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!
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3$\begingroup$ When $\kappa$ is an infinite cardinal, what is the cardinality of a $\mathbb{Q}$-vector space of dimension $\kappa$? $\endgroup$– Robin ChapmanCommented Jul 18, 2010 at 15:06
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2$\begingroup$ 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. $\endgroup$– Asaf Karagila ♦Commented Jul 18, 2010 at 15:19
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1$\begingroup$ Continuing with this Socratic approach, is there a countable $\mathbb{R}$-vector space? $\endgroup$– Donu ArapuraCommented Jul 18, 2010 at 15:20
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$\begingroup$ @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. $\endgroup$– Asaf Karagila ♦Commented Jul 18, 2010 at 15:26
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1$\begingroup$ 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. $\endgroup$– Donu ArapuraCommented Jul 18, 2010 at 15:53
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2 Answers
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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.
answered Jul 18, 2010 at 15:10
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$\begingroup$ Does the theory of $R$-modules get a constant per ring element? $\endgroup$ Commented Jul 18, 2010 at 15:14
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$\begingroup$ @Mariano: More precisely, one unary function per ring element. $\endgroup$ Commented Jul 18, 2010 at 15:15
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$\begingroup$ Ah, ok. And the ring operations are codified by axioms, I guess, $\endgroup$ Commented Jul 18, 2010 at 15:41
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$\begingroup$ The Lowenheim-Skolem theorem, as described above, is also called the Lowenheim-Skolem-Tarski theorem by some. $\endgroup$ Commented Jul 18, 2010 at 17:54
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$\begingroup$ 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? $\endgroup$ Commented Jul 19, 2010 at 15:03
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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.
answered Jul 19, 2010 at 10:23
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$\begingroup$ Or one can also just use large permutation groups. $\endgroup$ Commented Jan 19, 2012 at 1:10