By a theorem of Specker, the group $\mathrm{Hom}(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$ is isomorphic to $\bigoplus_{\aleph_0}\mathbb{Z}$ and is in particular a free abelian group. I wonder, if this generalizes to all cardinals:

Question: Is it true that $\mathrm{Hom}(\prod_{\kappa} \mathbb{Z},\mathbb{Z})$ is a free abelian group for each cardinal $\kappa$ ?

As the answer to such questions sometimes depends on the underlying set theory, I included the "set-theory" tag.


If there are no measurable cardinals, or just if there are no measurable cardinals $\leq\kappa$, then the answer to your question is yes, and in fact all homomorphisms from $\prod_\kappa\mathbb Z$ to $\mathbb Z$ are linear combinations of the $\kappa$ projection maps. If, on the other hand, $\kappa$ or some smaller cardinal is measurable, so $\kappa$ supports a non-trivial, countably complete ultrafilter $U$, then the "in fact" clause in the preceding sentence is false; a counterexample is given by the homomorphism sending any $f\in\prod_\kappa\mathbb Z$ to the value that $f$ takes at $U$-almost all elements of $\kappa$. I believe that Hom($\prod_\kappa\mathbb Z,\mathbb Z)$ may nevertheless be free (though with a more complicated base than just the projections), but this probably depends on detials of the structure of the countably complete ultrafilters on $\kappa$. (In a couple of days, I'll be back in Michigan, where I can look in my copy of the book "Almost Free Modules" by Eklof and Mekler, where the section on the Los-Eda theorem should give me a lot more information about this. Anyone who wants to look for themselves rather than waiting for me should take into account that "Los" here is really "{\L}o\'s".)

Edit: OK, I'm back in Michigan, and I'm looking at Corollary 3.6 of the Eklof-Mekler book. It's stated in more generality than the present question wants (using slender modules over general rings), but if I specialize the ring $R$ and all the modules $H$ and $M_i$ in this corollary to be $\mathbb Z$, it tells me that Hom($\prod_\kappa\mathbb Z,\mathbb Z)$ is freely generated by the homomorphisms that I described above, one homomorphism for each countably complete ultrafilter on $\kappa$ (including the principal ultrafilters, which give the projection homomorphisms).

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  • $\begingroup$ Andreas, thank you very much for your answer! Do you have a refernce for the "no measurable cardinals $\le \kappa$" case ? I have access to the Eklof-Mekler book and will have a look. But since I don't know much about ultrafilters I would appreciate very much if you will also have a look when you're back. Thanks again. $\endgroup$ – Ralph Aug 29 '12 at 1:25
  • $\begingroup$ Ralph, the non-measurable case is Corollary 3.3 of Eklof-Mekler, specialized as in the "Edit" that I added to my answer. The original sources are listed on page 80 of E-M; in particular, they attribute the (specialized) non-measurable case to a 1962 paper of Balcerzyk. $\endgroup$ – Andreas Blass Aug 30 '12 at 17:27
  • $\begingroup$ Shelah 904 seems somewhat relevant too, the abstract states: "Answering problem (DG) of [EM90], [EM02], we show that there is a reflexive group of cardinality >= first measurable."(shelah.logic.at/904_abs.html) $\endgroup$ – Asaf Karagila Aug 30 '12 at 17:30
  • $\begingroup$ @Andreas: Very nice, thank you very much (the moment when you posted your edit, I was just reading Cor. 3.7 (ii) that also states the freeness of the dual). BTW: The book of Eklof-Mekler seems to be quite interesting. Thanks for drawing my attention to it. $\endgroup$ – Ralph Aug 30 '12 at 17:39

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