Free modules over integers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:51:44Z http://mathoverflow.net/feeds/question/102900 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102900/free-modules-over-integers Free modules over integers truebaran 2012-07-22T21:57:54Z 2012-07-22T22:20:32Z <p>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 $\mathbb{Z}$): </p> <p>a) $\mathbb{Z}^{\infty}$-all sequences of integers,</p> <p>b) $\mathbb{Z}^{\mathbb{R}}$-all functions $f: \mathbb{R} \to \mathbb{Z}$,</p> <p>c) the set of all functions $f: \mathbb{R} \to \mathbb{Z}$ with at most countable support?</p> http://mathoverflow.net/questions/102900/free-modules-over-integers/102901#102901 Answer by Andreas Blass for Free modules over integers Andreas Blass 2012-07-22T22:20:32Z 2012-07-22T22:20:32Z <p>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.</p>