dec 20 '23 at 11:26
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I just wanted to point out that I don't think that the Axiom of Choice is necessary to construct arbitrarily large groups. Can't you for each set $X$ take the collection of all formal (finite) linear combinations of elements of $X$ over the rationals, which then becomes a $\mathbb Q$-vector space of dimension the size of $X$?
Note that without AC we probably don't know that (in the case that $X$ is infinite) there is a bijection between $X$ and this vector space.

Stefan Geschke
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