I believe it is known that if I is a set of non-measurable 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?

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    $\begingroup$ 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 $\endgroup$ May 28, 2013 at 2:18
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    $\begingroup$ en.wikipedia.org/wiki/Measurable_cardinal $\endgroup$ May 28, 2013 at 3:45
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    $\begingroup$ As you write it, the answer is obvious, since the target is $Z^1$. $\endgroup$ May 28, 2013 at 9:12
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    $\begingroup$ 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 $I$. $\endgroup$ May 28, 2013 at 12:59
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    $\begingroup$ Some of the older literature uses "measurable" to mean "supporting a non-principal 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". $\endgroup$ May 28, 2013 at 13:01

1 Answer 1


This runs under the name Łoś-Eda 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 non-principal 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 non-principal 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 Eklof-Mekler 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 non-principal countably complete ultrafilters on $I$ give rise to non-trivial (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/dual-of-zi-for-uncountable-i/… (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 Shelah-Strüngmann paper in the question you linked, is based on a non-measurable index set. However, it should be pointed out that in the measurable as well as in the non-measurable 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. $\endgroup$
    – Ralph
    May 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 Shelah-Strüngmann paper (it's freely available on degruyter.com/view/j/jgt.2013.16.issue-3/issue-files/…). 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 Shelah-Strüngmann consider homomorphisms from the free complete product of groups into the integers while I take the ... $\endgroup$
    – Ralph
    May 28, 2013 at 20:34

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