Welcome to MathOverflow

Question

1

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!

flag	asked Jul 18 2010 at 15:03 ashpool 1,419●4●16

3

When $\kappa$ is an infinite cardinal, what is the cardinality of a $\mathbb{Q}$-vector space of dimension $\kappa$? – Robin Chapman Jul 18 2010 at 15:06

2

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 2010 at 15:19

1

Continuing with this Socratic approach, is there a countable $\mathbb{R}$-vector space? – Donu Arapura Jul 18 2010 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 2010 at 15:26

1

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 2010 at 15:53

show 5 more comments

Joel David Hamkins · Accepted Answer · 2010-07-18 15:10:14Z UTC

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.

Stefan Geschke · Answer 2 · 2010-07-19 10:23:58Z UTC

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.

Group & modules of arbitrary cardinality

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.