# Homomorphisms from powers of Z to Z

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?

• 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
• en.wikipedia.org/wiki/Measurable_cardinal – Allen Knutson May 28 '13 at 3:45
• As you write it, the answer is obvious, since the target is $Z^1$. – Fernando Muro May 28 '13 at 9:12
• 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
• 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

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).

• 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