## Free modules over integers

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After the course of linear algebra I'm more familiar with vector spaces rather than modules so my question may seem to be silly but I think it's quite natural for someone who thinks of modules as 'vector spaces over a ring': which of the following is free module (free means: having a basis, all of them are over ring $\mathbb{Z}$):

a) $\mathbb{Z}^{\infty}$-all sequences of integers,

b) $\mathbb{Z}^{\mathbb{R}}$-all functions $f: \mathbb{R} \to \mathbb{Z}$,

c) the set of all functions $f: \mathbb{R} \to \mathbb{Z}$ with at most countable support?

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It is well known non of them is free. This is not at a research level. Actually, it is enough to check that a) is not free, as it is a direct summand of b) and c). – Fernando Muro Jul 22 at 22:23

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This raises an interesting question about "research level". The proofs are not difficult and are many decades old, yet I know from personal experience that there are very capable research mathematicians who not only didn't know that (a) isn't free but, when I mentioned the result, were sure I had made a mistake. – Andreas Blass Jul 22 at 22:27

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For me, a research question is a question arising in your research that you, an experienced mathematician (at least to some extent), have tried to solve on your own, looking up in the literature, etc. This question can be solved in one google search. – Fernando Muro Jul 22 at 23:17

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A module over the integers is precisely an abelian group. So, for example, the volumes by László Fuchs will tell you this story and of course much more. @Fernando Muro...what did you google?...one try for me landed in the world of photoshop freeware LOL. – David Feldman Jul 23 at 0:59

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This question and the accompanying answers deal with a very similar topic: mathoverflow.net/questions/10239 – Emerton Jul 23 at 5:40
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## 1 Answer

None of these are free. For (a), there is a theorem of Specker ("Additive Gruppen von Folgen ganzer Zahlen" Portugaliae Math. 1950) that covers this group and lots of its subgroups (the so-called monotone subgroups). I believe that for this particular group, the result may already be in a 1937 paper of Baer. Since the groups in (b) and (c) have subgroups isomorphic to the group in (a), and since subgroups of free abelian groups are free, it follows that (b) and (c) aren't free either.

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Here reh.math.uni-duesseldorf.de/~schroeer/… is a short self-contained proof that an infinite direct product of the integers is non-free. – SJR Jul 23 at 7:31
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