Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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 –  Gerhard Paseman May 28 '13 at 2:18
"non-measurable cardinality"? –  Greg Martin May 28 '13 at 3:12
As you write it, the answer is obvious, since the target is $Z^1$. –  Fernando Muro May 28 '13 at 9:12
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". –  Andreas Blass May 28 '13 at 13:01

1 Answer 1

up vote 10 down vote accepted

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 $$\phi: \bigoplus_{i \in I} Hom(\mathbb{Z},\mathbb{Z}) \to Hom(\prod_{i \in I}\mathbb{Z},\mathbb{Z}),\; (g_i)_i \mapsto \big(\; (m_i)_i \mapsto \sum_i g_i(m_i)\;\big)$$ 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 $$Hom(\prod_{i \in I}\mathbb{Z},\mathbb{Z}) \cong \bigoplus_D Hom(\mathbb{Z},\mathbb{Z})$$ where $D$ runs through all countably complete ultrafilters on $I$ (Cor. III.3.7).

share|improve this answer
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. –  Andreas Blass May 28 '13 at 13:06
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.) –  Todd Trimble May 28 '13 at 15:51
@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. –  Ralph May 28 '13 at 19:09
@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? –  Todd Trimble May 28 '13 at 19:35
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 ... –  Ralph May 28 '13 at 20:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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