I believe it is known that if I is a set of nonmeasurable cardinality, then any homomorphism $Z^I\to Z$ factors through a finite power. Here $Z$ is the group of integers. Can anyone give a reference for this?

1$\begingroup$ I remember a similar exercise in "Algebras, Lattices, and Varieties" by McKenzie, McNulty, and Taylor. (Ch 4.4, exercise 20.) I would guess it is a result of Specker, but that it just a guess. I can't tell from a glance at the bibliography where it came from. Maybe someone else can. Gerhard "Someone Take The Baton Now" Paseman, 2013.05.27 $\endgroup$– Gerhard PasemanMay 28, 2013 at 2:18

2$\begingroup$ en.wikipedia.org/wiki/Measurable_cardinal $\endgroup$– Allen KnutsonMay 28, 2013 at 3:45

3$\begingroup$ As you write it, the answer is obvious, since the target is $Z^1$. $\endgroup$– Fernando MuroMay 28, 2013 at 9:12

2$\begingroup$ Fernando is right; the intended conclusion is that any homomorphism factors through the standard projection to some finite power, i.e, simply restricting functions $I\to Z$ to a finite subset $F$ of $I$. $\endgroup$– Andreas BlassMay 28, 2013 at 12:59

3$\begingroup$ Some of the older literature uses "measurable" to mean "supporting a nonprincipal countably complete ultrafilter", which nowadays would be expressed as "greater than or equal to the first measurable cardinal". The same concept is, I believe, sometimes called "Ulam measurable". $\endgroup$– Andreas BlassMay 28, 2013 at 13:01
1 Answer
This runs under the name ŁośEda Theorem. A reference is the book Paul C. Eklof, Alan H. Mekler, Almost Free Modules (2002):
Call a set $I$ $\omega$measurable if its cardinality is greater or equal to the first measurable cardinal. This is equivalent to $I$ being uncountable and supporting a nonprincipal countably complete ultrafilter.
First note that $\mathbb{Z}$ is slender (Cor. III.2.4). Then, by Cor. III.3.6 (and the discussion before Lemma III.3.5), if $I$ is not $\omega$measurable, the natural map $$\phi: \bigoplus_{i \in I}\operatorname{Hom}(\mathbb{Z},\mathbb{Z}) \to \operatorname{Hom}(\prod_{i \in I}\mathbb{Z},\mathbb{Z}),\; (g_i)_i \mapsto \big((m_i)_i \mapsto \sum_i g_i(m_i)\big)$$ is an isomorphism.
Remarks: 1) If $I$ is $\omega$measurable, not all homomorphisms $\prod_I \mathbb{Z} \to \mathbb{Z}$ factor through a finite subset of $I$. For, let $D$ be a nonprincipal countably complete ultrafilter on $I$ and let $K_D = \lbrace x \in \prod_I \mathbb{Z} \mid I \setminus \sup(x) \in D\rbrace$. Then it's not hard to show that the composition $\prod_I \mathbb{Z} \twoheadrightarrow \prod_I \mathbb{Z}/K_D \cong \mathbb{Z}$ doesn't factor through a finite subset of $I$ (the latter isomorphism uses II.3.3).
2) Irrespective whether $I$ is $\omega$measurable or not, there is a canonical isomorphism $$\operatorname{Hom}(\prod_{i \in I}\mathbb{Z},\mathbb{Z}) \cong \bigoplus_D \operatorname{Hom}(\mathbb{Z},\mathbb{Z})$$ where $D$ runs through all countably complete ultrafilters on $I$ (Cor. III.3.7).

$\begingroup$ The EklofMekler book is also the first reference that I'd suggest. I believe, though, that the theorem the OP asked about is entirely due to Łoś. Eda's contribution concerned what happens when $I$ is greater than or equal to the first measurable cardinal. For such $I$ it's clear that nonprincipal countably complete ultrafilters on $I$ give rise to nontrivial (i.e., not factoring through projections to finite products) homomorphisms, but it takes some work to show that these together with the trivial homomorphisms actually generate all the homomorphisms. $\endgroup$ May 28, 2013 at 13:06

$\begingroup$ I'm having some trouble reconciling this answer with the answer given here: mathoverflow.net/questions/12586/dualofziforuncountablei/… (This is not to say that I disbelieve the present answer. I'm guessing that something might have gotten lost in the translation with the other answer.) $\endgroup$– Todd Trimble ♦May 28, 2013 at 15:51

$\begingroup$ @Todd: As explained by Andreas Blass, not all homs factor when I isn't measurable (I'll correct my remark above later). I guess the ShelahStrüngmann paper in the question you linked, is based on a nonmeasurable index set. However, it should be pointed out that in the measurable as well as in the nonmeasurable case $Hom(\prod_I \mathbb{Z},\mathbb{Z})\cong \bigoplus_D Hom(\mathbb{Z},\mathbb{Z})$ is a free abelian group with basis a set $D$ of ultrafilters on I. $\endgroup$– RalphMay 28, 2013 at 19:09

$\begingroup$ @Ralph: perhaps I'm being thick, but I'm having a hard time seeing how you're addressing my query. Martin's question (the page I linked to) asked about general uncountable $I$. For the moment, let's say that $I$ is less than the first measurable cardinal (where Andreas's remark would not apply). Then you seem be be asserting that the map $\phi$ is an isomorphism. Whereas Mariano gave the opposite answer. Am I missing something? $\endgroup$– Todd Trimble ♦May 28, 2013 at 19:35

$\begingroup$ Todd, if it seemed that I didn't take your comment/question seriously, I apologize. I just had a look into the ShelahStrüngmann paper (it's freely available on degruyter.com/view/j/jgt.2013.16.issue3/issuefiles/…). Unfortunately Mariano didn't give a precise reference within the paper where it is shown that $\phi$ (the map from my answer) fails to be an isomorphism for uncountable cardinals. But it seems to me that the point is that ShelahStrüngmann consider homomorphisms from the free complete product of groups into the integers while I take the ... $\endgroup$– RalphMay 28, 2013 at 20:34