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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?

Michael Barr
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    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 02:18
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    "non-measurable cardinality"? – Greg Martin May 28 '13 at 03:12
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    http://en.wikipedia.org/wiki/Measurable_cardinal – Allen Knutson May 28 '13 at 03:45
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    This is not exactly the statement. Indeed, if $\alpha$ is the smallest measurable cardinal and if $\beta$ is the next cardinal, then $\beta$ is not measurable. The correct statement is: "if $I$ admits no nonprincipal ultrafilter stable by countable intersections", or equivalently "if the cardinal $I$ is smaller than every measurable cardinal", or equivalently "every subset of $I$ is non-measurable". – YCor May 28 '13 at 08:16
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    As you write it, the answer is obvious, since the target is $Z^1$. – Fernando Muro May 28 '13 at 09:12
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    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 $I$. – Andreas Blass May 28 '13 at 12:59
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    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
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    I gave a proof at this answer https://mathoverflow.net/a/326379/14094 (proof of the lemma'), where "reasonable" means "non-Ulam-measurable", and granted the countable case. More precisely it's a proof of the fact that every homomorphism vanishing on the direct sum vanishes on the product. But then the conclusion (that every homomorphism vanishes on all but finitely many factors) is immediate from the countable case. – YCor Apr 02 '19 at 17:38

1 Answers1

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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).

user26857
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Ralph
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  • 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: http://mathoverflow.net/questions/12586/dual-of-zi-for-uncountable-i/12588#12588 (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 http://www.degruyter.com/view/j/jgt.2013.16.issue-3/issue-files/jgt.2013.16.issue-3.xml). 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
  • ... direct product. This may also be the reason why the title of their paper includes the words "non-commutative". In particular, they refer to the Eklof-Mekler book for generalizations of the Specker phenomenon on uncountable direct products (p. 420, before Def. 2.1). To summarize: 1. The isomorphism in my 1st comment is correct (reference: Eklof-Mekler, Cor. III.3.7). 2. $\phi$ is an isomorphism if $I$ is not $\omega$-measurable (references in my answer). – Ralph May 28 '13 at 21:03
  • @Ralph: Right, this is what I thought was going on (I say this even though I'm not sure what is meant by the "free complete product"). So this means a clarification is in order for the other question; I'll refer to your answer here. What is somewhat annoying is that the notation used in the Shelah-Strüngmann paper, referring to free complete products, looks identical to cartesian product notation, and this looks like a likely source of confusion. Thanks for your attention to this matter! – Todd Trimble May 28 '13 at 21:11