11
$\begingroup$

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.

$\endgroup$

1 Answer 1

16
$\begingroup$

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).

$\endgroup$
4
  • $\begingroup$ 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, 2012 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. $\endgroup$ Aug 30, 2012 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, 2012 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, 2012 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.